Chemistry Class 12 NCERT Solutions: Chapter 7 The p Block Elements Part 6

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Q: 39. Why do noble gases have comparatively large atomic sizes?

Answer:

Noble gases do not form molecules. In case of noble gases, the atomic radii corresponds to van der Waal’s radii. On the other hand, the atomic radii of other elements correspond to their covalent radii. By definition, van der Waal’s radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period.

Q: 40. List the uses of Neon and argon gases.

Answer:

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.

Text Solution:

Q: 1. Why are pentahalides more covalent than trihalides?

Answer

In pentahalides, the oxidation state is and in trihalides, the oxidation state is . Since the metal ion with a high charge has more polarizing power, pentahalides are more covalent than trihalides.

Q: 2. Why is the strongest reducing agent amongst all the hydrides of Group elements?

Answer:

As we move down a group, the atomic size increases and the stability of the hydrides of group elements decreases. Since the stability of hydrides decreases on moving from to , the reducing character of the hydrides increases on moving from to .

Q: 3. Why is less reactive at room temperature?

Answer

The two N atoms in are bonded to each other by very strong triple covalent bonds. The bond dissociation energy of this bond is very high. As a result, is less reactive at room temperature.

Q: 4. Mention the conditions required to maximise the yield of ammonia

Answer

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

Q: 5. How does ammonia react with a solution of ?

Answer:

acts as a Lewis base. It donates its electron pair and forms a linkage with metal ion.

Q: 6. What is the covalence of nitrogen in ?

Answer:

Q 6 structure of Dinitrogen pentoxide

Q 6 Structure of Dinitrogen Pentoxide

Q 6 structure of Dinitrogen pentoxide

From the structure of , it is evident that the covalence of nitrogen is .

Q: 7. Bond angle in is higher than that in . Why?

Answer:

In , P is hybridized. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the tetrahedral shape associated with bonding is changed to pyramidal. combines with a proton to form in which the lone pair is absent. Due to the absence of lone pair in , there is no lone pair-bond pair repulsion. Hence, the bond angle in is higher than the bond angle in .

Q 7 1 Three Orbitals are Involved in Bonding with Three Hydrogen

Q 7 1 Three Orbitals Are Involved in Bonding with Three Hydrogen

Q 7 1 Three Orbitals are Involved in Bonding with Three Hydrogen

Q: 8. What happens when white phosphorus is heated with concentrated solution in an inert atmosphere of ?

Answer:

White phosphorous dissolves in boiling solution (in a atmosphere) to give phosphine, .