Bihar Board - Class 12 Chemistry - Chapter 7: The P Block Elements Long Answer Question

Long Question Answer

Q. 1. Why does the reactivity of nitrogen differ from phosphorus ?

Ans⇒ Nitrogen molecule is a diatomic and the two nitrogen atoms are linked by triple bond (N=N). As the bond dissociation energy is very high (946 kJ mol^-1 ). It is not possible to cleave the triple bond so easily. Therefore the reactivity of nitrogen differs from phosphones.

Q.2. How is ammonia manufactured industrially ?

Ans⇒ Ammonia is manufactured industrially by Haber’s process.

N₂ (g) +3 H₂(g)⇌ 2NH₃ (g)

F=-46.1 kJ mol^-1 HO

A mixture of dry nitrogen and hydrogen gasses in the ratio of 1 : 3 by volume is compressed to about 200 to 300 atm and passed over iron mixed with aluminum oxide (Al₂O₃) and potassium oxide (K₂O) which act as promoters.

Ammonia being formed is continuously removed by liquefy it.

The optimum conditions for the production of ammonia are a pressure of about 200 atm., a temperature of -700 K and use of a catalyst such as iron oxide with small amounts of K₂O and Al₂O₃ to increase the rate of attainment of equilibrium.

Q. 3. How are XeO₃ are XeOF₄ prepared ?

Ans⇒ (i) Preparation of XeO₃ : It is prepared by the hydrolysis of XeF₄ and XeF₆ under controlled pH of medium.

6XeF₄+12H₂O₄Xe +2XeO₃+24HF+3O₂

XeF₆+3 H₂OXeO₃+6HF

(ii) Preparation of XeOF₄  :

Partial hydrolysis of XeF₆  gives XeOF₄

XeF₆+H₂OXeOF₄+2HF

Q.4. How is ammonia manufactured industrially? 

Ans⇒Ammonia is prepared on a large-scale by the Haber’s process. 

N₂ (g)+3H₂ (g)2NH₃ (g) f H=-46.1 kJ/mol

The optimum conditions for manufacturing ammonia are: 

Pressure ( around 200 105 Pa ) 

Temperature (4700 K)

Catalyst such as iron oxide with small amounts of Al₂O₃ and K₂O.

Q. 5. List the uses of neon and argon gasses.

Ans⇒ Uses of Neon : (i) Neon is mainly used in fluorescent lamps of tubes for advertising purposes. These are known as neon signs and can be seen at long distances even when there is a fog. Neon actually produces an orange red glow in the tube and on mixing with the vapors of other gasses, glows or signs of different colors can be obtained.

(ii) It is used in filling sodium vapor lamps.

(iii) It is used in safe devices for protecting certain electrical instruments (voltmeters, relays, rectifiers.)

Uses of Argon : (i) It is used in metal filament electric lamps since it increases the life of the tungsten filament by retarding its vapourisation.

(ii) A mixture of argon and mercury vapor is used in fluorescent tubes.

(iii) It is used to create an inert atmosphere for welding and for carrying certain chemical reactions.

Q. 6. Assign reason for each of the following :

(a) Noble gasses are mostly chemically inert.

(b) Bismuth is a strong oxidizing agent in a pentavalent state.

Ans⇒ (a) Noble gasses are mostly chemically inert due to the following reasons :

(i) Atoms of the noble gasses have stable closed shell electronic configuration.

(ii) Noble gasses have exceptionally high ionization energies.

(iii) Noble gasses have very low electron affinities.

(b) The stability of +5 oxidation state decreases down the group due two inert pair effects. So +5  oxidation state of Bi  is less stable than its +3  oxidation state.

Q. 7. NF3  does not have donor properties like ammonia. Explain.

Ans⇒ NF3 has a pyramidal shape with one lone-pair on N atom.

