Mahesh

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 7th Lesson d and f Block Elements & Coordination Compounds Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 7th Lesson d and f Block Elements & Coordination Compounds

Very Short Answer Questions

Question 1.
What are transition elements ? Give examples.
Answer:
Transition elements are the elements which contains partially filled d-subshells in their ionic state (or) in their elementary state.
Eg : Mn, Co, Ag etc.

Question 2.
Which elements of 3d, 4d, and 5d series are not regarded as transition elements and why ?
Answer:
Zn (3d-series), Cd (4d-series), Hg (5d-series) are not regarded as transition elements because these contains completely filled d-subshells.

Question 3.
Why are d-block elements called transition elements ?
Answer:
The name transition elements given to d-block is due to their properties which are transitioned between the electropositive s-block metals and electronegative p-block elements.

Question 4.
Write the general electronic configuration of transition elements.
Answer:
General electronic configuration of transition elements is (n – 1)d^1-10 ns^1-2.

Question 5.
In what way is the electronic configuration of transition elements different from non . transition elements ?
Answer:

• The general electronic configuration of transition elements is (n – 1) d^1-10 ns^1-2.
• The general electronic configuration of non-transition elements is (n – 1)d^1-10 ns².

Question 6.
Write the electronic configuration of chromium (Cr) and copper (Cu).
Answer:

• Electronic configuration of chromium (Cr) – [Ar]4s¹d⁵.
• Electronic configuration of copper (Cu) – [Ar]4s¹3d¹⁰.

Question 7.
Why do transition elements exhibits characteristic properties ?
Answer:
Due to the presence of partially filled d-orbitals transition elements exhibit characteristic properties such as variable oxidation states, colour property, magnetic property, complex tendency etc.

Question 8.
Scandium is a transition element. But Zinc is not. Why
Answer:
Scandium has electronic configuration [Ar] 4s²3d¹.
Zinc has electronic configuration [Ar] 4s²3d¹⁰.
Scandium has one unpaired d-electron where as Zinc has zero unpaired d-electrons so Scandium is transition element but Zinc is not.

Question 9.
Even though silver has d¹⁰ configuration, it is regarded as transition element. Why ?
Answer:
The outer electronic configuration of silver is – 4d¹⁰5s¹. It is having general electronic configuration of a transition element (n – 1) d^1-10 ns^1-2.
So silver is a transition element.

Question 10.
Write the electronic configuration of Co²⁺ and Mn²⁺.
Answer:

• The electronic configuration of Co⁺² is [Ar] 4s⁰ 3d⁷.
• The electronic configuration of Mn⁺² is [Ar] 4s⁰ 3d⁵.

Question 11.
Why are Mn2+ compounds more stable than Fe²⁺ towards oxidation to their +3 state ?
Answer:

• Mn⁺² has electronic configuration [Ar] 4s⁰ 3d⁵.
• Fe⁺² has electronic configuration [Ar] 4s¹ 3d⁶.
• Mn⁺² has half filled d-subshell which is more stable.
• Hence Mn⁺² compounds are more stable than Fe⁺² toward oxidation to their +3 state.

Question 12.
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why ?
Answer:
Copper exhibits +1 oxidation state most frequently because Cu⁺ has electronic configuration [Ar] 4s⁰3d¹⁰ which has a fulfilled d-subshell which is more stable.

Question 13.
Why do transition elements exhibit more than one oxidation state (variable oxidation states) ?
Answer:
Transition elements exhibits more than one oxidation state.

Reason:
The energy difference between (n -1) d subshell and ns subshell is very low. So both of these subshells compelete to lose the electrons.

Question 14.
Though Sc is a transition element, it does not exhibit variable oxidation state. Why ?
Answer:
Scandium has electronic configuration [Ar] 4s²3d¹. It has only one unpaired d-electron. So it does not exhibits variable oxidation state. It exhibits +3 stable oxidation state.

Question 15.
Why is it difficult to obtain M³⁺ oxidation state in Ni, Cu and Zn ?
Answer:

• Ni has electronic configuration [Ar] 4s²3d⁸.
• Ni+2 has electronic configuration [Ar] 4s⁰3d⁸
• It is difficult to remove an electron from 3d⁸. So, Ni⁺³ is difficult to obtain. (Ni has high negative enthalpy of hydration).
• Cu has electronic configuration [Ar] 4s¹3d¹⁰.
• Cu+ has electronic configuration [Ar] 3d¹⁰
• It is difficult to remove the electrons from 3d¹⁰ (stable). So, Cu⁺³ is difficult to obtain.
• Zn has electronic configuration [Ar] 4s²3d¹⁰
• Zn⁺² has electronic configuration [Ar] 4s⁰3d¹⁰
• It is difficult to remove the electron from 3d¹⁰ (stable). So Zn⁺³ is difficult to obtain.

Question 16.
Why is Cr²⁺ reducing and Mn³⁺ oxidizing even though both have the same d⁴ electronic configuration.
Answer:
Cr⁺² is reducing as its configuration changes from d⁴ to d³, the latter having a half filled t_2g level. On the other hand the change from Mn⁺² to Mn⁺³ results in the half filled (d5) configuration which has extra stability.

Question 17.
Although Cr, Mo and W belong to the same group (group 6) Cr (VI) is a strong oxidizing agent while Mo (VI) and W (VI) are not. Why ?
Answer:
In group 6, Mo (VI) and W (VI) are more stable than Cr (VI). Thus Cr (VI) in the form of dichromate in acidic medium is a strong oxidising agent. Where as MoO₃ and WO₃ are not.

Question 18.
What do you infer from the fact that M³⁺/M²⁺ standard electrode potential of Mn is comparatively higher and that of Fe is comparitively lower ?
Answer:
M³⁺/M²⁺ standard electrode potential of Mn is comparatively higher than that of Fe is comparatively lower. This is because much larger 3^rd ionisation energy of Mn (d⁵ to d⁴).

Question 19.
Transition elements have high melting points. Why ?
Answer:
Transition elements have high melting points because of involvement of greater number of electrons from (n – 1)d in addition to ns electrons in the interatomic metallic bonding of these metals.

Question 20.
Among the first transition series (3d series) Chromium has highest melting point. Why? Ans. Among first transition series (3d-series) chromium has highest melting point.
Reason : In chromium one unpaired electron per d-orbital is particularly favourable for strong inter atomic interactions.

Question 21.
Compared to s-block elements, the transition elements exhibit higher enthalpy of atomization. Why ?
Answer:
Because of large no. of unpaired electrons in their atoms transition elements have stronger inter atomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation.

Question 22.
Among the first transition series (3d series) zinc has lowest enthalpy of atomization. Why ?
Answer:
Among the first transition series zinc has lowest enthalpy of atomisation because zinc has no unpaired electrons in their atomic state or in ionic state.

Question 23.
How do you expect the density of transition elements to vary in a given series and why ?
Answer:
In a given series the density of transition elements increases.

Eg. : From Titanium to copper there is a significant increase in the density.
This is due to decrease in metallic radius coupled with increase in atomic mass.

Question 24.
How do the atomic and ionic sizes vary among transition metals in a given series ?
Answer:
In a given series of transition elements atomic and ionic sizes progressively decreased. This is due to entering of a new electron into d-orbital each time.

Question 25.
Why do Mn, Ni and Zn exhibit more negative E^⊖ values than expected ?
Answer:
E⁰ values of Mn, Ni and Zn are more negative than expected from general trend. This is due to the stabilities of half filled d-subshell (d⁵) in Mn⁺² and completely filled d-subshell (d¹⁰) in Zinc. For Nickel E⁰ value is related to the highest negative enthalpy of hydration.

Question 26.
Among the first transition series (3d series) only copper has positive E^⊖M²⁺/M value. Why?
Answer:
Among the first transition series only copper has positive E^⊖M²⁺/M value. This is due to high ∆_aH⁰ and low ∆_hydH⁰ values.

Question 27.
Cu⁺² forms halides like CuF₂, CuCl₂ and CuBr₂ but not CuI₂. Why ?
Answer:
Cu⁺² forms halides like CuF₂, CuCl₂ and CuBr₂ but not CuI₂ because Cu⁺²oxidises I^– to I₂
2Cu⁺² + 4I^– → Cu₂I₂ + I₂

Question 28.
The highest Mn fluoride is MnF₄ where as the highest oxide is Mn₂O₇ Why ?
Answer:
The ability of oxygen to stabilize the high oxidation states exceeds that of fluoride. Thus the highest Mn fluoride is MnF₄ where as highest oxide is Mn₂O₇.

Question 29.
In its fluoride or Oxide, in which a transition metal exhibits highest oxidation state and why ?
Answer:

• In fluorides highest oxidation numbers are achieved in TiF₄, VF₅ and CrF₆.
• The +7 state of Mn is represented in MnO₃F.
• The highest oxidation number in oxides acheived from Sc₂O₃ to Mn₂O₇.
• In Mn₂O₇, Mn oxidation state is +7.

Question 30.
Why Zn²⁺ is diamagnetic whereas Mn²⁺ is paramagnetic ? [T.S. Mar. 15]
Answer:

• Zn⁺² electronic configuration is [Ar] 4s⁰3d¹⁰. It has no unpaired electrons. So it is dia magnetic.
• Mn⁺² electronic configuration is [Ar] 4s⁰3d⁵. It has five unpaired electrons so it is paramagnetic.

Question 31.
Write ‘spin only’ formula to calculate the magnetic moment of transition metal ions.
Answer:
Spin only formula to calculate the magnetic moment of transition metal ions is
μ = \(\sqrt{n(n+2)}\) BM
BM = Bohr Magneton.

Question 32.
Calculate the ‘spin only’ magnetic moment of Fe²⁺_(aq) ion.
Answer:
Fe⁺² ion has electronic configuration [Ar] 4s⁰3d⁶
It has four unpaired electrons n = 4
Spin only magnetic moment μ = \(\sqrt{n(n+2)}\) BM = \(\sqrt{4(4+2)}\) = \(\sqrt{24}\) BM = 4.9 BM

Question 33.
What is meant by ‘disproportionation’ ? Give an example of disproportionation reaction in aqueous solution.
Answer:
The reactions in which only one element undergo both oxidation as well, as reduction are called disproportion reactions.

When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation.
Eg: Cu⁺ ion is not stable in aqueous solution because it undergo disproportionation in aqueous solution.
2Cu⁺_(aq) → Cu⁺²_(aq) + Cu_(s)

Question 34.
Aqueous Cu²⁺ ions are blue in colour, where as Aqueous Zn²⁺ ions are colourless. Why ?
Answer:

• Electronic configuration of Cu⁺² ion is [Ar] 4s⁰3d⁹. If contains one unpaired electron due to presence of this unpaired electron aq. Cu⁺² ions are blue in colour.
• Electronic configuration of Zn⁺² ion is [Ar] 4s⁰3d¹⁰. It contains no unpaired electrons, due to absence of unpaired electrons aq. Zn⁺² ions are colourless.

Question 35.
What are complex compounds ? Give examples.
Answer:
Complex compounds: Transition metal atoms or ions form a large number of compounds in which anions or neutral groups are bound to metal atom or ion through co-ordinate covalent bonds. Such compounds are called co-ordination compounds (or) complex compounds.
Eg.: [Fe(CN)₆]^4-, [Co(NH₃)₆]³⁺.

Question 36.
Why do the transition metals form a large number of complex compounds ?
Answer:
Transition elements (metals) forms a large number of complex compounds this is due to

1. Small size of these ions
2. High effective nuclear charge,
3. Possessing in completely filled d-orbitals.

Question 37.
How do transition metals exhibit catalytic activity ?
Answer:
Catalytic properties:

• Transition metals and their compounds form important catalysts in industry and in biological systems.
• The catalytic activity of transition metals depends on their ability to exist in different oxidation states of oxidation (or) to form co-ordination compounds.
  Eg : 1) V₂O₅ is used as catalyst in manufacturing of SO₃ from SO₂.
  
  2) Fe is used as a catalyst in manufacturing of NH₃.
  

Question 38.
Give two reactions in which transition metals or their compounds acts as catalysts.
Answer:
1) V₂O₅ is used as catalyst in manufacturing of SO₃ from SO₂.

2) Fe is used as a catalyst in manufacturing of NH₃.

Question 39.
What is an alloy ? Give example.
Answer:
Alloy : An intimate mixture having physical properties similar to that of the metal formed by a metal with other metals or metalloids or sometimes a non metal is called as an alloy.
Eg.: Invar – 64% Fe, 35% Ni, Mn 8cc in traces
Nichrome – 60% Ni, 25% Fe, 15% Cr.

Question 40.
Why do the transition metals readily form alloys ?
Answer:
Because of similar atomic or ionic radii and similar characterstic properties of transition elements alloys are readily formed by these elements.

Question 41.
How do the ionic character and acidic nature vary among the oxides of first transition series ?
Answer:

• As the oxidation number of a metal increases ionic character decreases in case of transition elements. Eg : Mn₂O₇ is a covalent green oil.
• In CrO₃ and V₂O₅ the acidic character is predominant.
• V₂O₅ is however amphoteric though mainly acidic V₂O₅ reacts with alkali as well as acids to give V0₄^-3 and VO⁺₄.
• CrO is basic
• Mn₂O₇gives HMnO₄ and CrO₃ gives H₂CrO₄ and H₂Cr₂O₇.

Question 42.
What is the effect of increasing pH on a solution of potassium dichromate ?
Answer:
On increasing pH of K₂Cr₂O₇ (orange) it changes into K₂CrO₄ (yellow)

Question 43.
Name the oxometal anions of the first series of transition metals in which the metal exhibits the oxidation state equal to its group number.
Answer:
V₄^-3 ion exhibits ‘+5’ oxidation state which is equal to the V group number
V₄^-3 + 4(-2) = -3, x = +5

Question 44.
Permanganate titrations are carried out in the presence of sulphuric acid but not in presence of hydrochloric acid. Why ?
Answer:
Per manganate titrations are carried out in the presence of sulphuric acid but not in presence of hydrochloric acid because hydrochloric acid is oxidised to chlorine.

Question 45.
What is lanthanoid contraction ? [T.S. Mar. 19]
Answer:
The slow decrease of atomic and ionic radii in lanthanides with increase in atomic number is called lanthanide contraction.

Question 46.
What are the different oxidation states exhibited by the lanthanoids ?
Answer:

• Lanthanoids exhibits +2, +3 states majorly. Sometimes +2 and +4 states exhibited in solid compounds.
• The common oxidation state of lanthanoids is +3.

Question 47.
What is mischmetall ? Give its composition and uses. [A.P. Mar. 19, 16]
Answer:

• Mischmetall is an alloy which consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al.
• It is used in Mg- based alloy to produce bullets, shell and lighter flint.

Question 48.
What is actinoid contraction ?
Answer:
The gradual decrease in the size of atoms or M⁺³ ions across, the actinoid series is called actinoid contraction.

Question 49.
What are co-ordination compounds ? Give two examples.
Answer:
Co-ordination compounds : Transition metal atoms or ions form a large number of compounds in which anions or neutral groups are bound to metal atom or ion through coordinate covalent bonds. Such compounds are called co-ordination compounds (or) complex compounds.
Eg.: [Fe(CN)₆]^4-, [Co(NH₃)₆]⁺³.

Question 50.
What is a coordination polyhedron ?
Answer:
The spatial arrangement of the ligands which are directly bonded to the central atom or ions defines the geometry about the central atom is called co-ordination polyhedron
Eg. : Octahedral, tetrahedral etc.

Question 51.
What is a double salt ? Give example.
Answer:
The salts which contains two cations and one anion are called double salts.
These dissociates into simple ions completely when dissolved in water.
Eg : KCl.MgCl₂.6H₃O -Camallite.

Question 52.
What is difference between a double salt and a complex compound ?
Answer:
Double salts dissociate into simple ions completely when dissolved in water while complex compounds dissociate to give complex ion and the counter ions.

Question 53.
What is a ligand ?
Answer:
Ligand : A co-ordinating entity which is bound to the central atom by donating electron pairs is called a ligand.
Eg.: Cl^–, NH₃, CN^– etc.

Question 54.
Give one example each for ionic and neutral ligands.
Answer:
Examples for ionic ligands – CN^–, I^–, Cl^–
Examples for Neutral ligands – NH₃, H₂O

Question 55.
How many moles of AgCl is precipitated when 1 mole of CoCl₃ is treated with AgNO₃, solution ?
Answer:
3 moles of AgCl is precipitated by the reaction of 1 mole of CoCl₃ with AgNO₃ solution
3 AgNO₃ + CoCl₃ → CO(NO₃)₃ +3 AgCl ↓

Question 56.
What is a chelate ligand ? Give example.
Answer:
The ligands which can form two co-ordinate covalent bonds through two donar atoms are called bidentate ligands. These bidentate ligands are also called chelate ligands.
Eg.: C₂O₄^-2, CO₃;^-2 etc.

Question 57.
What is an ambidentate ligand ? Give example. [A.P. Mar. 16]
Answer:
A unidentate ligand containing two possible donor atoms can co-ordinate through either of donor atoms. Such ligands are called ambidentate ligands.
Eg : NO₂^–

Question 58.
CUSO₄.5H₂O is blue in colour where as anhydrous CuSO₄ is colourless. Why ?
Answer:
CUSO₄.5H₂O is blue in colour whereas anhydrous CuSO₄ is colourless because in the absence of ligand, crystal field splitting does not occurs.

Question 59.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test for Fe²⁺ ion but CuS04 mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu²⁺ ion. why ?
Answer:

• FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1 : 1 molar ratio gives the test for Fe²⁺ because of formation of double salt (FeSO₄(NH₄)₂ SO₄ 6H₂O Mohr’s salt).
• CuSO₄ mixed with aq.ammonia in 1 : 4 molar ratio does not give the test of Cu⁺² ion because of formation of complex compound [Cu(NH₃)₄]SO₄.

Question 60.
How many geometrical isomers are possible in the following coordination entities ?

1. [Cr(C₂O₄)₃]^3-
2. [CO(NH₃)₃Cl₃]

Answer:

1. [Cr(C₂O₄)₃]^3- : Two geometrical isomers are possible i) cis-isomer, ii) Trans-isomer.
2. [CO(NH₃)₃Cl₃] : Two geometrical isomers are possible i) cis-isomer, ii) Trans-isomer.

Question 61.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution copper sulphate ? Why ?
Answer:
When excess of aq. KCN is mixed with aq.CuSO₄ a complex named potassium tetra cyano cuprate (II) is formal

Question 62.
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic Why ?
Answer:
[Cr(NH₃)₆]³⁺ is paramagnetic due to the presence of three unpaired electrons.
[Ni(CN)₄]^2- is diamagnetic due to the absence of unpaired electrons.

Question 63.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Why ?
Answer:

• In the complex [Ni(H₂O)₆]²⁺, H₂O molecules are weak ligands they do not cause pairing.
  So the complex has two unpaired electrons, d-d-transitions takes place due to absorption of red light radiation and emission of green colour occurs.
• In the complex [Ni(CN)₄]^2-, CN^– ions are strong ligands they cause pairing. So there is no unpaired electrons so, no d-d transition. Hence it is colourless.

Question 64.
[Fe(CN)₄]^2- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why ?
Answer:
In the given complexes Fe has + 2 oxidation state with 3d⁶ outer electronic configuration. It has four unpaired electrons in presence of weak ligand H₂O. But in presence of strong ligand CN^– the electrons are paired up. Due to the difference in the no. of unpaired electrons both complex have different colours in dilute solutions.

Question 65.
What is the oxidation state of cobalt in (i) K[Co(CO)₄] and (ii) [Co(NH₃)₆]³⁺ ?
Answer:

1. K[CO(CO)₄] : 1 + x + 4(0) = 0, x = -1 .
2. [CO(NH₃)₆]³⁺ : x + 6(0) = + 3, x = + 3

Short Answer Questions

Question 1.
Compared to 3d series the corresponding transition metals of 4d and 5d transition series show high enthalpy of atomization. Explain.
Answer:
Compared to 3d-series the corresponding transition metals of 4d and 5d transition series show high enthalpy of atomisation.

Reason : This is due to the occurrence of much more frequent metal-metal bonding in compounds of heavy transition metals.

Question 2.
Compared to the changes in atomic and ionic sizes of elements of 3d and 4d series, the change in radii of elements of 4d and 5d series is virtually the same Comment.
Answer:
Compared to the change in atomic and ionic sizes of elements 3d and 4d series, the change in radii of elements 4d and 5d series is virtually the same.

This is due to the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbitals results in a regular decrease in atomic radii called lanthanoid contraction. The consequence of lanthanoid contraction is that the 4d and 5d series exhibit same size of radii.

Question 3.
Account for the zero oxidation state of Ni and Fe in [Ni(CO)₄] and [Fe(CO)₅] respectively.
Answer:
In [Ni(CO)₄] and [Fe(CO)₅] the oxidation state of Ni and Fe is zero.
These low oxidation states found when the complex compound has ligands capable of π-acceptor character in addition to the σ-bonding.

Question 4.
Why do the transition metal ions exhibit characteristic colours in aqueous solution. Explain giving examples.
Answer:
Colour property of transition metal ions in aqueous solution is due to the presence of unpaired d-electrons. These unpaired d-electrons from a lower energy level of a metal ion in a complex is excited to a higher energy d-orbital of the same n value. The energy of excitation corresponds to the frequency of light absorbed and this frequency lies in visible region. The colour observed corresponds to the complementary colour of light obsorbed. The frequency of light absorbed is determined by the nature of the ligand. In aqueous solutions water molecules are ligands the colour of the ions observed are listed in the following table.
Colours of Some of the First Row (aquated) Transition Metal Ions

Question 5.
Explain the catalytic action of Iron(DI) in the reaction between I arid S₂O₈^2- ions.
Answer:
Transition metal ions can change their oxidation states and become more effective catalysts. Iron (III) Catalyses the reaction between iodide and per sulphate ions

Catalytic action explained as follows
2Fe⁺³ + 2I^– → 2Fe⁺² + I₂
2Fe⁺² + S₂O₈^-2 → 2Fe⁺³ + 2SO₄^-2

Question 6.
What are interstitial compounds ? How are they formed ? Give two examples.
Answer:
The compounds which are formed when small atoms like H, C or N are trapped inside the crystal lattices of metal are called interstitial compounds.
Eg : TiC, Fe₃H, VH_0.56 etc.

• These are non stoichiometric and are neither typically ionic nor covalent.
• They have high melting points, higher than of pure metals.
• They are very hard, some borides approach diamond in hardness.
• They retain metallic conductivity.
• They are chemically inert.

Question 7.
Write the characteristics of interstitial compounds.
Answer:
Characteristics of interstitial compounds :

• These are non stoichiometric and are neither typically ionic nor covalent.
• They have high melting points, higher than of pure metals.
• They are very hard, some borides approach diamond in hardness.
• They retain metallic conductivity.
• They are chemically inert.

Question 8.
Write the characteristic properties of transition elements. [A.P. Mar. 15]
Answer:
Transition elements exhibits typical characteristic properties.

• Electronic configurations.
• Para and ferro magnetic properties.
• Alloy forming ability
• Complex forming ability.
• Interstitial compounds.
• Variable oxidation states.
• Formation of coloured hydrated ions.
• Catalytic property.
• Metallic character.

Question 9.
Write down the electronic configuration of ‘

1. Cr³⁺
2. Cu⁺
3. Co²⁺
4. Mn²⁺

Answer:

1. Cr³⁺ electronic configuration is [Ar] 4s⁰3d³
2. Cu⁺ electronic configuration is [Ar] 4s⁰3d¹⁰
3. Co²⁺ electronic configuration is [Ar] 4s⁰3d⁷
4. Mn²⁺ electronic configuration is [Ar] 4s⁰3d⁵

Question 10.
What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms : 3d³ 3d⁵ 3d⁸ and 3d⁴ ?
Answer:

1. 3d³-stable oxidation states are +2,+3,+4 and+5 (V)
2. 3d⁵-stable oxidation states are +3, +4 and +6 (Cr)
3. 3d⁵-stable oxidation states are +2, +4, +6 and +7 (Mn)
4. 3d⁸-stable oxidation states are +2, +3 (Co)
5. 3d⁴-This configuration does not exist.

Question 11.
What is lanthanoid contraction ? What are the consequences of lanthanoid contraction?
Answer:
Lanthanoid contraction : The overall decrease in atomic and ionic radii from lanthanum to leutetium is observed in the lanthanoids. This phenomenon is called lanthanoid contraction. It is due to the fact that with every additional proton in the nucleus, the corresponding electron goes into a 4f-subshell. This is too diffused to screen the nucleus as effectively as the more localised inner shell. Hence, the attraction of the nucleus for the outermost electrons increases steadily with the atomic number.

Consequences of Lanthanoid Contraction : The important consequences of lanthanoid contraction are as follows :
i) Basic character of oxides and hydroxides : Due to the lanthanoid contraction, the covalent nature of La-OH bond increases and thus, the basic character of oxides and hydroxides decreases from La(OH)₃ to Lu(OH)₃.
ii) Similarity in the size of elements of second and third transition series : Because of
lanthanoid contraction, elements which follow the third transition series are considerably smaller than would otherwise be expected. The normal size increases from Sc → Y → La and disappears after lanthanides. Thus, pairs of elements such as Zr/Hf, Nb/Ta and Mo/w are almost identical in size.
Due to almost similar size, such pairs have very sijnilar properties which makes their separation difficult.

iii) Separation of lathanoids : Due to lanthanoid contraction, there is a difference in some properties of lanthanoid like solubility, degree of hydration and complex formation. These difference enable the separation of lanthanoids by ion exchange method.

Question 12.
How is the variability in oxidation states of transition metals different from that of the non transition metals ? Illustrate with examples.
Answer:

• In transition elements the oxidation states vary by unity (due to incomplete filling of d- orbitals)
  Eg : Mn exhibits +2, +3, +4, +5, +6 and +7 all differing by 1.
• In non-transition element, this variation is selective, always differing by 2.
  Eg : S exhibits 2, 4, 6 oxidation states. N exhibits 3, 5 etc.

Question 13.
Describe the preparation of potassium dichromate from iron chromite ore.
Answer:
Preparation of K₂Cr₂O₇ from chromite ore :
i) a) Potassium dichromate is obtained by the fusion of chromite ore (FeCr₂O₄) with sodium (or potassium) carbonate in the presence of excess of air.

b) The solution is filtered and treated with sulphuric acid.

c) Now sodium dichromate is treated with potassium chloride. As a result, potassium dichromate is produced.

ii) Effect of increasing pH on K₂Cr₂O₇ solution :

On increasing pH, K₂Cr₂O₇ changes into K₂CrO₄ (orange to yellow).

Question 14.
Describe the oxidising action of potassium dichromate and write the ionic equations for its.
With (i) iodide (ii) iron (II) solution (iii) H₂S and (iv) Sn(II)
Answer:
Potassium dichromate is a strong oxidising agent. In acidic solution, its oxidising action may be represented as :
Cr₂O₇^2- + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O (E° = 1.33 V)
Ionic reactions :

1. Reaction of K₂Cr₂O₇ with I^–
   Cr₂O₇^2- + 14H⁺ + 6I^– → 2Cr³⁺ + 3I₂ + 7H₂O
2. Reaction of K₂Cr₂O₇ with Fe²⁺ (aq)
   Cr₂O₇^2- + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
3. Reaction of K₂Cr₂O₇with H₂S
   Cr₂O₇^2- + 8H⁺ + 3H₂S → 2Cr⁺³ + 3S + 7H₂O

Question 15.
Describe the preparation of potassium permanganate.
Answer:
Preparation of KMnO₄ Potassium permanganate is prepared by the fusion of MnO₂ with an alkali metal hydroxide and an oxidising agent like KNO₃. It forms dark green, K₂MnO₄ which disproportionates in a neutral or acidic solution to give permanganate.

Question 16.
How does the acidified permanganate solution react with

1. iron (II) ions
2. SO₂ and
3. oxalic acid.

Write the ionic equations for the reactions.

