Question

How many unpaired electrons are there in ${\text{N}}{{\text{i}}^{{\text{2 + }}}}$ $\left( {{\text{Z = 28}}} \right)$ ?
A 0
B 8
C 2
D 4

Answer 1

From the atomic number of nickel, write its electronic configuration. Also write the electronic configuration of ${\text{N}}{{\text{i}}^{{\text{2 + }}}}$ cation. From the electronic configuration, determine the number of unpaired electrons.

Complete Step by step answer: The atomic number of nickel is 28. Its electronic configuration is $\left[ {{\text{Ar}}} \right]3{d^8}4{s^2}$
Nickel atoms lose two electrons to form ${\text{N}}{{\text{i}}^{{\text{2 + }}}}$cation. The electronic configuration of ${\text{N}}{{\text{i}}^{{\text{2 + }}}}$ cation is $\left[ {{\text{Ar}}} \right]3{d^8}$ .
8 electrons are present in 3d subshell.

Six electrons are present in lower ${t_{2g}}$ level and two electrons are present in upper ${e_g}$ level.
Thus, the number of unpaired electrons in ${\text{N}}{{\text{i}}^{{\text{2 + }}}}$ ion is 2.

Hence, the correct option is the option (C).

Additional information: When 8 electrons are present in the d orbitals, 6 electrons will pair and two remain unpaired. This is irrespective of if strong field ligand is present or weak field ligand is present. Due to presence of unpaired electrons, ${\text{N}}{{\text{i}}^{{\text{2 + }}}}$ ion shows paramagnetic behaviour. If all the electrons were paired, then the ion would have been diamagnetic.

Note: The five d orbitals of a metal are degenerate. They have the same energy level. This is true in absence of ligands. In the presence of an octahedral field of ligands, the five degenerate d orbitals of metal split into two energy levels. The lower energy level contains three d orbitals and is called ${t_{2g}}$ level. The upper energy level contains two d orbitals and is called ${e_g}$ level.