JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Coordination Compounds

In $[\mathrm{Ni}(\mathrm{CO}{)}_4]$ the metal is zero-valent $3d^{10}$ and the strong-field carbonyls give a tetrahedral, diamagnetic complex of zero moment. In $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$ the metal is $\mathrm{N}{\mathrm{i}}^{2+}$ $3d^8$ and the strong-field cyanide pairs the electrons, giving a square planar, diamagnetic ion of zero moment. In $[\mathrm{NiC}{\mathrm{l}}_4{]}^{2-}$ the weak-field chloride leaves two unpaired electrons in a tetrahedral geometry, with moment $\sqrt{8}\approx 2.8\text{BM}$. In $[\mathrm{MnB}{\mathrm{r}}_4{]}^{2-}$ the metal is $\mathrm{M}{\mathrm{n}}^{2+}$ $3d^5$ with five unpaired electrons in a tetrahedral field, moment $\sqrt{35}\approx 5.9\text{BM}$. The correct pairing is tetrahedral and zero for the carbonyl, square planar and zero for the cyanide, tetrahedral and 2.8 BM for the chloride, and tetrahedral and 5.9 BM for the bromide.