Answer:
Reactions of KMnO₄ in acidic medium
MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O …………. (i)
1) Iron (II) ions : Ferrous is oxidised to ferric.
[Fe²⁺ → Fe³⁺ + e] × 5 ………………. (ii)
From equation (i) and (ii) 5Fe²⁺ + MnO₄^– + 8H+ → Mn²⁺ + 4H₂O + 5Fe³⁺

2) SO₂ : It is oxidised to SO₄^2- by acidified KMnO₄.
SO₂ + 2H₂O → SO₄^2- + 4H⁺ + 2e^– ……………….(iii)
From equation (i) and (iii) .
5SO₂ + 2MnO₄^– + 2H₂O → 2Mn²⁺ + 4H⁺ + 5SO₄^2-

3) Oxalic acid : C₂O₄^2- → 2CO₂ + 2e^– ……………… (iv)
From equation (1) and (iv)
5C₂O₄^2- + 2MnO₄^– + 16H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂

Question 17.
Predict which of the ions Cu⁺, Sc³⁺, Mn²⁺, Fe²⁺ are coloured in aqueous solution ? Give reasons.
Answer:
Only those ions will be coloured which have incomplete d-orbitals. Ions which has complete or vacant d-orbitals are colourless.

As Sc³⁺ and Cu⁺ have 3d⁰ and 3d¹⁰ configuration in their valence shell so their aqueous solutions are colourless. All others, i.e., Ti³⁺, V³⁺, Mn²⁺ Fe⁺² and Co²⁺ are coloured in aqueous medium.

Question 18.
Compare the stability of +2 oxidation state of the elements of the first transition series.
Answer:
Element (+2 state) ₂₁Sc²⁺ Ti²⁺ _{22 22}V²⁺ ₂₄Cr²⁺ ₂₅Mn²⁺,
Electronic configuration 3d¹ 3d² 3d³ 3d⁴ 3d⁵
In all the elements listed, the removal of two 4s. Electrons (in Cr²⁺, i.e., from 4s and 1e^– from 3d), the 3d-orbitais get gradually occupied. Since, the number of empty d-orbitais decreases or the number unpaired electrons in 3d-orbitais increases with increase in atomic number of cations, so the stability of the cations (M²⁺) increases from Sc²⁺ to Mn²⁺.

Question 19.
Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion and calculate its magnetic moment on the basis of ‘spin-only’ formula.
Answer:
Ce(Z = 58) = [Xe] = 4f¹ 5d¹ 6s²
Ce³⁺ = [Xe]4f¹ (only one unpaired electron)
By ‘spin-only’ formula, Magentic moment of Ce³⁺(μ) = \(\sqrt{n(n+2)}\)
[∵ n = 1 (unpaired electron)] = \(\sqrt{1(1+2)}\) = \(\sqrt{3}\) = 1.73 BM

Question 20.
Write down the number of 3d electrons in each of the following ions : Ti²⁺, V²⁺ Cr³⁺ and Mn²⁺ Indicate how would you expect the five 3d orbitais to be occupied for these hydrated ions (octahedral).
Answer:

Question 21.
Explain Werner’s theory of coordination compounds with suitable examples. [A.P. Mar. 18; T.S. Mar. 18, 15] [Mar. 14]
Answer:
Werner’s theory:
Postulates :
1) Every complex compound has a central metal atom (or) ion.

2) The central metal shows two types of valencies namely primary valency and secondary valency.

A) Primary valency: The primary valency is numerically equal to the oxidation state of the
metal. Species or groups bound by primary valencies undergo complete ionization. These valencies are identical with ionic bonds and are non-directional. These valencies are represented by discontinuous lines (…..)
Eg- : CoCl₃ contains Co³⁺ and 3Cl^– ions. There are three Primary Valencies or three ionic bonds.

B) Secondary Valency : Each’ metal has a characteristic number of Secondary Valencies. They are directed in space around the central metal.
The number of Secondary Valencies is called Coordination numbe (C.N.) of the metal. These valencies are directional in Nature.
For example in CoCl₃. 6NH₃
Three Cl^– ions are held by primary Valencies and 6NH₃ molecules are held by Secondary Valencies. In CuSO₄ . 4NH₃ complex SO₄^2- ion is held by two Primary Valencies and 4NH₃ molecules are held by Secondary Valencies.

3) Some negative ligands, depending upon the complex, may satisfy both primary and secondary valencies. Such ligands, in a complex, which satisfy both primary as well as secondary valencies do not ionize.

4) The primary valency of a metal is known as its outer sphere of attraction or ionizable valency while the Secondary valencies are known as the inner sphere of attraction or coordination Sphere. Groups bound by secondary valencies do not undergo ionization in the complex.

Example to Clarify Werner’s Theory

Question 22.
Give the geometrical shapes of the following complex entities

1. [Co(NH₃)₆]³⁺
2. [Ni(CO)₄]
3. [Pt Cl₄]^2- and
4.  [Fe(CN)₈]^4-.

Answer:

1. Geometrical shape of [Co(NH₃)₆]³⁺ is octahedral
2. Geometrical shape of [Ni(CO)₄] is tetrahedral
3. Geometrical shape of [Pt Cl₄]^2- is square planar
4. Geometrical shape of [Fe(CN)₈]^4- is octahedral

Question 23.
Explain the terms
(i) Ligand
(ii) Coordination number
(iii) Coordination entity
(iv) Central metal atom/ion.
Answer:
i) Ligand : The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be
a) simple ions such as Cl^–
b) small molecules such as H₂O or NH₃
c) large molecules such as H₂NCH₂CH₂NH₂ or N(CH₂CH₂NH₂)₃ or even (d) macromolecules, such as proteins.
On the basis of the number of donor atoms available for coordination, the ligands can be classified as :
a) Unidentate : One donor atom, Eg.:
b) Bidentate : Two donor atoms, Eg. : H₂NCH₂CH₂NH₂
(ethane-1, 2-diamine or en), C₂O₄^2- (oxalate), etc.
c) Polydentate : More than two donor atoms, Eg. : N(CH₂CH₂NH₂)₃ EDTA, etc.

ii) Coordination number: The coordination number (CN) of metal ion in a complex can be defined as the number of ligands or donor atoms to which the metal is directly bonded.
Eg : In [PtCl₆]^2-, CN of Pt = 6, In [Ni(NH_{3/sub>)4]²⁺, CN of Ni 4.}

iii) Coordination entity: A central metal atoms or ion bonded to a fixed number of molecules or ions (ligands) is known as coordination entity. For example [CoCl₃(NH₃)₃. Ni(CO)_{4/sub>], etc.}

iv) Central metal atom/ion : In a coordination entity, the atom/ion to which a fixed no. of ions/groups are bound in a definite geometrical arrangement around it is called central metal atom or ion. Eg: K₄[Fe(CN)₆] ‘Fe’ is central metal.

Question 24.
Explain the terms (i) Unidentate ligand (ii) bidentate ligand (iii) polydentate ligand and (iv) ambidentate ligand giving one example for each.
Answer:
i) Unidentate : The negative ion or neutral molecule having only one donor atom is called unidentate ligand
Eg :
ii) Bidentate (or didentate) : The ions or molecules having two donor atoms are called bidentate ligands.
Eg:
iii) Polydentate ligands : Ligands having more than one donor atom in the coordinating group and are capable of forming two or more coordinate bonds with same central atom simultaneously are called poly dentate ligands.
Eg : C₂O₄^-2.
iv) Ambidentate: Ligand which can ligate through two different atoms is called ambidentate ligand. Eg: NO₂^–, SCN^–ions, NO₂^– ion can coordinate either through nitrogen or through oxygen to a central metal atom/ion. Similarly, SCN^– ion can coordinate through the sulphur or nitrogen atom.

Question 25.
What is meant by chelate effect ? Give example.
Answer:
When a didentate or a polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal ion, a 5-or 6-membered ring is formed, the effect is known as chelate effect. Example

Question 26.
Give the oxidation numbers of the central metal atoms in the following complex entities
(i) [Ni(CO)₄]
(ii) [CO(NH₃)₆]³⁺
(iii) [Fe(CN)₆]^4- and
(iv) [Fe(C₂O₄)₃]^3-
Answer:
i) [Ni(CO)₄] :
x + 4(0) = 0
x = 0
‘Ni’ oxidation state is zero.

ii) [CO(NH₃)₆]³⁺
x + 6(0) = + 3
x = +3
‘Co’ oxidation state is + 3.

iii) [Fe(CN)₆]^4-
x + 6(-1) = -4
x = + 2
‘Fe’ oxidation state is + 2.

iv) [Fe(C₂O₄)₃]^3-
x + 3(-2) = -3
x = + 3
‘Fe’ oxidation state is + 3.

Question 27.
Using IUPAC norms write the formulas for the following.

1. Tetrahydroxozincate (II)
2. Hexaamminecobalt (III) sulphate
3. Potassium tetrachloropalladate (II) and
4. Potassium tri(oxalato) chromate (III)

Answer:

1. Tetrahydroxozincate (II) – [Zn(OH)₄]^-2
2. Hexa ammine cobalt (III) sulphate – [Co(NH₃)₆]₂ (SO₄)₃
3. Potassium tetrachloropalladate – K₂[PdCl₄]
4. Potassium tri(oxalato) chromate (III) – K₃[Cr(C₂O₄)₃]

Question 28.
Using IUPAC norms write the systematic names of the following. [A.P. Mar 19]

1. [CO(NH₃)₆]Cl₃
2. [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
3. [Ti(H₂O)₆]³⁺ and
4. [NiCl₄]^2-

Answer:

1. Hexa ammine cobalt (III) chloride
2. Diammine chlorido (methyl amine) platinum (II) chloride
3. Hexa aquo titanium (III) ion
4. Tetra chloro nickelate (III) ion

Question 29.
Explain geometrical isomerism in Co-ordination compounds giving suitable examples. [A.P. Mar 19]
Answer:

• Geometrical isomerism arises in Co-ordination complexes due to different possible geometric arrangements of the ligands.
• This isomerism found in complexes with Co-ordination numbers 4 and 6.
• In a square planar complex of formula [MX₂L₂] the two ligands may be arranged in adjacent to each other in a cis isomer (or) opposite to each other in a trans isomer.
  
• Square planar complex of type [MAB XL] shows three isomers two-cis and one trans. (A, B, X, L are unidentate ligands is square planar complex).
• Geometrical isomerism is not possible in tetrahedral geometry.
• In octahedral complexes of formula [MX₂L₄] in which two ligands X may be oriented cis or trans to each other.
  
• Another type of geometrical isomerism occurs in octahedral Co-ordination compounds of type [Ma₃b₃] if three donar atoms of the same ligands occupy adjacent positions at the corners of an octahedral face then it is facial (fac) isomer. When the positions occupied are around the meridian of the octahedran then it is meridonial (mer). isomer.
  

Question 30.
What are homoleptic and heteroleptic complexes ? Give one example for each.
Answer:
Homoleptic complexes : These are the complexes in which a metal is bound by only ore kind of ligands.
eg.: [Co(NH₃)₆]³⁺
Heteroleptic complexes: These are the complexes in which a metal is bound by more than one kind of ligends. eg.: [Co(NH₃)₄Cl₂]⁺

Long Answer Questions

Question 1.
Explain giving reasons :
(i) Transition metals and many of their compounds show paramagnetic behaviour.
(ii) The enthalpies of atomisation of the transition metals are high.
(iii) The transition metals generally form coloured compounds.
(iv) Transition metals and their many compounds act as good catalysts.
Answer:
i) Transition elements have unpaired electrons. Each unpaired electron has a magnetic moment associated with its spin angular momentum and orbital angular momentum. This is the reason of paramagnetism in transition metals.

ii) The reason for the high enthalpy of atomisation is the presence of large number of unpaired electrons in their atoms. These atoms have strong interatomic interaction and hence, stronger bonding between them.

iii) Formation of coloured compounds by transition metals is due to partial adsorption of visible light. The electron absorbs the radiation of a particular frequency (of visible region) and jumps into next orbital.

iv) Catalysts, at the solid surface, involve the formation of bonds between reactants molecules and atoms of the surface of the catalyst (I row transition metals utilise 3d and 48- electrons for bonding). This has the effect of increasing the concentration of the reactants at the catalyst surface and also lowering of the activation energy.
Transition metal ions show variable oxidation states, so they are effective catalysts.

Question 2.
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with
(i) iron (II) ions
(ii) SO₂ and
(iii) oxalic acid ?
Write the ionic equations.
Answer:
Preparation of KMnO₄ Potassium permanganate is prepared by the fusion of MnO₂ with an alkali metal hydroxide and an oxidising agent like KNO₃. It forms dark green, K₂MnO₄ which disproportionates in a neutral or acidic solution to give permanganate.

Reactions of KMnO₄ in acidic medium
MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O …………. (i)
i) Iron (II) ions : Ferrous is oxidised to ferric.
[Fe²⁺ → Fe³⁺ + e] × 5 ………………. (ii)
From equation (i) and (ii) 5Fe²⁺ + MnO₄^– + 8H+ → Mn²⁺ + 4H₂O + 5Fe³⁺

ii) SO₂ : It is oxidised to SO₄^2- by acidified KMnO₄.
SO₂ + 2H₂O → SO₄^2- + 4H⁺ + 2e^– ……………….(iii)
From equation (i) and (iii) .
5SO₂ + 2MnO₄^– + 2H₂O → 2Mn²⁺ + 4H⁺ + 5SO₄^2-

iii) Oxalic acid : C₂O₄^2- → 2CO₂ + 2e^– ……………… (iv)
From equation (1) and (iv)
5C₂O₄^2- + 2MnO₄^– + 16H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂

Question 3.
Compare the chemistiy of actinoids with that of the lanthanoids with special reference to :
(i) electronic configuration (ii) oxidation state (iii) atomic and ionic sizes and (iv) chemical reactivity.
Answer:

Question 4.
How would you account for the following.
(i) Of the d4 species, Cr²⁺ is strongly reducing while manganese (HI) is strongly oxidising.
(ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised,
(iii) The d₁ configuration is very unstable in ions.
Answer:
i) E° value of Cr³⁺/Cr²⁺ is negative (-0.41 V) while that of Mn³⁺/Mn²⁺ is positive (+1.57 V). This means that Cr²⁺ ions can lose electrons to form Cr³⁺ ions and act as a reducing agent while Mn³⁺ ions can accept electrons and can act as an oxidising agent.

ii) Cobalt (III) ion has greater tendency to form complexes than cobalt (II) ion. Therefore, Co (II) ion, being stable in aqueous solution, changes to Co (III) ion, in the presence of complexing reagents and gets oxidised.

iii) Ions of transition metals with d1 configuration tend to lose one electron to acquire d⁰ configuration that is quite stable. Therefore, such ions (with d¹) undergo either oxidation or disproportionation hence unstable.

Question 5.
Give examples and suggest reasons for the following features of the transition medals.
(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.
Answer:
i) Acidic strength of oxides increases with the increase in oxidation state of the element Eg.: MnO(Mn²⁺) is basic whereas Mn₂O₇(Mn⁷⁺) is acidic in nature.

ii) Both oxygen and fluorine being highly electronegative can increase the oxidation state of a particular transition metal. In certain oxides, the element oxygen is involved in multiple bonding with the metal .and this is responsible for the higher oxidation state of the metal.

iii) This is also due to high electronegativity of oxygen eg.: chromium exhibits oxidation states of +6 in oxoanion [CrO₄]^2- and manganese shows oxidation state of +7 in oxoanion [MnO₄]^–

Question 6.
Compare the chemistry of the actinoids with that of lanthanoids with reference to : (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity
Answer:

Question 6.
Explain IUPAC nomenclature of Co-ordination compounds with suitable examples.
Answer:
IUPAC nomenclature : The formula of a compound is an abreviated description of the constitution of the compound. The fpllowing rules are prescribed by IUPAC for naming of Coordination compounds.
i) Positive ions are named first followed by negative ions,
eg.: Potassium hexacyanoferrate (II), K₄[Fe(CN)₆]

ii) Within the Co-ordination sphere ligands are named before the metal atom/ion. However, in formulae, metal ion is written first.
eg.: Tetraammine copper (II) sulphate [Cu(NH₃)₄]SO₄

iii) Prefixes are used to denote the number of same ligands that the present in the Co¬ordination sphere. Complex ions are denoted in paranthesis ( ) and prefixed by bis, tris etc.
Examples:

eg. : [CO(NH₂CH₂CH₂NH₂)Cl₂] Cl is named as dichloro bis (ethylenediamine) cobalt (III) chloride.

iv) Ligands are named in alphabetical order.
eg.: [PtCl₂(NH₃)₂ diammine dichloro platinum (II)

v) Anionic ligands are denoted by a suffix ‘O’ and neutral ligands are denoted by their original names.
eg.: Cl^– – Chloro, CN^– – Cyano
Exception for the above are indicated below.

vi) Oxidation state of the metal ion is indicated by Roman numerical in parenthesis, eg.: [Ag(NH₃)₂] [Ag(CN)₂] is named as diammine silver (I) dicyanoargentate (I).

vii) If the charge of the Co-ordination entity is negative, the name of the metal ends with a suffix-ate.
eg.: [CO(SCN)₄]^2- – tetrahiocyanato cobaltate (II)
Some metal ions are denoted by their names from which their symbols are derived
eg.: Fe – ferrate
Pb – plumbate
Sn – stannate .
Ag-argentate
Au-aurate

viii) Prefixes cis – and trans are used to designate adjacent and opposite geometric locations of the ligands, in a complex.
eg. :

ix) Bridging ligands between two metal ions in a Co-ordination entity are denoted by prefix p(greek letter’mu).
eg.: [(NH₃)₄ CO(OH) (NH₂)Co(NH₃)₄]⁺ is named as μ-amido-μ hydroxo bis (tetraammine) cobalt (IV) .

1. Tetrahydroxozincate (II) – [Zn(OH₄]^-2
2. Hexa ammine cobalt (III) sulphate – [Co(NH₃)₆]₂ (SO₄)₃
3. Potassium tetrachloropalladate – K₂[PdCl₄)
4. Potassium tri(oxalato) chromate (III) – K₃[Cr(C₂O₄)₃]

Question 7.
Explain different types of isomerism exhibited by Co-ordination compounds, giving suitable examples.
Answer:
Isomerism in Co-ordination compounds : Isomers are compounds that have the same chemical formula but different arrangement of atoms. Two principal types of isomerism are known among Co-ordination compounds namely stereo isomerism and structural isomerism.
a) Stereoisomengni: Stereoisomerism is a form of isomerism in which two substances have the same composition and structure but differ in the relative spatial positions of the ligands. This can be sub divided into two classes namely.

1. Geomethcal isomerism and
2. optical isomerism

b) Structural isomerism:

1. Linkage isomerism
2. Co-ordination isomerism
3. Ionisation isomerism
4. Hydrate isomerism

a) (1) Geometrical isomerism:

• Geometrical isomerism arises in Co-ordination complexes due to different possible geometric arrangements of the ligands.
• This isomerism found in complexes with Co-ordination numbers 4 and 6.
• In a square planar complex of formula [MX₂L₂] the two ligands may be arranged in adjacent to each other in a cis isomer (or) opposite to each other in a trans isomer.
  
• Square planar complex of type [MAB XL] shows three isomers two-cis and one trans. (A, B, X, L are unidentate ligands is square planar complex).
• Geometrical isomerism is not possible in tetrahedral geometry.
• In octahedral complexes of formula [MX₂L₄] in which two ligands X may be oriented cis or trans to each other.
  
• Another type of geometrical isomerism occurs in octahedral Co-ordination compounds of type [Ma₃b₃] if three donar atoms of the same ligands occupy adjacent positions at the corners of an octahedral face then it is facial (fac) isomer. When the positions occupied are around the meridian of the octahedran then it is meridonial (mer). isomer.
  

a(2) Optical isomerism : Optical isomerism arises when two isomers of a compound exist such that one isomer is a mirror image of the other isomer. Such isomers are called optical isomers or enantiomers. The molecules or ions that cannot be superimposed are called chiral. The two forms are called dextro (d) and laevo (1) depending upon the direction they rotate the plane of polarised light in a polarimeter (d rotates to the right, l to the left). Optical isomerism is common in octahedral complexes involving bidentate ligands.

b(1) Linkage isomerism: Linkage isomerism arises in Co-ordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocyanate ligand-NCS^–, which may bind through the nitrogen to give M-NCS or through sulphur to give M-SCN.
eg. : [Mn(CO)₅SCN] and [Mn(CO)₅NCS]

(2) Co-ordinate isomerism : This type isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex.
eg. : [CO(NH₃)₆] [Cr(CN)₆] and [Co(CN)₆] [Cr(NH₃)₆]

(3) Ionisation isomerism: This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion
eg. : [CO(NH₃)₅SO₄] Br and [Co(NH₃)₅Br]SO₄

(4) Hydrate isomerism : This form of isomerism is known as ‘hydrate isomerism since water is involved as a solvent. Hydrate isomers differ by whether or not a hydrate molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice.

Question 8.
Discuss the nature of bonding and magnetic behaviour in the following Co-ordination entities on the basis of valence bond theory.
(i) [Fe(CN₆)]^4-
(ii) [FeF₆]^3-
(iii) [Co(C₂O₄)₃]^3- and
(iv) [CoF₆]^3-
Answer:
i) Fe(CN₆)]^4- : In this complex Fe is present as Fe²⁺.
Fe = [Ar] 3d⁶4s²
Outer configuration of Fe²⁺ = 3d⁶4s⁰

CN being strong field ligand, pair up the unpaired d electrons Thus, two 3d-orbital are now available for CN^– ions.

Since, all the electrons are paired, the complex is diamagnetic. Moreover (n – 1) d- orbitals are involved in bonding, so, it is an inner orbital or low spin complex.

ii) [FeF₆]^3- : In this complex, the oxidation state of Fe is + 3.
Fe³⁺ = 3d⁵4s⁰

F- is not a strong field ligand. It is a weak field ligand, so no pairing occurs. Thus, 3d- orbitals are not available to take part in bonding.

Because of the presence of five unpaired electrons, the complex is paramagnetic. Moreover, nd-orbitals are involved in bonding, so it is an outer orbital or high spin complex.

iii) [Co(C₂O₄)₃]^3- : In this complex, the oxidation state of Co is + 3.
Outer configuration of Co = 3d⁷ 4s²

Oxalate ion being a strong field ligand pair up the 3d electrons, thus two out of the five 3d-orbitals are available for oxalate ions.

Since, all the electrons are paired, this complex is diamagnetic. It is an inner orbital complex because of the involvement of (n – 1) d-orbital for bonding.

iv) [CoF₆]^3- : In this complex, Co is present as Co³⁺.
Outer configuration of Co³⁺ = 3d⁶ 4s⁰

Because of the presence of four unpaired electrons, the complex is paramagnetic. Since, nd orbitals take part in bonding, it is an outer orbital complex or high spin complex.

Question 9.
Sketch the spliting of d orbitals in an octahedral crystal field.
Answer:

Question 10.
What is spectrochemical series ? Explain the difference between a weak field ligand and a strong field ligand.
Answer:
The arrangement of ligands in order of their increasing field strengths, i.e., increasing crystal field splitting energy (CFSE) values is called spectrochemical series.

The ligands with a small value of CFSE (∆₀) are called weak field ligands. For such ligands ∆₀ < P where P is the pairing energy. Whereas the ligands with a large value of CFSE are called strong field ligands. In case of such ligands ∆₀ > P.

When ligands approach a transition metal ion, the d-orbitals split into two sets (t_2g and e_g), one with lower energy and the other with higher energy. The difference of energy between the two sets of orbitals is called crystal field splitting energy, ∆₀ for octahedral field (and ∆_t for tetrahedral field)

If ∆₀ < P (pairing energy), the 4^th electron enters one of the e_g orbitals giving the configuration \(t_{2 g}^{3} e_{g}^{1}\) thereby forming high spin complex. Such ligands for which ∆₀ < P are known as weak field ligands.

If ∆₀ > P, the 4^th electron pairs up in one of the t_2g orbitals giving the configuration \(\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{0}\), thus forming low spin complexes. Such ligands for which ∆₀ > P are called strong field ligands.

Question 11.
Discuss the nature of bonding in metal carbonyls.
Answer:
The metal-carbon bond in metal carbonyls have both sigma (σ) and pi (π) characters. The metal-carbon σ-bond is formed by the donation of lone pair of electrons of the carbonyl carbon to a vacant orbital of the metal. The metal- carbon re-bond is formed by the donation of a pair of electrons from a filled d-orbital of metal into the vacant

Question 12.
Explain the applications of Co-ordination compounds in different fields.
Answer:
Applications of Co-ordination compounds : Co-ordination compounds are of great importance. These compounds are widely present in the mineral, plant and animal worlds and are known to play many important functions in the area of analytical chemistry, metallurgy, biological systems, industry and medicine. These are described below.

Co-ordination compounds find use in many qualitative and quantitative chemical analysis. The familiar colour reactions given by metal ions with a number of ligands (especially chelating ligands), as a result of formation of co-ordination enitites, form the basis for their detection and estimation by classical and instrumental methods of analysis. Examples of such reagents include EDTA, DMG (dimethylglyoxime), oc-nitroso-p-naphtol, cupron etc.

Hardness of water is estimated by simple titration with Na₂EDTA. The Ca²⁺ and Mg²⁺ ions form stable complexes with EDTA. The selective estimation of these ions can be done due to difference in the stability constants of calcium and magnesium complexes.

Some important extraction processes of metals, like those of silver and gold, make use of complex formation. Gold for example, combines with cyanide in the presence of oxygen and water to form the co-ordination entity [Au(CN)₂]^– in aqueous solution. Gold can be separated in metallic from this solution by the addition of zinc.

Similarly, purification of metals can be achieved through formation and subsequent decomposition of their coordination compounds. For example, impure nickel is converted to [Ni(CO)₄], which is decomposed to yield pure nickel.

Co-ordination compounds are of great importance in biological systems. The pigment responsible for photosynthesis, chlorophyll, is a co-ordination compound of magnesium. Haemoglobin, the red pigment of blood which acts as oxygen carrier is a co-ordination compound of iron. Vitamin B₁₂, cyanocobalamine, the anti-pernicious anaemia factor, is a coordination compound of cobalt. Among the other compounds of biological importance with coordinated metal ions are the enzymes like, carboxypeptidase A and carbonic anhydrase (catalysts of biological systems).

Co-ordination compounds are used as catalysts for many industrial processes. Examples include rhodium complex, [(Ph₃P)₃RhCl], a Wilkinson catalyst, is used for the hydrogenation of alkenes.

Articles can be electroplated with silver and gold much more smoothly and evenly fr6m solutions of the complexes, [Ag(CN)₂]^– and. [Au(CN)₂]^– than from a solution of simple metal ions.

In black and white photography, the developed film is fixed by washing with hypo solution which dissolves the undecomposed AgBr to form a complex ion, [Ag(S₂O₃)₂]^3-.

There is growing interest in the use of chelate therapy in medicinal chemisty. An example is the treatment of problems caused by the presence of metals in toxic proportions in plant/animal systems. Thus, excess of copper and iro are removed by the chelating ligands D-penicillamine and desferrioxime B via the formation of coordination compounds. EDTA is used in the treatment of lead poisoning. Some co-ordination compounds of platinum effectively inhibit the growth of tumours. Examples are : cis-platin and related compounds.

Textual Examples

Question 1.
On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not ?
Solution:
On the basis of incompletely filled 3d orbitals in case of scandium atom in its atomic state (3d¹), it is regarded as a transition element. On the other hand, zinc atom has completely filled d orbitals (3d¹⁰) in its ground state as well as in its oxidised state, hence it is not regarded as a transition element.

Question 2.
Why do the transition elements exhibit higher enthapies of atomisation ?
Solution:
Because of large number of unpaired electrons in their atoms they have stronger interatomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation.

Question 3.
Name at transition element which does not exhibit variable oxidation states.
Solution:
Scandium (Z = 21) does not exhibit variable oxidation states.

Question 4.
Why is Cr²⁺ reducing and Mn³⁺ oxidising when both have d4 configuration ?
Solution:
Cr2²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g level. On the other hand; the change from Mn²⁺ to Mn³⁺ results in the half-filled (d<sup5) configuration which has extra stability.

Question 5.
How would you account for the increasing oxidising power in the series VO₂⁺ < Cr₂O₇^2- < MnO₄^–?
Solution:
This is due to the increasing stability of the lower species to which they are reduced.

Question 6.
For the first row transition metals the E^⊖ values are

Explain the non-regularity in the above values.
Solution:
The E^⊖ (M²+/M) values are not regular which can be explained from the non-regular variation of ionisation enthalpies (∆_iH₁ + ∆_iH₂) and also the sublimation enthalpies which are relatively much less for manganese and vanadium.

