FSc Chemistry 12th Ch 4 Group VA and Group VIA Elements Exercise Short Questions
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Exercise Short Questions
Q4 (i) How does nitrogen differ from other elements of its group?
(ii) Why does aqua regia dissolve gold and platinum?
(iii) Why the elements of group VI-A other than oxygen show more than two oxidation states?
(iv) Write down a comparison of the properties of oxygen and sulphur.
(v) Write the equation for the reaction between conc. H2SO4 and copper and explain the type of reaction.
Ans.
(i) How does nitrogen differ from other elements of its group?
- Nitrogen is a gas while other members are solids.
- Nitrogen has no allotropes while P, Sb, Bi has allotropes.
- Nitrogen does not use its d-orbital for bond formation.
- Nitrogen forms multiple bond with nitrogen atom (N ≡ N) and with the atoms of other elements.
- Nitrogen exists in diatomic form (N2) while phosphorous occurs as P4.
All these properties are not shown by other elements of V-A group. Therefore these properties make nitrogen different from rest of its group members.
(ii) Why does aqua regia dissolve gold and platinum?
Aqua Regia is three parts of concentrated HCl and one part of concentrated HNO3. Metals like gold and platinum can dissolve in aqua regia by the formation of their chlorides.
3HCl(conc.) + HNO3(conc.)→ NOCl(aq) + Cl2(g) + 2H2O(l)
2NOCI → 2NO + Cl2
This liberated chlorine gas converts noble metals to their chlorides.
2Au + 3Cl2 → 2AuCl3
Over all reaction is as follows:
2Au + 3HCI + HNO2 → AuCl3 + NO + 2H2O
(iii) Why the elements of group VI-A other than oxygen show more than two oxidation states?
Except oxygen, the other members of group VI-A show the valences of +2, +4 or+6 e.g.
+2 oxidation state is shown due to unpaired electrons in the p-orbital.
+4 oxidation state is shown when 1 electron from p-orbital is promoted to the next vacant d-orbital.
+6 oxidation state is shown when another electron from s-orbital is also promoted to the next vacant d-orbital.
(iv) Write down a comparison of the properties of oxygen and sulphur.
Comparison of Oxygen and Sulphur
Similarities:
- Both oxygen and sulphur have same outer electronic configuration of ns2 np4.
- Both oxygen and sulphur are usually divalent.
- Both oxygen and sulphur exhibit allotropic forms.
- Both have polyatomic molecules. Oxygen has diatomic O2, while sulphur S2 and S8 molecules.
- Both combine with metals in the form of O-2 and S-2 with oxidation sates -2.
- Both combine with non-metals and form covalent compounds, e.g, H2O and H2S, CO2 and CS2, etc.
- Both are typical non-metals.
- Both are found in free and combined states on earth.
Dissimilarities:
| Oxygen | Sulphur |
| 1. There are two allotropic forms of Oxygen O2 and O3. | There are 3 allotropic forms of sulphur, rhombic, monoclinic and plastic |
| 2. It is gas at ordinary temperature. | It is solid at ordinary temperature. |
| 3. Oxygen is sparingly soluble in water. | Sulphur is not soluble in water. |
| 4. Oxygen helps in combustion. | Sulphur is itself combustible. |
| 5. It is paramagnetic in nature. | It is diamagnetic in nature. |
| 6. It does not react with water. | When steam is passed through boiling sulphur a little hydrogen sulphide and sulphur dioxide are formed. |
| 7. It does not react with acids. | It is readily oxidized by conc. sulphuric acid or nitric acid. |
| 8. It does not react with alkalies. | It reacts with alkali solution and forms sulphides and thiosulphate. |
| 9. It shows -2 oxidation state. | It shows oxidation states of -2, +2, +4 and +6. |
(v) Write the equation for the reaction between conc. H2SO4 and copper and explain the type of reaction.
Cu+2H2SO4 → CuSO4 + SO2 +2H2O
(Cupric sulphate)
It is a redox reaction. Cu acts as reducing agent and H2SO4 acts as an oxidizing agent.
Q5. (a) Explain the Birkeland and Eyde’s process for the manufacture of nitric acid.
