Identify the reaction which does not liberate hydrogen

We first recall that molecular hydrogen, $$H_2,$$ is produced in a reaction only when two hydrogen atoms come together and are released as a gas. We shall test every option one by one, write the balanced equation, and look carefully whether $$H_2$$ appears on the product side.

Option A  -  Reaction of lithium hydride with diborane

Diborane, $$B_2H_6,$$ is an electron-deficient Lewis acid. Lithium hydride, $$LiH,$$ can supply a hydride ion $$H^-.$$ The accepted laboratory equation is

$$B_2H_6 + 2\,LiH \rightarrow 2\,LiBH_4$$

On the right-hand side we obtain lithium borohydride, $$LiBH_4,$$ and we notice that no extra $$H_2$$ is produced. All hydrogens remain bonded inside the borohydride ion $$BH_4^-.$$ Therefore, this reaction does not liberate hydrogen gas.

Option B  -  Electrolysis of acidified water using Pt electrodes

During electrolysis water splits. The cathode reaction is

$$2\,H_2O + 2\,e^- \rightarrow H_2 + 2\,OH^-$$

We clearly see $$H_2$$ appears; thus hydrogen is liberated here.

Option C  -  Reaction of zinc with aqueous alkali

In alkaline medium zinc forms sodium zincate (if the alkali is sodium hydroxide) and releases hydrogen:

$$Zn + 2\,NaOH \rightarrow Na_2ZnO_2 + H_2$$

Again $$H_2$$ is on the product side, so hydrogen is liberated.

Option D  -  Allowing a solution of sodium in liquid ammonia to stand

Freshly prepared blue solutions contain solvated electrons. On standing they disproportionate and yield sodium amide and hydrogen:

$$2\,Na + 2\,NH_3 \rightarrow 2\,NaNH_2 + H_2$$

Here too, $$H_2$$ gas is set free.

Comparing all four possibilities, only the first reaction produces no hydrogen gas. The others all have $$H_2$$ among their products.

Hence, the correct answer is Option A.