Question 7.
Why is the E^⊖ value for the Mn³⁺/Mn²⁺ couple much more positive than that for Cr³⁺/ Cr²⁺ or Fe³⁺/Fe²⁺ ? Explain.
Solution:
Much larger third ionisation energy of Mn (where the required change is d⁵ to d⁴) is mainly responsible for this. This also explains why the +3 state of Mn is of little importance.

Question 8.
Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25. [A.P. Mar. 16]
Solution:
With atomic number 25, the divalent ion in aqueous solution will have d⁵ configuration (five unpaired electrons). The magnetic moment, μ is μ = \(\sqrt{5(5+2)}\) = 5.92 BM

Question 9.
What is meant by ‘disproportionation’ of an oxidation state ? Give an example.
Solution:
When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation. For example, manganese (VI) becomes unstable relative to manganese (VII) and manganese (TV) in acidic solution.
3Mn^VIO₄^2- + 4H⁺ → 2Mn^VIIO₄^– + Mn^IVO₂ + H₂O

Question 10.
Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
Solution:
Cerium (Z = 58)

Question 11.
On the basis of the following observations made with aqueous solutions, assign secondary valencies to metals in the following compounds.

Solution:
i) Secondary 4 ii) Secondary 6 iii) Secondary 6 iv) Secondary 6 v) Secondary 4

Question 12.
Write the formulas for the following Co-ordination compounds
a) Tetraammineaquachloro cobalt (III) chloride
b) Potassium tetrahydroxozincate (II)
c) Potassium trioxalatoaluminate (III)
d) Dichlorobis (ethane-1, 2-diamine) cobalt (III)
e) Tetracarbonylnickel (0)
Solution:
a) [Co(NH₃)₄(H₂O)Cl]Cl₂
b) K₂[Zn(OH)₄]
c) K₃[Al(C₂O₄)₃]
d) [CoCl₂(en)₂]⁺
e) [Ni(CO)₄]

Question 13.
Write the IUPAC names of the following coordination compounds.
a) [Pt(NH₃)₂Cl(NO₂)]
b) K₃[Cr(C₂O<sub4)₃]
c) [CoCl₂(en)₂]Cl
d) [Co(NH₃)₅(CO₃)]Cl
e) Hg[Co(SCN)₄]
Solution:
a) Diamminechloronitrito-N-platinum (II)
b) Potassium trioxalatochromate (III)
c) Dichlorobis (ethane-1, 2-diamine) cobalt (III) chloride
d) Pentaairuninecarbonatocobalt (III) chloride
e) Mercury tetrathiocyanocobaltate (III)

Question 14.
Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands Co-ordinated with the central metal ion ?
Solution:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other.

Question 15.
Draw structures of geometrical isomers of [Fe(NH₃)₂(CN)₄]^–
Solution:

Question 16.
Out of the following two Co-ordination entities which is chiral (optically active) ?
a) cis-(CrCl₂(ox)₂]^3-
b) trans-[CrCl₂(ox)₂]^3-
Solution:

Out of the two, (a) cis-(CrCl₂(ox)₂]^3- is chiral (optically active).

Question 17.
The spin only magnetic moment of [MnBr₄]^2- is 5.9 BM. Predict the geometry of the complexion?
Solution:
Since the Co-ordination number of Mn²⁺ ion in the complex ion is 4, it will be either tetrahedral (sp3 hybridisation) or square planar (dsp² hybridisation). But the fact that the magnetic moment of the complex ion is 5.9 BM, it should be tetrahedral in shape rather than square planar because of the presence of five unpaired electrons in the d-orbitals.

Intext Questions

Question 1.
Silver atom has completely filled d-orbitals (4d¹⁰) in its ground State. How can you say that it is a transition element ?
Solution:
Silver (atomic no. = 47), in its +1 oxidation state exhibits 4d¹⁰5s⁰ configuration. But in some compounds it also shows +2 oxidation state, i.e., 4d⁹5s⁰ configuration. Here d-orbital is not completely filled. Therefore, silver is a transition element.

Question 2.
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol^-1, why ?
Solution:
In zinc (3d¹⁰4s¹) d-orbital is complete so the electrons of d-orbital are not involved in bonding. So, metallic bond is weaker than the other elements of the series, where the electrons of d-orbitals are involved in metallic bonds. So, enthalpy of atomisation of zinc is lowest in its transition series.

Question 3.
Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why ?
Solution:
Manganese (atomic no. = 25) has electronic configuration [Ar] 3d⁵4s². It shows maximum number of oxidation states, i.e., from +2 to +7 (+2, +3, +4, +5, +6, +7) in its compounds.

Question 4.
The E°(M²⁺/M) value for copper is positive (+0.34 V). What is possibly the reason for this ?
Answer:
E° (M²⁺/M) value for any metal depends on three factors : •
i) ∆_aH (Enthalpy of atomisation); M(s). + ∆_aH → M(g)
ii) ∆_iH (Enthalpy of ionisation); M(g) + ∆_iH → M²⁺(g)
iii) ∆_hydH (Hydration enthalpy); M²⁺(g) + (aq → M²⁺(aq)
Copper has high value of enthalpy of atomisation and low value of enthalpy of hydration. It means that ∆_iH required is not compensated by the energy released. Therefore, E⁰_(Cu^2+/Cu) is positive.

Question 5.
How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements ?
Solution:
In the first transition series, there is irregular variation of ionisation enthalpies because the stability of 3d configuration differs to some extent. Generally the ionisation enthalpy increases with increase in effective nuclear charge, Thus, it is lower for Cr due to absence of any change in the d-configuration while high for Zn as it represents an jonisation from the 4s level. Second ionisation energy, shows the removal of second. The configurations like d⁵ and d¹⁰ are exceptionally stable. So, they have high ionisation enthalpies.

Question 6.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only ?
Solution:
Oxygen as well as fluorine both have high values of electronegativity. So, in their compounds (oxides and fluorides) they can oxidise the metal to their highest oxidation states.

Question 7.
Which is the stronger reducing agent Cr²⁺ or Fe²⁺ and why ?
Solution:
Cr²⁺ is a stronger reducing agent than Fe²⁺. The E° values are
E°_{Cr³⁺/Cr²⁺} = -0-41 V and E⁰Fe³⁺/Fe²⁺ = 0.77 V

Therefore, Cr²⁺ is stronger reducing agent (itself gets oxidised easily) than Fe²⁺.

Question 8.
Calculate the ‘spin only’ magnetic moment of M²⁺(aq) ion (Z = 27).
Solution:
Electronic configuration of M(Z = 27) is [Ar] 3d⁷ 4s²
Electronic configuration of M²⁺ = [Ar]3d⁷ or
Three unpaired electrons are present in M²⁺ (aq) ion. i.e., n = 3. Applying ‘spin only1 formula for magnetic moment.
μ = \(\sqrt{n(n+2)}\) = \(\sqrt{3(3+2)}\) = \(\sqrt{15}\) BM = 3.87 BM
Note : Unit of magnetic moment is Bohr Magnetons (BM).

Question 9.
Explain why Cu⁺ ion is not stablein aqueous solutions ?
Solution:
In aqueous solution, Cu⁺(aq) undergoes disproportionation reaction as follows.
2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)
The higher stability of Cu²⁺ (aq) in aqueous solution is due to higher negative enthalpy of hydration ∆_hydH° than that of Cu⁺(aq). It compensates the second IE of Cu involved in the formation of Cu²⁺ ion. Hence, Cu⁺ ion changes to more stable Cu²⁺ ion in aqueous medium.

Question 10.
Actinoid contraction is greater from element to element than lanthanoid contraction ?
Solution:
The decrease in atomic (or ionic) radii in actinoid elements (actinoid contraction) is greater than lanthanoid contraction because 5f-electrons have poor shielding effect as compared to 4f- electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

Question 11.
Write the formulae for the following Co-ordination compounds
i) Tetraamminediaquacobalt (III) chloride,
ii) Potassium tetracyanonickelate (II)
iii) Tris-(ethane-l, 2-diamine) chromium (m) chloride
iv) Amminebromidochloridonitrito-N-platinate (II)
v) Dichlorido bis-(ethane-1,2-diamine) platinum (IV) nitrate
Solution:
i)

To find the value of x, we have to find the charge on the complex.
III
[CO(NH₃)₄ (H₂O)₂]^x+ + 3 + 4 × 0 + 2 × 0 = x, × = +3
o, the formula of the complex is [CO(NH₃)₄ (H₂O)₂]Cl₃.

ii)

To find the value of x, find the charge on the complex.
[Ni(CN)₄]^x- (as K+ is positive)
+2 + (-1) x 4 = – x
-x = -2 or x = + 2
So, the formula of the complex is K₂[Ni(CN)₄].
Similarly
iii) [Cr(en)₃]Cl₃
iv) [Pt(NH₃) BrCl(NO₂)]^–
v) [PtCl₂(en)₂] (NO₃)₂
vi) Fe₄[Fe(CN)₆]₃

Question 12.
Write the IUPAC names of the following Co-ordination compounds.
(i) [Co(NH₃)₆]Cl₃
(ii) [Co(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆]
(iv) K₃[Fe(C₂O₄)₃]
(v) K₂[PdCl₄]
(vi) [Pt(NH₃)₂Cl(NH₂CH₃)Cl.
Solution:
i)

Let the oxidation state of Co be x.
x + (0)6 + (-1) × 3 = 0, x + 0 – 3 = 0, x = +3
So, the name of this complex is hexaamminecobalt (III) chloride,

ii) [CO(NH₃)₅Cl]Cl₂
x + (0) × 5 + (-1) × 1 + (-1) × 2 = 0, x + 0 – 3 = 0, x =.+ 3
So, the name of the complex is pentaamminechloridocobalt (III) chloride
[K3[Fe(CN)6]

iii)

Let the oxidation state of Fe is x.
(+ 1) 3 + x + (-1) 6 = 0, + 3 + x – 6 = 0, x = + 3
So, the name of the complex is potassium hexacyanoferrate (III) .

iv)

Let the oxidation state of Fe is x.
(+ 1) 3 + x + (-2) 3 = 0 [∵ Oxalate ion (C₂O₄^2-) bears – 2 charge.]
3 + x- 6 = 0, x = +3
So, the name of the complex is potassium trioxalateferrate (III).

v)

Let the oxidation state of Pd is x.
(+ 1) 2 + x + (-1) 4 = 0, 2 + x – 4 = 0, x = + 2 .
So, the name of the complex is potassium tetrachloridopalladate (II).

vi)

Let the oxidation state of Pt is x.
x + (0) 2 + (-1) × 1 + 0 + (-1) × 1 = 0
x + 0- 1 + 0 -1 = 0, x – 2 = 0, x = + 2
So, the name of the complex is diamminechlorido (methylamine) platinum (II) chloride.

Question 13.
Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers :
(i) K[Cr(H₂O)₂(C₂O₄)₂]
(ii) [Co(en)₃]Cl₃
(iii) [Co(NH₃)₅(NO₂)](NO₃)₂
(iv) [Pt(NH₃)(H₂O)Cl₂]
Solution:
i) [Cr(H₂O)₂(C₂O₄)₂] or K[Cr(H₂O)₂(ox)₂] :
(where, ox = oxalate ion)
a) It exists as geometrical isomers : cis and trans forms.
(In cis form, the same groups occupy adjacent positions while in trans form, they are present at alternate positions.)

b) The cis isomer can also exist as a pair of optical isomers (i.e., d-and 1- forms) (due to absence of plane of symmetry).

ii) [Co(en)₃]Cl₃ : It has two optical isomers (i.e. d-and 1-forms)

iii) [Co(NH₃)₅(NO₂)](NO₃)₂ : It can exist as a pair of ionization isomers as well as linkage iosmers.
Ionization isomers :
[Co(NH₃)₅(NO₂)](NO₃)₂ and [Co(NH₃)₅(NO₃)](NO₂) (NO₃) as they give different ions on ionization.

Linkage isomers : [Co(NH₃)₅(NO₂)](NO₃)₂ and [CO(NH₃)₅ONO](NO₃)₂

iv) [Pt(NH₃)(H₂O)Cl₂] : It can exist as two geometrical iosmers. .

Question 14.
Give evidence that [Co(NH₃)₅Cl]SO₄ and [CO(NH₃)₅SO₄]Cl are ionization isomers.
Solution:
We dissolve both the compounds in water in separate test tubes. To the both test tube
I Step or I Test: Add BaCl₂ solution One compound give white ppt., indicating the presence of SO₄^2- ions.
The other compound does not give white ppt, indicating the absence of SO₄^2- ions.

II Test: Add AgNO₃ solution to both the Compounds in separate test tubes :
Only [II] compound gives white ppt, not the [I] one, due to absence of Cl^– as counter ion.

These two tests prove that the given two compounds are a pair of ionization isomers.

Question 15.
Explain on the basis of valence bond theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and the [NiCl₄] ion with tetrahedral geometry is paramagnetlc.
Solution:

(Cl^– being weak field ligand, cannot cause pairing.)

[NiCl₄]^2- ion is paramagnetic in nature as it has two unpaired electrons.

Question 16.
[NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral.
Solution:
In the complex [NiCl₄]^2- Ni is in +2 oxidation state and has the configuration 3d⁸4s⁰. The cr ion being a weak field ligand cannot pair the two unpaired electrons present in 3d-orbitals. This means that 3d-orbitals are not involved in hybridization. Thus, the complex is sp3 hybridized (tetrahedral) and.is paramagnetic in nature.

In the complex [Ni(CO)₄], the oxidation state of nickel is zero and electronic configuration is 3d⁸4s². In the presence of the ligand CO, the 4s-electrons shift to the two half-filled 3d-orbitals and make all the electrons paired. The valence 4s and 4p-orbitals are involved in hybridization. Thus, the complex is tetrahedral but diamagnetic in nature.
Outer configuration of Ni atom = 3d⁸ 4s².

Question 17.
[Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly paramagnetic Explain.
Solution:
Outer configuration of ₂₆Fe = 3ds⁶4s². In both the complexes Fe is present as Fes³⁺ ion.
Fes³⁺ ion = 3d⁵4s⁰

In the presence of CN^– (a strong field ligand) the 3d-electrons pair up leaving only one unpaired electron. The hybridization is d²sp³ forming an inner orbital complex.

In the presence of H₂O, (a weak field ligand), 3d-electrons do not pair up. The hybridization is sp³d² forming an outer orbital complex containing five unpaired electrons. Hence, it is strongly paramagnetic.

Question 18.
Explain [Co(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex.
Solution:
In [Co(NH₃)₆]³⁺, CO is present as Co³⁺ arid has 3d⁰ configuration.
In the presence of NH₃, the 3d-electrons pair up leaving two d-orbitals empty to the involved in d²sp³ hybridization forming inner orbital complex in the case of [Co(NH₃)₆]³⁺.

Since, (n – 1) d-orbitals are not available but the nd-orbitals are used in bond formation i.e., in hybridization, the complex is called outer orbital complex.

Question 19.
Predict the number of unpaired electrons in the square planar [Pt(CN)₄]^2- ion.
Solution:
₇₈Pt is present in group 10 of the Periodic Table. Its outer configuration is 5d⁹6s¹.

For square plannar shape, the hybridization is dsp². Hence, the unpaired electrons in 5d- orbital pair up to make one d-orbital empty for dsp² hybridization. Thus, there is no unpaired electron.

Question 20.
The hexaquo manganese (II) ion contains five unpaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using crystal field theory.
Solution:
Mn(II) ion has 3d⁵ configuration. In the presence of H₂O molecules (acting as weak field ligands), the distribution of these five electrons is \(\mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^2\) all the electrons remain unpaired to form a high spin complex.

However, in the presence of CN^– (acting as strong field ligands), the distribution of these electrons is \(\mathrm{t}_{2 \mathrm{~g}}^5 \mathrm{e}_{\mathrm{g}}^0\), i.e., two t_2g orbitals contain paired electrons while the third t_2g orbital contains one unpaired electron. The complex formed is a low spin complex.

Question 21.
Calculate the overall complex dissociation equilibrium constant for the Cu(NH₃)₄²⁺ ions, given that β₄ for this complex is 2.1 × 10¹³.
Solution:
Overall complex dissociation equilibrium constant
= \(\frac{1}{\beta_4}\) = \(\frac{1}{2.1 \times 10^{13}}\)
= 4.7 × 10^-14

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 6(d) Group-18 Elements Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 6(d) Group-18 Elements

Very Short Answer Questions

Question 1.
What inspired Bartlett for carrying out reaction between Xe and PtF₆ ?
Answer:

1. At first Bartlett prepared a red compound \(\mathrm{O}_2^{+}+\mathrm{PtF}_6^{-}\)
2. As the first Ionisation Enthalpy of molecular oxygen is identical with that the Xe.
3. He made efforts to prepare same type of compound with Xe and was successful in preparing another red colour compound Xe⁺ \(\mathrm{PtF}_6^{-}\) by mixing PtF₆ and Xe.

Question 2.
Which of the following does not exist ?
a) XeOF₄
b) NeF₂
c) XeF₂
d) XeF₆
Answer:
Given compounds are XeOF₄, NeF₂, XeF₂, XeF₆.
NeF₂ does not exist among the above compounds.

Reason : Due to small size and high I.E. of ‘Ne’. It does not form chemical compounds.

Question 3.
Why do noble gases have comparatively large atomic sizes ?
Answer:
Noble gases have comparatively large atomic sizes as they have vander waals radii which is larger than both the ionic and covalent radii.

Question 4.
List out the uses of Neon. [A.P. Mar. 18]
Answer:
Uses of Ne:

1. Ne is used in discharge tubes and fluor escent bulbs for advertisement display purpose.
2. ‘Ne’ – bulbs are used in botanical gardens and in green houses.

Question 5.
Write any two uses of argon.
Answer:
Uses of Ar :

1. ‘Ar’ is used to create inert atmosphere in high temperature net allurgical process.
2. ‘Ar’ is ued in filling electric bulbs.

Question 6.
In modern diving apparatus, a mixture of He and O₂ is used – Why ? [A.P. Mar. 16]
Answer:
In modem diving apparatus, a mixture of He and O₂ is used because He is very low soluble in blood.

Question 7.
Helium is heavier than hydrogen. Yet helium is used (instead of H₂) in filling baloons for meteorological observations – Why ?
Answer:
‘He’ is a non-inflammable and light gas. Hence it is used in filling baloons for meterological observations.

Question 8.
How is XeO₃ prepared ?
Answer:
XeF₆ on hydrolysis produce XeO₃
XeF₆ + 3H₂O → XeO₃ + 6HF

Question 9.
Give the preparation of
a) XeOF₄ and
b) XeOaF₂
Answer:
Partial hydrolysis of XeF₆ gives oxy fluorides XeOF₄ and XeO₂F₂
XeF₆ + H₂O → XeOF₄ + 2HF
XeF₆ + 2H₂O → XeO₂F₂ + 4HF

Question 10.
Explain the structure of XeO₃. [T.S. Mar. 16]
Answer:
Structure of XeO₃

1. Central atom is ‘Xe’
2. ‘Xe’ under goes sp³ hybridisation in 3^rd excited state.
   
3. ‘Xe’ forms 3σ-bonds and 3π-bonds with three oxygens.
4. Shape of molecule is pyramidal with bond angle 103°.

Question 11.
Noble gases are inert – explain.
Answer:
Noble gases are chemically inert:

1. Noble gases have stable electronic configuration coctet configuration except He)
2. Noble gas have high Ionisation energy values and have large positive values of electron gain Enthalpy.

Question 12.
Write the name and formula of the first noble gas compound prepared by Bertlett.
Answer:
The first noble gas compound prepared by Bertlett is XePtF6. Name of the compound is xenon hexa fluoro platinate.

Question 13.
ExplaIn the shape of XeF₄ on the basis of VSEPR theory
Answer:
Shape of XeF₄:

1. Central atom in XeF₄ is ‘Xe’.
2. Xe undergoes sp³d² hybridisation in it’s 2^nd excited. state.
   
3. Shape of the molecule is square planar with bond angle 90° and bond length 1 .95Å.
   
4. Xe – forms four o-bonds by the overlap of sp³d² – 2p_z(F) orbitals.

Question 14.
Give the outer electronic configuration of noble gases.
Answer:
The outer electronic configuration of noble gases is ns²np⁶ (except He (1s²)

Question 15.
Why do noble gases form compounds with fluorine and oxygen only ?
Answer:
Noble gases form compounds with flourine and oxygen only.
Reason : Oxygen and Fluorine are most electronegative elements.

Question 16.
How is XeOF₄ prepared ? Describe its molecular shape.
Answer:
Partial hydrolysis of XeF₆ gives XeOF₄

XeF₆ + H₂O → XeOF₄ + 2HF
XeOF₄ is a colourless volatile liquid it has a square pyramidal shape.

Question 17.
What is the major source of helium ?
Answer:
The major source of helium is natural gas.

Question 18.
Which noble gas is radioactive ? How is it formed ?
Answer:
Radon (Rn) is radio active noble gas. Radon is obtained as decay product of ₈₆Ra²²⁶
₈₆Ra²²⁶ → ₈₆Rn²²² + ₂He⁴

Question 19.
Name the following :
a) most abundant noble gas in atmosphere
b) radioactive noble gas
c) noble gas with least boiling point
d) noble gas forming large number of compounds
e) noble gas not present in atmosphere
Answer:
a) Argon is the most abundant noble gas in atmosphere.
b) Radon is the radio active noble gas.
c) Helium has lowest boiling point (4.2K)
d) Xenon forms large number of compounds.
e) Radon is not present in atmosphere.

Short Answer Questions

Question 1.
How are Xenon fluorides XeF₂, XeF₄ and XeF₆ obtained?
Answer:
Xenon forms the binary fluorides XeF₂, XeF₄, XeF₆ as follows. These are formed by direct combination of Xe and F₂.

Question 2.
How are XeO₃ and XeOF₄ prepared? [Mar. 14]
Answer:
XeF₆ on hydrolysis produce XeO₃
XeF₆ + 3H₂O → XeO₃ + 6HF
Partial hydrolysis of XeF₆ gives XeOF₄
XeF₆ + H₂O → XeOF₄ + 2HF

Question 3.
Give the formulae and describe the structures of a noble gas species, isoelectronic with
a) ICl₄
b) IBr₂
c) BrO^–₃
Answer:
a) ICl₄ is also electronic with XeF₄ and it has square planar shape.
b) IBr^–₂ is also electronic with XeF₂ and it has linear shape.
c) BrO^–₃ is also electronic with XeO₄ and it has tetrahedral shape.

Question 4.
Explain the reaction of the following with water.
a) XeF₂
b) XeF₄
c) XeF₆
Answer:
a) XeF₂ is hydrolysed to form Xe, HF and O₂
2XeF₂ + 2H₂O → 2Xe + 4HF + O₂

b) XeF₄ is hydrolysed to give XeO₃
6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24 HF + 3O₂

c) XeF₆ undergo hydrolysis to form XeO₃
XeF₆ + 3H₂O → XeO₃ + 6HF
XeF₆ undergo partial hydrolysis to form XeOF₄ + XeO₂F₂
XeF₆ + H₂O → XeOF₄ + 2HF
XeF₆ + 2H₂O → XeO₂F₂ + 4HF

Question 5.
Explain the structures of [A.P. Mar. 18] [Mar. 14]
a) XeF₂ and
b) XeF₄
Answer:
a) Structure of XeF₂ :

1. In XeF₂ central atom is ‘Xe’.
2. ‘Xe’ undergoes sp³d hybridisation in it’s 1^st excited state
   
3. Shape of molecule is linear
4. Xe form two σ – bonds with two fluorines (sp³ – 2p_z overlap)
   

b) Structure of XeF₄ :

1. Central atom in XeF₄ is ‘Xe’.
2. Xe undergoes sp³d² hybridisation in it’s 2^nd excited, state.
   
3. Shape of the molecule is square planar with bond angle 90° and bond length 1.95Å.
   
4. Xe – forms four a – bonds by the overlap of sp³d² – 2p_z(F) orbitals.

Question 6.
Explain the structures of
a) XeF₆ and
b) XeOF₄
Answer:
a) Structure of XeF₆ :

1. Central atom in XeF₆ is ‘Xe’
2. Xe under goes sp³d³ hybridisation in it’s 3^rd excited state.
   
3. Shape of molecule is distorted octahedral.
   

b) Structure of α XeOF₄

1. In XeOF₄ molecule ‘Xe’ under goes sp³d² hybridisation.
2. Shape of the molecule is square pyramid.
3. There is one Xe-O double bond containing
   pπ = dπ overlaping.
   Partial hydrolysis of XeF₆ gives XeOF₄
   XeF₆ + H₂O → XeOF₄ + 2HF
   XeOF₄ is a colourless volatile liquid it has a square pyramidal shape.
   

Question 7.
Complete the following.
a) XeF₂ + H₂O →
b) XeF₂ + PF₅ →
c) XeF₄ + SbF₅ →
d) XeF₆ + ASF₅ →
e) XeF₄ + O₂F₂ →
f) NaF + XeF₆ →
Answer:
a) 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂
b) XeF₂ + PF₅ → [XeF]⁺ + A[F₆]^–
c) XeF₄ + SbF₅ → [XeF₃]⁺ [SbF₆]^–
d) XeF₆ + ASF₅ → [Xe₂F₁₁]⁺ [ASF₆]^–
e) XeF₄ + O₂F₂ → XeF₆ + O₂
f) NaF + XeF₆ → Na⁺ [XeF₇]^–

Question 8.
How are XeF₂ and XeF₄ prepared ? Give their structures. [T.S. Mar. 18]
Answer:
Xenon forms the binary fluorides XeF₂, XeF₄, XeF₆ as follows. These are formed by direct combination of Xe and F₂.

Structure of XeF₂

1. In XeF₂ central atom is Xe’.
2. ‘Xe’ undergoes sp³d hybridisation in it’s 1^st excited state
   
3. Shape of molecule is linear.
4. Xe form two σ bonds with two fluorines.

b) Structure of XeF₄ :

1. Central atom in XeF₄ is ‘Xe’.
2. Xe undergoes sp³d² hybridisation in it’s 2^nd excited state.
   
3. Shape of the molecule is square planar with bond angle 90° and bond length 1.95Å.
   
4. Xe – forms four σ-bonds by the overlap of sp³d² – 2p_z (F) orbitals.

Long Answer Question

Question 1.
How are XeF₂, XeF₄ and XeF₆ prepared ? Explain their reaction with water. Discuss their structures. [A.P. Mar. 15]
Answer:
Xenon forms the binary fluorides XeF₂, XeF₄, XeF₆ as follows. These are formed by direct combination of Xe and F₂.

Reaction with water:

1. XeF₂ is hydrolysed to form Xe, HF and O₂
   2XeF₂ + 2H₂O → 2Xe + 4HF + O₂
2. XeF₄ is hydrolysed to give XeO₃
   6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24 HF + 3O₂
3. XeFg undergo hydrolysis to form XeO₃
   XeF₆ + 3H₂O → XeOF₃ + 6HF
4. XeF₆ undergo partial hydrolysis to form XeOF₄ and XeO₂F₂
   XeF₆ + H₂O → XeOF₄ + 2HF
   XeF₆ + 2H₃O → XeO₂F₂ + 4HF

Structure of XeF₂:

1. In XeF₂ central atom is ‘Xe’.
2. ‘Xe’ undergoes sp³d hybridisation in it’s 1^st excited state
   
3. Shape of molecule is linear.
4. Xe form two σ-bonds with two fluorines.
   

b) Structure of XeF⁴ :

1. Central atom in XeF⁴ is ‘Xe’.
2. Xe undergoes sp³d² hybridisation in it’s 2^nd excited state.
   
3. Shape of the molecule is square planar with bond angle 90° and bond length 1.95Å.
   
4. Xe – forms four σ-bonds by the overlap of sp³d² – 2p_z (F) orbitals.

Structure of XeF₆ :

1. Central atom in XeF₆ is ‘Xe’
2. Xe under goes sp³d³ hybridisation in it’s 3^rd excited state.
   
3. Shape of molecule is distorted octahedral.
   

Textual Examples

Question 1.
Why are the elements of group 18 known as noble gases ?
Solution:
The elements present in Group 18 have their valence shell orbitals completely filled and, therefore, react with a few elements only under certain conditions. Therefore, they are now known as noble gases.