(b) Which metals evolve hydrogen upon reaction with nitric acid? Illustrate along with chemical equation.
(c) What is meant by fuming nitric acid?
Ans.
(a) Explain the Birkeland and Eyde’s process for the manufacture of nitric acid.
Birkeland and Eyde’s process
This process consists of the following steps:
(i) Formation of nitric oxide
Atmospheric nitrogen and oxygen are combined to give nitric oxide in an electric arc (3000oC).
N2(g) + O2(g) –3000oC→ 2NO(g)
NO formed is cooled quickly to 1000oC at which it does not decompose.
(ii) At 600oC, NO combines with O2 to form NO2.
2NO(g) + O2(g) → 2NO2(g)
(iii) Nitrogen dioxide is absorbed in water to give dilute HNO3 alongwith nitrous acid.
2NO2(g) + H2O → HNO3(aq) + HNO2(aq)
(iv) Nitrous acid is oxidized to nitric acid and nitric oxide which is re-oxidized to NO2.
3HNO2(g) →HNO3(aq)+ 2NO(g) + H2O(l)
(b) Magnesium and manganese react with HNO3 and evolve hydrogen gas:
Mn + 2HNO3 → Mn(NO3)2 + H2
Mg + 2HNO3 → Mg(NO3)2 + H2
(c) What is meant by fuming nitric acid?
Concentrated HNO3 is a colourless volatile liquid and it fumes strongly in air and to evolve NO2 gas. Therefore mixture of HNO3+ NO2 is called as fuming nitric acid.
Q6. (a) Sulphuric acid is said to act as an acid, an oxidizing agent and a dehydrating agent, describe two reactions in each case to illustrate the truth of this statement.
(b) Give the advantages of contact process for the manufacture of sulphuric acid.
Ans. (a) Sulphuric acid is said to act as an acid, an oxidizing agent and a dehydrating agent, describe two reactions in each case to illustrate the truth of this statement.
(i) As an acid:
H2SO4(ag) + NaOH(aq) → NaHSO4(aq) + H2O(l)
NaHSO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l)
(ii) As on oxidizing agent:
C(s) + 2H2SO4(aq) → CO2(g) + 2SO2(g) + 2H2O(g)
H2S + H2SO4(aq) → S(s) + SO2(g) + 2H2O
(iii) As dehydration agent:
HCOOH(aq) –Conc. H2SO4→ CO(g) + H2O(l)
C2H5OH –Conc. H2SO4/100°C→ C2H4 + H2O
(b) Give the advantages of contact process for the manufacture of sulphuric acid.
- Solid catalysts V2O5, is used and there is no chance of nitrogen oxide impurities.
- In this process 100% H2SO4 is produced.
Q7. (a) Describe the chemistry of the Industrial preparation of sulphuric acid from sulphur by the contact process.
(b) Why is SO2 dissolved in H2SO4 and not in water?
(c) Explain the action of sulphuric acid on metals along with chemical equations.
Ans.
(a) Describe the chemistry of the Industrial preparation of sulphuric acid from sulphur by the contact process.
Manufacture of Sulphuric Acid
Sulphuric acid is being manufactured commonly by contact process.
Contact Process
This method was developed by Knietsch in Germany. Basically, it involves the catalytic combination of sulphur and oxygen to form SO2 which is then dissolved in water to form H2SO4.
Principle
SO2 obtained by burning sulphur or iron pyrites is oxidized to SO3 in the presence of V2O5 which acts as a catalyst.
The best yield of SO3 can be obtained by using excess of oxygen or air and keeping the temperature between 400-500oC
SO3 formed is absorbed in concentrated H2SO4 and “Oleum” (H2S2O7) formed can be converted to sulphuric acid of any strength by mixing adequate quantities of water.
The process is completed in the steps given below.
a. Suphur Burners
Sulphur or iron pyrites are burnt in excess of air to produce SO2.
b. Purifying Unit
SO2 is purified from impurities like dust and arsenic oxide, to avoid poisoning of the catalyst. Purifying unit consists of the following parts.
- Dust Remover: Steam is injected to remove dust particles from the gases.
- Cooling Pipes: The gases are passed through lead pipes to cool them to 100oC.
