Chemistry Class 11 NCERT Solutions: Chapter 9 Hydrogen Part 3

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Q: 10. Do you expect the carbon hydrides of the type to act as ‘Lewis’ acid or base? Justify your answer:


For carbon hydrides of type , the following hydrides are possible for

For a hydride to act as a Lewis acid i.e., electron accepting, it should be electron- deficient. Also, for it to act as a Lewis base i.e., electron donating, it should be electron-rich.

Taking as an example, the total number of electrons are 14 and the total covalent bonds are seven. Hence, the bonds are regular centered bonds.

Hence, hydride has sufficient electrons to be represented by a conventional Lewis structure. Therefore, it is an electron-precise hydride, having all atoms with complete octets. Thus, it can neither donate nor accept electrons to act as a Lewis acid or Lewis base.

Q: 11. What do you understand by the term “non-stoichiometric hydrides”? Do you expect this type of the hydrides to be formed by alkali metals? Justify your answer.


Non-Stoichiometric hydrides are hydrogen-deficient compounds formed by the reaction of dihydrogen with d-block and f-block elements. These hydrides do not follow the law of constant composition. For example: etc.

Alkali metals form stoichiometric hydrides. These hydrides are ionic in nature. Hydride ions have comparable sizes with alkali metal ions. Hence, strong binding forces exist between the constituting metal and hydride ion. As a result, stoichiometric hydrides are formed. Alkali metals will not form non-stoichiometric hydrides.

Q: 12. How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.


Metallic hydrides are hydrogen deficient, i.e., they do not hold the law of constant composition. It has been established that in the hydrides of Ni, Pd, Ce, and Ac, hydrogen occupies the interstitial position in lattices allowing further absorption of hydrogen on these metals. Metals like Pd, Pt, etc. have the capacity to accommodate a large volume of hydrogen. Therefore, they are used for the storage of hydrogen and serve as a source of energy.

Q: 13. How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes? Explain.


Atomic hydrogen atoms are produced by the dissociation of dihydrogen with the help of an electric arc. This releases a huge amount of energy This energy can be used to generate a temperature of 4000 K, which is ideal for welding and cutting metals. Hence, atomic hydrogen or oxy-hydrogen torches are used for these purposes. For this reason, atomic hydrogen is allowed to recombine on the surface to be welded to generate the desired temperature.

Q: 14. Among and HF, which would you expect to have highest magnitude of hydrogen bonding and why?


The extent of hydrogen bonding depends upon electronegativity and the number of hydrogen atoms available for bonding. Among nitrogen, fluorine, and oxygen, the increasing order of their electronegativities are . Hence, the expected order of the extent of hydrogen bonding is . But, the actual order is

Although fluorine is more electronegative than oxygen, the extent of hydrogen bonding is higher in water. There is a shortage of hydrogens in HF, whereas there are exactly the right numbers of hydrogens in water. As a result, only straight chain bonding takes place. 0n the other hand, oxygen forms a huge ring-like structure through its high ability of hydrogen bonding.

In case of ammonia, the extent of hydrogen bonding is limited because nitrogen has only one lone pair. Therefore, it cannot satisfy all hydrogens.

Q: 15. Saline hydrides are known to react with water violently producing fire. Can , a well-known fire extinguisher, be used in this case? Explain.


Saline hydrides (i.e., , etc.) react with water to form a base and hydrogen gas. The chemical equation used to represent the reaction can be written as:

The reaction is violent and produces fire.

is heavier than dioxygen. It is used as a fire extinguisher because it covers the fire as a blanket and inhibits the supply of dioxygen, thereby dousing the fire. can be used in the present case as well. It is heavier than dihydrogen and will be effective in isolating the burning surface from dihydrogen and dioxygen.