Question 2.
Noble gases have very low boiling points. Why ?
Solution:
Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.

Question 3.
Does the hydrolysis of XeF₆ lead to a redox reaction ?
Solution:
No, the products of hydrolysis are XeOF₄ and XeO₂F₂ where the oxidation states of all the elements remain the same as it was in the reacting state.

Intext Questions

Question 1.
Why is helium used in diving apparatus ?
Solution:
Helium is used in diving apparatus due to its very low solubility in blood.

Question 2.
Balance the following equation :
XeF₆ + H₂O → XeO₂F₂ + HF
Solution:
XeF₆ + 2H₂O → XeO₂F₂ + 4HF

Question 3.
Why has it been difficult to study the chemistry of radon ?
Solution:
Radon is a radioactive element with very short half-life of 3.82 days. That is why, the study of chemistry of radon is a difficult task.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 6(b) Group-16 Elements Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 6(b) Group-16 Elements

Very Short Answer Questions

Question 1.
Why is dioxygen a gas but sulphur a solid?
Answer:
Dioxygen is a gas but sulphur a solid.
Explanation:

• Due to small atomic size and high electronegativity oxygen forms Pπ – Pπ multiple bond and exists as O₂ molecules held together by weak vander waal’s forces. Thus oxygen exists as a gas at room temperature.
• Due to large atomic size and less electronegativity sulphur forms strong S—S single bonds and exists as S₈ molecules with puckered ring structure. Hence sulphur is a solid at room temperature.

Question 2.
What happens when
a) KClO₃ is heated with MnO₂
b) O₃ is passed through KI soultion
Answer:
a) When KClO₃ is heated with Mn02 and liberates oxygen gas.

b) O₃ is passed through KI solution I₂ is liberated
2KI + O₃ + H₂O → 2KOH + I₂ + 2O₂

Question 3.
Give two examples each for amphoteric oxides and neutral oxides.
Answer:

• Examples of amphoterc oxides – Al₂O₃, SiO₂, PbO.
• Examples of neutral oxides – CO, NO and N₂O.

Question 4.
Oxygen generally exhibits an oxidation state of -2 only while the other members of the group show oxidation states of +2, +4 and +6 also – explain.
Answer:

• Oxygen exhibits -2 oxidation state due to its high electronegativity. The tendency of exhibiting -2 oxidation state decrease down the group.
• Due to decrease of electronegativity down the group the other elements exhibit +2, +4 and +6 oxidation states also.

Question 5.
Write any two compounds, in which oxygen shows an oxidation state different from -2. Give the oxidation states of oxygen in them.
Answer:
OF₂ and O₂ F₂ are two compounds in which oxygen shows an oxidation state different from-2.

• In OF₂ the oxidation state of oxygen is +2
• In O₂F₂ the oxidation state of oxygen is +1.

Question 6.
Oxygen molecule has the formula O₂ while sulphur has S₈ – explain.
Answer:
Due to small atomic size and high electronegativity oxygen forms Pπ – Pπ multiple bond and exists as O₂ molecule.
Due to large atomic size and less electronegativity sulphur forms strong S – S single bonds and exists S₈ molecule.

Question 7.
Why is H₂O a liquid while H₂S is a gas ?
Answer:
H₂O is liquid due to the presence of intermolecular hydrogen bonding. While H₂S is gas because it is not having such type of bonding.

Question 8.
H₂O is neutral while H₂S is acidic – explain.
Answer:
H₂O is neutral while H₂S is acidic.
Reason: The O-H bond dissociation Enthalpy is greater than the S – H bond dissociation Enthalpy.

Question 9.
Name the most abundant element present in earth’s crust.
Answer:
The most abundant element present in earth’s crust is oxygen (about 46.6%).

Question 10.
Which element of group-16 shows highest catenation ?
Answer:
Sulphus shows highest catenation among group – 16 elements and exists as S₉ molecule with puckered ring structure.

Question 11.
Among the hydrides of chalcogens, which is most acidic and which is most stable ?
Answer:

• Among hydrides of chalcogens, H₂Te is most acidic.
• Among hydrides of chalcogens, H₂O is most stable.

Question 12.
Give the hybridization of sulphur in the following.
a) SO₂
b) SO₃
c) SF₄
d)SF₆
Answer:
a) Hybridisation of’S’ in SO₂ is Sp²
b) Hybridisation of ‘S’ in SO₃ is Sp²
c) Hybridisation of ‘S’ in SF₄ is Sp³d
d) Hybridisation of ‘S’ in SF₆ is Sp³d²

Question 13.
Write the names and formulae of any two oxyacids of sulphur. Indicate the oxidation state of sulphur in them.
Answer:

• Peroxy mono sulphuric acid – H₂SO₅ ‘S’ oxidation state +6 .
• Peroxy di sulphuric acid – H₂S₂O₈ ‘S’ oxidation state +6

Question 14.
Explain the structures of SF₄ and SF₆.
Answer:
Structure of SF₄:

• It SF₄ ‘S’ undergoes sp³d hybndisation.
• It has trigonal bipyramidal structure in which one of the equitorial positions is occupied by a lone pair of electrons. This geometry is also known as see – saw geometry.

Structure of SF₆:

• In SF₆ ‘S’ undergoes sp³d² hybridisation.
• It has octahedral structure.

Question 15.
Give one example each for
a) a neutral oxide
b) a peroxide
c) a super oxide
Answer:
a) CO, N₂O are neutral oxides.
b) Na₂O₂, BaO₂ are pernxides.
c) KO₂, RbO₂ are super oxides.

Question 16.
What is tailing of mercury? How is it removed? [A.P. & T.S. (Mar. 15)]
Answer:
Mercury loses it’s lustreness, meniscus and consequently sticks to the walls of glass vessel when it reacts with ozone. This phenomenon is called tailing of mercury.
2Hg + O₃ → Hg₂O + O₂
It is removed by shaking it with water which dissolves Hg₂O.

Question 17.
Write the principle involved in the quantitative estimation of ozone gas.
Answer:
When ozone reacts with an excess of Kl solution buffered with a borate buffer (pH 9.2) iodine is liberated which can be titrated against a standard solution of sodium thiosulphate. In this way O₃ is estimated quantitatively.

Question 18.
Write the structure of ozone.
Answer:
Structure of Ozone:

• O₃ is angular molecule with bond angle 117°
• O – O bond length is 128 pm.

Question 19.
SO₂ can be used as an anti-chior. Explain.
Answer:
SO₂ gas is used as anti-chlor. Anti-chior means the substance which removes the excess of on clothes. SO₂ reacts with chlorine in presence of charcoal to give suiphuryl chloride
SO_2(g) + Cl_2(g) → SO₂Cl_2l

Question 20.
How is ozone detected ?
Answer:
Ozone is a pale blue gas, dark blue liquid and violet black solid and it has characterstic smell.

• It is detected by the reaction with ‘Hg’ which is called as tailing of mercury.
  2Hg + O₃ → Hg₂O + O₂
• Ozone turns benzidene paper to brown colour.

Question 21.
How does ozone react with Ethylene ?
Answer:
Ethylene reacts with ozone to form Ethylene ozonoid followed by the hydrolysis to form formaldehyde.

Question 22.
Out of O₂ and O₃, which is paramagnetic ?
Answer:

• O₂ is paramagnetic due to presence of unpaired electrons
• O₃ (gaseous) is diamagnetic due to absence of unpaired electrons.

Question 23.
Between O₃ and O₂, ozone is a better oxidizing agent – why ?
Answer:
Ozone is next to fluorine in the oxidising capacity. It is best oxidising agent than O₂. (Fluorine is the powerful oxidising agent). Ozone liberates nascent oxygen easily.

Question 24.
Write any two uses each for O₃ and H₂SO₄.
Answer:
Uses of O₃:

• Ozone is used in sterilisation of water.
• Ozone is used in manufacture of artificial silk and camphor etc.
• Ozone is used to identify unsaturation in carbon compounds.

Uses of H₂SO₄:

• H₂SO₄ is used in the manufacture of fertilisers.
• H₂SO₄ is used in petrol refining.
• H₂SO₄ is used in detergent industry.

Question 25.
Which form of sulphur shows paramagnetism ?
Answer:
In vapour state sulphur partly exists as S₂ molecule which has two unpaired electrons in the antibonding π (π*)orbitals like O₂. Hence exhibits paramagnetism.

Question 26.
How is the presence of SO₂ detected ?
Answer:
SO₂ has a pungent odour SO₂ presence can be detected by the following tests.

1. SO₂ changes the colour of acidified potassium dichromate solution from orange to green.
   
2. SO₂ decolourises acidified KMnO₄ solution.
   

Question 27.
Why are group – 16 elements called chalcogens ?
Answer:
Chalcogens means mineral forming (or) ore forming elements. Most of elements exist in earth crust as oxides, sulphides, selinides, telurids etc. So Group – 16 elements are called as chalcogens.

Question 28.
Among chalcogens, which has highest eletronegativity and which has highest electron gain enthalpy?
Answer:

• Among chalcogens Oxygen has high electronegativity.
• Among chalcogens Sulphur has high electron gain Enthalpy.

Question 29.
Which hydride of group – 16 has highest boiling point and weakest acidic character ?
Answ:

• Among group – 16 hydrides water (H₂O) has high bioling point.
• Among group – 16 hydrides water (H₂O) has weakest acidic, character.

Short Answer Questions

Question 1.
Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation states and hydride formation.
Answer:
1) Electronic configurations :
Oxygen (O) – [He] 2s² 2p²
Sulphur (S) – [Ne] 3s² 3p²
Selenium (Se) – [Ar] 3d¹⁰ 4s²4p⁴
Tellurium (Te) – [Kr] 4d¹⁰ 5s² sp⁴
Polonium (Po) – [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁴
All the above elements has general outer electronic configuration ns²np⁴.

2) Oxidation states :
All the gives elements (chalcogens) exhibits common oxidation state of-2
O^-2, S^-2, Se^-2 etc.

3) Hydride formation:
All these elements (chalcogens) forms hydrides of type EH₂(E = chalcogen)
Eg : H₂O, H₂S, H₂Se, H₂Te, H₂Po.
The above mentioned concepts evident that the elements O, S, Se, Te and Po are present in the same group of periodic table.

Question 2.
Describe the manufacture of H₂SO₄ by contact process.
Answer:
Manufacture of H₂SO₄ by contact process:
Manufacturing of H₂SO₄ involves three main steps.
Step – 1.
SO₂ production : The required SO₂ for this process is obtained by burning S(or) Iron pyrites in oxygen.
S + O₂ → SO₂
4FeS₂ + 15O₂ → 2Fe₂O₃ + 8SO₃
Step – 2
SO₃ formation: SO₂ is oxidised in presence of catalyst with atmosphric air to form SO₃

Step – 3
Formation of H₂SO₄: SO₃ formed in the above step absorbed in 98% H₂SO₄ to get oleum (H₂S₂O₇). This oleum is diluted to get desired concentration of H₂SO₄.
SO₃ + H₂SO₄ → H₂S₂O₇
H₂S₂O₇ + H₂O → 2H₂SO₄

Question 3.
How is ozone prepared ? How does it react with the following? [Mar. 14]
a) PbS
b) KI
c) Hg
d) Ag
Answer:
Preparation of Ozone:
A slow dry stream of oxygen under silent electric discharge to form ozone (about 10%). The product obtained is known as ozonised oxygen.
3O₂ → 2O₃ ∆ H° = 142kJ/mole

• The formation of ozone is an endothermic reaction.
• It is necessary to use silent electric discharge in the preparation of O₃ to prevent its decomposition

a) Reaction with PbS : Black lead suiphide oxidised to white lead sulphate in presence of ozone.
PbS + 4O₃ → PbSO₄ + 4O₂

b) Reaction with KI: Moist Kl is oxidised to Iodine in presence of ozone.
2KI + H₂O + O₃ -→ 2KOH + I₂ + O₂

c) Reaction with Hg : Mercury loses it’s lustreness, meniscus and consequently sticks to the walls of glass vessel when it reacts with ozone. This phenomenon is called tailing of mercury.
2Hg + O₃ → Hg₂ O + O₂
It is removed by shaking it with water which dissolves Hg₂O.

d) Reaction with Ag : Ag metal oxidised to Ag₂O (Ag metal is tarnished):
2Ag + O₃ → Ag₂O + O₂

Question 4.
Write a short note on the allotropy of sulphur.
Answer:
The important allotropes of sulphur are
a) yellow rhombic (α. sulphur)
b) Monoclinic (β – sulphur)

• The stable from is α-sulphur (at room temperature)

Rhombic sulphur (α – Sulphur):

• Colour : Yellow. ,
• Melting point: 385.8K.
• Specific gravity : 2.06.
• It is insouble in water and partially soluble in alcohol, benzene etc. and readily soluble in CS₂.

Monoclinic sulphur (β – Sulphur):

• Melting point: 392K
• Specific gravity : 1.98.
• It is soluble in CS₂.
• Rhombic sulphur transforms to monoclinic sulphur by heating above 369K. This temperature is called transition temperature.

Question 5.
How does SO₂ react with the following ?
a) Na₂SO₃(aq)
b) Cl₂
c) Fe⁺³ ions
d) KMnO₄
Answer:
a) Sodium sulphite (aq) reacts with So₂ to form sodium hydrogen sulphite.
Na₂SO₃ + H₂O + SO₂ → 2NaHSO₃

b) SO₂ as reacts with chlorine gas in the presence of charcoal to form sulphuryl chloride.
SO_2(g) + Cl_2(g) → SO₂Cl_2(l)

c) Fe⁺³ ions are reduced to Fe⁺² ions by SO₂.
2Fe⁺³ + SO₂ + 2H₂O → 2Fe⁺² + SO^-2₄ + 4H⁺

d) SO₂ gas decolourises acidified potassium permanganate (VII) solution.
5SO₂ + 2MnO^–₄ + 2H₂O → 5SO^-2₄ + 4H⁺+ 2Mn⁺²

Question 6.
Starting from elemental sulphur, how is H₂SO₄ prepared ?
Answer:
Manufacture of H₂SO₄ by contact process :
Manufacturing of H₂SO₄ involves three main steps.
Step – 1.
SO₂ production : The required SO₂ for this process is obtained by burning S(or) Iron pyrites in oxygen.
S + O₂ → SO₂
4FeS₂ + 15O₂ → 2Fe₂O₃ + 8SO₃
Step – 2
SO₃ formation: SO₂ is oxidised in presence of catalyst with atmosphric air to form SO₃

Step – 3
Formation of H₂SO₄: SO₃ formed in the above step absorbed in 98% H₂SO₄ to get oleum (H₂S₂O₇). This oleum is diluted to get desired concentration of H₂SO₄.
SO₃ + H₂SO₄ → H₂S₂O₇
H₂S₂O₇ + H₂O → 2H₂SO₄

Question 7.
Describe the structures (shapes) of SO^-2₄ and SO₃.
Answer:
Structure of SO₃:

• In SO₃ sulphur undergoes sp² hybridisation.
• Shape : Trigonal planait s
• Bondangle: 120°.
• S -0 bond length: 143 pm.

Structure of SO^-2₄:

• In SO^-2₄ sulphur undergo sp³ hybridisation.
• Shape : Tetrahedral.
• It has several resonance structures.
• In this two Pπ – dπ bonds are present.

Question 8.
Which oxide of sulphur can act as both oxidizing and reducing agent? Give one example each.
Answer:
Sulphur dioxide (SO2) acts as both oxidising as well as reducing agent.
SO₂ as Oxidising agent:
Sodium suiphide oxidises to hypo with SO₂.
2Na₂S + 3SO₂ → 2Na₂S₂O₂ + S
SO₂ as Reducing agent:
SO₂ reduces Fe⁺³ ions to Fe⁺² ions.
2Fe⁺³ + SO₂ + 2H₂O → 2Fe⁺² + SO^-2₄ + 4H⁺

Question 9.
Explain the conditions favourable for the formation of SO₃ from SO₂ in the contact process of H₂SO₄.
Answer:
Le Chatlier’s principle – Application to produce SO₃:
The oxidation of SO₂ to SO₃ in the presence of a catalyst is a reversible reaction. The thermochemical equation for the conversion is written as
2SO_2(g) + O_2(g) ⇌ 2SO_3(g); ∆H = -196 kJ
The equation reveals the following points :

1. 3 volumes of the reactants convert into 2 volumes of SO₃. i.e., a decrease of volume accompanies the reaction.
2. the reaction is an exothermic change.
3. the catalyst may be present to increase the SO₃ yields.

According to Le Chatlier’s principle,
i) a decrease in volume of the system is favoured at high pressures. But in practice only about 2 bar pressure is used. The reason for not using high pressures is acid resisting towers which can withstand high pressures cannot be built.

ii) exothermic changes are favoured at low temperatures. It is not always convenient in the industry to work at low temperatures. In such situations an optimum temperature is maintained. At this temperature considerable amounts of the product are obtained. In the manufactrue of H₂SO₄, the optimum temperature suitable for the conversion of SO₂ into SO₃ is experimentally found to be 720K.

iii) The rate of formation of SO₃ is enhanced by the use of a catalyst. (V₂ O₅ (or) Pt – asbestos).
Favourable Conditions :
Temperature : 720K
Pressure : 2 bar
Catalyst: V₂O₅ (or) platinized asbestos.

Question 10.
Complete the following
a) KCl + H₂SO₄ (Conc) →
b) Sucrose
c) Cu + H₂SO₄ (Conc) →
d) C + H₂SO₄ (Cone) →
Answer:
a) 2KCl + Cone. H₂SO₄ → 2HCl + K₂SO₄
b) C₁₂H₂₂O₁₁ 12C + nH₂O
c) Cu + 2H₂SO₄(Conc) → CuS0₄ + SO₂ + 2H₂O
d) C + 2H₂SO₄(Conc) → CO₂ + 2SO₂ + 2H₂O

Question 11.
Which is used for drying ammonia ?
Answer:
For drying ammonia quick line (CaO) is used.
For drying ammonia cone. H₂SO₄, P₄O₁₀ and anhydrous CaCl₂ cannot be used because they react with ammonia and forms (NH₄)₂SO₄, (NH₄)₃PO₄ and CaCl₂. 8NH₃
H₂SO₄ + 2NH₃ → (NH₄)₂ SO₄

CaCl₂ + 8NH₃ → CaCl₃8NH₃.

Question 12.
Why cone H₂SO₄, P₄O₁₀ and anhydrous CaCl₂ cannot be used to dry ammonia ?
(Hint: ammonia reacts with them forming (NH₄)₂ SO₄; (NH₄)₃ PO₄ and CaCl₂, 8NH₂)
Answer:
For drying ammonia quick line (CaO) is used.
For drying ammonia cone. H₂SO₄, P₄O₁₀ and anhydrous CaCl₂ cannot be used because they react with ammonia and forms (NH₄)₂SO₄, (NH₄)₃PO₄ and CaCl₂. 8NH₃
H₂SO₄ + 2NH₃ → (NH₄)₂ SO₄

CaCl₂ + 8NH₃ → CaCl₃8NH₃.

Long Answer Questions

Question 1.
Explain in detail the manufacture of sulphuric acid by contact process. [T.S. Mar. 18, 16]
Answer:
Manufacture of H₂SO₄ by contact process :
Manufacturing of H₂SO₄ involves three main steps.
Step – 1.
SO₂ production : The required SO₂ for this process is obtained by burning S(or) Iron pyrites in oxygen.
S + O₂ → SO₂
4FeS₂ + 15O₂ → 2Fe₂O₃ + 8SO₃
Step – 2
SO₃ formation: SO₂ is oxidised in presence of catalyst with atmosphric air to form SO₃

Le Chatlier’s principle – Application to produce SO₃ :
The oxidation of SO₂ to SO₃ in the presence of a catalyst is a reversible reaction. The thermochemical equation for the conversion is written as
2SO_2(g) + O_2(g) ⇌ 2SO_3(g); ∆H = -196 kJ
The equation reveals the following points : .

1. 3 volumes of the reactants convert into 2 volumes of SO₃. i.e., a decrease of volume accompanies the reaction.
2. the reaction is an exothermic change.
3. the catalyst may be present to increase the SO₃ yields.

According to Le Chatlier’s principle, .
i) a decrease in volume of the system is favoured at high pressures. But in practice only about 2 bar pressure is used. The reason for not using high pressures is acid resisting towers which can withstand high pressures cannot be built.

ii) exothermic changes are favoured at low temperatures. It is not always convenient in the industry to work at low temperatures. In such situations an optimum temperature is maintained. At this temperature considerable amounts of the product are obtained. In the manufactrue of H₂SO₄, the optimum temperature suitable for the conversion of SO₂ into SO., is experimentally found to be 720K.

iii) The rate of formation of SO₃ is enhanced by the use of a catalyst. (V₂ O₅ (or) Pt – asbestos).
Favourable Conditions :
Temperature : 720K .
Pressure : 2 bar
Catalyst: V₂O₅ (or) platinized asbestos.

Step – 3 :
Formation of H₂SO₄: SO₃ formed in the above step absorbed in 98% H₂SO₄ to get oleum (H₂S₂O₇). This oleum is diluted to get desired concentration of H₂SO₄.
SO₃ + H₂SO₄ → H₂S₂O₇
H₂S₂O₇ + H₂O → 2H₂SO₄

Question 2.
How is ozone prepared from oxygen ? Explain its reaction with [A.P. Mar. 19, 18, 17, 16] [Mar. 14]
a) C₂H₄
b) KI
c) Hg
d) PbS.
Answer:
Preparation of Ozone :
A slow dry stream of oxygen under silent electric discharge to form ozone (about 10%). The product obtained is known as ozonised oxygen.
3O₂ → 2O₃; ∆H° = 142kJ/mole

• The formation of ozone is an endothermic reaction.
• It is necessary to use silent electric discharge in the preparation of O₃ to prevent its decomposition

a) Reaction with C₂H₃ : Ethylene reacts with ozone to form Ethylene ozonoid followed by the hydrolysis to form formaldehyde.

) Reaction with KI: moist KI is oxidised to Iodine in presence of ozone.
2KI + H₂O + O₃ → 2KOH + I₂ + O₂

c) Reaction with Hg : Mercury loses it’s lustreness, meniscus and consequently sticks to the walls of glass vessel when it reacts with ozone. This phenomenon is called tailing of mercury.
2Hg + O₃ → Hg₂O + O₂
It is removed by shaking it with water which dissolves Hg₂O.

d) Reaction with PbS : Black lead sulphide oxidised to white lead sulphate in presence of ozone.
PbS + 4O₃ → PbSO₄ + 4O₂.

Intext Questions

Question 1.
List the important sources of sulphur.
Answer:
Occurrence or sources of sulphur in the earth’s crust, percentage of sulphur is only 0.03 to 0. 1%. In combined state, it occurs

1. In the form of sulphates, eg., gypsum (CaSO₄ . 2H₂O). epsom salt (MgSO₄ . 7H₂O), baryte
   (BaSO₄).
2. In the form of sulphides, eg., galena (PbS), zinc blende (ZnS), Copper pyrites (CuFeS₂). In volcanoes, traces of sulphur occur as H2S. In organic materials such as eggs, proteins, garlic, onion,mustard, hair and wool, sulphur is present in trace amounts.

Question 2.
Write the order of thermal stability of the hydrides of group 16 elements.
Answer:
The thermal stability of hydrides of group 16 elements is directly proportional to the bond dissociation enthalpy of H – E bond. On moving down the group, bond dissociation energy decreases because bond length increases.
Thus, the order of bond dissociation energy is
H₂O > H₂S > H₂Se > H₂Te > H₂Po
This is also the order of thermal stability.

Question 3.
Why is H₂O a liquid and H₂S a gas ?
Answer:
The difference in electronegativity values of 0(3.5) and H(2.1) is more than the difference between the electronegativity values of H(2.1) and S(2.5) i.e., O – H bond is more polar than S – H bond. That is why H-bonding is present among water molecules but absent in H₂S. Thus,
strong intermolecular interactions causes water to exist as a liquid but due to weak van der waals’ force H₂S exists as a gas.

Question 4.
Which of the following does not react with oxygen directly ? Zn, Ti, Pt, Fe.
Answer:
Platinum (Pt).

Question 5.
Complete the following reactions.
i) C₂H₄ + O₂ →
ii) 4Al + 3O₂ →
Answer:

Question 6.
Why does O₃ act as a powerful oxidising agent ?
Answer:
Because ozone liberates nascent oxygen very easily.

Question 7.
How is Oa estimated quantitatively ?
Answer:
When ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated which can be titrated against a standard solution of sodium thiosulphate. In this way O₃ can be estimated quantitatively.

Question 8.
What happens when sulphur dioxide is passed through an aqueous solution of Fe (III) salt ?
Answer:
When SO₂ is passed through an aqueous solution of Fe (III), i.e., ferric salt, it is reduced to Fe (II) i,e., ferrous salt. Here, SO₂ acts as a reducing agent.

Question 9.
Comment on the nature of two S – O bonds formed in SO₂ molecule. Are the two S – O bonds in this molecule equal ?
Answer:
The two S – O bonds in SO₂ molecule are covalent in nature. These are equal with bond length = 143 pm. The resonating structures of SO₂ are as follows.

SO₂ is a resonance hybrid of these two canonical forms.

Question 10.
How is the presence of SO₂ detected ?
Answer:
SO₂ has a pungent odour. Two tests to detect the presence of SO₂ are as follows :

1. SO₂ decolourises acidified KMnO₄ solution.
   
2. SO₂ Changes the colour of acidified potassium dichromate solution from orange to green.
   

Question 11.
Mention three areas in which H2S04 plays an important role.
Answer:
Uses of sulphuric acid.

1. It is used in the manufacture of pigments, paints and dyestuff intermediate.
2. It is used in petroleum refining.
3. It is also used in fertilizer industry.

Question 12.
Write the conditions to maximize the yield of H₂SO₄ by contact process.
Answer:
The key step in the manufacture of H₂SO₄ is catalytic oxidation of SO₂ to produce SO₃ in presence of V₂O₅.

The reaction is exothermic, reversible and the forward reaction results in the decrease in volume. Thus according to Le-Chatelier’s principle, the forward reaction should be favoured by low temperature and high pressure. But the temperature should not be very low otherwise the. rate of reaction will become very slow.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 6(a) Group-15 Elements Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 6(a) Group-15 Elements

Very Short Answer Questions

Question 1.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:
Nitrogen gas exists as diatomic molecule. Dut to the presence of triple bond between Base N – Atoms bond dissociation energy is high (941.4 KJ /mol). Hence nitrogen is inert and unreactive.
Phosphorus is a tetra atomic molecule and P-P single bond is weaker than N ≡ N. P – P bond dissociation energy is 213 KJ/mole. Hence phosphorus is more reactive than Nitrogen.

Question 2.
How is nitrogen prepared in the laboratory ?
Write the chemical equations of the reactions involved.
Answer:
Preparation of di Nitrogen :

• Very pure nitrogen is obtained by the thermal decomposition of sodium (or) barium a-zide.
  Ba(N₃)₂ → Ba + 3N₂
• In the laboratory dinitrogen is prepared by treating an aqueous solution of NH₄Cl with NaNO₂
  NH₄Cl_(aq) + NaNO_2(aq) → N_2(g) + 2H₂O_(I) + NaCl_(aq)
• Nitrogen can also be obtained by the thermal decompostion of ammonium dichromate.
  (NH₄)₃ Cr₂O₇ N₂ + 4H₂O + Cr₂O₃

Question 3.
Nitrogen exists as diatomic molecule and phosphorus as P₄ – Why ?
Answer:
Nitrogen exists as diatomic molecule :

• Nitrogen has small size and high electronegativity and nitrogen atom forms Pπ – Pπ multiple bonds with it self (triple bond). So it exists as a discrete diatomic molecule in elementary state.

Phosphorus exists as tetra atomic molecule :

• Phosphorus has large size and less electronegative and it forms P-P single bonds. So it exists as tetra atomic i.e., P₄.

Question 4.
Why does nitrogen show catenation properties less than phosphorus ?
Answer:
Explanation :
The single N-N bond is weaker than the single P-P bond due to high inter electronic repulsion of the non-boriding electrons in N₂ because of small bond length. Therefore the catenation property is weaker in nitrogen as compared to phosphorus.

