Statement (I): A solution of is green in colour.

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Solution

Q.
Given below are two statements: [2024]

Statement (I): A solution of ${[Ni {(H_{2} O)}_{6}]}^{2 +}$ is green in colour.
Statement (II): A solution of ${[Ni {(CN)}_{4}]}^{2 -}$ is colourless.

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

1 Both Statement I and Statement II are incorrect

2 Both Statement I and Statement II are correct

3 Statement I is incorrect but Statement II is correct

4 Statement I is correct but Statement II is incorrect

Ans.
(2)

${[Ni {(H_{2} O)}_{6}]}^{2 +}$

Oxidation number of Ni = +2

${Ni}^{2 +} = {}_{18}{[Ar] 3 d^{8}}$

Since coordination number of the complex is 6, it is an octahedral complex. Octahedral splitting of d orbitals is shown below:

Irrespective of the ligand, $d^{8}$ complexes always split into $t_{2 g}^{6}, e_{g}^{2}$ . From electronic configuration, it is clear that d-d transition can take place as $e_{g}$ orbitals are vacant. Because of this, ${[Ni {(H_{2} O)}_{6}]}^{2 +}$ is coloured. Magnitude of $Δ_{o}$ is such that red colour is absorbed during d-d transition of electron from $t_{2 g}$ to $e_{g}$ . Thus, complementary colour green is observed.

${[Ni {(CN)}_{4}]}^{2 -}$

Oxidation number of Ni = +2

${Ni}^{2 +} : {}_{18}{[Ar] 3 d^{8}}$

$C N^{-}$ is a strong field ligand and hence causes pairing of $3 d^{8}$ electrons.
As there is no vacant orbital, d-d transition cannot occur. Hence, this is a colourless complex.

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