- Scrubbers: The cooled gases are washed by a spray of water, as SO2 is not soluble in water at high temperature.
- Drying Tower: The moisture of gases is removed by concentrated H2SO4 trickling down through the coke filled in this tower.
- Arsenic Purifier: Arsenic oxide is then removed by passing the gases through a chamber provided with shelves packed with freshly prepared ferric hydroxide.
- Testing box: In this box a beam of light is introduced which indicates the presence or absence of solid particles. If present the gases are sent back for further purification.
c. Contact Tower
Preheated gases at 400-500oC are passed through vertical iron columns packed with the catalyst V2O5. Here SO2 is oxidized to SO3.
2SO2(g) + O2(g) –400-500oC/V2O5→ 2SO3(g)
The reaction is highly exothermic so no heating is required once the reaction is started.
d. Absorption Unit
The SO3 obtained from the contact tower is dissolved in 98% H2SO4 to form pyrosulphuric acid (oleum), (H2S2O7). It can be diluted with water to get any required concentration of sulphuric acid.
(b) Why is SO2 dissolved in H2SO4 and not in water?
In the process of the formation of H2SO4, SO3 is dissolved in H2SO4 to form oleum (H2S2O7) which is diluted to get pure H2SO4.
While if SO3 is dissolved in water, the reaction will be highly exothermic and metallic fog is formed which impure the product (H2SO4).
(c) Explain the action of sulphuric acid on metals along with chemical equations.
Sulphuric acid reaction with metals
(a) Cold dilute acid reacts with almost all metals to produce hydrogen gas and sulphate salts.
Fe + H2SO4 → FeSO4 + H2
Zn + H2SO4 → ZnSO4 + H2
Mg + H2SO4 → MgSO4 + H2
Sn + H2SO4 → SnSO4 + H2
(b) Cold concentrated H2SO4 does not react with most of the metals like Cu, Ag, Hg, Pb, Au.
(c) With certain metals hot concentrated sulphuric acid gives metal sulphates, water and SO2.
Cu(aq) + 2H2SO4(aq) → CuSO4(aq) + 2H2O(l) + SO2(aq)
2Ag(aq) + 2H2SO4(aq) → Ag2SO4(aq) + 2H2O(l) + SO2(aq)
Hg(l) + 2H2SO4(aq) → HgSO4(aq) + 2H2O(l) + SO2(aq)
Q8. Describe the preparation of NO2 gas. Also give its reactions.
Ans. Preparation of NO2 gas
1. It can be prepared in small quantities by heating lead nitrate.
2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g)
2. It can also be prepared by reacting conc. HNO3 with copper.
Cu(s) + 4HNO3(conc.) → Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g)
Nitrogen is a reddish brown gas with a pungent smell. It dissolved readily in water to form a blue acidic solution.
Reactions of NO2 gas
1. On cooling, NO2 is converted into a yellow liquid which can be frozen to a colorless solid dinitrogen tetraoxide (N2O4). If this solid is heated to 140oC, the mixture contains NO2 and N2O4 but above 140oC NO2 is converted to NO and O2 molecules which are colorless. This decomposition is complete at 620oC.
NO2(g) → NO2(g) → N2O4(l) ⇌140oC⇌ 2NO2(g) ⇌620oC⇌ 2NO(g) + O2
2. Elements like phosphorus, potassium and carbon continue burning in NO2 as it yields O2 on decomposition.
2NO2(g) → 2NO(g) + O2
2P(s) + 5NO2(g) → P2O5(s) + 5NO(g)
3. In the absence of air, it dissolved in water to form nitric and nitrous acids.
2NO2(g) + H2O(l) → HNO3(aq) + HNO2(aq)
However in the presence of air or oxygen, nitric acid is the final product.
