Given below are two statements :

Statement (I) : A solution of $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ is green in colour.

Statement (II) : A solution of $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is colourless.

In the light of the above statements, choose the most appropriate answer from the options given below :

<p>In order to determine the correctness of the given statements, we need to analyze the nature of the mentioned complexes.</p> <p><b>Statement (I)</b> asserts that a solution of $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ is green in color. This complex involves a nickel(II) cation coordinated with six water molecules. Nickel(II) has the electronic configuration $[Ar]3d^8$. In an octahedral field created by water ligands, the d-orbitals split into two sets, $t_{2g}$ and $e_g$, due to crystal field splitting. The presence of unpaired electrons allows for d-d transitions when light is absorbed, which is responsible for the characteristic color of the complex. Since typical $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ complexes are indeed green in color due to such transitions, Statement (I) is correct.</p> <p><b>Statement (II)</b> states that a solution of $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is colorless. In this complex, nickel(II) is surrounded by four cyanide ligands. Cyanide is a strong field ligand, which leads to a large crystal field splitting that exceeds the pairing energy of the electrons. Hence, all the electrons in the nickel(II) ion are paired, and there are no unpaired electrons available for d-d transitions. As there are no d-d transitions to absorb light in the visible region, the complex appears colorless. Therefore, Statement (II) is also correct.</p> <p>Given that both statements are correct, the most appropriate answer would be :</p> <p>Option B : Both Statement I and Statement II are correct.</p>