JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Coordination Compounds

The spin-only moment $\mu =\sqrt{n(n+2)}$ B.M. increases with the number of unpaired electrons $n$, so the order of moments is the order of $n$. In $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}$, $\mathrm{F}{\mathrm{e}}^{3+}$ is $d^5$ and the strong field of $\mathrm{C}{\mathrm{N}}^{-}$ gives $t_{2g}^5$, so $n=1$. In $[\mathrm{Mn}(\mathrm{CN}{)}_6{]}^{3-}$, $\mathrm{M}{\mathrm{n}}^{3+}$ is $d^4$ low-spin, $t_{2g}^4$, so $n=2$. In $[\mathrm{Co}{\mathrm{F}}_6{]}^{3-}$, $\mathrm{C}{\mathrm{o}}^{3+}$ is $d^6$ and the weak field of ${\mathrm{F}}^{-}$ gives high-spin $t_{2g}^4{e}_g^2$, so $n=4$. In $[\mathrm{MnB}{\mathrm{r}}_4{]}^{2-}$, $\mathrm{M}{\mathrm{n}}^{2+}$ is $d^5$, tetrahedral with weak-field $\mathrm{B}{\mathrm{r}}^{-}$, so $n=5$. The increasing order of magnetic moment is $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}<[\mathrm{Mn}(\mathrm{CN}{)}_6{]}^{3-}<[\mathrm{Co}{\mathrm{F}}_6{]}^{3-}<[\mathrm{MnB}{\mathrm{r}}_4{]}^{2-}$.