4NO2(g) + H2O(l) + O2(g) → 4HNO3(aq)
4. A mixture of nitrate and nitrite is formed when NO → is passed through strong alkalies.
2NaOH(aq) + 2NO2(g) → NaNO3(aq) + NaNO2(aq) + H2O(l)
2KOH(aq) + 2NO2(g) → KNO3(aq) + KNO2(aq) + H2O(l)
5. It is a strong oxidizing agent and oxidized H2S to sulphur, ferrous sulphate to ferric sulphate and KI to I2.
H2S(g) + NO2(g) → H2O(l) +S(s) + NO(g)
2FeSO4(aq) + H2SO4(aq) + NO2(g) → H2O(l) +Fe2(SO4)3(s) + NO(g)
2KI(aq) + 2NO2(g) → 2KNO3(aq) + I2(s)
Q9. How PCl3 and PCl5 can be used for the preparation of other chemical compounds?
Ans.
PCl3 Reactions
1. It combines with chlorine to form phosphorus pentachloride.
PCl3(l) + Cl2(g) → PCl5(s)
2. It combines with atmospheric oxygen slowly to form phosphorus oxychloride.
2PCl3(l) + O2(g) → 2POCl3(s)
3. It is soluble in organic solvents, but readily reacts with water to form phosphorus acid.
PCl3(l) + 3H2O(l) → H3PO3(aq) + 3HCl(aq)
4. It reacts with alcohols and carboxylic acids forming the respective chloro derivatives.
3CH3OH(l) + PCl3(l) → 3CH3Cl(l) + H3PO3(l)
3CH3COOH(l) + PCl3(l) → 3CH3COCl(l) + H3PO3(l)
PCl5 Reactions
1. It decomposes on heating producing PCl3 and chlorine.
PCl5(s) → PCl3(l) + Cl2(g)
2. It gets decomposed by water forming phosphorus oxychloride which further reacts will water to produce orthophosphoric acid.
PCl5(s) + H2O(l) → POCl3(l) + 2HCl(aq)
POCl3(l) + 3H2O(l) → H3PO4(aq) + 3HCl(aq)
PCl5(s) + 4H2O(l) → H3PO4(aq) + 3HCl(aq)
3. It converts metals into their chlorides.
Zn(s) + PCl5(s) → ZnCl2(s) + PCl3(l)
Q10. Answer the following questions:
(i) Describe “Ring test” for the confirmation of the presence of nitrate ions in solution.
(ii) NO2 as a strong oxidizing agent. Prove the truth of this statement giving example.
(iii) Write down the chemical equations and names of the products formed as a result of the reaction of HNO3 with arseniec and antimony.
(iv) Give the methods of preparation of PCl5.
(v) P2O5 is a powerful dehydrating agent. Prove giving example.
Ans. (i) Ring Test: This test is used to confirm NO3-1 The nitrate is dissolved in H2O and freshly prepared FeSO4 is added in it. Conc. H2SO4 is added along the walls of the test tube. Brown ppts. of FeSO4. NO are formed.
FeSO4 +NO → FeSO2. NO
(ii) NO2 is Strong Oxidizing Agent:
2KI+ 2NO2 → 2KNO2+ I2
H2S(g) + NO2(g) → S(s) + NO(g) + H2O(l)
2FeSO4 + H2SO4+ NO2 → Fe2(SO4)3 + NO + H2O
(iii) Reactions of HNO3 on As, Sb:
As + 5HNO3 → H3AsO4+ 5NO2 + H2O
Sb + 5HNO3 → H3SbO4 + 5NO2 + H2O
(iv) Preparation of PCl5:
(i) By passing Cl2 through PCI3 at 0°C:
PCl3 + C2 –0°C→ PCl5
(ii) By passing Cl2 in cold solution of phosphorous in CS:
(v) P2O5 is Dehydrating Agent:
The substance which is used to remove water from a substance is called dehydrating agent.
H2SO4 + P2O5 → SO3 + 2HPO3
2HNO3 + P2O5 → N2O5 + 2HPO3
C2H5OH + P2O5 → C2H4 + 2HPO3
2CH3COOH(aq) + P2O5 → (CH3CO)2O(l) + 2HPO3(aq)
Q11. Complete and balance the following chemical equations:
Ans. (i) 2P + 5NO2 → P2O5 +5NO
(ii) 2NO+ Cl2 → 2NOCl
(iii) H2S+ 2NO → H2O+ N2O +S
(iv) 2Pb(NO3)2 → 2PbO + 4NO2 + O2
(v) 2NO2 + H2O → HNO3 + HNO2
(vi) 2HNO2 + 2HI → 2H2O + 2NO + I2
(vii) HNO2+ NH3 → 2H2O + N2
(viii) 2HNO2 + CO(NH2)2 → 2N2 + CO2+ 3H2O
(ix) 2KNO3 + H2SO4 → K2SO4 + 2HNO3
Q12. Describe the methods of preparation of phosphours pentaoxide and explain its reaction.
