Question

The number of unpaired electrons calculated in $\left[Co\right(N{H}_3{)}_6{]}^{3+}$ and $[Co{F}_6{]}^{3-}$ are:

• A. 4 and 4
• B. 0 and 2
• C. 2 and 4
• D. 0 and 4

Answer 1

The correct option is D 0 and 4

$N{H}_3$ is a strong field ligand and $F^{-}$ is a weak field ligand.
In $\left[Co\right(N{H}_3{)}_6{]}^{3+}$, $Co$ is in +3 oxidation state.
So the elctronic configuration will be $\left[Ar\right]3d^6$
So accroding to crystal field theory the configuration is $t_{2g}^{4}e{g}^2$
But since $N{H}_3$ is a strong field ligand, thus the splitting of $t_{2g}$ and $e_g$ will be more and pairing enrgy will be less than that. Thus configuration would be $t_{2g}^6{e}_g^0$.
Thus number of unpaired electrons in $\left[Co\right(N{H}_3{)}_6{]}^{3+}$ is 0.

Again, in $[Co{F}_6{]}^{3-}$, $Co$ is in +3 oxidation state. but here since $F^{-}$ is a weak field ligand., thus the pairing will not take place and hence number of unpaired electron will be 4 ($t_{2g}^4{e}_g^2$)