Question

Generally, transition elements and their salts are colored due to the presence of unpaired electrons in metal ions. Which of the following compounds are colored?

• A. $Ce(S{O}_4{)}_2$
• B. $TiC{l}_4$
• C. $C{u}_{2}C{l}_2$
• D. $KMn{O}_4$

Answer 1

Answer: (A), (D)

$KMn{O}_4$

Analyzing the given options :

Option $\left(A\right)$

The oxidation state of $Mn$ in $KMn{O}_4$ is $+7.$

Electronic configuration of $Mn$ in its $+7$ state is:

$M{n}^{7+}=\left[Ar\right]3d^{0}4s^0.$

Since, no unpaired electron is present, it will not show $d-d$ transition. But $KMn{O}_4$ is a colored $\left(\text{purple}\right)$ compound because it shows momentary charge transfer from ligand $O$ to $Mn.$

$KMn{O}_4$ shows colour due to ligand to metal charge transfer. The charge transfers from $2p$ of oxygen $\left(\text{Ligand}\right)$ to $3d\left(\text{Metal}\right)$ of $Mn.$ Therefore, we can conclude that due to $L\to M$ charge transfer, it is coloured.

Option$\left(B\right)$

The oxidation state of $Ce$ in $Ce(S{O}_4{)}_2$ is $+4.$

Electronic configuration of $Ce$ in its $+4$ state is:

$C{e}^{4+}=\left[Xe\right]4f^{0}5d^{0}6s^0$

It has no unpaired electron yet $Ce(S{O}_4{)}_2$ is a colored $\left(\text{yellow}\right)$ compound because it shows charge transfer and not $f-f$ transition.

Option$\left(C\right)$

In $TiC{l}_4,$ Ti is in $+4$ oxidation state.

Electronic configuration of $Ti=\left[Ar\right]3d^{2}4s^2$

Electronic configuration of $T{i}^{4+}=\left[Ar\right]$

All its orbitals are fully-filled. So, it does not contain any unpaired electron. It does not show charge transfer spectra either.

Hence, $TiC{l}_4$ is colorless.

Option$\left(D\right)$

$C{u}_{2}C{l}_2$ has $Cu$ in $+1$ oxidation state.

Electronic configuration of $C{u}^{+}=\left[Ar\right]3d^{10}$

It does not have any unpaired electron, also does not show any charge transfer.

Thus, it is also colorless.

So correct answers are options $\left(A\right)$ and $\left(B\right).$