Analyze Statement (I):

The complex is $[\mathrm{Ni}(\mathrm{H_2O})_6]^{2+}$

Here, nickel is in the oxidation state $+2$.

$\mathrm{Ni}^{2+}$ has electronic configuration: $[\mathrm{Ar}]\,3d^8$

Since $\mathrm{H_2O}$ is a weak field ligand, the complex is octahedral and shows $d$-$d$ transitions. Such transitions absorb visible light, so the complex is coloured. The hexaaquanickel(II) ion is known to be green in aqueous solution.

Hence, Statement (I) is correct.

Analyze Statement (II):

The complex is $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$

Oxidation state of Ni: $x + 4(-1) = -2 \Rightarrow x=+2$

So again, nickel is $\mathrm{Ni}^{2+}$, i.e. $3d^8$.

Now $\mathrm{CN^-}$ is a strong field ligand. For a $d^8$ nickel(II) complex with coordination number 4, the complex becomes square planar due to pairing of electrons.

In the square planar arrangement, all the electrons are paired and the crystal field splitting is such that there is no suitable low-energy $d$-$d$ transition in the visible region. Therefore, the complex appears colourless.

Hence, Statement (II) is also correct.

Evaluate options:

• A: Statement I incorrect, Statement II correct $\rightarrow$ False
• B: Both Statement I and Statement II are correct $\rightarrow$ True
• C: Both incorrect $\rightarrow$ False
• D: Statement I correct, Statement II incorrect $\rightarrow$ False

Final Answer:

$\boxed{\text{B}}$

Comparison with stored answer:

Stored correct answer = B

My derived answer = B

Therefore, they agree.