CBSE Class 12 Chemistry Chapter 7 Important Questions – Free PDF Download
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Important Questions Class 12 Chemistry Chapter 7 – The p-Block Elements
CBSE Class 12 Chemistry Important Questions Chapter 7 – The p-Block Elements
3 Marks Questions
1. Write three differences between white and red phosphorous.
Ans.
| White phosphorous | Red phosphorous |
| 1. It is translucent waxy solid | 1. It has iron grey Lustre. |
| 2. It is poisonous and glows in dark | 2. It is non poisonous and does not glow in dark. |
| 3. It is less stable and more reactive. | 3. It is more stable and less reactive. |
2. Why does 
dimerise?
Ans. contains odd number of valence electrons. It behaves as a typical odd molecule. On demerisation, it is converted to stable 
molecule with even number of electrons.
3. Mention the conditions required to maximise the yield of ammonia.
Ans. Ammonia is prepared using the Haber’s process. The yield of ammonia can be maximized under the following conditions:
(i) High pressure 
(ii) A temperature of 
(iii) Use of a catalyst such as iron oxide mixed with small amounts of 
and 
4. What happens when H3PO3is heated?
Ans. 
,on heating, undergoes disproportionation reaction to form 
and. The oxidation numbers of P in ,, and are +3, – 3, and +5 respectively. As the oxidation number of the same element is decreasing and increasing during a particular reaction, the reaction is a disproportionation reaction.
5. How is 
estimated quantitatively?
Ans. Quantitatively, ozone can be estimated with the help of potassium iodide. When ozone is made to react with potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated. This liberated iodine can be titrated against a standard solution of sodium thiosulphate using starch as an indicator. The reactions involved in the process are given below.
6. How is the presence of 
detected?
Ans. 
is a colourless and pungent smelling gas.
It can be detected with the help of potassium permanganate solution. When is passed through an acidified potassium permanganate solution, it decolonizes the solution as it reduces 
ions to 
ions.
7. Mention three areas in which 
plays an important role.
Ans. Sulphuric acid is an important industrial chemicaland is used for a lot of purposes. Some important uses of sulphuric acid are given below.
(i) It is used in fertilizer industry. It is used to make various fertilizers such as ammonium sulphate and calcium super phosphate.
(ii) It is used in the manufacture of pigments, paints, and detergents.
(iii) It is used in the manufacture of storage batteries.
8. Write the conditions to maximize the yield of by Contact process.
Ans. Manufacture of sulphuric acid by Contact process involves three steps.
1. Burning of ores to form
2. Conversion of to 
by the reaction of the former with 
(
is used in this process as a catalyst.)
3. Absorption of in 
to give oleum 
The key step in this process is the second step. In this step, two moles of gaseous reactants combine to give one mole of gaseous product. Also, this reaction is exothermic. Thus, in accordance with Le Chatelier’s principle, to obtain the maximum amount of gas, temperature should be low and pressure should be high.
9. Why is 
for in water?
Ans. 
It can be noticed that 
This is because a neutral has a much higher tendency to lose a proton than the negatively charged 
. Thus, the former is a much stronger acid than the latter.
10. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of 
and 
.
Ans. Fluorine is a much stronger oxidizing agent than chlorine. The oxidizing power depends on three factors.
1. Bond dissociation energy
2. Electron gain enthalpy
3. 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.
11. Illustrate how copper metal can give different products on reaction with 
.
Ans. Concentrated nitric acid is a strong oxidizing agent. It is used for oxidizing most metals. The products of oxidation depend on the concentration of the acid, temperature, and also on the material undergoing oxidation.
12. Give the resonating structures of 
and
.
Ans.
(1)
(2)
13. The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of 
hybridisation in 
and only s-p bonding between hydrogen and other elements of the group].
Ans. Hydride 
H-M-H angle
The above trend in the H-M-H bond angle can be explained on the basis of the electronegativity of the central atom. Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes greater repulsion between the electron pairs around nitrogen, resulting in maximum bond angle. We know that electronegativity decreases on moving down a group. Consequently, the repulsive interactions between the electron pairs decrease, thereby decreasing the H-M-H bond angle.
14. Explain why is basic while 
is only feebly basic.
Ans. is distinctly basic while is feebly basic.
Nitrogen has a small size due to which the lone pair of electrons is concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density. Hence, the electron donating capacity of group 15 element hydrides decreases on moving down the group.
15. Nitrogen exists as diatomic molecule and phosphorus as 
. Why?
Ans. Nitrogen owing to its small size has a tendency to form 
multiple bonds with itself. Nitrogen thus forms a very stable diatomic molecule, 
. On moving down a group, the tendency to form bonds decreases (because of the large size of heavier elements). Therefore, phosphorus (like other heavier metals) exists in the state.
16. Why is dioxygen a gas but sulphur a solid?
Ans. Oxygen is smaller in size as compared to sulphur. Due to its smaller size, it can effectively form 
bonds and form 
(O==O) molecule. Also, the intermolecular forces in oxygen are weak van der Wall’s, which cause it to exist as gas. On the other hand, sulphur does not form 
molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid.
17. Knowing the electron gain enthalpy values for 
and 
as -141 and 
respectively, how can you account for the formation of a large number of oxides having 
species and not 
?
(Hint: Consider lattice energy factor in the formation of compounds).
Ans. Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be.
Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving 
ion is much more than the oxide involving O-ion. Hence, the oxide having ions are more stable than oxides having . Hence, we can say that formation of is energetically more favourable than formation of .
18. Why are halogens strong oxidising agents?
Ans. The general electronic configuration of halogens is 
, where n = 2-6. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron. Hence, they act as strong oxidizing agents.
19. How are xenon fluorides 
and 
obtained?
Ans. and are obtained by a direct reaction between Xe and 
. The condition under which the reaction is carried out determines the product.
20. Arrange the following in the order of property indicated for each set:
(i) 
increasing bond dissociation enthalpy.
(ii) HF, HCl, HBr, HI – increasing acid strength.
(iii) 
– increasing base strength.
Ans. (i) Bond dissociation energy usually decreases on moving down a group as the atomic size increases. However, the bond dissociation energy of is lower than that of 
and 
. This is due to the small atomic size of fluorine. Thus, the increasing order for bond dissociation energy among halogens is as follows:
(ii) HF < HCl < HBr < HI
The bond dissociation energy of H-X molecules where X = F, Cl, Br, I, decreases with an increase in the atomic size. Since H-I bond is the weakest, HI is the strongest acid.
(iii) 
On moving from nitrogen to bismuth, the size of the atom increases while the electron density on the atom decreases. Thus, the basic strength decreases.
21. List the uses of Neon and argon gases.
Ans. Uses of neon gas:
(i) It is mixed with helium to protect electrical equipments from high voltage.
(ii) It is filled in discharge tubes with characteristic colours.
(iii) It is used in beacon lights.
Uses of Argon gas:
(i) Argon along with nitrogen is used in gas-filled electric lamps. This is because Ar is more inert than N.
(ii) It is usually used to provide an inert temperature in a high metallurgical process.
(iii) It is also used in laboratories to handle air-sensitive substances.
