Chapter 9 Hydrogen Solutions
Question - 11 : - What do you understand by the term ‘non-stoichiometric hydrides’? Do you expect this type of hydrides to be formed by alkali metals? Justify your answer.
Answer - 11 : - Those hydrides which do not have fix composition are called non-stoichiometric hydrides, and the composition varies with temperature and pressure. This type of hydrides are formed by d and /block elements. They cannot be formed by alkali metals because alkali metal hydrides form ionic hydrides.
Question - 12 : - How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.
Answer - 12 : - In metallic hydrides, hydrogen is adsorbed as H-atoms. Due to the adsorption of H atoms the metal Lattice expands and become unstable. Thus, when metallic hydride is heated, it decomposes to form hydrogen and finely divided metal. The hydrogen evolved can be used as fuel.
Question - 13 : - How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes ? Explain.
Answer - 13 : - When hydrogen is burnt in oxygen the reaction is highly exothermic, it produces very high temperature nearly 4000°C which is used for cutting and welding purposes.
Question - 14 : - Among NH3 H2O and HE, which would you expect to have highest magnitude of hydrogen bonding and why?
Answer - 14 : - HF is expected to have highest magnitude of hydrogen bonding since, ‘F’ is most electronegative. Therefore, HF is the most polar.
Question - 15 : - Saline hydrides are known to react with water violently producing fire. Can C02, a well known fire extinguisher, be used in this case? Explain.
Answer - 15 : - No. Because if saline hydrides react with water the reaction will be highly exothermic thus the hydrogen evolved in this case can catch fire. C02 cannot be used as fire extinguisher because C02 will get absorbed in alkali metal hydroxides.
Question - 16 : - Arrange the following:
(i) CaH2, BeH2 and TiH2 in order ofincreasing electrical conductance.
(ii) LiH, NaH and CsH in order of increasing ionic character.
(iii) H-H, D—D and F—F in order of increasing bond dissociation enthalpy.
(iv) NaH, MgH2 andH2O in order ofincreasing reducing property.
Answer - 16 : -
(i) BeH2< TiH2 < CaH2
(ii) LiH < NaH < CsH
(iii) F—F < H—H < D—D
(iv) H2O < MgH2 < NaH
Question - 17 : - Compare the structures of H2O and H2O2
Answer - 17 : -
In water, O is sp3 hybridized. Due to stronger lone pair-lonepair repulsions than bond pair-bond pair repulsions, the HOH bond angledecreases from 109.5° to 104.5°. Thus water molecule has a bent structure.
H2O2 has a non-planar structure. The O—H bondsare in different planes. Thus, the structure of H2O2 is like an open book.
Question - 18 : - What do you understand by the term ‘auto-protolysis’ of water? what is its significance?
Answer - 18 : - Auto-protolysis means self-ionisation of water. It may be represented as
Due to auto-protalysis water is amphoteric in nature, i.e., it can act as an acid as well as base.
Answer - 19 : -
2F2(ag) + 2H20(l)—————> O2(g) + 4H+(aq) + 4F(aq)
In this reaction water acts as a reducing agentand itself gets oxidised to O2 while F2 acts as an oxidising agent and hence itselfreduced to F– ions.
Question - 20 : - Complete the following chemical reactions.
(i) PbS(s) + H2O2 (aq) ————->
(ii) MnO4– (aq) + H2O2 (aq)————->
(iii) CaO(s) + H2O(g)————->
(iv) AlCl3(g) + H2O(l)————->
(v) Ca3N2(S) + H2O(l) ————->
Classify the above into (a) hydrolysis, (b) redox and (c) hydrationreactions.
Answer - 20 : -
(i) PbS(s) +4H2O2(aq) ————-> PbSO4(s) + 4H2O(l)
(ii) 2MnO4– (aq) +H2O2(aq) + 6H+(aq) ————-> 2Mn (aq) + 8H2O(l) + 5O2(g)
(iii) CaO(s) + H2O(g) ————->Ca(OH)2(aq)
(iv) AlCl3(aq) + 3H2O(l) ————-> Al(OH)3(S) + 3HCl (aq)
(v) Ca3N2(s) + H2O(l) ————->3Ca(OH)2(aq) + 2NH3(aq)
(a) Hydrolysis reactions, (iii) (iv) and (v)
(b) Redox reactions (i) and (ii)