Find the electronic configuration of $Ni$ and $Ni^{2+}$

Nickel has atomic number $28$.

$Ni: [Ar] \, 3d^8 4s^2$

For $Ni^{2+}$, two electrons are removed from the $4s$ orbital first:

$Ni^{2+}: [Ar] \, 3d^8$

Determine the number of unpaired electrons

In aqueous solution, $Ni^{2+}$ exists as a hydrated ion such as $[Ni(H_2O)_6]^{2+}$.

Since $H_2O$ is a weak field ligand, the configuration remains high spin for $d^8$ octahedral ion.

For octahedral $d^8$:

$t_{2g}^6 e_g^2$

The two $e_g$ electrons occupy separate orbitals first, so there are:

$n = 2 \text{ unpaired electrons}$

Use the spin-only magnetic moment formula

$\mu = \sqrt{n(n+2)} \; \mu_B$

Substituting $n=2$:

$\mu = \sqrt{2(2+2)} = \sqrt{8} = 2\sqrt{2} \approx 2.84 \; \mu_B$

Match with the options

$2.84 \; \mu_B$

So, the correct option is A.

Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

Hence, they agree.