Chemistry - Coordination Chemistry Question with Solution | TestHub

To determine the order of spin-only magnetic moments, we need to find the number of unpaired electrons ($n$) for each complex. The spin-only magnetic moment is given by $\mu =\sqrt{n(n+2)}$ B.M. Higher $n$ means higher $\mu$.

1. For $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}$: Fe is in +3 oxidation state ($d^5$). CN⁻ is a strong field ligand, so it's low spin. Electron configuration: $t_{2g}^5{e}_g^0$. Number of unpaired electrons $n=1$.

2. For $[\mathrm{Mn}(\mathrm{CN}{)}_6{]}^{3-}$: Mn is in +3 oxidation state ($d^4$). CN⁻ is a strong field ligand, so it's low spin. Electron configuration: $t_{2g}^4{e}_g^0$. Number of unpaired electrons $n=2$.

3. For $[{\mathrm{CoF}}_6{]}^{3-}$: Co is in +3 oxidation state ($d^6$). F⁻ is a weak field ligand, so it's high spin. Electron configuration: $t_{2g}^4{e}_g^2$. Number of unpaired electrons $n=4$.

4. For $[{\mathrm{MnBr}}_4{]}^{2-}$: Mn is in +2 oxidation state ($d^5$). Br⁻ is a weak field ligand (tetrahedral complexes are usually high spin). Electron configuration: $e^2{t}_2^3$. Number of unpaired electrons $n=5$.

Ordering by increasing number of unpaired electrons (and thus increasing magnetic moment):

$n=1$ for $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}$

$n=2$ for $[\mathrm{Mn}(\mathrm{CN}{)}_6{]}^{3-}$

$n=4$ for $[{\mathrm{CoF}}_6{]}^{3-}$

$n=5$ for $[{\mathrm{MnBr}}_4{]}^{2-}$

Therefore, the correct order is:

$[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}<[\mathrm{Mn}(\mathrm{CN}{)}_6{]}^{3-}<[{\mathrm{CoF}}_6{]}^{3-}<[{\mathrm{MnBr}}_4{]}^{2-}$