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

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.