Q.Which of the following reagents can be used to convert nitrobenzene to aniline?

Complete reduction of nitrobenzene to aniline needs a strong acidic reducing system such as $Sn/HCl$ (or equivalently $Fe/HCl$ or catalytic $H_2/Ni$); Zn/NH$_4$Cl only partially reduces the nitro group.

Nitrobenzene reduction proceeds through several stages: $C_6H_5NO_2 \rightarrow C_6H_5NO \rightarrow C_6H_5NHOH \rightarrow C_6H_5NH_2$. Tin with concentrated hydrochloric acid ($Sn/HCl$) supplies enough reducing power to carry the reduction all the way to aniline ($C_6H_5NH_2$) — the standard laboratory method for this conversion. Zn dust with neutral $NH_4Cl$ solution (option b) is a much milder reducing system that arrests the sequence at the N-phenylhydroxylamine stage rather than going to completion. Zn(Hg)/NaOH (option a, Clemmensen-type conditions) is used to reduce a ketone/aldehyde carbonyl to $-CH_2-$, not to reduce a nitro group. Since (a) and (b) do not achieve full reduction to aniline, (d) 'all of these' is also incorrect.

Zn/NaOH is a strongly reducing alkaline system that reduces nitrobenzene completely through nitroso- and azoxy-/azo- intermediates to hydrazobenzene, $C_6H_5-NH-NH-C_6H_5$.

The reduction of nitrobenzene proceeds through a well-defined sequence of intermediates, $C_6H_5NO_2 \rightarrow C_6H_5NO \rightarrow C_6H_5NHOH \rightarrow$ (coupling) $\rightarrow C_6H_5-N=N(\to O)-C_6H_5$ (azoxybenzene) $\rightarrow C_6H_5-N=N-C_6H_5$ (azobenzene) $\rightarrow C_6H_5-NH-NH-C_6H_5$ (hydrazobenzene), and which product is isolated depends entirely on the strength and pH of the reducing system used: Zn/$NH_4Cl$ (neutral, mild) stops at phenylhydroxylamine; Sn/HCl or Fe/HCl (acidic, strong) goes all the way to aniline; Zn dust in strongly alkaline NaOH solution is the classical reagent that carries the reduction through to the fully reduced coupled product, hydrazobenzene. So among the options, azoxybenzene (option d) and phenylhydroxylamine-type intermediates are only partial-reduction stages under milder/different conditions, not the product of the strongly alkaline Zn/NaOH reduction asked about here.

The reducing agent used on nitrobenzene changes the final product: acidic reduction (Sn/HCl or Fe/HCl) gives aniline directly; alkaline reduction (Zn dust + $NaOH$) proceeds through a coupling pathway to give hydrazobenzene, via successive intermediates.

Under alkaline conditions, nitrobenzene is first reduced to nitrosobenzene and phenylhydroxylamine, which then condense with each other to give azoxybenzene ($C_6H_5-N(O)=N-C_6H_5$); further reduction (with excess Zn/NaOH) reduces this to azobenzene ($C_6H_5-N=N-C_6H_5$), and yet further reduction under the same alkaline conditions reduces the $N=N$ double bond to give hydrazobenzene ($C_6H_5-NH-NH-C_6H_5$). Azobenzene (a) and azoxybenzene (d) are intermediate stages, not the final product with excess Zn/NaOH. Aniline (b) is the product of acidic (not alkaline) reduction.