AP Inter 2nd Year Chemistry Important Questions Chapter 6(b) Group-16 Elements

Students get through AP Inter 2nd Year Chemistry Important Questions Lesson 6(b) Group-16 Elements which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions Lesson 6(b) Group-16 Elements

Very Short Answer Questions

question 1.
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 2.
Why H₂O a liquid while H₂S Is a gas? ( IPE May – 2014)
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 3.
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 4.
Explain the structures of SF₄ and SF₆.
Answer:
Structure of SF₄:

• In SF₄ ‘S’ undergoes sp³d hybridisation.
• 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 5.
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 peroxides.
c) KO₂, RhO₂ are super oxides.

Question 6.
What is tailing of mercury? How is it removed? (IPE Mar – 2015 (TS))
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 7.
How does ozone react with Ethylene?
Answer:
Ethylene reacts with ozone to form Ethylene ozonoid followed by the hydrolysis to form formaldehyde.

Question 8.
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 it exhibits paramagnetism.

Question 9.
Why are group — 16 elements called chalçogens ?
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 10.
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.

Short Answer Questions

Question 1.
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 : 3858K.
• Specific gravity: 2.06.
• It is insoluble 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 2.
Which is used for drying ammonia?
Answer:
For drying ammonia quick lime (CaO) is used.

• For drying ammonia conc. H₂SO₄ , P₄O₁₀ and anhydrous CaCl₂ cannot be used because they react with ammonia and forms (NH₄)₂SO₄, (NH₄)₃PO₄ and CaCl₂. 8NH₃

Question 3.
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

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 Chattier’s principle,

i. a decrêase 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 manufacture 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. (V₂O₅ (or) Pt – asbestos).
Favourable Conditions:
Temperature: 720K
Pressure :2 bar
Catalyst : V₂O₅ (or) platinized asbestos.

Question 4.
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 sulphide oxidises to hypo with SO₂.
2Na₂S + 3SO₂ → 2Na₂S₂O₃ + S
SO₂ as Reducing agent:
SO₂ reduces Fe⁺³ ions to Fe⁺² ions.
2Fe⁺³ + SO₂ + 2H₂O → 2Fe²⁺ + \(\mathrm{SO}_4^{-2}\) + 4H⁺

Long Answer Questions

Question 1.
Explain in detail the manufacture of sulphuric acid by contact process. ( IPE 2016 (TS))
Answer:
Manufacture of H₂SO₄ by contact process:
Manufacturing of H₂SO₄ involves three main steps.

Step-I:
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 atmospheric air to form SO₃.

Le Chatliers 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

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,

1. 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.
2. 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 manufacture of H₂SO₄, the optimum temperature suitable for the conversion of SO₂ into SO₃ is experimentally found to be 720K.
3. 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 → H₂SO₄

Question 2.
How is ozone prepared from oxygen? Explain its reaction with
a) C₂H₄
b) KI
c) Hg
d) PbS. (T.S. & A.P. Mar. ’17) (A.P.Mar. ’16) (Mar. ’14)
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 ΔH° = 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 C₂H₄ : Ethylene reacts with ozone to form Ethylene ozonoid followed by the hydrolysis to for formaldehyde.

b) 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.

Question 3.
Write the structures of oxoacids of sulphur. (IPE 2016 (TS))
Answer:
Oxoacids of sulphur: Sulphur forms a number of oxoacids such as H₂SO₃. H₂S₂O₃, H₂S₂O₄, H₂S₂O₅, H₂S₂O₆ (x = 2 to 5), H₂SO₄, H₂S₂O₇, H₂SO₅, H₂S₂O₈.

Question 4.
Write any two oxidation and any two reduction properties of ozone with equations.
Answer:
Oxidation properties:

1. Ozone oxidises moist potassium idodide and liberates I₂.
   2KI + H₂O + O₃ → 2KOH + I₂ + O₂
2. Ozone oxidises black lead sulphide to white lead sulphate. PbS + 4O₃ → PbSO₄ + 4O₂

Reduction properties:

1. Ozone reduces H₂O₂ to H₂O.        H₂O₂ + O₃ → H₂O + 2O₂
2. Ozone reduces Ag₂O to Ag.         Ag₂O + O₃ → 2Ag + 2O₂