Question 5.
Nitrogen molecule is highly stable – Why ?
Answer:
Nitrogen Molecule is more stable because in between two nitrogen atoms of N₂, a triple bond is present. To break this triple bond high energy is required (941.4KJ/mole).

Question 6.
Why are the compounds of bismuth more stable in +3 oxidation state ?
Answer:
Bismuth compounds are more stable in +3 oxidation state because ‘Bi’ exhibits +3 stable oxidation state instead of +5 due to inert pair effect.

Question 7.
What is allotropy ? Explain the different allotropic forms of phosphorus.
Answer:
Allotropy: The existance of an element in different physical forms having similar chemical properties is called allotropy.
Allotropes of ‘P’: → White ‘P’ (or) Yellow ‘P’.

• Red ‘P’
• Scarlet ‘P’
• Violet ‘P’
• α – black ‘P’
• β – black ‘P’.

White phosphorus :

• It is poisonous and insoluble in water and soluble in carbon disulphide and glows in dark. It is a translucent white waxy solid. ‘
•  It dissolves in boiling NaOH solution and gives PH₃.
  P₄ + 3NaOH + 3H₂ → PH₃ + 3NaH₂PO₂. (sodium hypo phosphite)
•  It is more reactive than other solid phases.
• Bond angle is 60° and it readily catches fire.

Red phosphorus :

• Red ‘F possesses iron grey lustre.
• In is odour less, non poisonous and insoluble in water as well as CS₂.
• Red F’ is much less reactive than white ‘P’.
  Black ‘P’:
• α – Black ‘P’: It is formed when red ‘P’ is heated in a sealed tube 803K.
• β – Black F : It is prepared by heating white P’ at 473 K under high pressure.

Question 8.
How do you account for the inert character of dinitrogen ?
Answer:
Di Nitrogen is chemically inert. .
Explanation : In nitrogen molecule there exists a triple bond between two nitrogen atoms. To break this triple bond high energy (bond dissociation energy) is required (941.4 KJ/mole).

Question 9.
Explain the difference in the structures of white and red phosphorus.
Answer:
White ‘P’ molecule has tetrahedral structure (discrete molecule). Discrete ‘P’ molecules are held by vander waal’s forces

White phosphorus
Red ‘P’ is polymeric consisting of chains of P₄ tetrahedron linked together through covalent bonds.

Question 10.
How is α – black phosphorus prepared from red phosphorus?
Answer:
α – Black ‘P: It is formed when red? is heated in a sealed tube 803K.

Question 11.
Write the difference between the properties of white phosphorus and red phosphorus.
Answer:
White ‘P’

• It is white waxy solid (translucent).
• It is insolable in water and solable in CS₂.
• It is more reactive.
• It is poisonous.

Red ‘P’

• It possesses iron grey lustre.
• It is in soluble in CS₂ as cool water.
• It is more reactive than white ‘P’.
•  It is non – poisonous.

Question 12.
What is inert pair effect ?
Answer:
Inert pair effect: The reluctance of ns pair of electrons to take part in bond formation is called inert pair effect.
Bi exhibits +3 oxidation state instead of +5 due to inert pair effect.

Question 13.
Explain why is NH₃ basic while BiH₃ is only feebly basic.
Answer:
NH₃ is basic while BiH₃ is only feebly basic.

Explanation :
Due to small atomic size of nitrogen, the electron density on nitrogen atom is greater than that on Bi atom. So electron releasing tendency is greater in NH₃.
Hence NH₃ is basic and BiH₃ is feebly basic.

Question 14.
Arrange the hydrides of group – 15 elements in the increasing order of basic strength and decreasing order of reducing character.
Answer:

1. Increasing order of basic strength of Group – 15 elements hydrides is
   BiH₃ < SbH₃ < ASH₃ < PH₃ < NH₃.
2. Decreasing order of reducing character of Group – 15 elements hydrides is
   BiH₃ > SbH₃ > ASH₃ > PH₃ > NH₃.

Question 15.
PH₃ is a weaker base than NH₃ – Explain.
Answer:
PH₃ is a weaker base than NH₃.

• In NH₃ nitrogen atom undergoes SP³ hybridisation and due to small size it has high electron density than in ‘P’ of PH₃.
• Due to large size of ‘P’ atom and availability of large surface area, lonepair of electron spread in PH₃. Hence PH₃ is weaker base then NH₃.

Question 16.
A hydride of group -15 elements dissolves in water to form a basic solution. This solution dissolves the AgCl precipitate. Name the hydride. Write the chemical equations involved.
Answer:
Given a hydride of group – 15 elements dissolves in water to form a basic solution;This solution dissolves the Agcl precipitate.

The given hydride is ammonia. It forms basic solution when dissolved in water due to formation of OH^– ions.
NH_3(g) + H_xO_l ⇌ \(\mathrm{NH}_4^{+}(\mathrm{aq})\) + OH^–_(aq)
This solution dissolves AgCl. Ppt due to formation of complex compound.
AgCl_(s) (white Ppt) + 2NH_3(aq) → [Ag(NH³)²]Cl_(aq) (colourless)

Question 17.
What happens when white phosphorus is heated with cone. NaOH solution in an inert atmosphere of CO₂? [A.P. Mar. 19, 15]
Answer:
When white phosphorus heated with con. NaOH solution in an inert atomosphere of CO₂ forms PH₃.
P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂.

Question 18.
NH₃ forms hydrogen bonds but PH₃ does not – Why ?
Answer:
NH₃ forms hydrogen bonds but PH₃ does not.
Reason : Ammonia forms hydrogen bonds because it it a polar molecule and N-H bond is highly polar. Nitrogen has more electronegativity than phosphrous. In case of PH₃ P-H bond polarity decreases.

Question 19.
The HNH angle is higher than HPH, HAsH and HSbH angles – Why?
Answer:
The central atom ε (where, ε = N, P, As, Sb, Bi) in all given hydrides is Sp³ hybridized. However, its electronegativity decreases and atomic size increase on moving down the group. Therefore is a gradual decrease in the force of repulsion between the shared electron pairs around the central atom. Thus, bond angle decreases as we move down the group.

Question 20.
How do calcium phosphide and heavy water react?
Answer:
Calcium phosphide reacts with heavy water to form Deutero phosphine.
Ca₃P₂ + 6D₂O → 3 Ca (OD)₂ + 2PD₃

Question 21.
Ammonia is a good complexing agent – Explain with an example. [Mar. 14]
Answer:
NH₃ is a lewis base and it donates electron pair to form dative bond with metal ions. This results in the formation of complex compound.

Question 22.
A mixture of Ca₃P₂ and CaC₂ is used in making Holme’s signal – Explain. [A.P. Mar. 16]
Answer:
A Mixture of Ca₃P₂ and CaC₂ is used in Holme’s signal.This Mixture containing containers are pierced and thrown in the sea, when the gas is evolved bum and serve as a signal.
The spontaneous combustion of PH₃ is the technical use of Holme’s signal.

Question 23.
Which chemical compound is formed in the brown ring test of nitrate ions ?
Answer:
In the brown ring test of nitrate salts a brown ring is formed. It’s formula is [Fe(H₂O)₅N0]⁺².

Question 24.
Give the resonating structures of NO₂ and N₂O₅.
Answer:
Resonance structure of NO₂:

Resonance structure of N₂O₅

Question 25.
Why does R₃P = O exist but R₃N = O does not (R = alkyl group) ?
Answer:
R₃P = O exist but R₃N = O does not.
Explanation :
Nitrogen does not form dπ – Pπ multiple bond with oxygen because of lack of d – orbitals in Nitrogen atom. But in case of R₃N = O the value of nitrogen should be 5. So these compounds do not exist where as in case of ‘P’ atom d-orbitals are available. So P-atom can able to form dπ – Pπ multiple bonds hence R₃P = O exist.

Question 26.
How is nitric oxide (NO) prepared ?
Answer:
The catalytic oxidation of NH₃ by atmospheric oxygen gives nitric oxide.

Question 27.
Give one example each of normal oxide and mixed oxide of nitrogen.
Answer:

• Nitric oxide (NO) is an example of normal oxide of Nitrogen.
• Dinitrogen trioxide (N₂O₃) is an example of mixed oxide of nitrogen.

Question 28.
NO is paramagnetic in gaseous state but diamagnetic in liquid and solid states – Why ?
Answer:
In gaseous state NO₂ exists as a Monomer and contains one unpaired electron but in solid state it dimerises to N₂O₄ so it doesnot contain unpaired electron.
Hence NO₂ is parce magnetic is geseous state but diamagnetic in solid state.

Question 29.
Give an example of
a) acidic oxide of phosphorus
b) neutral oxide of nitrogen.
Answer:
a) P₂O₅ (or) P₄O₁₀, phosphorus pentoxide is an example of acidic oxide of phosphorus
b) Nitrous oxide (N₂O) and Nitric oxide (NO) are neutral oxides of nitrogen.

Question 30.
Explain the following
a) reaction of alkali with red phosphorus.
b) reaction between PCl₃ and H₃PO₃.
Answer:
a) Red phosphorus reacts with an alkali to form hypophosphoric acid (H₄P₂O₆)
b) PCl₃ undergoes hydrolysis to form H₃PO₃
PCl₃ + 3H₂O → H₃PO₃ + 3HCl

Question 31.
How does PCl₃ react with
a) CH₃COOH
b) C₂H₅OH and
c) water.
Answer:
a) PCl₃ reacts with CH₃COOH and form phosphorous acid, acetyl chloride.
3CH₃COOH + PCl₃ → 3CH₃COCl + H₂PO₃

b) PCl₃ reacts with phosphorous acid and forms phosphorus acid, Ethyl chloride
3C₂H₅OH + PCl₃ → 2C₂H₅Cl₃ + H₃PO₃

c) PCl₃ reacts with water (hydrolysis) to form phosphorus acid. It undergo hydrolysis in presence of moisture.
PCl₃ + 3H₂O → H₃PO₃ + 3HCl

Question 32.
PCl₃ can act as an oxidizing as well as a reducing agent – Justify.
Answer:
PCl₃ acts as reducing agent. It is evidented by the following reaction.

PCl₃ acts as oxidising agent. It is evidented by the following reaction.

Question 33.
Which of the following are not known ?
PCl₃, AsCl₃, SbCl₃, NCl₅, BiCl₅, PH₅
Answer:
NCl₅, BiCl₅, PH₅ are not known in the given compounds.

Question 34.
Which of the following is more covalent – SbCl₅ or SbCl₃ ?
Answer:
SbCl₅ (Penta halide) is more covalent than SbCl₃ (Tri halide). Because Sb in the higher oxidation state exert more polarising power.

Question 35.
Write the oxidation states of phosphorus in solid PCl₅.
Answer:

• In solid state PCl₅ exists as an ionic solid [PCl₄)⁺ [PCl₆]^–
• ‘P’ exhibits +5 oxidation state in [PCl₄)⁺ [PCl₆]^–

Question 36.
Illustrate how copper metal can give different products on reaction with HNO₃.
Answer:
Reaction of Cu metal with dilute HNO₃.

Reaction of Cu Metal with cone. HNO₃

Question 37.
Which oxide of nitrogen has oxidation number of N same as that in nitric acid ?
Answer:

• In HNO₃, ‘N’ has oxidation state +5.
• Among oxides of nitrogen N₂O₅ exhibits ‘+5’ oxidation state.

Question 38.
Write the chemical reactions that occur in the manufacture of nitric acid.
Answer:
Chemical reactions involved in the manufacturing of HNO₃.

• Oxidation of NH₃
  
• NO₂ formation :
  2NO + O₂ ⇌ 2NO₂
• Formation of HNO₃:
  3NO₂ + H₂O → 2HNO₃ + NO

Question 39.
Iron becomes passive in cone. HNO₃ – Why ?
Answer:
Iron becomes passive in cone. HNO₃ due to formation of a passive film of oxide on the surface of iron.

Question 40.
Give the uses of
a) nitric acid and
b) ammonia.
Answer:
Use of HNO₃:

• HNO₃ is used in the manufacture of ammonium nitrate for fertilisers and other nitrates for use in explosives and pyrotechnics.
• HNO₃ is used in the pickling of stainless steel.
• HNO₃ is used as oxidiser in rocket fuels.

Question 41.
What are the oxidation states of phosphorus in the following ?
i) H₃PO₃
ii) PCl₃
iii) Ca₃P₂
iv) Na₃PO₄
v) POF₃
Answer:
i) H₃PO₃
3(1) + x + 3(-3) = 0
x = +3

ii) PCl₃
x + 3(-1) = 0
x = 3

iii) Ca₃P₂
3(+2) + 2x = 0
x = -3

iv) Na₃PO₄
3(1) + x + 4(-2) = 0
x = +5

v) POF₃
x + (-2) + 3(-1) = 0
x = +5

Question 42.
H₃PO₃ is diprotic while H₃PO₂ is monoprotic – Why ?
Answer:
H₃PO₃ is diprotic :
Structure of H₃PO₃:

H₃PO₂ is monoprotic :
Structure of H₃PO₂ :

Question 43.
Give the disproportionation reaction of H₃PO₃.
Answer:
Orthophosphoric acid (H₃PO₃) on heating disproportionates to give orthophosphoric acid and phosphine.
4H₃PO₃ → 3H₃PO₄ + PH₃.

Question 44.
H₃PO₂ is a good reducing agent – Explain with an example.
Answer:
In H₃PO₂, two H-atoms are bonded directly to P-atom which imparts reducing character to the acid.

Question 45.
Draw the structures of
a) Hypo phosphoric acid
b) Cyclic meta phosphoric acid.
Answer:
a) Structure of hypo phosphoric acid (H₄P₂O₆) :

b) Structure of cyclic meta phosphoric acid (HPO₃)₃

Short Answer Questions

Question 1.
Discuss the general characteristics of Group – 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionization enthalpy and electronegativity.
Answer:
1) Electronic Configuration : All these elements of this group have s²p³ configuration in their respective outmost orbits.
Nitrogen (N) : [He] 2s²2p³; phosphorus (P) : [Ne] 3s²3p³, Arsenic (As) : [Ar] 3d¹⁰4s²4p³
Atimony (Sb) : [Kr] 4d¹⁰ 5s² 5p³ and Bismuth (Bi) = [Xe] 4f¹⁴ 5d¹⁰ 6s²6p³

2) Oxidation Spates : As per their configurations, these elements may utilize either the three electrons iti the p-orbitals or the five electrons present in both s -and p -orbitals in the exhibition of oxidation states. Consequently, they exhibit common oxidation states +III and +V.

3) Atomic size: In Group-15 elements atomic size increase down the group. A considerable increase in covalent radius observed from N to P and from As to Bi only a small increase in atomic size observed.

4) Ionisation enthalpy : In group – 15 elements ionisation enthalpy decreases down the group due to gradual increase in atomic size.

5) Electronegativity : In group-15 elements electronegativity values decrease down the group with increase in atomic size.

Question 2.
Discuss the trends in chemical reactivity of group – 15 elements.
Answer:
Nitrogen gas exists as diatomic molecule. Dut to the presence of triple bond between Base N – Atoms bond dissociation energy is high (941.4 KJ /mol). Hence nitrogen is inert and unreactive.

Phosphorus is a tetra atomic molecule and P-P single bond is weaker than N ≡ N. P – P bond dissociation energy U 213 KJ/mole. Hence phosphorus is more reactive than Nitrogen.

Explanation :
The single N-N bond is weaker than the single P-P bond due to high inter electronic repulsion of the non-bonding electrons in N₂ because of small bond length. There fore the catenation property is weaker in nitrogen as compoud to phosphorares.
R₃ P = O exist but R₃N = O does not.

Explanation :
Nitrogen does not form dπ – Pπ multiple bond with oxygen because of lack of d – orbitals in Nitrogen atom. But in case of R₃N = 0 the value of nitrogen should be 5. So these compounds do not exist where as in case of P’ atom d-orbitals are available. So P-atom can able to form dπ – Pπ multiple bonds hence R₃P = 0 exist.

1. Reactivity towards hydrogen: Group -15 elements forms EH₃ type hydrides (E = Group – 15 elements) .
   Eg : PH₃, NH₃, AsH₃, BiH₃, SbH₃.
   • Among above hydrides NH₃ is mild reducing agent while BiH₃ is strong reducing agent.
   • Stability of hydrides decreases from NH₃ to BiH₃.
   • Basicity of hydrides decreases as follows.
     NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃.
2. Reactivity towards Oxygen : These forms two types of oxides E₂O₃ and E<sub2O₅.
   Eg : P₂O₃ > N₂O₅, N₂O₃, P₂O₅.
   • Acidic character of oxides decrease down the group.
   • E₃O₃ of N and P are acidic, As and Sb are amphoteric while ‘Bi’ is basic.
3. Reactivity towards halogens : These elements forms two types of halides EX₃ and EX₅.
   ‘N’ – does not form penta halides due to lack of the d-orbitals.
   Penta halides are more covalent than tri halides because the elements in higher oxidation state have more polarising power.
4. Reactivity towards metals : All these elements react with metals to form their binary compounds containing – 3 oxidation state.
   Eg. : Ca₃N₂, Ca₃P₂ etc.

Question 3.
How does P₄ react with the following ?
a) SOCl₂
b) SO₂Cl₂
Answer:
a) When P₄ reacts with SOCl₂ to form phosphorus trichloride.
P₄ + 8 SOCl₂ → 4 PCl₃ + 4 SO₂ + 2 S₂Cl₂
b) When P₄ reacts with SOCl₂ to form phosphorus pentachloride.
P₄ + 10 SOCl₂ → 4 PCl₅ + 10 SO₂

Question 4.
Explain the anomalous nature of nitrogen in group -15.
Answer:
Anomalous properties of Nitrogen : Nitrogen differs from the remaining elements of this group due to its small size, high electronegativity, high ionisation enthalpy and non-availability of d-orbitals.

Nitrogen gas exists as diatomic molecule. Due to the presence of triple bond between base N – Atoms bond dissociation energy is high (941.4 KJ /mol). Hence nitrogen is inert and un-reactive.

Explanation :

• The single N-N bond is weaker than the single P-P bond due to high inter electronic repulsion of the non-bonding electrons in N₂ because of small bond length. Therefore the catenation property is weaker in nitrogen as compound to phosphorus.
• Nitrogen does not form penta halides due to lack of d-orbitals.
  R₃ P = O exist but R₃N = O does not.

Explanation :
Nitrogen does not form dπ – Pπ multiple bond with oxygen because of lack of d – orbitals in Nitrogen atom. But in case of R₃N = O the value of nitrogen should be 5. So these compounds do not exist where as in case of ‘P atom d-orbitals are available. So P-atom can able to form dπ – Pπ multiple bonds hence R₃P = O exist.

Question 5.
Complete the following reactions.
a) Ca₃P₂ + HaO →
b) P₄ + KOH →
c) CuSO₄ + NH₃ →
d) Mg + N₂ →
d) (NH₄)₂ + Cr₂O₇
f) Decomposition of nitrous acid →
Answer:
a) Ca₃P₂ + 6 H₂O → 3 Ca(OH)₂ + 2PH₃
b) P₄ + 3 KOH + 3 H₂O → PH₃ + 3KH₂ PO₂
c) CuSO_4(aq) (blue) + 4NH₃ → [Cu (NH₃)₄] SO₄ (deep blue)
d) 3 Mg + N₂ → Mg₃N₂
e) (NH₄)₂ + Cr₂O₇ N₂ + 4 H₂O + Cr₂O₃
f) 3HNO₂ → HNO₃ + 2 NO + H₂O

Question 6.
How does PCl₅ react with the following?
a) Water
b) C₂H₅OH
c) CH₃COOH
d) Ag
Answer:
a) PCl₅ undergo hydrolysis to form phosphoric acid.
PCl₅ + H₂O → POCl₃ + 2HCl
POCl₃ + 3H₂O → H₃PO₄ + 3 HCl

b) PCl₅ reacts with C₂H₅OH to form Ethyl chloride.
C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl

c) PCl₅ reacts with CH₃COOH to form acetyl chloride.
CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl

d) PCl₅ reacts with Ag to form PCl₃ and AgCl
PCl₅ + 2 Ag → 2 AgCl + PCl₃

Question 7.
Complete the following.
a) NH₄NO₃
b) HNO₃ + P₄O₁₀ →
c) Pb(NO₃)₂
d) Zn + dil.HNO₃ →
e) P₄ + conc.HNO₃ →
f) HgCl₂ + PH₃ →
Answer:
a) NH₄NO₃ N₂O (Nitrous oxide) + 2 H₂O
b) 12 HNO₃ + P₄O₁₀ → 6 N₂O₅ + 4 H₃PO₄
c) 2 Pb(NO₃)₂ 2 PbO + 4 NO₂ + O₂
d) 4 Zn + 10 HNO₃ (dil) → 4 Zn (NO₃)₂ + 5 H₂O + N₂O
e) P₄ + 20 HNO₃ → 4H₃PO₄ + 20 NO₂ + 4 H₂O
f) 3 HgCl₂ + 2 PH₃ → Hg₃P₂ + 6 HCl

Long Answer Questions

Question 1.
How is ammonia manufactured by Habes process? Explain the reactions of ammonia with [A.P. Mar. 18]
a) ZnSO_4(aq)
b) CuSO_4(aq)
c) AgCl_(s)
Answer:
In Haber process ammonia is directly synthesised from elements (nitrogen and hydrogen). The principle involved in this is
N₂ (g) + 3H₂O (g) ⇌ 2NH₃ + 92.4kJ
This is a reversible exothermic reaction.
According to Le Chatelier’s principle favourable conditions for the better yield of ammonia are low temperature and high pressure. But the optimum conditions are
Temperature : 720k
Pressure : 200 atmospheres
Catalyst : Finely divided iron in the presence of molybdenum (Promoter).
Procedure : A mixture of nitrogen and hydrogen in the volume ratio 1 : 3 is heated to 725 – 775K at a pressure of 200 atmospheres is passed over hot finely divided iron mixed with small amount of molybdenum as promotor. The gases coming out of the catalyst chamber consists of 10 – 20% ammonia gas are cooled and compressed, so that ammonia gas is liquified, and the uncondensed gases are sent for recirculation.

b) Aq.CuSO₄ reacts with ammonia to form a deep blue complex,

c) Solid AgCl reacts with ammonia to form a colourless complex.

Question 2.
How is nitric acid manufactured by Ostwald’s process ? How does it react with the following ? [T.S. Mar. 19]
a) Copper
b) Zn
c) S8
d) P4
Answer:
Ostwald’s process : Ammonia, mixed with air in 1 : 7 or 1 : 8, when passed over a hot platinum gauze catalyst, is oxidised to NO mostly (about 95%).
The reaction is

The liberated heat keeps the catalyst hot. The ‘NO’ is cooled and is mixed with oxygen to give the dioxide, in large empty towers (oxidation chamber). The product is then passed in to warm water, under pressure in the presence of excess of air, to give HNO₃.
4NO₂ + O₂ + 2H₂O → 4HNO₃.
The acid formed is about 61% concentrated.

a) Copper reacts with dil. HNO₃ and cone. HNO₃ and liberates Nitric Oxide and Nitrogen dioxide respectively
3 Cu + 5 HNO₃ (dil) → 3 Cu (NO₃)₂ + 2 NO + 4 H₂O
Cu + 4 HNO₃ (cone.) → Cu (NO₃)₂ + 2 NO₂ + 2 H₂O

b) Zn reacts with dil. HNO₃ and Cone. HNO₃ and liberates Nitrous oxide and Nitrogen dioxide respectively.
4 Zn + 10 HNO₃ (dil) → 4 Zn (NO₃)₂ + 5 H₂O + N₂O
Zn + 4 HNO₃ (cone.) → Zn (NO₃)₂ + 2 H₂ + 2 NO₂

c) S₈ reacts with cone, nitric acid to form Sulphuric acid, NO₂ gas.
S₈ + 48 HNO₃ → 8 H₂SO₄ + 48 NO₂ + 16 H₂O
d) P₄ reacts with cone, nitric acid to form phosphoric acid and NO₂ gas.
P₄ + 20 HNO₃ → 4 H₃PO₄ + 20 NO₂ + 4 H₂O

Textual Examples

Question 1.
Though nitrogen exhibits +5 oxidation state, it does not form peniahalide. Give reason.
Answer:
Nitrogen with n = 2, has s and p orbitals only. It does not have d orbitals to expand its covalence beyond four. That is why it does not form pentahalide.

Question 2.
PH₃ has lower boiling point than NH₃. Why ? [T.S. Mar. 16]
Answer:
Unlike NH₃. PH₃ molecules are not associated through hydrogen bonding in liquid state. That is why the boiling point of PH₃ is lower than NH₃.

Question 3.
Write the reaction of thermal decomposition of sodium azide.
Answer:
Thermal decomposition of sodium azide gives dinitrogen gas.
2NaN₃ → 2Na + 3N₂

Question 4.
Why does NH₃ act as a Lewis base ?
Answer:
Nitrogen atom in NH₃ has one lone pair of electrons with is available for donation. Therefore, it acts as a Lewis base.

Question 5.
Why does NO₂ dimerise ?
Answer:
NO₂ contains odd number of valence electrons. It be haves as a typical odd molecule. On dimerisation. It is converted to stable N₂O₄ molecule with even number of electrons.

Question 6.
In what way can it be proved that PH₃ is basic in nature ?
Answer:
PH₃ reacts with acids like HI to form PH₄I which shows that it is basic in nature.
PH₃ + HI →PH₄I
Due to lone pair on phosphorus atom, PH₃ is acting as a Lewis base in the above reaction.

Question 7.
Why does PCl₃ fume in moisture ?
Answer:
PCl₃ hydrolyses in the presence of moisture giving fumes of HCl.
PCl₃+ 3H₂O → H₃PO₃ + 3HCl

Question 8.
Are all the five bonds in PCl₅ molecule equivalent ? Justify your answer.
Answer:
PCl₅ has a trigonal bipyramidal structure and the three equatorial P-Cl bonds are equivalent. While the two axial bonds are different and longer than equatorial bonds.

Intext Questions

Question 1.
Why are pentahalides more covalent than trihalides ?
Answer:
Higher the positive oxidation state of central atom, more will be its polarising power, which in turn increases the covalent character of the bond formed between the central atom and the halogen atom. In pentahalides, the central atom is in +5 oxidation state while in trihalides, it is in +3 oxidation state. Therefore, pentahalides are more covalent than trihalides.

Question 2.
Why is BiH₃ the strongest reducing agent amongst all the hydrides of Group 15 elements ?
Answer:
Among the 15 group elements, the size of Bi atom is largest and hence, the Bi-H bond length is largest or Bi-H bond dissociation energy is lowest. That’s why Bi-H bond dissociates more readily than the other hydrides of the group and hence, BiH₃ is the strongest reducing agent.

Question 3.
Why is N₂ less reactive at room temperature ?
Answer:
Dinitrogen is inert of less reactive because the bond enthalpy of N ≡ N bond is very high.

Question 4.
Mention the conditions required to maximise the yield of ammonia.
Answer:
Ammonia is produced by Haber’s process as.

Yield of ammonia is favoured by high pressure according to Le-Chatelier’s principle, Other conditions, that favour the production of ammonia are as follows :

1. Temperature – approximately 700 K
2. Pressure – 200 atm or 200 × 10⁵ Pa
3. Catalyst – Iron oxide
4. Promotor – Molybdenum of K₂O and Al₂O₃

Question 5.
How does ammonia react with a solution of Cu²⁺?
Answer:
When ammonia (aqueous solution is ammonium hydroxide) reacts with a solution of Cu²⁺, a deep blue solution is obtained due to the formation of tetraamine copper (II) ion.

Question 6.
What is the covalence of nitrogen in N₂O₅?
Answer:
Structural formula of N₂O₅

Since, N atom has 4 shared electron pairs; the valence of N is 4.

Question 7.
Bond angle in PH₄⁺ is higher than that in PH₃. Why ?
Answer:
In both PH₄⁺ and PH₃, phosphorus atom is sp³ hybridized. In PH₄⁺, all the four orbitals are bonded whereas in PH₃, there is a lone pair of electrons too. Due to lone pair-bonded pair repulsion in PH₃, the bond angle is less than 109.5°.

Question 8.
What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂? .
Answer:
White phosphorus dissolves in boiling NaOH in an inert atmosphere of CO₂ producing phosphine (PH₃) gas and sodium hypophosphite (NaH₂PO₂).

