JEE Main · 2004 · 24 Apr 2004 · Offline · d and f Block Elements

Spin-only moments follow $\mu =\sqrt{n(n+2)}$, so count the unpaired electrons. In $[\mathrm{MnCl}{}_4]{}^{2-}$ manganese is $d^5$ with weak-field chloride, five unpaired, giving $\mu \approx 5.92\text{BM}$. In $[\mathrm{CoCl}{}_4]{}^{2-}$ cobalt is $d^7$ with chloride, three unpaired, $\mu \approx 3.87\text{BM}$. In $[\mathrm{Fe}(\mathrm{CN}){}_6]{}^{4-}$ iron is $d^6$ with strong-field cyanide, fully paired, $\mu =0$. The order is therefore $[\mathrm{MnCl}{}_4]{}^{2-}>[\mathrm{CoCl}{}_4]{}^{2-}>[\mathrm{Fe}(\mathrm{CN}){}_6]{}^{4-}$.