The total number of reagents from those given below, that can convert nitrobenzene into aniline is _________. (Integer answer)
I. Sn - HCl
II. Sn - NH$$_4$$OH
III. Fe - HCl
IV. Zn - HCl
V. H$$_2$$ - Pd
VI. H$$_2$$ - Raney Nickel

We begin with the basic fact that reduction of the nitro group $$\left( -\,\mathrm{NO_2}\right)$$ to the amino group $$\left( -\,\mathrm{NH_2}\right)$$ requires a source of nascent hydrogen or a catalytic hydrogenation system. A reagent or a reagent‐combination capable of supplying such hydrogen will convert nitrobenzene $$\left(\mathrm{C_6H_5NO_2}\right)$$ into aniline $$\left(\mathrm{C_6H_5NH_2}\right).$$

Now we examine each option one by one.

I. $$\mathrm{Sn + HCl}$$    In the acidic medium, metallic tin reacts to give $$\mathrm{Sn^{2+}}$$ ions and nascent hydrogen: $$\mathrm{Sn + 2\,HCl \;\rightarrow\; SnCl_2 + 2\,[H]}.$$ The nascent hydrogen reduces the nitro group stepwise $$\bigl(-\mathrm{NO_2}\;\rightarrow\;-\mathrm{NHOH}\;\rightarrow\;-\mathrm{NH_2}\bigr)$$ so that $$\mathrm{C_6H_5NO_2 \xrightarrow[HCl]{Sn} C_6H_5NH_2}.$$ Hence this reagent works.

II. $$\mathrm{Sn + NH_4OH}$$    Tin needs an acidic medium to generate nascent hydrogen. In the alkaline medium furnished by $$\mathrm{NH_4OH},$$ tin does not produce the necessary hydrogen; therefore the nitro group is not reduced all the way to aniline. So this combination fails for the required conversion.

III. $$\mathrm{Fe + HCl}$$    Exactly analogous to Sn/HCl, iron in acid gives nascent hydrogen: $$\mathrm{Fe + 2\,HCl \;\rightarrow\; FeCl_2 + 2\,[H]}.$$ So we get $$\mathrm{C_6H_5NO_2 \xrightarrow[HCl]{Fe} C_6H_5NH_2}.$$ Thus this reagent is effective.

IV. $$\mathrm{Zn + HCl}$$    Zinc in hydrochloric acid behaves similarly: $$\mathrm{Zn + 2\,HCl \;\rightarrow\; ZnCl_2 + 2\,[H]},$$ followed by $$\mathrm{C_6H_5NO_2 \xrightarrow[HCl]{Zn} C_6H_5NH_2}.$$ Hence this reagent also works.

V. $$\mathrm{H_2}\;-\;\mathrm{Pd}$$    This is catalytic hydrogenation. The general statement is: “Hydrogen gas in the presence of a transition-metal catalyst (Pd, Pt, Ni, Raney Ni) reduces $$\mathrm{-NO_2}$$ to $$\mathrm{-NH_2}$$.” Therefore $$\mathrm{C_6H_5NO_2 + 3\,H_2 \xrightarrow{Pd} C_6H_5NH_2 + 2\,H_2O}.$$ So it is a valid reagent.

VI. $$\mathrm{H_2}\;-\;\text{Raney Ni}$$    Raney nickel is another well-known hydrogenation catalyst, hence $$\mathrm{C_6H_5NO_2 + 3\,H_2 \xrightarrow{Raney\;Ni} C_6H_5NH_2 + 2\,H_2O}.$$ Thus this reagent also accomplishes the conversion.

Counting all the successful cases we have:

Sn/HCl},\;$$\text{Fe/HCl}$$,\;$$\text{Zn/HCl}$$,\;$$\text{H}_2\text{/Pd}$$,\;$$\text{H}_2\text{/Raney$$ Ni

Total $$= 5.$$

Hence, the correct answer is Option 5.