Question 9.
What happens when PCl₅ is heated ?
Answer:
In PCl₃, there are 5P – Cl bonds, out of which three are equatorial (longer) and two are axial (shorter). When PCl₅ is heated strongly, two less stable axial bonds break and phosphorus trichloride (PCl₃) is formed.

Structure of PCl₅ showing axial and equatorial bonds.

Question 10.
Write a balanced equation for the hydrolytic reaction of PCl₅ in heavy water.
Answer:
Reaction of PCl₅ in heavy water (D₂O)

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 3(a) Electro Chemistry Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 3(a) Electro Chemistry

Very Short Answer Questions

Question 1.
What is a galvanic cell or a voltaic cell? Give one example. [T.S. Mar. 19]
Answer:
Galvanic cell: A device that converts chemical energy into electrical energy by the use of spontaneous redox reaction is called a Galvanic cell (or) voltaic cell.
Eg.: Daniell cell.

Question 2.
Write the Chemical reaction used in the construction of the Daniell cell together with the half-cell reactions.
Answer:
The chemical reactions used in the construction of the Daniell cell.

Question 3.
Name the two half-cell reactions that are taking place in the Daniell cell.
Answer:
The two half cell reactions that are takes place in the Daniell cell are oxidation and reduction.
Zn → Zn⁺² + 2e^– (Oxidation)
Cu⁺² + 2e^– → Cu (Reduction)

Question 4.
How is a galvanic cell represented on paper as per IUPAC convention ? Give one example.
Answer:
Representation of a Galvanic cell:
As per IUPAC convention

• Oxidation half cell represented on the left side.
• Reduction half cell represented on the right side.
• Oxidation half cell and reduction half cell are connected by a salt bridge indicated by two vertical parallel lines.
  Eg.: Cu_(s) | \(\mathrm{Cu}_{(\mathrm{aq})}^{+2}\) || \(\mathrm{Ag}_{(\mathrm{aq})}^{+}\) | Ag_(s)

Question 5.
Write the cell reaction taking place in the cell
Cu_(s) | \(\mathrm{Cu}_{(\mathrm{aq})}^{+2}\) || \(\mathrm{Ag}_{(\mathrm{aq})}^{+}\) | Ag_(s)
Answer:
Given cell is Cu_(s) | \(\mathrm{Cu}_{(\mathrm{aq})}^{+2}\) || \(\mathrm{Ag}_{(\mathrm{aq})}^{+}\) | Ag_(s)
Cu → Cu⁺² + 2e^– (Oxidation)
2Ag⁺ + 2e^– → 2Ag (Reduction)

Question 6.
What is standard hydrogen electrode ?
Answer:
The electrode whose potential is known as standard electrode (or) standard hydrogen electrode.

To determine the potential of a single electrode experimentally it combine with standard hydrogen electrode and the EMF of cell so constructed is measured with potentiometer.

Question 7.
Give a neat sketch of standard hydrogen electrode.
Answer:

Standard Hydrogen Electrode (SHE)

Question 8.
What is Nernst equation ? Write the equation for an electrode with electrode reaction Mⁿ⁺(aq) + ne^– ⇌ M(s).
Answer:
The electrode potential at any concentration measured with respect to standard hydrogen electrode is represented by Nernst equation.
Nernst equation is
E = E⁰ + \(\frac{\mathrm{RT}}{\mathrm{nF}}\) ln[Mⁿ⁺] [Metal Electrodes]
Given electrode reaction is \(\mathrm{M}_{(\mathrm{aq})}^{\mathrm{n}+}\) + ne^– ⇌ M_(s)
For the above electrode reaction Nemst equation is
E_(Mⁿ⁺/M) = \(\mathrm{E}_{\left(\mathrm{M}^{\mathrm{n}+} / \mathrm{M}\right)}^0+\frac{\mathrm{RT}}{\mathrm{nF}} \ln \left[\mathrm{M}^{\mathrm{n}+}\right]\)
Here E_(Mⁿ⁺/M) = Electrode potential
E⁰E_(Mⁿ⁺/M) = Standard Electrode potential
R = gas constant = 8.314 J/k.mole
F = Faraday = 96487 c/mole
T = temperature
[Mⁿ⁺] = concentration of species Mⁿ⁺

Question 9.
A negative E⁰ indicates that the redox couple is _______________ reducing couple than H⁺/H₂, couple, (powerful or weak)
Answer:
A negative E⁰ value indicates that the redox couple is powerful reducing couple than H⁺/H₂ couple.

Question 10.
A positive E⁰ indicates that the redox couple is a weaker _____________ couple than H⁺/H₂ couple. (oxidising or reducing)
Answer:
A positive E⁰ value indicates that the redox couple is as weaker reducing couple than H⁺/H₂ couple.

Question 11.
Write the Nernst equation for the EMF of the cell .
Ni_(s) |\(\mathrm{Ni}_{(\mathrm{aq})}^{2+}\) || \(\mathrm{Ag}_{(\mathrm{aq})}^{+}\) | Ag
Answer:
Given cell is Ni_(s) |\(\mathrm{Ni}_{(\mathrm{aq})}^{2+}\) || \(\mathrm{Ag}_{(\mathrm{aq})}^{+}\) | Ag
Nernst equation for the cell is
E_cell = \(\mathrm{E}_{\text {cell }}^0+\frac{\mathrm{RT}}{\mathrm{nF}} \ln \frac{\left[\mathrm{Ag}^{+}\right]}{[\mathrm{Ni}]^{+2}}\)

Question 12.
Write the cell reaction for which E_cell = \(\mathrm{E}_{\text {cell }}^0-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2}\)
Answer:
Given E_cell = \(\mathrm{E}_{\text {cell }}^0-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2}\)
The cell reaction is Mg_(s)/\(\mathrm{Mg}_{(\mathrm{aq})}^{2+}\) || \(\mathrm{Ag}_{(\mathrm{aq})}^{+}\) | Ag_(s)

Question 13.
How is E⁰ cell related mathematically to the equilibrium constant K_c of the cell reaction?
Answer:
Relation between E₀ cell and equilibrium constant K_C of the cell reaction
\(\mathrm{E}_{\text {cell }}^0=\frac{2.303 \mathrm{RT}}{\mathrm{nF}}\) log K_C
n = number of electrons involved
F = Faraday 96500 C mol^-1,
T = Temperature
R = gas constant

Question 14.
How is Gibbs energy (G) related to the cell emf (E) mathematically ?
Answer:
Relation between Gibb’s energy (G) and emf (E) mathematically
∆G⁰ = -nFE_(cell)
∆G = change in Gibb’s energy
n = number of electrons involved
F = Faraday = 96500 C mol^-1

Question 15.
Define conductivity of a material. Give its SI units.
Answer:
The reciprocal of specific resistance (or) resistivity is called conductivity.
It is represented by (κ)
(Or)
The conductance of one unit cube of a conductor is also called conductivity.
SI units : ohm^-1 m^-1 (or) Sm^-1 S = Siemen

Question 16.
What is cell constant of a conductivity cell ?
Answer:
Resistance R = ρ . \(\frac{l}{\mathrm{~A}}\) = \(\frac{l}{\kappa \cdot \mathrm{A}}\)
\(\frac{l}{\mathrm{~A}}\) = cell constant
Cell constant G = \(\frac{l}{\mathrm{~A}}\) = R × κ
The cell constant of a conductivity cell is the product of resistance and specific conductance.

Question 17.
Define molar conductivity ∧_m and how is it related to conductivity (κ) ?
Answer:
Molar conductivity: The conductivity of a volume of solution containing one gram molecular weight of the electrolyte placed between two parallel electrodes separated by a distance of unit length of 1 meter is called molar conductivity (∧_m).
Relation between conductivity and molar conductivity:
∧_m = \(\frac{\kappa}{\mathrm{c}}\);
∴ c = constant

Question 18.
Give the mathematical equation which gives the variation of molar conductivity with ∧_m the molarity (c) of the solution ?
Answer:
The mathematical equation which gives the variation of molar conductivity ∧_m with the molarity (c) of the solution is
∧_m = \(\frac{\kappa}{1000\left(\text { lit } / \mathrm{cm}^3\right) \times \text { molarity (moles/lit) }}=\frac{\kappa \times 1000\left(\mathrm{~cm}^3 / \mathrm{lit}\right)}{\text { molarity (mole/lit) }}\)

Question 19.
State Kohlrausch’s law of independent migration of ions.
Answer:
Kohlrausch’s law of independent migration of ions: The limiting molar conductivity of an electrolytes can be represented as the sum of the individual contributions of the anion and the cation of the electrolytes.
\(\lambda_{\mathrm{m}(\mathrm{AB})}^0=\lambda_{\mathrm{A}^{+}}^0+\lambda_{\mathrm{B}^{-}}^0\)
\(\lambda_m^0\) = Limiting molar conductivity
\(\lambda_A^0\) = Limiting molar conductivity of cation
\(\lambda_B^0\) = Limiting molar conductivity of anion

Question 20.
State Faraday’s first law of electrolysis. [A.P. Mar. 18, 16] [Mar. 14]
Answer:
The amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of current passing through the electrolyte.
(Or)
The mass of the substance deposited at an electrode during the electrolysis of electrolyte is directly proportional to quantity of electricity passed through it.
m ∝ Q; m ∝ c × t
m = ect; m = \(\frac{\text { Ect }}{96,500}\)
e = electrochemical equivalent
c = Current in amperes
t = time in seconds
E = Chemical equivalent

Question 21.
State Faraday’s second law of electrolysis. [T.S. Mar. 18] [Mar. 14]
Answer:
The amounts of different substances liberated when the same quantity of electricity is passed through the electrolytic solution are proportional to their chemical equivalent weights.
m ∝ E

Qustion 22.
What are the products obtained at the platinum anode and the platinum cathode respectively in the electrolysis of fused or molten NaCl ?
Answer:
In the electrolysis of fused (or) molten NaCl with platinum anode and platinum cathode, sodium metal obtained at cathode and chlorine gas at anode
2 NaCl → 2 Na⁺ + 2Cl^–
2 Cl^– → Cl₂ + 2e^– (anode)
2 Na⁺ + 2e^– → 2 Na (cathode)

Question 23.
Give the products obtained at the platinum electrodes (cathode and anode) when aqueous solution of K₂SO₄ is electrolysed.
Answer:
During the electrolysis of aq.K₂SO₄ using pt-anode and pt-cathode hydrogen gas liberated at cathode and oxygen gas at anode.
K₂SO₄ → 2K⁺ + \(\mathrm{SO}_4^{-2}\)

Question 24.
Write the chemical equation corresponding to the oxidation of H₂O_(l) at the platinum anode.
Answer:
The chemical equation corresponding to the oxidation of H₂O_(l) at the platinum anode is
2H₂O_(l) → O_2(g) + 4H⁺_(aq) + 4e^–

Question 25.
Give the chemical equation that represents the reduction of liquid water H₂O_(l)at the platinum cathode.
Answer:
The chemical equation that represents the reduction of liquid water H₂O_(l) at the platinum cathode is H₂O_(l) + e^– → \(\frac{1}{2}\)H_2(g) + OH^–

Question 26.
What is a primary battery ? Give one example. [A.P. Mar. 17]
Answer:
The batteries which after their use over a period of time, becomes dead and the cell reaction is completed and this cannot be reused again are called primary batteries.
Eg: Leclanche cell, dry cell.

Question 27.
Give one example for a secondary battery. Give the cell reaction.
Answer:
Lead storage battery is an example of secondary battery.
The cell reactions when the battery is in use are
Pb_(s) + \(\mathrm{SO}_4^{-2} \text { (aq) }\) → PbSO_4(s) + 2e^– (Anode)
PbO_2(s) + \(\mathrm{SO}_4^{-2} \text { (aq) }\) → \(4 \mathrm{H}_{(\mathrm{aq})}^{+}\) + 2e^– → PbSO_4(s) + 2 H₂O_(l) (Cathode)
Overall cell reaction is .
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(s) + H₂O_(l)

Question 28.
Give the cell reaction of nickel-cadmium secondary battery.
Answer:
The cell reaction of nickel-cadmium secondary battery is
Cd_(s) + 2Ni(OH)_3(s) → CdO_(s) + 2 Ni(OH)_2(s) + H₂O_(l)

Question 29.
What is a fuel cell ? How is it different from a conventional galvanic cell ?
Answer:
A fuel cell is a galvanic cell in which the chemical energy of fuel-oxidant system is converted directly into electrical energy.

• Conventional Galvanic cell converts chemical energy into electrical energy by spontaneous redox reactions.
• Fuel cell convert energy of combustion of fuels like hydrogen, methane etc., into electrical energy. These cause less pollution.

Question 30.
Give the electrode reactions occurijig at the anode and at the cathode in H₂, O₂, fuel cell.
Answer:
At cathode : O_2(g) + 2 H₂, O₂ + 4e^– → 4OH^–_(aq)
At anode : 2H_2(g) + 4 OH^–_(aq) → 4H₂O_(l) + 4e^–
Overall reaction : 2H_2(g) + O_2(g) → 2H₂O_(l)

Question 31.
What is metallic corrosion ? Give one example. [T.S. Mar. 17; A.P. Mar. 15]
Answer:
Metallic corrosion: The natural tendency of conversion of a metal into its mineral compound form i n interaction with the environment is known as metallic corrosion.
Eg : 1) Iron converts itself into its oxide [Rusting] (Fe₂O₃)
2) Silver converts itself into it’s sulphate
[tarnishing] [Ag₂S]

Question 32.
Give the electro-chemical reaction that represents the corrosion or rusting of iron.
Answer:
Rusting of iron: Iron converts itself into its oxixe Fe₂O₃
The electrochemical reaction that represents the corrosion or rusting of iron is
Anode : 2 Fe_(s) → 2 Fe²⁺ + 4e^– \(\mathrm{E}_{\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right)}^0\) = -0.44V
Cathode : O_2(g) + 4H⁺_(aq) + 4e^– → 2H₂O_(l) \(\mathrm{E}_{\mathrm{H}^{+} / \mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\Theta}\) = 1.23V
Overall reaction: 2 Fe_(s) + O_2(g) + 4H⁺_(aq) → 2Fe⁺²_aq + 2 H₂O_(l) \(\mathrm{E}_{\text {(cell) }}^{\Theta}\) = 1.67 V

Short Answer Questions

Question 1.
What are galvanic cells ? Explain the working of a galvanic cell with a neat sketch taking Daniell cell as example. [T.S. & A.P. Mar. 18]
Answer:
Galvanic cell: A device which converts chemical energy into electrical energy by the use of spontaneous redox reaction is called Galvanic cell (or) voltaic cell.
Eg : Daniell cell.

Daniell cell: It is a special type of galvanic cell. It contains two half cells in the same vessel. The vessel is devided into two chambers. Left chamber is filled with ZnSO₄ (aq) solution and Zn – rod is dipped into it. Right chamber is filled with aq. CuSO₄ solution and a copper rod is dipped into it. Process diaphragm acts as Salt bridge. The two half cell’s are connected to external battery.

Cell reactions :
Ion Zn/ZnSO₄ half cell, oxidation, reaction occurs
Zn → Zn²⁺ + 2e^–
Ion Cu/CuSO₄ half cell, reduction reaction occurs.
The net cell reaction is
Zn + Cu⁺² ⇌ Zn⁺² + Cu
Cell is represented as Zn / Zn⁺² || Cu⁺² / Cu

Question 2.
Give the construction and working of a standard hydrogen electrode with a neat diagram.
Answer:
To determine the potential of a single electrode experimentally, it is combined with a standard hydrogen electrode (electrode whose potential is known) and the EMF of the cell. So constructed is measured with a potentiometer. Standard hydrogen electrode is constructed and is used as standard electrode or reference electrode. Standard Hydrogen Electrode (SHE) or Normal Hydrogen Electrode (NHE).

Pure hydrogen gas is bubbled into a solution of 1M HCl along a platinum electrode coated with platinum block. A platinum block electrode placed in the solution at atmospheric pressure.

Generally the electrode is fitted into the tube. The tube will have two circular small holes. This tube is immersed in the acid solution such that one haf of the circular hole is exposed to air and another half in the solution.
The following equilibrium exists at the electrode.
\(\frac{1}{2}\) H_2(g) (1_atm) ⇌ \(\mathrm{H}_{(\mathrm{aq})}^{+}\) (1M) + e^–

Question 3.
State and explain Nernst equation with the help of a metallic electrode and a non-metallic electrode.
Answer:
The electrode potential at any concentration measured with respect to standard hydrogen electro represented by Nernst equation.
Nernstst equation is
E = E⁰ + \(\frac{\mathrm{RT}}{\mathrm{nF}}\) In [Mⁿ⁺] [Metal Electrodes]
Given electrode reaction is
\(\mathrm{M}_{(\mathrm{aq})}^{\mathrm{n}+}\) + ne^– ⇌ M_(s)
For the above electrode reaction Nernst equation is
E_{(Mⁿ⁺ / M)} = \(\mathrm{E}_{\left(\mathrm{M}^{\mathrm{n}+} / \mathrm{M}\right)}^0+\frac{\mathrm{RT}}{\mathrm{nF}} \ln \left[\mathrm{M}^{\mathrm{n}+}\right]\)
Here E_{(Mⁿ⁺ / M)} = Electrode potential
E⁰_{(Mⁿ⁺ / M)} = Standard Electrode potential
R = gas constant = 8.314 J/k.mole
F = Faraday = 96487 c/mole
T = temperature
[Mⁿ⁺] = concentration of species Mⁿ⁺
For non-metal electrodes :
E = E⁰ – \(\frac{\mathrm{RT}}{\mathrm{nF}}\) ln C ; C = concentration
Example for metal electrode:
Given cell is Ni_(s) | \(\mathrm{Ni}_{(a q)}^{2+}\) || \(\) | Ag
Nernst equation for the cell is
E_cell = \(\mathrm{E}_{\text {cell }}^0+\frac{\mathrm{RT}}{\mathrm{nF}} \ln \frac{\left[\mathrm{Ag}^{+}\right]}{[\mathrm{Ni}]^{+2}}\)
Example for non-metal electrode: Pt, Cl₂/Cl^–
E = E⁰ – \(\frac{\mathrm{RT}}{\mathrm{nF}}\) log C
E = E⁰ – \(\frac{\mathrm{RT}}{\mathrm{F}}\) log C

Question 4.
Explain with a suitable example the relation between the Gibbs energy of chemical reaction (G) and the functioning of the electrochemical cell.
Answer:
Electrical work done in one second in an electrochemical cell is equal to the electrical potential multiplied by the total charge passing. If we want to obtain maximum work from a galvanic cell then the charge has to be passed reversibly. The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy and therefore, if the emf is E and nF is the amount of charge passing and ∆_rG = – nFE_[cell]
It may be remembered that E_[cell] is an intensive parameter but ∆_rG is an extensive thermodynamic property and the value depends on n. Thus, if we write the reaction
Zn_[s] + \(\mathrm{Cu}_{[\mathrm{aq}]}^{2+}\) → \(\mathrm{Zn}_{\text {[aq] }}^{2+}\) + Cu_[s]
∆_rG = -2FE_[cell]
but when we write the reaction
2 Zn_[S] + 2 \(\mathrm{Cu}_{[\mathrm{aq}]}^{2+}\) → \(\mathrm{Zn}_{[\text {aq] }}^{2+}\) + 2 Cu_[s]
∆_rG = -4FE_[cell]
If the concentration of each of the reacting species is unity, then E_[cell] = \(\mathrm{E}_{\text {[cell] }}^{\Theta}\) and we have
∆_rG^⊖ -nF\(\mathrm{E}_{\text {[cell] }}^{\Theta}\)
We can calculate equilibrium constant using the equation
∆_rG^⊖ = – RTln K_c

Question 5.
On what factors the electrical conductance of an aqueous solution of electrolyte depends?
Answer:
The electrical conductance of an aqueous solution of electrolyte depends on

1. The nature of the electrolyte
2. Size and solvation of the ions formed in the dissociation of the electrolyte.
3. The nature and viscosity of the solvent.
4. Concentration of the electrolyte
5. Temperature

Question 6.
How is molar conductivity of an aqueous electrolyte solution measured experimentally ?
Answer:
Molar conductivity: The conductivity of a volume of solution containing one gram molecular weight of the electrolyte placed between two parallel electrodes separated by a distance of unit length of 1 meter is called molar conductivity (∧_m)
Relation between conductivity and molar conductivity:
∧_m = \(\frac{\kappa}{\mathrm{c}}\);
∴ c = constant .
The conductance of a solution measured in a conductivity cell.
By using the conductivity cells
Resistance (R) = \(\frac{l}{\kappa \times A}\)
l = distance between electrodes; A = Area of cross section; K = conductivity
G* = \(\frac{l}{\mathrm{~A}}\) = cell constant
-4 Cell constant is measured by measuring the resistance of the cell containing a solution whose conductivity is known.
-4 Cell constant determined is used for measuring the resistance (or) conductivity.
∴ molar conductivity ∧_(m) = \(\frac{\kappa}{\mathrm{c}}\)
∧_(m) = \(\frac{\kappa\left(\mathrm{Scm}^{-1}\right)}{1000\left(\mathrm{~L}^{-3}\right) \text { molarity (moles/lit) }}\)

Question 7.
Explain the variation of molar conductivity with the change in the concentration of the electrolyte. Give reasons.
Answer:
Molar conductivity of a solution at a given concentration is the conductance of the volume of solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length.

Molar conductivity ∧_m = \(\frac{\mathrm{\kappa A}}{l}\)
∧_m = κ × V (∵ l = 1, A = V)
V = Volume containing 1 mole of electrolyte
Molar conductivity increase with decrease in concentration of electrolyte (or) increase in dilution. This is due to total volume V of solution containing one mole of electrolyte also increases.

When concentration approaches to zero, the molar conductivity is known as limiting molar conductivity and it is represented by
For strong electrolytes molar conductivity is given by \(\wedge_m^0\)
∧_m = \(\wedge_m^0\) – Ac^1/2
For weak electrolytes limiting’ molar conductivity is given by
λ_AB⁺ = λ_A⁺ + λ_B^– (Kohlrausch law)

Question 8.
State and explain Kohlrausch’s law of independent migration of ions. [T.S. Mar. 16; A.P. Mar. 15]
Answer:
Kohlrausch’s law of independent migration of ions: The limiting molar conductivity of an electrolytes can be represented as the sum of the individual contributions of the anion and the cation of the electrolytes.
\(\lambda_{\mathrm{m}(\mathrm{AB})}^0=\lambda_{\mathrm{A}^{+}}^0+\lambda_{\mathrm{B}^{-}}^0\)
\(\lambda_{\mathrm{m}}^0\) = Limiting molar conductivity
\(\lambda_{\mathrm{A}^{+}}^0\) = Limiting molar conductivity of cation
\(\lambda_{\mathrm{B}^{-}}^0\) = Limiting molar conductivity of anion

Applications:

1. Kohlrausch’s law is used in the calculation of the limiting molar conductivity of weak electrolytes.
   Eg : λ_CH3COOH = λ_CH3COONa + λ_HCl – λ_NaCl
   = λ_CH3COO^– + λ_Na⁺ + λ_H⁺
2. This law is used in the calculation of degree of dissociation of a weak electrolyte.
3. This law is used in the calculation of solubility of sparingly soluble salts like AgCl, BaSO₄ etc.

Question 9.
What is electrolysis ? Give Faraday’s first law of electrolysis. [A.P. & T.S. Mar. 15]
Answer:
Electrolysis : The decomposition of a chemical compound in the molten state or in the solution state into its constituent elements under the influence of an applied EMF is called electrolysis.
The amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of current passing through the electrolyte.
(Or)
The mass of the substance deposited at an electrode during the electrolysis of electrolyte is directly proportional to quantity of electricity passed through it.
m ∝ Q; m ∝ c × t
m = ect; m = \(\frac{\text { Ect }}{96,500}\)
e = electrochemical equivalent
t = time in seconds
c = Current in amperes
E = Chemical equivalent

Question 10.
What are the products obtained at the cathode and anode during the electrolysis of the following when platinum electrodes are used in the electrolysis
a) Molten KCl
b) Aq. CuSO₄ solution
c) Aq. K₂SO₄ solution
Answer:
a) During the electrolysis of molten KCl using platinum electrodes. Potassium is obtained at cathode and chlorine at anode.
2 KCl → 2K⁺ + 2Cl^–
2 Cl^– → Cl₂ + 2e^– (anode)
2K⁺ + 2e^– → 2K (cathode)

b) During the electrolysis of aq.CuS04 solution using platinum electrodes, O₂ gas liberated at anode and Cu deposited at cathode.
2 CuSO₄ → 2 Cu⁺² + 2 SO^-2₄
2 Cu⁺² + 4e^– → 2 Cu (cathode)
2H₂O – 4e^– → O₂ + 4H⁺ (anode)

c) During the electrolysis of aq.k₂SO₄ using pt-anode and pt-cathode hydrogen gas liberated at cathode and oxygen gas at anode.

Question 11.
What are primary and secondary batteries ? Give one example for each.
Answer:
The batteries which after their use over a period of time’ becomes dead and the cell reaction is completed and this cannot be reused again are called primary batteries.
Eg : Leclanche cell, dry cell.

Secondary battery: A secondary battery is the battery in which after it’s use can be recharged and can be used again.
A good secondary battery undergoes large no. of discharging and charging cycles.
Lead storage battery is an example of secondary battery. ,
The cell reactions when the battery is in use are
Pb_(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → PbSO₄(s) + 2e^– (Anode)
PbO_2(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → \(4 \mathrm{H}_{\text {(aq) }}^{+}\) + 2e → PbSO_4(s) + 2 H₂O_(l) (Cathode)
Overall cell reaction is
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(aq) + 2H₂O_(l)

Question 12.
What are the fuel cells ? How are they different from galvanic cells ? Give the construction of H₂, O₂ fuel cell? [T.S. Mar. 17]
Answer:
A fuel cell is a galvanic cell in which the chemical energy of fuel-oxidant system is converted directly into electrical energy.

• Conventional Galvanic cell converts chemical energy into electrical energy by spontaneous redox reactions.
• Fuel cell convert energy of combustion of fuels like hydrogen, methane etc., into electrical energy. These cause less pollution.

H₂ – O₂ fuel cell: In this cell, hydrogen and oxygen are bubbled through porous carbon electrodes into Cone. NaOH solution. Electrodes are embedded with suitable catalysts. The electrode reactions are:

Overall reaction : 2H_2(g) + O_2(g) → 2 H₂O_(l)
The cell functions as long as the reacting gases are in supply. The heat of combustion is directly converted into electrical energy.

Question 13.
What is metallic corrosion ? Explain it with respect to iron corrosion.
Answer:
Metallic corrosion: The natural tendency of conversion of a metal into its mineral compound form on interaction with the environment is known as metallic corrosion.
Eg.: 1) Iron converts itself into its oxide [Rusting] (Fe₂O₃)
2) Silver converts itself into it’s sulphate [tarnishing] [Ag₂S]
Rusting of iron : Iron converts it self into its oxixe Fe₂O₃
The electrochemical reaction that represents the corrosion or rusting-of iron is
Anode : 2 Fe_(s) → 2 Fe²⁺ + 4e^– \(\mathrm{E}_{\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right)}^0\) = -0.44 V
Cathode : O_2(g) + \(4 \mathrm{H}_{\text {(aq) }}^{+}\) + 4e^– → 2₂O_(l) \(\) = 1.23 V
Overall reaction : 2 Fe_(s) + O_2(g) + \(4 \mathrm{H}_{\text {(aq) }}^{+}\) → \(2 \mathrm{Fe}_{\mathrm{aq}}^{+2}\) + 2 H₂O_(l) \(\mathrm{E}_{(\text {cell) }}^{\Theta}\) = 1.67 V

Long Answer Questions

Question 1.
What are electro chemical cells ? How are they constructed ? Explain the working of the different types of galvanic cells ?
Answer:
Electrochemical cell is a device which makes use of a spontaneous redox reaction for the generation of electrical energy.
Eg : Galvanic cell, Daniell cell etc.
Galvanic cell: A device which converts chemical energy into electrical energy by the use of spontaneous redox reaction is called Galvanic cell (or) voltaic cell.
Eg : Daniell cell.

