Question

The hexaquo manganese$\left(II\right)$ ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.

Answer 1

$\left[Mn\right(H_{2}O{)}_6{]}^{2+}$

In this compound, the oxidation state of $\text{Mn}$ is $+2$ and its outer electronic configuration is $3d^5.$

As $H_{2}O$ is a weak field ligand and crystal field is octahedral so the arrangement of electrons will be $t_2^{3}ge{g}^2.$

Hence, $\left[Mn\right(H_{2}O{)}_6{]}^{2+}$ have $5$ unpaired electrons.

$\left[Mn\right(CN{)}_6{]}^{4-}$

In this compound, the oxidation state of $9\text{Mn}$ is also $+2$ and its outer electronic configuration is also $3d^5.$

As $C{N}^{-}$ is a strong field ligand and crystal field is octahedral so the arrangement of electrons will be $t_{2g}^{5}e{g}^0.$

Hence, $\left[Mn\right(CN{)}_6{]}^{4-}$ have one unpaired electron.