Ans. Preparation
It is prepared by burning phosphorus in excess of dry air.
P4(s) + 5O2(g) → 2P2O5(s)
Properties of Phosphorus Pentaoxide
It is a white hygroscopic powder having a faint, garlic like odour due to the presence of traces of P2O3. It sublimes at 360oC.
Reactions
1. With cold water phosphorus pentaoxide forms metaphosphoric acid.
P2O5(s) + H2O(l) → 2HPO3(aq)
With hot water, it forms orthophosphoric acid
P2O5(s) + 3H2O(l) → 2H3PO4(aq)
2. It is a powerful dehydrating agent, thus, with HNO3, H2SO4, CH3COOH and C2H5OH, it gives N2O5, SO3, (CH3CO)2O and C2H4, respectively.
P2O5(s) + 2HNO3(aq) → N2O5(g) + 2HPO3(aq)
P2O5(s) + H2SO4(aq) → SO3(g) + 2HPO3(aq)
P2O5(s) + 2CH3COOH(aq) → (CH3CO)2O(l) + 2HPO3(aq)
P2O5(s) + C2H5OH(l) → C2H4(g) + 2HPO3(aq)
Q13. Discuss the trends in physical properties of group VI-A elements.
Ans. All the elements of group VIA are non-metals except Po which is a radioactive metal. Atomic radii, density, melting points generally increase with increase in atomic number down the group. Ionization energies of the group members are very high which shows their reluctance to lose electrons. Oxygen is the most electronegative element after fluorine. All these elements show the property of allotropy. Oxygen has two allotropic forms (O2 and O3), sulphur has 3(α, β, γ), Se has two (red and grey), Te has two (metallic and non-metallic). They also show the property of catenation. This property decreases down the group. All the elements are polymeric in nature (they form poly-atomic molecules). They attain the electronic configuration of the nearest noble gas by gaining 2 electrons forming O-2, S-2, Se-2, etc. Except oxygen the other members of the group show a covalency of +2, +4, and +6, for example SCl2, SCl4, SCl6. +2 oxidation state is shown due to 2 unpaired electrons in the p orbitals. +4 oxidation state is shown when 1 electron from p-orbital is promoted to the next vacant d-orbital, while +6 oxidation state is shown when another electron from s-orbital is also promoted to the next vacant d-orbital.
| Physical Properties | Oxygen O | Sulphur S | Selenium Se | Tellurium Te | Polonium Po |
| Atomic number | 8 | 16 | 34 | 52 | 84 |
| Electronic configuration | [He] 2s2 2p4 | [Ne] 3s2 3p4 | [Ar] 4s2 4p4 | [Kr] 5s2 5p4 | [Xe] 6s2 6p4 |
| Ionization energy (kJ/mol) | 1314 | 1000 | 941 | 869 | 813 |
| Electron affinity (kJ/mol) | 141.1 | 200.42 | -195 | -183 | -180 |
| Electronegativity | 3.5 | 2.5 | 2.4 | 2.1 | 2.0 |
| Atomic radius (pm) | 66 | 104 | 117 | 137 | 152 |
| Ionic radius 2-ion (pm) | 140 | 184 | 198 | 221 | — |
| Melting points (oC) | -218 | 113 | 217 | 450 | 254 |
| Boiling points (oC) | -183 | 444.6 | 684 | 990 | 962 |
| Density (g/cm3) | 0.00143 | 2.06 | 4.8 | 6.25 | 9.4 |
MCQs
Additional Short Questions
Q1. Write down the equation for the reaction between concentrated HNO3 and copper. Explain what type of reaction it is?