The lone-pair on N is in opposite direction to the N-F bond moments and therefore it has very low dipole moment Gout 0.234 D. Thus it does not show donor properties. But ammonia has a high dipole moment because the lone pair is in the same direction as the N-H . bond moments. Thus it has donor properties

Q. 8. Describe the chief uses of fluorine, chlorine and their compounds.

Ans⇒ Uses of fluorine :

(i) It is used in the preparation of fluoro carbon which are non-inflammable and chemically inert and are used as solvents, lubricants and insulators.

(ii) It is used as a refrigerant in many of the cooling processes.

(iii) It finds considerable use as DDFT, which similar to DDT, is extremely efficient as a fungicide and fumigant.

(iv) In nuclear physics and higher voltage electricity, fluorine as SF, finds great use e.g., in the separation of isotopes of uranium.

Uses of chlorine :

(i) It is used in sterilization of drinking water.

(ii) Large quantities of chlorine are used for bleaching paper pulp and textiles.

(iii) It is used in the manufacture of inorganic chemicals such HCl . Sodium hypochlorite (NaOCl), bleaching powder (CaOCl₂), phosphorus trichloride (PCl₃) , phosphorus pentachloride(PCl₅) etc.

(iv) It is used in the manufacture of vinyl chloride which is the starting material for the plastic polyvinyl chloride (PVC) .

(v) It is used in the manufacture of insecticides like DDT. germicides, dyes and drugs.

Q. 9. Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionization enthalpy and electronegativity.

Ans⇒ Electronic configuration :

All the elements of Group 15 have ns²np³ (=5 electrons) electronic configuration in their valence shells. The s-subshell is completely filled and p-subshell is

exactly half-filled (P,x P,y P, z). This imparts extra stability to their electronic configuration.

Nitrogen =N=7 = [He] 2s²p³

Phosphorus =P=15= [Ne] 3s²3p³

Arsenic =As=33=[Ar] 3d¹⁰ 4s² 4p³

Antimony = Sb =51=[Kr] 4d¹⁰ 5s² 5p³

Bismuth = Bi=83= [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p³

Oxidation States : The common oxidation states of these are -3,+3 and +5. The tendency to show -3 oxidation state decreases from top to bottom ase in size and metallic character. N does oxidation state because it has only 4 orbitals

(One s and three p) for bonding. Br hardly forms any compound in the -3 oxidation state. The stability of +5 oxidation state decreases down the group and that of +3 state increases (due to inert pair effect). N exhibits +1 , +2 and +4 oxidation states when it reacts with oxygen. P also shows +1 and +4 oxidation states in some oxoacids.

Atomic size : Atomic size increases from N to Bi due to the addition of one orbit each time. There is a considerable increase in covalent radius from N to P. However from As to Bi, only a small increase in covalent radius is observed due to the presence of completely filled d and / or f orbitals in heavier members.

Ionization Enthalpy : The ionization enthalpy of group 15 elements is much greater than that of Group 14 elements due to the extra-stability of half-filled p orbitals and smaller size. It decreases from top to bottom due to gradual increase in atomic size.

Electronegativity : Electronegativity decreases from top to bottom in the group. The difference of electronegativity is not much pronounced amongst the heavier elements.

Q. 10. Write three uses of Dinitrogen and Ammonia.

Ans⇒ Three uses of Dinitrogen :

(i) For providing an inert atmosphere during many industrial processes where presence of air or O₂ is to be avoided.

(ii) For manufacture of HNO₃.

(iii) For manufacture of nitronium.

Three uses of Ammonia :

(i) Used as a refrigeration fluid.

(ii) For remaining grease because NH₄OH dissolves grease.

(iii) As laboratory reagent.

Q. 11. White two uses of HNO₃ and P₄O₁₀

Ans⇒Two uses of HNO₃ :

(i) It is also used for the preparation of nitroglycerin, trinitrotoluene and other organic nitro compounds.

(ii) The magic use of nitric acid is in the manufacture of ammonium-nitrate for fertilizers.

Two uses of P₄O₁₀:

(i) For drying acidic gasses.

(ii) For the preparation of SO₃ and N₂O₅.

Q. 12. Write two uses of H₂O₂ and H₂S.

Ans⇒ Two uses of H₂O₂ :

(i) In bleaching of delicate materials such as sill wolf, cotton, ivory etc.

