Question

1) Give the electronic configuration of the following complex on the basis ofCrystal Field Splitting theory.$[Co{F}_6{]}^{3-}$

2) Give the electronic configuration of the following complex on the basis of Crystal Field Splitting theory.$\left[Fe\right(CN{)}_6{]}^{4-}$

3) Give the electronic configuration of the following complex on the basis ofCrystal Field Splitting theory.
$\left[Cu\right(N{H}_3{)}_6{]}^{2+}$

Answer 1

(i) $[Co{F}_6{]}^{3-}$

Electronic configuration of 𝐶𝑜 $\left(27\right)is\left[Ar\right]3d^{7}4s^2$ but in $[Co{F}_6{]}^{3-},Coisin+3$ oxidation state, i.e., it exists as $C{o}^{3+}$.

So, electronic configuration of $C{o}^{3+}$ will be $3d^{6}4s^0$ and on the basis of crystal field splitting theory it will be $t_{2g}^4{e}_g^2$.

Fluoride is a weak field ligand so it will not cause the pairing of electrons.

Crystal field splitting diagram of $C{o}^{3+}$ ion in presence of weak field ligand



(ii)

In $\left[Fe\right(CN{)}_6{]}^{4-}$, Fe will be in +2 oxidation state i.e., as $F{e}^{2+}$.

Electronic configuration of $Fe\left(26\right)is\left[Ar\right]3d^{6}4s^2$, so electronic configuration of $F{e}^{2+}willbe\left[Ar\right]3d^{6}4s^2$

As ${CN}^{-}$ is a strong field, it will cause pairing of electrons:

Crystal field splitting diagram of $F{e}^{2+}$ in presence of strong field ligand



So, the electronic configuration of $\left[Fe\right(CN{)}_6{]}^{4-}is:t_{2g}^6{e}_g^0$.

(iii)

In $\left[Cu\right(N{H}_3{)}_6{]}^{2+}$, 𝐶𝑢 will be in $+2$ oxidation state i.e., as $C{u}^{2+}$.

Electronic configuration of $Cu\left(29\right)is\left[Ar\right]3d^{10}4s^1$, so electronic configuration of $C{u}^{2+}$ will be $\left[Ar\right]3d^{9}4s^0$.

Here field strength of the ligand does not matter as only one configuration is possible.

Crystal field splitting diagram of $C{u}^{2+}$



So, the electronic configuration of $\left[Cu\right(N{H}_3{)}_6{]}^{2+}is{t}_{2g}^6{e}_g^3$.