Question

Identify the reaction which does not liberate hydrogen?
A. Reaction of lithium hydride with ${B_2}{H_6}$
B. Electrolysis of acidified water using Pt electrodes
C. Reaction of zinc with aqueous alkali
D. Allowing a solution of sodium in liquid ammonia to stand

Answer 1

In the options given above one of them does not liberate hydrogen and to find this reaction we need to know every reaction product. We can also try some assumptions for example in the process of electrolysis with various compounds hydrogen gas is released in many reactions.

Complete step by step answer:
To answer this question we need to look into each reaction carefully.
A. Reaction of lithium hydride with ${B_2}{H_6}$
Lithium Hydride is an inorganic compound with formula $LiH$ . This alkali metal hydride is a colourless solid but the commercial samples are grey. It has characteristics of salt-like hydride and has a high melting point. It is not soluble in water but is reactive with all organic and protic solvents. Diborane is a chemical compound consisting of boron and hydrogen. It is colourless, pyrophoric gas with a repulsively sweet odour.
Lithium Hydride or $LiH$ reacts with diborane or ${B_2}{H_6}$ in diethyl ether forming lithium borohydride.
$2LiH + {B_2}{H_6} \to 2Li[B{H_4}]$

B. Electrolysis of acidified water using Pt electrodes
Electrolysis is a way of splitting up the compound using electrical energy. In the process of electrolysis the electrolyte used is dilute sulphuric acid which is also referred to as acidified water which during hydrolysis split into hydrogen and oxygen gases. The reaction is as given below:
$2{H_2}O \to {O_2} + 4H + 4{e^ - }$
$4{H_2}O + 4{e^ - } \to 2{H_2} + 4O{H^ - }$

C. Reaction of zinc with aqueous alkali
Zinc reacts with aqueous alkalis like sodium or potassium hydroxide to form zincates. Given below is the reaction of zinc with Sodium hydroxide. In this reaction sodium will replace hydrogen and hydrogen gas would be evolved.
$2NaOH + Zn \to N{a_2}Zn{O_2} + {H_2}$

D. Allowing a solution of sodium in liquid ammonia to stand
When a solution of sodium reacts with liquid ammonia and dissolves in it to form a blue solution due to the presence of solvated electrons in liquid ammonia. Solvated electron is a free electron in a solution and is the smallest possible anion. Also, hydrogen gas is liberated in this reaction. The reaction is as given below.
$2N{H_3} + 2Na \to 2NaN{H_2} + {H_2}$
From the above reactions we can see that hydrogen is released in all the reactions except in the first reaction of lithium hydride with diborane.

**Therefore the correct answer is option A. Reaction of lithium hydride with ${B_2}{H_6}$

Note: **
Hydrogen gas is released in many reactions like electrolysis or reaction of zinc with aqueous alkali. But hydrogen gas is not released in reactions like the reaction of lithium hydride with diborane.