Daniell cell: It is a special type of galvanic cell. It contains two half cells in the same vessel.
The vessel is devided into two chambers. Left chamber is filled with ZnSO₄ (aq) solution and Zn – rod is dipped into it, Right chamber is filled with aq. CuSO₄ solution and a copper rod is dipped into it. Process diaphragm acts as Salt bridge. The two half cell’s are connected to external battery.

Cell reactions :
Ion Zn/ZnSO₄ half cell, oxidation, reaction occurs
Zn → Zn²⁺ + 2e^–
Ion Cu/CuSO₄ half cell, reduction reaction occurs.
Cu⁺² + 2e^– → Cu
The net cell reaction is
Zn + Cu⁺² ⇌ Zn⁺² + Cu
Cell is represented as Zn / Zn⁺² // Cu²⁺ / Cu

• The two vertical lines (||) represents a salt bridge.
• It is U shaped tube filled with KCl and agar agar gel salt bridge maintains the electrical neutrality and flow of current.

The batteries which after their use over a period of time, becomes dead and the cell reaction is completed and this cannot be reused again are called primary batteries.
Eg : Leclanche cell, dry cell.

Secondary battery: A secondary battery is the battery in which after it’s use can be recharged and can be used again.
A good secondary battery undergoes large no. of discharging and charging cycles.
Lead storage battery is an example of secondaiy battery.
The cell reactions when the battery is in use are
Pb_(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → PbSO₄(s) + 2e^– (Anode)
PbO_2(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → \(4 \mathrm{H}_{\text {(aq) }}^{+}\) + 2e → PbSO_4(s) + 2 H₂O_(l) (Cathode)
Overall cell reaction is
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(aq) + 2H₂O_(l)
In batteries also the principle involved same as in the galvanic cell.

Question 2.
What is electrical conductace of a solution? How ¡s it measured experimentally?
Answer:
The reciprocal of resistance (R) is known as conductance (c)
\(\frac{l}{\mathrm{R}}=\kappa \times \frac{1}{l / \mathrm{A}}\)
c = \(\kappa \times \frac{1}{l / \mathrm{A}}\)
The conductance or the current conducting capacity of an electrolytic solution can be expressed as .

1. Specific conductance (κ),
2. Molar conductivity (∧_m)

1. Specific conductance : The conductance of the solution enclosed between two parallel electrodes of unit area of cross section separated by unit distance is called specific conductance (κ).

2. Molar conductivity (∧_m) : The conductivity of a volume of solution containing one gram molecular weight of the electrolyte placed between two parallel electrodes separated by a distance of unit length of 1 meter is called molar conductivity (∧_m)
Relation between conductivity and molar conductivity :
∧_m = \(\frac{\kappa}{\mathrm{c}}\); ∴ c = constant
Measurement of electrical conductance :

• The resistance of a metallic wire can be measured with a Wheatstone bridge.
•  In measuring the resistance of an electrolytic solution two problems are identified.
  1. On passing direct current (DC) through the solution changes the composition of the solution due to electrolysis.
  2. A solution cannot be connected to the measuring bridge like a metallic wire.
• First problem solved by using AC instead of DC. .
• Second problem solved by using specially designed vessel called conductivity cell.
  
  By using above cells,
  Resistance R = \(\frac{l}{\kappa \times \mathrm{A}}\)
  l = distance between electrodes; A = Area of cross section; κ = conductivity
  \(\frac{l}{\mathrm{~A}}\) = cell constant = G*
  ∴ G* = \(\frac{l}{\mathrm{~A}}\) = R × κ
  The cell constant (G*) is determined by measuring the resistance of the cell containing a solution whose conductivity is already known
• Cell constant once determined, used for measuring the resistance (or) conductivity.
• The set up for the measurement of the resistance is shown in following figure.

Question 3.
Give the applications of Kohlrausch’s law of independent migration of ions. [T.S. Mar. 16]
Answer:
Kohlrausch’s law of independent migration of ions: The limiting molar conductivity of an electrolytes can be represented as the sum of the individual contributions of the anion and the cation of the electrolytes.
\(\lambda_{\mathrm{m}(\mathrm{AB})}^0=\lambda_{\mathrm{A}^{+}}^0+\lambda_{\mathrm{B}^{-}}^0\)
\(\lambda_{\mathrm{m}}^0\) = Limiting molar conductivity
\(\lambda_{\mathrm{A}^{+}}^0\) = Limiting molar conductivity of cation
\(\lambda_{\mathrm{B}^{-}}^0\) = Limiting molar conductivity of anion

Applications:
1) Calculation of limiting molar conductance at infinite dilution (∧_∞) of weak electrolytes :
Limiting molar conductance of weak electrolytes at infinite dilution- cannot be obtained graphically by extrapolation method. However the application of Kohlrausch’s law enables indirect evaluations in such cases.

Eg: Limiting molar conductance of acetic acid at infinite dilution can be calculated from the limiting molar conductances at infinite dilution of hydrochloric acid, sodium chloride and sodium acetate, as illustrated below:

2) Calculation of degree of dissociation of a weak electrolyte :
Degree of dissociation (α) = \(\frac{\lambda_{\mathrm{v}}}{\lambda_{\infty}}\)
Where λ_v = Limiting molar conductivity at a given concentration or dilution.
λ_∞ = Limiting molar conductivity of given electrolyte at infinite dilution.

3) Calculation of the solubility (s) of sparingly soluble salts like AgCl, BaSO₄ etc.
S = \(\frac{\kappa \times 1000}{\wedge_{\infty}}\)
where K = specific conductance
∧_∞ = Limiting molar conductance at infinite dilution.

Question 4.
Give the different types of batteries and explain the construction and working of each type of battery. [A.P. Mar. 19]
Answer:
The batteries which after their use over a period of time, becomes dead and the cell reaction is completed and this cannot be reused again are called primary batteries.
Eg.: Leclanche cell, dry cell.

Secondary battery:
A secondary battery is the battery in which after it’s use can be recharged and can be used again.
A good secondary battery undergoes large no. of discharging and charging cycles.
Lead storage battery is an example of secondary battery.
The cell reactions when the battery is in use are
Pb_(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → PbSO₄(s) + 2e^– (Anode)
PbO_2(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → \(4 \mathrm{H}_{\text {(aq) }}^{+}\) + 2e → PbSO_4(s) + 2 H₂O_(l) (Cathode)
Overall cell reaction is
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(aq) + 2H₂O_(l)

Construction and working of primary batteries :

Dry cell:

1. This is a modification of Leclanche cell. The liquid state electrolytes (in Leclanche cell) are replaced by paste electrolytes.
2. A cylindrical ‘Zn’ vessel is covered with a cardboard. This is sealed with pitch. Zn vessel acts as negative electrode. A carbon rod is introduced at the centre of the Zn vessel. This carbon rod acts as positive electrode.
3. Carbon rod is surrounded by a paste of (C + MnO₂). The remaining space is filled with (NH₄Cl + ZnCl₂ paste. The two pastes are separated by a porous sheet.
4. These cells are easy to handle and used in radios, watches, torch lights etc. The cell potential is approximately 1.5 V
   
5. Electrode reactions:
   Cathode : MnO₂ + \(\mathrm{NH}_4^{+}\) + e^Θ → MnO (OH) + NH₃
   Anode : Zn + 2MnO₂ + 2H₂O → Zn²⁺ + 2OH^Θ + 2MnO (OH)
   Some of the secondary reactions are
   2 NH₄Cl + 2OH^Θ → 2NH₃ + 2Cl^– + 2H₂O
   Zn²⁺ + 2NH₃ + 2Cl^– → [Zn (NH₃)₂] Cl₂

Secondary batteries – construction and working : A secondary cell after its use can be recharged and can be used again. A good secondary cell undergoes a large number of discharging and charging cycles. The most important secondary cell in use is the lead storage battery (fig.(b)). This is commonly used in automobiles and invertors. It consists of a lead anode and a grid of lead packed with lead dioxide (PbO₂) as cathode. A 38% solution of sulphuric acid is used as electrolyte.
The cell reactions when the battery is in use (discharging) are :
Anode : Pb_(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → PbSO_4(s) + 2e^–
Cathode: PbO_2(s) + \(\mathrm{SO}_4^{-2}(\mathrm{aq})\) → \(4 \mathrm{H}_{\text {(aq) }}^{+}\) + 2e → PbSO_4(s) + 2 H₂O_(l)
i.e., overall cell reaction consisting of cathode and anode reactions is
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(aq) + 2H₂O_(l)
These reactions occur during discharge i.e., during use of the battery.
On charging the discharged battery the above reaction is reversed and PbSO₄ (s) on anode and cathode is converted into Pb and PbO₂, respectively.

Problems

Question 1.
The standard potentials of some electrodes are as follows. Arrange the metals in an increasing order of their reducing power.
1) K⁺/K = -2.93 V
2) Ag⁺/Ag = 0.8OEV
3) Cu²⁺/Cu = 0.34 V
4) Mg²⁺/Mg = -2.37 V
5) Cr³⁺/Cr = – 0.74V
6) Fe²⁺/Fe = —0.44 V
Solution:
Given
E₀ of cells K⁺/K = —2.93
Mg⁺²/Mg = -2.37 V
Ag⁺/Ag = 0.80 V
Cr⁺³/Cr = -0.74 V
Cu⁺²/Cu = 0.34 V
Fe⁺²/Fe = -0.44 V
Lower the reduction potential values indicates high reducing power and high value indicates the high oxidising power.
So, increasing order of the given cells reducing power is
Ag < Cu < Fe < Cr < Mg < K.

Question 2.
Calculate the emf of the cell at 25°C
Cr | Cr³⁺ (0.1 M) || Fe²⁺ (0.01M) | Fe, given that \(\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^0\) = – 0.74V and \(\mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^0\) = 0.44 V.
Solution:
Given cell is
Cr | \(\mathrm{Cr}_{(0.1 \mathrm{M})}^{+3}\) || \(\mathrm{Fe}_{(0.01 \mathrm{M})}^{+2}\) | Fe
E⁰ of Cr⁺³/Cr = -0.74 V
E⁰ of Fe⁺²/Fe = -0.44 V
E_Cr⁺³/Cr = E⁰ + \(\frac{0.059}{3}\) log10 [Cr⁺³]
= -0.74 + \(\frac{0.059}{3}\) log 0.1
= -0.76 V
E_Fe⁺²/Fe = -0.44 + \(\frac{0.059}{2}\) log 0.01
= -0.44 – 0.059 = – 0.499 V
EMF of cell = E_RHS – E_LHS
= (- 0.499) – (-0.76) = 0.26 1 V .

Question 3.
Calculate the potential of a Zn – Zn²⁺ electrode in which the molarity of Zn²⁺ is 0.001 M. Given that \(\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^0\) = – 0.76 V
R = 8.314 JK^-1 mol^-1; F = 96500 C mol^-1.
Solution:
Given electrode Zn | Zn⁺²(0.001m) \(\mathrm{E}_{\mathrm{Zn}^{+2} / \mathrm{Zn}}^0\) = -0.76V
Nemst equation is
E = E⁰ + \(\frac{2.303 \mathrm{RT}}{\mathrm{nF}}\) log [Mⁿ⁺]
Given R = 8.314 J/K. mole
F = 96500 c/mole
E = E⁰ + \(\frac{0.059}{\mathrm{n}}\) log C
= -0.76 + \(\frac{0.059}{2}\) log 0.001
= -0.76 – \(\frac{0.059}{2}\) × 3
= – 0.76 – 0.0295 × 3
= -0.76 – 0.0885
= -0.8485 V

Question 4.
Determine ∆G⁰ for the button cell used in the watches. The cell reaction is
Zn_(s) + Ag₂O_(s) + H₂O_(l)→ \(\mathrm{Zn}_{(a q)}^{2+}\) + 2 Ag_(s) + 2 \(\mathrm{OH}_{(a q)}^{-}\)
\(\mathrm{E}_{\mathrm{Ag}^+ / \mathrm{Ag}}^0\) = + 0.80 V; \(\mathbf{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^0\) = – 0.76 V.
Solution:
Given E⁰ of Ag⁺/Ag = 0.80 V
E⁰ of Zn⁺²/Zn = – 0.76 V
Cell representation Zn/Zn⁺²|| Ag⁺ / Ag
EMF = E_RHS – E_LHS = 0.80 – (-0.76) = 1.56 V
∆G = -nFE⁰ = -2 × 96500 × 1.56 = -301.08 kJ/mole

Question 5.
Calculate the emf of the cell consisting the following half cells
Al/Al³⁺(0.001 M), Ni/Ni²⁺ (0.50 M). Given that \(\mathrm{E}_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^0\) = – 0.25 V \(\mathrm{E}_{\mathrm{Al}^{3+} / \mathrm{Al}}^0\) = – 1.66 v (log 8 × 10^-6 = – 5.0969).
Solution:
Given E⁰ of Al⁺³/Al =-1.66 V
E⁰ of Ni⁺²/Ni = – 0.25 V
Cell representation Al/\(\mathrm{Al}_{(0.001 \mathrm{M})}^{+3}\) || \(\mathrm{Ni}_{(0.50 \mathrm{M})}^{+2}\) | Ni
Nemst equation is
E_Al⁺³/Al = \(\mathrm{E}_{\mathrm{Al}^{+3} / \mathrm{A} l}^0\) + \(\frac{0.059}{n}\) log [Al⁺³]
= -1.66 + \(\frac{0.059}{3}\) log 0.001 = – 1.66 – 0.059 = – 1.719V
E_Ni⁺²/Ni = \(\mathrm{E}_{\mathrm{Ni}^{+2} / \mathrm{Ni}}^0+\frac{0.059}{\mathrm{n}}\) log [Ni⁺²]
= -0.25 + \(\frac{0.059}{2}\) log [0.5]
= -0.25 – 0.0295 × 0.3010 = -0.25 – 0.0088795 = -0.25888
EMF of cell = E_RHS – E_LHS
= -0.25888 – (-1.719) = 1.46012 V

Question 6.
Determine the values of Kc for the following reaction
Ni_(s) + 2 \(\mathrm{Ag}_{\text {(aq) }}^{+}\) → \(\mathrm{Ni}_{\text {(aq) }}^{2+}\) + 2 Ag_(s) E⁰ = 1.05 V.
Solution:
Given
Ni_(s) + 2 \(\mathrm{Ag}_{\text {(aq) }}^{+}\) → \(\mathrm{Ni}_{\text {(aq) }}^{2+}\) + 2 Ag_(s)
E⁰ = 1.05 V.
Nernst Equation is E = E⁰ – \(\frac{0.059}{\mathrm{n}}\) log K_c
at equilibrium E = 0
E⁰ = \(\frac{0.059}{\mathrm{n}}\) log K_c
1.05 = \(\frac{0.059}{2}\) log K_c
log K_c = \(\frac{2.1}{0.059}\) = 35.59
K_c = 3.42 × 10³⁵

Question 7.
Calculate the potential of the half-cell containing 0.1 M
K₂Cr₂O_7(aq), 0.2 M \(\mathrm{Cr}_{\mathrm{aq}}^{3+}\) and 1.0 × 10^-4 M \(\mathrm{H}_{(\mathrm{aq})}^{+}\). The half-reaction \(\mathrm{Cr}_2 \mathrm{O}_{7(\mathrm{aq})}^{2-}+14 \mathrm{H}_{\text {(aq) }}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+7 \mathrm{H}_2 \mathrm{O}_{(l)}\) (E⁰ of \(\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}\) = 1.33 V)
Solution:
Given half reaction

Question 8.
Calculate K for the reaction at 298 K
\(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{aq})}^{2+} \rightleftharpoons \mathrm{Zn}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(\mathrm{s})}\)
\(\mathrm{E}_{\mathrm{Zn}^{2+/ Z n}}^0\) = -0.76 V; \(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0\) = + 0.34 V.
Solution:
Given \(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{aq})}^{2+} \rightleftharpoons \mathrm{Zn}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(\mathrm{s})}\)
E⁰ of Zn⁺²/Zn = -0.76 V
E⁰ of Cu⁺²/Cu = -0.34 V
E⁰ of cell = E_RHS – E_LHS = 0.34 – (- 0.76) = 1.1 V
∆G⁰ = – RTlnK_c = – 2.303 RT log K_c
-212300 = -2.303 × 8.314 × 298 × log K_c
log K_c = \(\frac{212300}{2.303 \times 8.314 \times 298}\) = 37.207
K_c = 1.6 × 10³⁷

Question 9.
Calculate the emf of the cell at 298 K
Sn_(s) | Sn²⁺ (0.05 M) || \(\mathrm{H}_{(\mathrm{aq})}^{+}\) (0.02 M) | H₂ 1 atm. Pt
Given that \(\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^0\) = -0.144 V
Solution:

Question 10.
Calculate the concentration of silver ions in the cell constructed by using 0.1 M concentration of Cu²⁺ and Ag⁺ ions. Cu and Ag metals are used as electrodes. The cell potential is 0.422 V.
[\(\mathrm{E}_{\mathrm{Ag}^{2+} / \mathrm{Ag}}^0\) = 0.80 V; \(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0\) = +0.34 V]
Solution:
Given E⁰ of Ag⁺/Ag = 0.80 V .
E⁰ of Cu⁺²/Cu = 0.34 V
E⁰ of cell = E_RHS – E_LHS = 0.80 – 0.34 = 0.46 V

Question 11.
Calculate the emf of the cell with the cell reaction
Ni_(s) + 2 Ag⁺ (0.002M) → Ni²⁺ (0.160 M) + 2 Ag_(s) ; \(\mathrm{E}_{\text {cell }}^0\) = 1.05 V.
Solution:
From the given cell reaction and Nernst equation

Question 12.
Cu²⁺ + 2e^– ⇌ Cu; E⁰ = +0.34 V
Ag⁺ + e^– ⇌ Ag; E⁰ = + 0.80 V
For what concentration of Ag⁺ ions will the emf of the cell be zero at 25°C. The concentration of Cu²⁺ is 0.1 M. (log 3.919 = 0.593).
Solution:
Cell is Cu|Cu⁺² | |Ag⁺|Ag
E⁰ of cell = E_RHS – E_LHS = 0.80 – 0.34 = 0.46 V
Nernst equation is E = E⁰ – \(\frac{0.059}{2} \log \frac{\left[\mathrm{Cu}^{+2}\right]}{\left[\mathrm{Ag}^{+}\right]^2}\)
EMF of cell = 0
E⁰ = \(\frac{0.059}{2} \log \frac{0.1}{\left[\mathrm{Ag}^{+}\right]^2}\)

Question 13.
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate molar conductance.
Solution:
Conductivity (K) = 0.0248 S cm^-1 = 0.0248 ohm^-1 cm^-1
Molar concentration [c] = 0.20 mol L^-1
= \(\frac{[0.2 \mathrm{~mol}]}{\left[1000 \mathrm{~cm}^3\right]}\) = 2.0 × 10^-4 mol cm^-3
Molar conductivity [∧_m] = \(\frac{\kappa}{\mathrm{c}}=\frac{\left[0.0248 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\right]}{\left[2.0 \times 10^{-4} \mathrm{~mol} \mathrm{~cm}^{-3}\right]}\)
= 124 ohm^-1 mol^-1 cm² or 1245 mol^-1 cm²
Molar conductivity = 124 mol^-1 cm²

Question 14.
Calculate the degree of dissociation (α) of CH₃COOH at 298 K.
Given that \(\wedge_{\mathrm{CH}_3 \mathrm{COOH}}^{\infty}\) = 11.75 cm² mol^-1
\(\wedge_{\mathrm{CH}_3 \mathrm{COO}^{-}}^0\) = 40.95 cm² mol^-1
\(\wedge_{\mathrm{H}^{+}}^0\) = 349.15 cm² mol^-1
Solution:
Given that \(\wedge_{\mathrm{CH}_3 \mathrm{COOH}}^{\infty}\) = 11.75 cm² mol^-1
\(\wedge_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\infty}\) = 40.95 cm² mol^-1
\(\wedge_{\mathrm{H}^{+}}^{\infty}\) = 349.15 cm² mol^-1
\(\wedge_{\mathrm{CH}_3 \mathrm{COOH}}^{\infty}\) = \(\wedge_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\infty}\) + \(\wedge_{\mathrm{H}^{+}}^{\infty}\) = 40.95 + 349.15 = 390.1
Degree of dissociation (α) = \(\frac{\lambda^\alpha}{\lambda^{\infty}}=\frac{11.75}{390: 1}\) = 0.03 = 3 × 10^-2
∴ α = 3× 10^-2

Textual Examples

Question 1.
Represent the cell in which the following reaction takes place Mg(s) + 2Ag⁺ (0.0001M) → Mg²⁺ (0.130M) + 2 Ag(s)
Calculate its E_(cell) if \(\mathrm{E}_{(\text {cell) }}^{\Theta}\) = 3.17 V
Solution:
The cell is represented as Mg |Mg²⁺ (0.130M) || Ag⁺(0.0001M) | Ag

Question 2.
Calculate the equilibrium constant of the reaction:
Cu_(s) + 2 \({Ag}_{\text {(aq) }}^{+}\) → \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) + 2 Ag_(s) ; \(\mathrm{E}_{(\text {cell) }}^{\Theta}\) = 0.46 V
Solution:
\(\mathrm{E}_{(\text {cell) }}^{\Theta}\) = \(\frac{0.059V}{2}\) log K_c = 0.46 V or
l0g K_c = \(\frac{0.46 \mathrm{~V} \times 2}{0.059 \mathrm{~V}}\) = 156
K_c = 3.92 × 10¹⁵

Question 3.
The standard emf of Daniell cell is 1.1V. Calculate the standard Gibbs energy for the cell reactions:
Zn_(s) + \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) → \(\mathrm{Zn}_{(\mathrm{aq})}^{2+}\) + Cu_(s)
Solution:
∆_rG^Θ = -nF\(\mathrm{E}_{(\text {cell) }}^{\Theta}\)
n in the above equation is 2, F = 96487 C mol^-1 and \(\mathrm{E}_{(\text {cell) }}^{\Theta}\) = 1.1 V
Therefore ∆_rG^Θ = -2 × 1.1V × 96487 C mol^-1
= – 21227 J mol^-1 =-212.27 kJ mol^-1

Question 4.
Resistance of a conductivity cell filled with 0.1 mol L^-1 KCl solution is 100 Ω. If .the resistance of the same cell when filled with 0.02 mol L^-1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L^-1 KCl solution. The conductivity of 0.1 mol L^-1 KCl solution is 1.29 S/m.
Solution:
The cell constant is given by the equation :
Cell constant = G* = conductivity × resistance
= 1.29 S/m 100 Ω = 129 m^-1 = 1.29 cm^-1
Conductivity of 0.02 mol L^-1 KCl solution = cell constant / resistance
= \(\frac{\mathrm{G}^*}{\mathrm{R}}=\frac{129 \mathrm{~m}^{-1}}{520 \Omega}\) = 0.248 S m^-1
Concentration = 0.02 mol L^-1
= 1000 × 0.02 mol m^-3 = 20 mol m^-3
Molar conductivity = ∧_m = \(\frac{\kappa}{\mathrm{c}}\)
= \(\frac{248 \times 10^{-3} \mathrm{Sm}^{-1}}{20 \mathrm{~mol} \mathrm{~m}^{-3}}\) = 124 × 10^-4 Sm² mol^-1
Alternatively, κ = \(\frac{1.29 \mathrm{~cm}^{-1}}{520 \Omega}\) = 0.248 × 10^-2 S cm^-1
and ∧_m = κ × 1000 cm³L^-1 molarity^-1
= \(\frac{0.248 \times 10^{-2} \mathrm{~S} \mathrm{~cm}^{-1} \times 1000 \mathrm{~cm}^3 \mathrm{~L}^{-1}}{0.02 \mathrm{~mol} \mathrm{~L}^{-1}}\)
= 124 S cm² mol^-1

Question 5.
The electrical resistance of a column 0.05 mol L^-1 NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 10³ ohm. Calculate its resistivity, conductivity and molar conductivity.
Answer:
A = πr² = 31.4 × 0.5² cm² = 0.785 cm2 = 0.785 × 10^-4 m²
l = 50 cm = 0.5 m

Question 6.
Calculate \(\wedge_m^0\) for CaCl₂ and MgSO₄ from the data given in Table 3.4.
Solution:
We know from Kohlrausch law that
\(\wedge_{\mathrm{m}\left(\mathrm{CaCl}_2\right)}^0=\wedge_{\mathrm{Ca}^{2+}}^0+2 \lambda_{\mathrm{Cl}}^0\) = 119.0 S cm² mol ^-1 + 2(76.3) S cm² mol^-1
= (119.0 + 152.6) Scm² mol^-1
= 271.6 S cm² mol^-1
\(\wedge_{\mathrm{m}\left(\mathrm{MgSO}_4\right)}^0=\lambda_{\mathrm{Mg}^{2+}}^0+\lambda_{\mathrm{SO}_4^{2-}}^0\) =106.0 S cm² mol^-1 + 160.0 S cm² mol^-1
= 266 S cm² mol^-1

Question 7.
\(\wedge_m^0\) for NaCl, HCl and NaAc are 126.4,425.9 and 91.0 S cm² mol^-1 respectively. Calculate ∧⁰ for HAc.
Solution:
\(\wedge_{\mathrm{m}(\mathrm{HAc})}^0=\lambda_{\mathrm{H}^{+}}^0+\lambda_{\mathrm{Ac}^0}^0=\lambda_{\mathrm{H}^{+}}^0+\lambda_{\mathrm{Cl}^{-}}^0+\lambda_{\mathrm{Na}^{+}}^0-\lambda_{\mathrm{Cl}^{-}}^0-\lambda_{\mathrm{Na}^{+}}^0\)
= \(\wedge_{\mathrm{m}(\mathrm{HCl})}^0+\wedge_{\mathrm{m}(\mathrm{NaAc})}^0-\wedge_{\mathrm{m}(\mathrm{NaCl})}^0\)
= (425.9 + 91.0 – 126.4) S cm² mol^-1
= 390.5 S cm² mol^-1

Question 8.
The conductivity of 0.001028 mol L^-1 acetic acid is 4.95 × 10^-5 S cm^-1. Calculate its dissociation constant if \(\wedge_m^0\) for acetic acid is 390.5 S cm² mol^-1.
Solution:

Question 9.
A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of eopper deposited at the cathode ? [A.P. & T.S. Mar. 15] [Mar. 15]
Solution:
t = 600 s charge = current × time = 1.5A × 600 s = 900 C
According to the reaction :
\(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) + 2e^– = Cu_(s)
We require 2F or 2 × 96487 C to deposit 1 mol or 63 g of Cu.
For 900 C, the mass of Cu deposited
= (63 g mol^-1 × 900 C) / (2 × 96487 C mol^-1) = 0.2938 g

Intext Questions

Question 1.
How would you determine the standard electrode potential of the system Mg²⁺/Mg ?
Solution:
E° value of Mg²⁺/Mg electrode is determined by setting up an electrochemical cell. For this purpose, a Mg electrode is dipped in 1 M MgSO₄ solution, which acts as one half cell i.e., oxidation half-cell.

In the same way, the standard hydrogen electrode acts as the other half-cell i.e., reduction half cell. The deflection of voltmeter placed in the cell circuit is towards the Mg electrode which indicates the direction of flow of current.
The cell may be represented as : Mg | Mg²⁺ (1M) || H⁺ (M) | H₂ (1 atm). Pt The reading as given by voltmeter gives \(\mathrm{E}_{\text {cell }}^0\)

Question 2.
Can you store copper sulphate solutions In a zinc pot?
Solution:
No, zinc pot cannot store copper sulphate solutions because the standard electrode potential (E⁰) value of zinc is less than that of copper. So, zinc is stronger reducing agent then copper.
\(\mathrm{Zn}_{(\mathrm{aq})}^{2+}\) + 2e^– → Zn(s); E⁰ = -0.76 V
\(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) + 2e^– → Cu(s); E⁰ = – 0.34 V
So, zinc will loss electrons to Cu²⁺ ions and redox reaction will occur as follows.
Zn(s) + \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) → \(\mathrm{Zn}_{(\mathrm{aq})}^{2+}\) + Cu(s)

Question 3.
Consult the table on standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Solution:
Fe²⁺ (ferrous) ions gets oxidised to Fe³⁺ (ferric) ion as follows
Fe²⁺ → Fe³⁺ + e^– \(\mathrm{E}_{0 \mathrm{x}}^0\) = -0.77V
Only those substances can oxidise Fe²⁺ ions to Fe³⁺ ions which can accept electrons released during oxidation or are placed above iron in electrochemical series. Three such substances are: Cl₂(g), Br₂(g) and F₂(g).