(ii) As oxidizing agent in rocket fuel.

Two uses of H₂S :

(i) As a laboratory reagent for the detection of basic radicals in qualitative analysis.

(ii) As a reducing agent.

Q. 13. Explain why NH₃  is basic while BiH₃  only feebly basic.

Ans⇒Nitrogen atoms have the smallest size and a very high electron density. Therefore, its electron releasing tendency or the basic strength is the maximum. Down the group, there is a gradual increase in atomic size and decrease in the electron density on the central atom. Consequently, the electron releasing tendency or basic strengths of the hydrides decrease in the order as given below.

                         NH₃>PH₃>AsH₃>SbH₃>BiH₃

Therefore, NH₃  is basic while BiH₃  is only feebly basic.

Q. 14. Nitrogen exists as a diatomic molecule but phosphorus as P₄. Why ?

Ans⇒ Due to large size, phosphorus has less tendency to form p-p bonds. So in order to gain stability, it polymerizes and exist as P4 solid while nitrogen completes its octet via  p-p bond formation and N₂ is a discrete molecule. In N₂, the intermolecular attraction are weak Van Der Waals forces and hence N₂ is a gas.

Q. 15. Bleaching of flowers by chlorine is permanent but that by sulfur dioxide is temporary.

Ans⇒ In presence of mixture, chlorine acts as an oxidizing agent and a bleaching agent.

Cl₂+H₂OHClHOCl (unstable)

HOClHCl+O (Nascent oxygen)

Coloured matter to Colorless matter

                      The bleaching action is permanent due to nascent oxygen. Bleaching action of SO₂ the presence of moisture is due to nascent hydrogen.

2SO₂+2H₂OH₂SO₄+2H

Coloured matter+H  Color matter So, bleaching action of SO₂ is temporary.

Q. 16. Why does the reactivity of nitrogen differ from phosphorus ?

Ans⇒ The reactivity of Nitrogen differs from Phosphorus due to its

(i) small size

(ii) high ionization enthalpy and electronegativity

(iii) non-availability of d-orbitals with N

(iv) Nitrogen has the unique ability to form p-p multiple bonds with itself and with other elements having small size and high electronegativity.

Q. 17. Discuss the trends in chemical reactivity of group 15 elements.

Ans⇒ N differs from other members of the group in its chemical reactivity. Some important trends in the chemical reactivity of group 15 elements are as follows.

Hydrides: All the elements of Group 15 form volatile hydrides of the formula EH3where E=N , P , As , Sb ,Bi. The lighter elements are also from hydrides of the formula E₂H . N forms HN3.

Oxides : N has a strong tendency to form p-p multiple bonding with N and Oatoms, whereas other elements of this group do not. Actually N form five oxides with oxidation states ranging from +1 to +5 (N₂O , NO , N₂O₃, NO₂/N₂O₄,N₂O₅ respectively) while other elements form oxides only in +3 and +5 oxidation states only slike P₄O₆ and P₄O₁₀]

Halides : The elements of Group 15 form halide to two types Trihalides and Pentahalides : EX₃ where E=N . P,  As , Sb ,Bi and X=F , Cl , Br and I.EX5 where E=P,  As , Sb ,Bi

Q.18. How is SO₂ an air pollutant?

Ans⇒ The environment is harmed by sulfur dioxide in many ways:

Sulphuric acid is formed when it is combined with water vapor present in the atmosphere. This causes acid that damages plants, soil, buildings (those made of marble are more prone), etc.

SO₂  causes irritation in the respiratory tract, throat, and eyes and can also affect the larynx to cause breathlessness.

The color of the leaves of the plant gets faded when it is exposed to sulfur dioxide for a long time. This defect is known as chlorosis. The formation of chlorophyll is affected by the presence of sulfur dioxide.