Ans. Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O
It is a redox reaction in which HNO3 acts as oxidizing agent and Cu acts as reducing agent. Oxidation state of copper changes from zero to +2. Oxidation state of nitrogen changes from +5 to +4 and it is reduced.
Q2. Why is SO3 dissolved in H2SO4 and not in water in contact process?
Ans. In contact process, SO3 is dissolved in conc. H2SO4 to form oleum.
SO3 + H2SO4 → H2S2O7
A measured quantity of water can be added to liquid oleum to change it into H2SO4.
H2S2O7 + H2O → 2H2SO4
When SO3 is directly dissolved in water, the reaction is highly exothermic and metallic fog is formed in the chamber.
Q3. Give one reaction of HNO2 in which it behave as oxidising agent and one reaction in which it behave as reducing agent?
Ans. HNO2 acts as oxidizing agent as well as reducing agent. HNO2 When reacts with HI and oxidation state of N changes from +4 → +2. HNO2 acts as oxidising agent.
H2S + 2HNO2 → S + 2NO + 2H2O
2HNO2 + 2HI → 2H2O+ 2NO + I2
HNO2 when reacts with aqueous bromine, the oxidation state of nitrogen changes from +4 to +5. In this reaction, HNO2 acts as reducing agent.
HNO2 + Br2 + H2O → HNO3+ 2HBr
2KMnO4 + 3H2SO4 + 5HNO2 → K2SO4 + 2MnSO4 + 5HNO3 + 3H2O
Q4. Why white phosphorus is more reactive than red phosphorus?
Ans. White phosphorus consists of individual P, molecules. The bond angle of P.P.P. in P4 molecule is of 60°, Due to smaller angle it has strain and more reactive because bond can be broken down easily, Red phosphorus is less reactive. It is macromolecule formed by the combination of P4 molecules.
Q5. H2SO4 acts as dehydrating agent. Give three reactions to support this statement.
Ans. C6H12O6 –Conc.H2SO4→ 6C + 6H2O
C2H5OH –Conc.H2SO4→ C2H4 + H2O
HCOOH –Conc.H2SO4→ CO+ H2O
Q6. H2SO4 acts as oxidizing agent. Give some examples?
Ans. C + 2H2SO4 → CO2 + 2SO2 + 2H2O
S + 2H2SO4 → 3SO2 + 2H2O
H2S + H2SO4 → S + SO2 + 2H2O
Q7. Sulphuric acid is dibasic acid, Explain it.
Ans. The acid which furnishes two protons when dissolved in water is called dibasic acid. H,SO, HS and oxalic acid are examples.
H2SO4 + H2O → H3O+ + HSO4–
HSO4– + H2O → H3O+ + SO4-2
Q8. What is poisoning of catalyst? How V2O5 is prevented from poisoning in contact process?
Ans. The phenomenon in which catalytic activity of a substance is reduced is called poisoning of catalyst. Vanadium pentoxide is poisoned by arsenic. Before going to contact chamber, the gases are passed through arsenic purifier which contains Fe(OH)3 to remove arsenic.
Q9. Which gases cannot be dried over sulphuric acid?
Ans. H2SO4 is good dehydrating agent and can be used to dry many gases. However, H2S, SO3 and NH3 cannot be dried with H2SO4 because these gases react with conc. H2SO4.
H2SO4 + SO3 → H2S2O7
H2SO4 + 2NH3 → (NH4)2SO4
H2SO4 + H2S → S + SO2 + 2H2O
Q10. Why HNO2 and H2SO4 acts as oxidizing agents and never reducing agents? Give reason for it.
Ans. Oxidation of N in HNO3 is +5 and S in H2SO4 is +6. During redox reaction charge of S or N always reduces or decreases and they always acts as oxidizing agent. Maximum oxidation state of the element is equal to its group No. Nitrogen belongs to group VA and sulphur belongs to group VIA.
Q11. What is the effect of temperature on NO2?
Ans. There is equilibrium between NO2 and N2O4. When temperature of the gas is increased, then it is in the NO2 state. At lower temperature it is in the N2O4 state.
N2O4 ⇌ 140oC⇌ 2NO2 ⇌ 620oC⇌ 2NO +O2
Color less Reddish brown Colorless