Question 4.
Calculate the potential of hydrogen electrode placed in a solution of pH 10.
Solution:

Question 5.
The cell in which the following cell reaction occurs,
\(2 \mathrm{Fe}_{(\mathrm{aq})}^{3+}+2 \mathrm{I}_{(\mathrm{aq})}^{-} \longrightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+\mathrm{I}_{2(\mathrm{~s})}\) has \(\mathrm{E}_{\text {cell }}^0\) = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Solution:
Two half reactions for the given redox reaction may be written as :
\(2 \mathrm{Fe}_{(\mathrm{aq})}^{3+}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{2+} ; 2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_2+2 \mathrm{e}^{-}\)
2 moles of electrons are involved in the reaction, so n = 2
∆_rG⁰ = – nF\(\mathrm{E}_{\text {cell }}^0\) = (-2 mol) × (96500 mol^-1) (0.236 V)
= -45548 CV = -45548 J
∆_rG⁰ = – 45.55 kJ
log K_c = \(\frac{\Delta \mathrm{G}^0}{2.303 \mathrm{RT}}\)
= \(\frac{(-45.55 \mathrm{~kJ})}{2.303 \times\left(8.314 \times 10^{-3} \mathrm{kJK}^{-1}\right) \times(298 \mathrm{~K})}\) = 7.983
K_c = Antilog (7.983) = 9.616 × 10⁷
K_c = 9.616 × 10⁷

Question 7.
Why does the conductivity of a solution decrease with dilution ?
Solution:
The conductivity of a solution is related with the number of ions present per unit volume of the solution. When the solution is diluted, the number of ions decreases. Hence, conductivity or specific conductance of the solution also decreases.

Question 8.
Suggest a way to determine the \(\wedge_{\mathrm{m}}^0\) value of water.
Solution:
The molar conductance of water at infinite dilution can be obtained from the knowledge of molar conductances at inifite dilution of sodium hydroxide, hydrochloric acid and sodium chloride (all strong electrolytes). This is in accordance with Kohlrausch’s law.
\(\wedge_{\mathrm{m}\left(\mathrm{H}_2 \mathrm{O}\right)}^0=\wedge_{\mathrm{m}(\mathrm{NaOH})}^0+\wedge_{\mathrm{m}(\mathrm{HCl})}^0-\wedge_{\mathrm{m}(\mathrm{NaCl})}^0\)

Question 9.
The molar conductivity of 0.025 mol L^-1 methanoic acid is 46.1 S cm² mol^-1. Calculate its degree of dissociation and dissociation constant.
Given, λ⁰(H⁺) = 349.6 S cm^-1 mol^-1 and λ⁰ (HCOO^–) = 54.6 S cm² mol^-1
Solution:
Step I: Calculation of degree of dissociation (a) of HCOOH

= \(\frac{\left(3.249 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\right)}{(0.886)}\)
= 3.67 × 10^-4 mol L^-1
α = 0.114
K_a = 3.67 × 10^-4 mol L^-1

Question 10.
If a current of 0.5 ampere flows through a metallic wire for 2h, then how many electrons would flow through the wire ?
Solution:
Quantity of charge (Q) passed = Current (C) × Time (t)
= (0.5 A) × (2 × 60 × 60 s)
= (3600) Ampere sec = 3600 C ,
Number of electrons flowing through the wire on passing charge of one Faraday (96500 C) = 6.022 × 10²³
Number of electrons flowing through the wire on passing a charge of 3600 C
= \(\frac{6.022 \times 10^{23} \times(3600 \mathrm{C})}{(96500 \mathrm{C})}\)
= 2.246 × 10²²
Number of electrons = 2.246 × 10²²

Question 11.
Suggest a list to metals that are extracted electrolytically.
Solution:
Those metals which are highly reactive and have large negative E⁰ values can be exerted electrolytically. This type of metals are powerful reducing agents.
Eg: Sodium, Potassium, Calcium, Magnesium etc.

Question 12.
Consider the reaction,
\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O
What is the quantity of electricity in coulombs nefeded to reduce 1 mole of \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) ?
Solution:
1 mole \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) require 6 Faraday electricity (6 F)
= 6 × 96500 C = 5.79 × 10⁵ C
Quantity of electricity = 5.79 × 10⁵ Coulomb

Question 13.
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Solution:
When lead storage battery is recharged, electrical energy is supplied to the cell from the external source : i.e., the cell operates as an electrolytic cell during recharging. All the chemical reactions, which take place during the use of battery, are reversed at this time.

Question 14.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Solution:
Methane (CH₄) and methanol (CH₃OH)

Question 15.
Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Solution:
In corrosion, a metal is oxidised by loss of electrons to oxygen with the formation of oxides. So, an electrochemical cell is set up.
Eg: Rusting of iron involves the following steps :
i) The water layer present on the surface of iron dissolves acidic oxides from air like CO₂ and forms acid to produce H⁺ ions.
H₂O + CO₂ H₂CO₃ ⇌ 2H⁺ + \(\mathrm{CO}_3^{2-}\)

ii) In the presence of H⁺ ions, iron starts losing electrons at some spot to form ferrous ions. This spot behaves as anode.

iii) The electrons released at anode move to another spot, where H⁺ ions and the dissolved oxygen gain these electrons. This spot becomes a cathode.

iv) Overall reaction, i.e., redox reaction is :

v) Ferrous ions are further oxidised by the atmospheric oxygen to ferric ions which combine with water molecules to form hydrated ferric oxide, Fe₂O₃. xH₂O. (Rust).

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Differential Equations Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B Differential Equations Important Questions

Question 1.
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^{x – y} + x² e^-y [Mar. 06; May 05]
Solution:

Question 2.
x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – y = 2x² sec²2x [May 11]
Solution:

Question 3.
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y tan x = sin x. [T.S. Mar. 16]
Solution:
I.F. = \(e^{\int \tan x d x}\) = e^{log sec x} = sec x
y.sec x = \(\int\) sin x . sec x dx = \(\int\) tan x dx
= log sec x + c

Question 4.
cos x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y sin x = sec²x [Mar. 14]
Solution:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + tan x . y = sec³x
I.F. = e^{\(\int\)tan x dx} = e^{log sec x} = sec x
y . sec x = \(\int\) sec⁴x dx = \(\int\) (1 + tan² x) sec² x
dx = tan x + \(\frac{\tan ^{3} x}{3}\) + c

Question 5.
(x + y + 1)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 1.
Solution:
\(\frac{\mathrm{dx}}{\mathrm{dy}}\) = x + y + 1
\(\frac{\mathrm{dx}}{\mathrm{dy}}\) = x + y + 1
I.F. = e^{\(\int\) -dy} = e^-y
x . e^-y = \(\int\) e^-y (y + 1)dy
= – (y + 1) . e^-y + \(\int\) e^-y . dy
= – (y + 1) e^-y – e^-y
= – (y + 2) e^-y + c
x = – (y + 2) + c. e^-y

Question 6.
Find the order and degree of r
\(\frac{\mathrm{d}^{3} \mathrm{~y}}{\mathrm{dx}^{3}}\) – 3 (\(\frac{\mathrm{dy}}{\mathrm{dx}}\)) – e^x = 4. [Mar. 14]
Solution:
The equation is a polynomial in \(\frac{\mathrm{dy}}{\mathrm{dx}}\) and \(\frac{\mathrm{d}^{3} \mathrm{~y}}{\mathrm{dx}^{3}}\).
The exponent of \(\frac{\mathrm{d}^{3} \mathrm{~y}}{\mathrm{dx}^{3}}\) is 2.
Hence the degree is 2.
\(\frac{\mathrm{d}^{3} \mathrm{~y}}{\mathrm{dx}^{3}}\) is the highest order derivative occuring in the equation.
Order of the equation is 3.

Question 7.
x^{\(\frac{1}{2}\)}(\(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\))^{\(\frac{1}{3}\)} + x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = 0 has order 2 and degree 1. Prove. [T.S. Mar. 15]
Solution:
The given equation can be written as

∴ The order of the equation is 2 and its degree is 1.

Question 8.
Find the order and degree of \(\left(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}\right)^{\frac{6}{5}}\) = 6y [Mar. 16; May 11]
Solution:
Given equation is \(\left(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}\right)^{\frac{6}{5}}\) = 6y
i.e., \(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\) + (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))³ = (6y)^{\(\frac{5}{6}\)}
Order = 2, degree = 1

Question 9.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x(2 \log x+1)}{\sin y+y \cos y}\) [Mar. 08]
Solution:
Given equation can be written as
(sin y + y cos y) dy = x(2 log x + 1) dx
\(\int\) sin y dy + \(\int\) y cos y dy = \(\int\) 2x log x dx + \(\int\) x dx
\(\int\) sin y dy + y sin y – \(\int\) sin y dy = x² log x – \(\int\) x² . \(\frac{1}{x}\) dx + \(\int\) x dx + c
y sin y = x² log x + c

Question 10.
(xy² + x) dx + (yx² + y) dy = 0. [A.P. Mar. 15, 07]
Solution:
(xy² + x) dx + (yx² + y) dy = 0
x(y² + 1) dx + y (x² + 1) dy = 0
Dividing with (1 + x²) (1 + y²)
\(\frac{x d x}{1+x^{2}}\) + \(\frac{y d x}{1+y^{2}}\) = 0
Integrating
\(\int \frac{x d x}{1+x^{2}}+\int \frac{y d y}{1+y^{2}}=0\)
\(\frac{1}{2}\)[(log (1 + x²) + log (1 + y²)] = log c
log (1 + x²) (1 + y²) = 2 log c = log c²
Solution is(1 + x²) (1 + y²) = k when k = c².

Question 11.
sin^-1 (\(\frac{\mathrm{dy}}{\mathrm{dx}}\)) = x + y [Mar. 07]
Solution:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sin (x + y)
x + y = t
1 + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{dt}}{\mathrm{dx}}\)
\(\frac{\mathrm{dt}}{\mathrm{dx}}\) – 1 = sin t
\(\frac{\mathrm{dt}}{\mathrm{dx}}\) = 1 + sin t
\(\frac{d t}{1+\sin t}\) = dx
Integrating both sides we get
\(\int \frac{d t}{1+\sin t}=\int d x\)
\(\int \frac{1-\sin t}{\cos ^{2} t} d t=x+c\)
\(\int\) sec² t dt = \(\int\) tan t . sec t dt = x + c
tan t – sec t = x + c
⇒ tan (x + y) – sec (x + y) = x + c

Question 12.
(x² – y²) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = xy [May 11]
Solution:

= log y + c
\(\frac{-x^{2}}{2 y^{2}}\) = (log y + c)
-x² = 2y² (c + log y)
⇒ Solution is x² + 2y² (c + log y) = 0.

Question 13.
Solve : x dy = (y + x cos² \(\frac{y}{x}\)) dx.
Solution:

Question 14.
(2x + y + 1) dx + (4x + 2y – 1) dy = 0 [T.S. Mar. 15]
Solution:

2v + log (v – 1) = 3x + c
2v – 3x + log (v – 1) = c
2(2x + y) – 3x + log (2x + y – 1) = c
4x + 2y – 3x + log (2x + y – 1) = c
Solution is x + 2y + log (2x + y – 1) = c

Question 15.
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y tan x = cos³x [May 11]
Solution:
I.F. = e^{\(\int\) tan x dx} = e^{log sec x} = sec x
y . sec x = \(\int\) sec x. cos³ x dx
= \(\int\) cos²x dx
= \(\frac{1}{2}\) \(\int\) (1 + cos 2x) dx
= \(\frac{1}{2}\) (x + \(\frac{sin2x}{2}\)) + c
\(\frac{2 y}{\cos x}\) = x + sin x . cos x + c
Solution is 2y = x cos x + sin x . cos² x + c . cos x^

Question 16.
(1 + x²) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = e^{tan-1 x} [May 07] [A.P. Mar. 16] [T.S. Mar. 15]
Solution:

Question 17.
Solve (1 + y²)dx = (tan^-1y – x)dy. [A.P. Mar. 15]
Solution:
Given \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\tan ^{-1} y-x}{1+y^{2}}\)

Question 18.
(x² – y²)dx – xy dy = 0 [May 06]
Solution:
(x² – y²)dx – xy dy = 0
(x² – y²)dx = xy . dy

–\(\frac{1}{4}\) [log (x² – 2y²) – log x²] = log x + log c
–\(\frac{1}{4}\) log (x² – 2y²) + \(\frac{1}{4}\) . 2 log x = log x + log c
–\(\frac{1}{4}\) log (x² – 2y²) = \(\frac{1}{2}\) log x + log c
– log (x² – 2y²) = – 2 log x – 4 log c
log (x² – 2y²) = – 2 log x + log k where
k = \(\frac{1}{c^{4}}\) = log \(\frac{\mathrm{k}}{\mathrm{x}^{2}}\)
x² – 2y² = \(\frac{\mathrm{k}}{\mathrm{x}^{2}}\)
Solution is x² (x² – 2y²) = k

Question 19.
\(\frac{d y}{d x}=\frac{3 y-7 x+7}{3 x-7 y-3}\) [T.S. Mar. 16]
Solution:


= 3ln (v – 1) – 3ln (v + 1) – 7ln (v + 1) – 7ln (v – 1)
14ln x – ln c = – 10 ln (v + 1) – 4 ln (v – 1)
ln (v + 1)⁵ + ln (v – 1)² + ln x⁷ = ln c
(v + 1)⁵ . (v – 1)² . x⁷ = c
(\(\frac{y}{x}\) + 1)⁵ (\(\frac{y}{x}\) – 1)² . x⁷ = c
(y – x)² (y + x)⁵ = c
[y – (x – 1)]² (y + x – 1)⁵ = c
Solution is [y – x + 1]² (y + x – 1)⁵ = c.

Question 20.
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) (x²y³ + xy) = 1 [Mar. 11]
Solution:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = xy + x²y³
This is Bernoulli’s equation
x^-2 . \(\frac{\mathrm{dx}}{\mathrm{dy}}\) – \(\frac{1}{x}\) . y = y³

Question 21.
Form the differential equation corresponding to y = A cos 3x + B sin 3x, where A and B are parameters. [AP Mar. 15]
Solution:
We have y = A cos 3x + B sin 3x
Differentiating w.r.to x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -3A sin 3x + 3B cos 3x
Differentiating again w.r.to. x
\(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\) = -9A cos 3x – 9B sin 3x
= – 9(A cos 3x + B sin 3x)
= -9y .
is \(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\) + 9y = 0.
Alternate method:
Eliminating A, B from the equation
y = A cos 3x + B sin 3x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = – 3A sin 3x + 3B sin cos 3x

This is the required differential equation.

Question 22.
Solve (x² + y²) dx = 2xy dy [A.P. Mar. 16]
Solution:
Given equation can be written as

log cx(1 – v² = log 1
cx (1 – v²) = 1
cx (1 – \(\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}\)) = 1
c(x² – y²) = x is the required solution.

Question 23.
Give the solution of x sin² \(\frac{y}{x}\) dx = y dx – x dy which passes through the point (1, \(\frac{\pi}{4}\)). [Mar. 14]
Solution:
The given equation can be written as

The given curve passes through (1, \(\frac{\pi}{4}\))
cot \(\frac{\pi}{4}\) = log 1 + c
1 = 0 + c ⇒ c = 1
Solution is cot \(\frac{y}{x}\) = log x + 1

Question 24.
Find the order and degree of the differential equation \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) = – p²y.
Solution:
The given equation is a polynomial equation in \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\)
Hence the degree is 1
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) is the highest order derivative occuring in the equation.
Its order is 2.

Question 25.
Find the order and degree of
(\(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\))² – 3 (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))² – e^x = 4 [Mar. 14]
Solution:
The equation is a polynomial in \(\frac{\mathrm{dy}}{\mathrm{dx}}\) and \(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\).
The exponent of \(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\) is 2.
Hence the degree is 2.
\(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\) is the highest õrder derivative occuring in the equation.
Order of the equation is 3.

Question 26.
x^{\(\frac{1}{2}\)}(\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\))^{\(\frac{1}{3}\)} + x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = 0 has order 2 and degree 1. Prove. [T.S. Mar. 15]
Solution:
The given equation can be written as
x^{\(\frac{1}{2}\)}(\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\))^{\(\frac{1}{3}\)} = -[x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y]
Cubing both sides
x^{\(\frac{3}{2}\)}(\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\)) = -[x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y]³
∴ The order of the equation is 2 and its degree is 1.

Question 27.
Find the order and degree of \(\left(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}\right)^{\frac{6}{5}}\) = 6y [A.P. Mar. 16; May 11]
Solution:
Given equation is \(\left(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}\right)^{\frac{6}{5}}\) = 6y
i.e., \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) + (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))³ = (6y)^{\(\frac{5}{6}\)}
Order = 2; degree = 1

Question 28.
Find the order of the differential equation corresponding to y = c(x – c)², where c is an arbitrary constant.
Solution:
The given differential equation is
y = c(x – c)²
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2c(x – c)
∴ Order of the differential equation is 1.

Question 29.
Find the order of the differential equation corresponding to y = Ae^x + Be^3x + Ce^5x; (A, B, C being parameters) is a solution.
Solution:
Required differential equation is obtained by eliminating A, B, C from y,
\(\frac{\mathrm{dy}}{\mathrm{dx}}\), \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\), \(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\)
Highest order deviation = \(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\)
Order of the differential equation = 3.

Question 30.
Form the differential equation corresponding to y = cx – 2c², where c is a parameter.
Solution:
Given y = cx – 2c² ………………. (1)
Differentiating (1) w.r.to
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = c
Substituting in (1), required differential equation is
y = x . (\(\frac{\mathrm{dy}}{\mathrm{dx}}\)) – 2(\(\frac{\mathrm{dy}}{\mathrm{dx}}\))²

Question 31.
Form the differential equation corresponding to y = A cos 3x + B sin 3x, where A and B are parameters. [A.P. Mar. 15]
Solution:
We have y = A cos 3x + B sin 3x
Differentiating w.r.to x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -3A sin 3x + 3B cos 3x
Differentiating again w.r.to. x
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) = -9A cos 3x – 9B sin 3x
= -9(A cos 3x + B sin 3x)
= -9y
is \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) + 9y = 0

Alternate Method:
Eliminating A, B from the equation
y = A cos 3x + B sin 3x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -3A sin 3x = 3B sin cos 3x
\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) = -9A cos 3x – 9B sin 3x
We get \(\left|\begin{array}{ccc}
y & -\cos 3 x & -\sin 3 x \\
\left(\frac{d y}{d x}\right) & 3 \sin 3 x & -3 \cos 3 x \\
\left(\frac{d^{2} y}{d x^{2}}\right) & 9 \cos 3 x & 9 \sin 3 x
\end{array}\right|\) = 0
y(27 sin² 3x + 27 cos² 3x) – (-9 sin 3x. cos 3x + 9 cos 3x. sin 3x) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + (3 cos² 3x + 3 sin² 3x) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) = 0
= 27y + 3 .\(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) = 0 or \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) + 9y = 0
This is the required differential equation.

Question 32.
Form the differential equation corresponding to the family of circles of radius r given by (x – a)² + (y – b)² = r², where a and b are parameters.
Solution:
We have (x – a)² + (y – b)² = r² ………………… (1)
Differentiating (1) w.r.to x
2(x – a) + 2(y – b) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0 ……………….. (2)
Differentiating (2) w.r.to. x
1 + (y – b) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) + (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))² = 0 ……………… (3)
From (2) (x – a) = -(y – b) \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
Substituting in (1), we get

i.e., \(r^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}=\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{3}\)
Which is the required differential equation.

Question 33.
Form the differential equation corresponding to the family of circles passing through the origin and having centres on Y-axis.
Solution:
The equation of the family of all circles passing through the origin and having centres on Y—axis is
x² + y² + 2hy = 0 …………………. (1)
Where h is a parameter
Differentiating (1) w.r. to x
2x + 2y . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2h . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
or x + y . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + h . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
-(x + y . \(\frac{\mathrm{dy}}{\mathrm{dx}}\)) = h. \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
h = \(\frac{-\left(x+y \cdot \frac{d y}{d x}\right)}{\frac{d y}{d x}}\)
Substituting in (1)
We get x² + y² – 2y \(\frac{\left(x+y \cdot \frac{d y}{d x}\right)}{\frac{d y}{d x}}\) = 0
x² . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y² . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 2xy – 2y² . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
or (x² – y²) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 2xy = 0
This is the required differential equation.

Question 34.
Express the following differential equations in the form f(x) dx + g(y) dy = 0.
i) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1+\mathrm{y}^{2}}{1+\mathrm{x}^{2}}\)
Solution:
⇒ \(\frac{d y}{1+y^{2}}=\frac{d x}{1+x^{2}}\)
\(\frac{d x}{1+x^{2}}-\frac{d y}{1+y^{2}}\) = 0

ii) y – x \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = a (y² + \(\frac{\mathrm{dy}}{\mathrm{dx}}\))
Solution:
y – x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = ay² + a. \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
y – ay² = (x + a) . \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
\(\frac{d x}{x+a}=\frac{d y}{y-a y^{2}}\)

iii) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^x-y + x² e^-y
Solution:
Multiplying in the e^y
e^y . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^x + x²
e^y . dy = (e^x + x²) dx
(e^x + x²) dx – e^y . dy = 0

iv) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + x² = x² e^3y
Solution:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x² . e^3y – x² = x²(e^3y – 1)
\(\frac{d y}{e^{3 y}-1}\) = x² dx ⇒ x² dx – \(\frac{d y}{e^{3 y}-1}\) = 0
or x²dx + \(\frac{1}{\left(1-e^{3 y}\right)}\) . dy = 0

Question 35.
Find the general solution of
x + y\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0.
Solution:
Given equation is x + y . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
x . dx + y . dy = 0
Integrating \(\frac{x^{2}}{2}\) + \(\frac{y^{2}}{2}\) = c
or x² + y² = 2 c = c’

Question 36.
Find the general solution of \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^x+y.
Solution:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^x+y = e^x . e^y
\(\frac{d y}{e^{y}}\) = e^x dx
\(\int\) e^-y dy = \(\int\) e^x dx
e – e^-y = e^x
or e^x + e^-y = c is the required solution.

Question 37.
Solve y² – x \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = a(y + \(\frac{\mathrm{dy}}{\mathrm{dx}}\))
Solution:

Question 38.
Solve \(\frac{d y}{d x}=\frac{y^{2}+2 y}{x-1}\)
Solution:

Question 39.
Solve \(\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}\) [Mar. 08]
Solution:
Given equation can be written as
(sin y + y cos y) dy = x(2 log x + 1) dx
\(\int\) sin y dy + \(\int\) y cos y dy = \(\int\) 2x log x dx + \(\int\) x dx
\(\int\) sin y dy + y sin y – \(\int\) sin y dy = x² log x – \(\int\) x² . \(\frac{1}{x}\) dx + \(\int\) x dx + c
y sin y = x² log x + c

Question 40.
Find the equation of the curve whose slope, at any point, (x, y) is \(\frac{y}{x^{2}}\) and which satisfies the condition y = 1 when x = 3.
Solution:

Question 41.
Solve y(1 + x) dx + x(1 + y) dy = 0
Solution:
The given equation can be written as
\(\frac{(1+x)}{x}\) dx + \(\frac{(1+y)}{y}\) . dy = 0
\(\int\) (1 + \(\frac{1}{x}\))dx + \(\int\) (1 + \(\frac{1}{y}\)) dy = 0
x + log x + y + log y = c
x + y + log (xy) = c is the required solution.

Question 42.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sin (x + y) + cos (x + y)
Solution:
Put x + y = t

x = log (1 + tan \(\frac{t}{2}\)) + c
But t = x + y
Solution is x = log (1 + tan \(\frac{x+y}{2}\)) + c

Question 43.
Solve (x – y)² \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = a²
Solution:
Put x – y = t

Question 44.
Solve \(\sqrt{1+x^{2}} \sqrt{1+y^{2}}\) dx + xy dy = 0
Solution:
Given equation can be written as

Question 45.
Solve \(\frac{d y}{d x}=\frac{x-2 y+1}{2 x-4 y}\)
Solution:
Put x – 2y = t

Question 46.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\sqrt{y-x}\)
Solution:
Put y – x = t²

Question 47.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 1 = e^x+y
Solution:
Put t = x + y
\(\frac{\mathrm{dt}}{\mathrm{dx}}\) = 1 + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^t
\(\int \frac{d t}{e^{t}}=\int d x\)
\(\int\) e^-t dt = \(\int\) dx
-e^-t = x + c
x + e^-t + c = 0
Solution is x + e^-(x+y) + c = 0

Question 48.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (3x + y + 4)²
Solution:
Put t = 3x + y + 4
\(\frac{\mathrm{dt}}{\mathrm{dx}}\) = 3 + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3 + t²
\(\frac{\mathrm{dt}}{\mathrm{t}^{3}+3}\) = dx
\(\int \frac{d t}{t^{2}+3}=\int d x\)
\(\frac{1}{\sqrt{3}}\) tan^-1 (\(\frac{t}{\sqrt{3}}\)) = x + c
Solution is \(\frac{1}{\sqrt{3}}\) tan^-1 (\(\frac{3 x+y+4}{\sqrt{3}}\)) = x + c

Question 49.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) – x tan (y – x) = 1
Solution:
Put y – x = t
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 1 = \(\frac{\mathrm{dt}}{\mathrm{dx}}\)
\(\frac{\mathrm{dt}}{\mathrm{dx}}\) = x tan t + 1 – 1 = x tan t
\(\frac{\mathrm{dt}}{\tan \mathrm{t}}\) = x dx
\(\int\) cot dt = \(\int\) x dx
log |sin | = \(\frac{x^{2}}{2}\) + c
Solution is log |sin (y – x)| = \(\frac{x^{2}}{2}\) + c

Question 50.
Show that f(x, y) = 1 + e^x/y is a homogeneous function of x and y.
Solution:
f(kx, xy) = 1 + e^kx/ky = 1 + e^x/y = f(x, y)
f(x, y) is a homogeneous function degree 0.

Question 51.
Show that f(x, y) = x\(\sqrt{x^{2}+y^{2}}\) – y² is a homogeneous function of x and y.
Solution:
f(kx, ky) = kx\(\sqrt{k^{2} x^{2}+k^{2} y^{2}}\) – k²y².
= k² (x\(\sqrt{x^{2}+y^{2}}\) – y²) = k² f(x, y)
f(x, y) is a homogeneous function of degree 2.

Question 52.
Show that f(x, y) = x – y log y + y log x is a homogeneous function of x and y.
Solution:
f(kx, ky) = kx – ky. log ky + ky log (kx)
= k(x – y log (ky) + y log kx)
= k(x – y log k – y log y + y log k + y log x)
= k(x – y log y + y log x)
= k. f(x, y)
f(x, y) is a homogeneous function of degree 1.

Question 53.
Express (1 + e^x/y)dx + e^x/y (1 – \(\frac{x}{y}\)) dy = 0 in the form \(\frac{\mathrm{dx}}{\mathrm{dy}}\) = F (\(\frac{x}{y}\))
Solution:

Question 54.
Express (x\(\sqrt{x^{2}+y^{2}}\) – y²) dx + xy dy = 0 in the form \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = F (\(\frac{x}{y}\))
Solution:
Given equation is

Question 55.
Express \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{y}{x+y e^{\frac{-2 x}{y}}}\) in the form \(\frac{\mathrm{dx}}{\mathrm{dy}}\) = F (\(\frac{x}{y}\))
Solution:

Question 56.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{y^{2}-2 x y}{x^{2}-x y}\)
Solution:
The given equation is a homogeneous equation.
Put y = vx

\(\log v \sqrt{2 v-3}=-3 \log \frac{x}{c}=\log \frac{c^{3}}{x^{3}}\)

Question 57.
Solve (x² + y²) dx = 2xy dy [A.P. Mar. 16]
Solution:
Given equation can be written as
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x^{2}+y^{2}}{2 x y}\)
This is a homogeneous function
Put y = vx

-log (1 – v²) = log x + log c
= log cx
log cx + log (1 – v²) = 0
log cx(1 – v²) = log 1
cx (1 – v²) = 1
cx (1 – \(\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}\)) = 1
c(x² – y²) = x is the required solution.