Q.19. What are the oxidation states of phosphorus in the following:

( a ) H₃PO₃

( b ) PCl₃

( c ) Ca₃P₂ 

( d ) Na₃PO₄

( e ) POF₃ 

Ans⇒ Let the oxidation state of phosphorous be x

(a) H₃PO₃ 

3 + x+3 (-2)=0

x-3=0

x=3

(b) PCl₃ 

x + 3(-1)=0

x=3  

(c) Ca3P2 

3 (2)+2(x)=0

2x=-6 

x = -3  

(d) Na₃PO₄ 

3 (1)+x+4(-2)=0

x-5=0

x=5

(e) POF₃ 

x + (-2)+3 (-1)=0

x - 5 = 0

x = 5  

Q.20. Arrange the following in the order of property indicated for each set:

(i) F ₂ , Cl₂ , Br₂ , I₂ - increasing bond dissociation enthalpy.

(ii) HF , HCl , HBr , HI-increasing acid strength.

(iii) NH₃ , PH₃ , AsH₃ , SbH₃ ,BiH₃- increasing base strength.

Ans⇒

(1) Bond dissociation energy normally lowers on moving down a group because of increase in the atomic size. However, F ₂ has a lower bond dissociation energy than Cl₂ and Br₂. This is because the atomic size of fluorine is very small.
Therefore, the increasing order for bond dissociation enthalpy is:

I₂<F ₂<Br₂ < Cl₂

(2)  Bond dissociation energy of  a H-X molecule ( where X=F , Cl , Br , I ) lowers with an increase in the size of an atom. As H-I bond is the weakest, it will be the strongest acid.
Therefore, the increasing order acidic strength is :

HF < HCl < HBr < HI

(3) BiH₃ < SnH₃ < AsH₃ < PH₃ < NH₃

On moving from nitrogen to bismuth, the atomic size increases but the electron density of the atom decreases. Hence, the basic strength lowers.

Q.21. Write main differences between the properties of white phosphorus and red phosphorus. 

Ans⇒

Q.22. How is the presence of SO₂ detected? 

Ans⇒Sulphur dioxide (SO2) is a colorless and pungent smelling gas. 

It can be detected with the help of potassium permanganate solution. When  SO2 is passed through an acidified potassium permanganate solution, it decolorizes the solution by reducing MnO4- ions to Mn2+ ions. The reaction taking place is as follows:

5 SO₂+2 MnO₄^-  + 2 H₂O₅SO₄₂^-+4H⁺+2Mn₂⁺

Q.23. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidizing power of F₂ and Cl₂ .

Ans⇒ Fluorine is a much stronger oxidizing agent than chlorine. The oxidizing power depends on three factors:

Bond dissociation energy 

Electron gain enthalpy 

Hydration enthalpy 

The electron gain enthalpy of chlorine is more negative than that of fluorine. However, the bond dissociation energy of fluorine is much lesser than that of chlorine Also, because of its small size, the hydration energy of fluorine is much higher than that of chlorine Therefore, the latter two factors more than compensate for the less negative electron gain enthalpy of fluorine. Thus, fluorine is a much stronger oxidizing agent than chlorine. 

Q.24. Why does O₃ act as a powerful oxidizing agent? 

Ans⇒ Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free radical, is very reactive.

O₃                 O₂⁺                [O]

Ozone               oxygen           nascent oxygen

Therefore, ozone acts as a powerful oxidizing agent 

Q.25. Describe the manufacture of H₂SO₄ by contact process? 

Ans⇒ Sulphuric acid is manufactured by the contact process. It involves the following steps: 

Step (i): Sulphur or sulfide ores are burnt in air to form SO₂ .

Step (ii): By a reaction with oxygen,SO₂ is converted into SO₃ in the presence of V₂O₅ as a catalyst. 

2SO₂ (g)+ O₂ (g)  2SO₃ (g)

Step (iii): SO₃ produced is absorbed on H₂SO₄ to give H₂S₂O₇ (oleum) 

SO₃ + H₂SO₄ H₂S₂O₇

This oleum is then diluted to obtain H₂SO₄ of the desired concentration. 

In practice, the plant is operated at 2 bar (pressure) and 720 K(temperature). The Sulphuric acid thus obtained is 96-98 %  pure.