Question

A first row transition metal in its +2 oxidation state has a spin-only magnetic moment value of 3.86 BM. The atomic number of the metal is

• A. 25
• B. 26
• C. 22
• D. 23

Answer 1

Answer: (D)

23

Answer 2

$22\text{Ti}^{2+} \rightarrow [\text{Ar}]3d^2 \quad 23\text{V}^{2+} \rightarrow [\text{Ar}]3d^3 \quad 25\text{Mn}^{2+} \rightarrow [\text{Ar}]3d^5 \quad 26\text{Fe}^{2+} \rightarrow [\text{Ar}]3d^6$
The spin-only magnetic moment is given by:
$\mu_s = \sqrt{n(n+2)} \, \text{BM},$
where $n$ is the number of unpaired electrons.
For $\mu_s = 3.86 \, \text{BM}$:
$3.86 = \sqrt{n(n+2)}.$
Squaring both sides:
$3.86^2 = n(n+2) \implies 14.9 \approx n(n+2).$
Solving for $n$,
we find: $n = 3.$
The element with $n = 3$ unpaired electrons in its $+2$ oxidation state can be identified as follows: Configuration in $+2$ state:
$22\text{Ti}^{2+} \rightarrow [\text{Ar}]3d^2 \, (n=2), \quad 23\text{V}^{2+} \rightarrow [\text{Ar}]3d^3 \, (n=3), \quad 25\text{Mn}^{2+} \rightarrow [\text{Ar}]3d^5 \, (n=5), \quad 26\text{Fe}^{2+} \rightarrow [\text{Ar}]3d^6 \, (n=4).$
Thus, the element is $V$ (Vanadium) with atomic number 23.