To understand why the correct answer is option 'B', let's examine the reduction reactions of nitrobenzene using different reagents:

(I) H2/Ni:
The reaction between nitrobenzene and H2/Ni is a catalytic hydrogenation reaction. In this reaction, the nitro group (-NO2) is reduced to an amino group (-NH2) by the addition of hydrogen gas (H2). However, this reaction does not stop at the formation of aniline (C6H5NH2). Further reduction can occur, leading to the formation of cyclohexylamine.

(II) Sn/HCl:
The reaction between nitrobenzene and Sn/HCl is a reduction using tin and hydrochloric acid. In this reaction, the nitro group is reduced to an amino group by the tin (Sn) metal. However, this reaction also does not stop at the formation of aniline. It can further reduce the aromatic ring, leading to the formation of cyclohexylamine.

(III) Zn/NaOH:
The reaction between nitrobenzene and Zn/NaOH is a reduction using zinc and sodium hydroxide. In this reaction, the nitro group is reduced to an amino group by the zinc (Zn) metal. However, this reaction also does not stop at the formation of aniline. It can further reduce the aromatic ring, leading to the formation of cyclohexylamine.

(IV) LiAlH4:
The reaction between nitrobenzene and LiAlH4 is a powerful reduction using lithium aluminum hydride. In this reaction, the nitro group is reduced to an amino group by the hydride ion (H-) provided by LiAlH4. This reaction selectively stops at the formation of aniline and does not further reduce the aromatic ring.

Therefore, the correct answer is option 'B' (I, II, and IV) because only the combination of H2/Ni (I), Sn/HCl (II), and LiAlH4 (IV) can be used to reduce nitrobenzene to aniline without further reducing the aromatic ring. Option 'B' provides a combination of reagents that selectively convert nitrobenzene to aniline while minimizing the formation of unwanted byproducts.