Question

Which of the following has the maximum number of unpaired d-electrons?
$\mathrm{A}) \mathrm{Fe}^{2+}$
$\mathrm{B}) \mathrm{Cu}^{+}$
$\mathrm{C}) \mathrm{Zn}^{2+}$
$\mathrm{D}) \mathrm{Ni}^{3+}$

Answer 1

Hint: Let’s look into the atomic configuration of the molecules. Then we will be able to answer this question.

Complete step by step answer:
The outer shell electronic configuration of $\mathrm{Fe}$ is $3 \mathrm{d}^{6} 4 \mathrm{s}^{2}$.Now we know as $4 \mathrm{s}$ is less in energy compared to $3 \mathrm{d}$, the electrons will be knocked out first from $4 \mathrm{s}$ orbital. $\mathrm{Fe}^{+2}$ will be now $3 \mathrm{d}^{6}$. Now each d orbitals can contain 10 electrons in 5 subshells maintaining the Hund's Rule.

Hence, there are 4 unpaired d-electrons.
$\mathrm{Cu}^{+}$ will be now $3 \mathrm{d}^{10} .$ Now each $\mathrm{d}$ orbitals can contain 10 electrons in 5 subshells maintaining the Hund's Rule. Thus, there are no unpaired electrons in d- orbital.

$\mathrm{Zn}^{2+}$ will be $\mathrm{d}^{10} .$ Now each d orbitals can contain 10 electrons in 5 subshells maintaining the
Hund's Rule. Thus, there are no unpaired electrons in d- orbital.
$\mathrm{Zn}^{2+} \uparrow \downarrow \uparrow \downarrow \uparrow \downarrow \uparrow \downarrow \uparrow \downarrow$

In $\mathrm{Ni}^{3+},$ the configuration gives us $3 \mathrm{d}^{7} .$ Now we know as $4 \mathrm{s}$ is less in energy compared to $3 \mathrm{d},$ the electrons will be knocked out first from 4 s orbital. $\mathrm{Ni}^{3+}$ will be now $3 \mathrm{d}^{7}$. Now each $\mathrm{d}$ orbitals can
contain 10 electrons in 5 subshells maintaining the Hund's Rule.
Thus, the maximum number of d-electrons unpaired is found in $$
\mathrm{Fe}^{+2}
$$
 i.e. 4
So, the correct answer is “Option A”.

Note: Always first find out the electronic configuration of the neutral molecule. Then find out the electronic configuration of the ions of the molecules to find out the unpaired electrons.