Complete reduction of benzene-diazonium chloride with Zn/HCl gives:

We begin with benzene-diazonium chloride, whose formula is written as $$C_6H_5N_2Cl$$. In the reaction mixture we add metallic zinc in the presence of dilute hydrochloric acid, symbolised as $$Zn/HCl$$. Zinc is easily oxidised from $$Zn$$ to $$Zn^{2+}$$, and during this oxidation it releases electrons. Simultaneously, the acid furnishes protons $$H^+$$. The combination of electrons from zinc and protons from the acid produces what chemists call “nascent hydrogen”, often represented simply as $$[H]$$. Thus, overall, the pair $$Zn/HCl$$ behaves as a reducing agent, supplying hydrogen atoms to whatever group is capable of being reduced.

Now, a diazonium group $$-N_2^+$$ attached to an aromatic ring is a very good leaving group because molecular nitrogen $$N_2$$ is extremely stable. When nascent hydrogen attacks the diazonium cation, two elementary steps occur in sequence:

First, one hydrogen atom neutralises the positive charge on nitrogen, and molecular nitrogen departs:

$$C_6H_5N_2^+Cl^- + [H] \;\rightarrow\; C_6H_5\!-\!N(H)\!-\!N \;+\; Cl^-$$

Second, a second hydrogen atom cleaves the $$N\!-\!N$$ bond and attaches to the nitrogen still bonded to the ring, while the other nitrogen leaves as $$N_2$$ gas:

$$C_6H_5\!-\!N(H)\!-\!N + [H] \;\rightarrow\; C_6H_5NH_2 + N_2 \uparrow$$

Combining both elementary steps in one overall balanced equation we obtain

$$C_6H_5N_2Cl + 2[H] \;\rightarrow\; C_6H_5NH_2 + N_2 \uparrow + HCl$$

We see that the product attached to the benzene ring is $$C_6H_5NH_2$$, which is called aniline. No $$N\!-\!N$$ bond remains in the final product; instead we have converted the diazonium group completely into a primary amine group $$-NH_2$$. This fits the phrase “complete reduction of benzene-diazonium chloride”.

A brief check of the other choices confirms that phenylhydrazine $$C_6H_5NHNH_2$$, azobenzene $$C_6H_5N{=}NC_6H_5$$, and hydrazobenzene $$C_6H_5NHNH C_6H_5$$ all still contain an $$N\!-\!N$$ linkage, so they cannot be the fully reduced product under these conditions. Only aniline lacks that linkage and matches the reaction we have just derived.

Hence, the correct answer is Option A.