Given below are two statements: Statement (I): A solution of [Ni(H2O)6]2+ is green in colour. Statement (II): A…

🧠 Two Independent Claims About Ni(II) Colour

Statement I: $[\mathrm{Ni(H_2O)_6}]^{2+}$ is green.

Ni²⁺ d⁸ in octahedral $\mathrm{H_2O}$ field is HS ($t_{2g}^6 e_g^2$). The d–d transition $t_{2g}\to e_g$ absorbs in the red (~700 nm) → transmits green. The aqueous solution is famously emerald green.

✓ Statement I is correct.

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

Ni²⁺ in CN⁻ is square planar, d⁸ low-spin (all 8 electrons paired in $d_{xy}, d_{xz}, d_{yz}, d_{z^2}$ with $d_{x^2-y^2}$ empty). The lone d–d transition is to high-energy $d_{x^2-y^2}$ — absorbed in the UV/near-UV, leaving no significant absorption in the visible.

The complex is described as colourless (or very faintly yellow) in standard JEE/NCERT treatment.

✓ Statement II is correct.

🗺️ Why Geometry Changes Everything for Ni(II)

Same metal, same d-count — but two completely different colour outcomes:

| Complex | Geometry | Spin | Visible absorption | |---|---|---|---| | $[\mathrm{Ni(H_2O)_6}]^{2+}$ | Oct | HS d⁸ | red → green colour | | $[\mathrm{Ni(CN)_4}]^{2-}$ | Sq. planar | LS d⁸ (diamag) | UV → colourless |

⚡ Field Strength + Geometry Together

CN⁻ is a strong-field ligand → forces Ni²⁺ into square planar (large Δ between $d_{x^2-y^2}$ and the rest). Weak-field $\mathrm{H_2O}$ leaves Ni²⁺ octahedral with smaller Δ → visible-region absorption.

⚠️ Don't Generalise the Colourless Claim

Many d⁸ square planar complexes (e.g. $[\mathrm{Ni(dmg)_2}]$ with dmg = dimethylglyoxime) are intensely coloured — that colour is from charge-transfer, not d–d. The "colourless" claim for $[\mathrm{Ni(CN)_4}]^{2-}$ is a special case driven by its electronic structure and lack of CT bands.

$\boxed{\text{Answer: (2) Both Statement I and Statement II are correct}}$