Question

Which one is more stable in the following pairs of complexes? Give reasons.
i) $\left[Co\right(N{O}_2{)}_6{]}^{4-}$ and $\left[Co\right(N{O}_2{)}_6{]}^{3-}$
ii) $K_4\left[Fe\right(CN{)}_6]$ and $K_3\left[Fe\right(CN{)}_6]$.
(iii)$\left[CO\right(H_{2}O{)}_6{]}^{2+}$ and $\left[CO\right(N{H}_3{)}_6{]}^{2+}$

Answer 1

i) Valence shell electronic configuration of $Co$ is $4s^{2}3d^7$

Oxidation state of $Co$ in $\left[Co\right(N{O}_2{)}_6{]}^{4-}$ is

$x+6\times (-1)=-4$ and $x=+2$ i.e. $3d^7$

and in $\left[Co\right(N{O}_2{)}_6{]}^{3-}$ is:

$x+6\times (-1)=-3$ and $x=+3$ i.e. $3d^6$

Since $N{O}_2^{-}$ is a strong field ligand, therefore it will completely fill the $t_{2g}$ orbital with $6$ electrons and then fill $e_g$ orbital.

Therefore $\left[Co\right(N{O}_2{)}_6{]}^{3-}$ will be more stable than $\left[Co\right(N{O}_2{)}_6{]}^{4-}$ complex as all electrons are in low energy orbital.

ii) Valence shell electronic configuration of $Fe$ is $4s^{2}3d^6$

Oxidation state of $Fe$ in $K_4\left[Fe\right(CN{)}_6]$ is

$x+6\times (-1)+4=0$ and $x=+2$ i.e. $3d^6$

and in $K_3\left[Fe\right(CN{)}_6]$. is:

$x+6\times (-1)+3=0$ and $x=+3$ i.e. $3d^5$

Since $C{N}^{-}$ is a strong field ligand, therefore it will completely fill the $t_{2g}$ orbital with $6$ electrons and then fill $e_g$ orbital.

Therefore $K_4\left[Fe\right(CN{)}_6]$ will be more stable than $K_3\left[Fe\right(CN{)}_6]$ complex as it has $6$ paired electrons and each stabilizes the complex by $0.4{\Delta }_{oh}$ in low energy orbital.