Question

Which of the following transition metal ions is colourless?
A) ${{\text{V}}^{2 + }}$
B) ${\text{C}}{{\text{r}}^{3 + }}$
C) ${\text{Z}}{{\text{n}}^{2 + }}$
D) ${\text{T}}{{\text{i}}^{3 + }}$

Answer 1

The elements or ions exhibit colours due to the presence of unpaired electrons in the d-orbital. To solve this we must know whether the given metal ions have unpaired electrons in their d-orbital. Determine the unpaired electrons in their d-orbital from the electronic configuration.

Complete solution:
We are given four transition metal ions ${{\text{V}}^{2 + }}$, ${\text{C}}{{\text{r}}^{3 + }}$, ${\text{Z}}{{\text{n}}^{2 + }}$ and ${\text{T}}{{\text{i}}^{3 + }}$. The elements or ions exhibit colours due to the presence of unpaired electrons in the d-orbital.
We know that the atomic number of vanadium $\left( {\text{V}} \right)$ is 23. Thus, the electronic configuration of vanadium is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^3}$
When we say ${{\text{V}}^{2 + }}$ or vanadium ion, two electrons are removed from the valence 4s orbital of vanadium. Thus, the electronic configuration of ${{\text{V}}^{2 + }}$ is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^3}$
From the electronic configuration, we can see that the electrons in the valence d-orbital of ${{\text{V}}^{2 + }}$ are unpaired. Thus, ${{\text{V}}^{2 + }}$ ion is not colourless.
We know that the atomic number of chromium $\left( {{\text{Cr}}} \right)$ is 24. Thus, the electronic configuration of chromium is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}3{d^5}$
When we say ${\text{C}}{{\text{r}}^{3 + }}$ or chromium ion, one electron is removed from the valence 4s orbital and one electron is removed from the valence 3d orbital of chromium. Thus, the electronic configuration of ${\text{C}}{{\text{r}}^{3 + }}$ is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^3}$
From the electronic configuration, we can see that the electrons in the valence d-orbital of ${\text{C}}{{\text{r}}^{3 + }}$ are unpaired. Thus, ${\text{C}}{{\text{r}}^{3 + }}$ ion is not colourless.
We know that the atomic number of zinc $\left( {{\text{Zn}}} \right)$ is 30. Thus, the electronic configuration of zinc is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}$
When we say ${\text{Z}}{{\text{n}}^{2 + }}$ or zinc ion, two electrons are removed from the valence 4s orbital of zinc. Thus, the electronic configuration of ${\text{Z}}{{\text{n}}^{2 + }}$ is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^{10}}$
From the electronic configuration, we can see that the electrons in the valence d-orbital of ${\text{Z}}{{\text{n}}^{2 + }}$ are paired. Thus, ${\text{Z}}{{\text{n}}^{2 + }}$ ion is colourless.
We know that the atomic number of titanium $\left( {{\text{Ti}}} \right)$ is 22. Thus, the electronic configuration of titanium is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^2}$
When we say ${\text{T}}{{\text{i}}^{3 + }}$ or titanium ion, three electrons are removed from the valence 4s orbital and one electron is removed from the valence 3d orbital of titanium. Thus, the electronic configuration of ${\text{Z}}{{\text{n}}^{2 + }}$ is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^1}$
From the electronic configuration, we can see that the electrons in the valence d-orbital of ${\text{T}}{{\text{i}}^{3 + }}$ are unpaired. Thus, ${\text{T}}{{\text{i}}^{3 + }}$ ion is not colourless.
Thus, the transition metal ions that are colourless are ${\text{Z}}{{\text{n}}^{2 + }}$.

Thus, the correct option is (C) ${\text{Z}}{{\text{n}}^{2 + }}$.

Note: The d-orbital has five sub-orbitals each occupying two electrons. Thus, the maximum capacity of d-orbital is 10 electrons. If the d-orbital contains 10 electrons then it is completely filled and has all paired electrons. If the number of electrons is less than 10 then we can say that the d-orbital is incomplete.