CBSE 2025, Set 1

Question

Why is a solution of $[	ext{Ni(H}_2	ext{O)}_6]^{2+}$ green while a solution of $[	ext{Ni(CN)}_4]^{2-}$ is colourless? (At. No. of Ni = 28)

Edvaya · Accepted Answer

The correct answer is Subjective. Ni: [Ar]3d$^8$4s$^2$. Ni$^{2+}$: [Ar]3d$^8$. **$[	ext{Ni(H}_2	ext{O)}_6]^{2+}$ is green:** H$_2$O is a **weak field ligand** → small $\Delta_o$ → high-spin configuration → Ni$^{2+}$ retains **2 unpaired electrons**. These unpaired electrons can undergo **d–d electronic transitions** by absorbing specific wavelengths of visible light. The complementary colour (green) is transmitted/reflected, giving the solution a green colour. **$[	ext{Ni(CN)}_4]^{2-}$ is colourless:** CN$^-$ is a **strong field ligand** → large $\Delta_o$ → square planar geometry → **all 8 d-electrons are paired** (low-spin, $d^8$ square planar). With no unpaired electrons and no suitable vacant d-orbital for transitions, **d–d transitions are not possible**. No visible light is absorbed → the solution is **colourless**.

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Chemistry > Inorganic Chemistry > Coordination Compounds

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