Question

What is value of spin only magnetic moment of $\text{Ni}\ (Z = 28)$ in +2 oxidation state?

• A. 3.1 BM
• B. 0.0 BM
• C. 2.8 BM
• D. 1.7 BM

Hint:

Here is an incredible time-saving shortcut for magnetic moments: the digit before the decimal point in the final answer is always exactly equal to the number of unpaired electrons ($n$)! Since $n = 2$, the answer must start with 2 (i.e., 2.8 BM).

Answer 1

The Correct Option is C

Step 1: Recall the spin-only formula.
The spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. So our whole task reduces to finding $n$.
Step 2: Write the configuration of neutral Ni.
Nickel has $Z = 28$, giving $[\text{Ar}]\,3d^8\,4s^2$.
Step 3: Form the Ni$^{2+}$ ion.
A transition metal loses its 4s electrons first, so Ni$^{2+}$ is $[\text{Ar}]\,3d^8$.
Step 4: Distribute the 3d$^8$ electrons.
Filling five d-orbitals by Hund's rule: the first five electrons go in singly, then three more pair up. This leaves two orbitals still singly occupied. So $n = 2$.
Step 5: Substitute into the formula.
$\mu = \sqrt{2(2+2)} = \sqrt{2 \times 4} = \sqrt{8}$ BM.
Step 6: Evaluate the square root.
Since $\sqrt{9} = 3$, $\sqrt{8}$ is just under 3, namely about $2.83$ BM, rounding to $2.8$ BM. This is option (3).
\[ \boxed{\mu \approx 2.8\ \text{BM}} \]