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

The Correct Option is D

\[ 22\text{Ti}^{2+} \rightarrow [\text{Ar}]3d^2 \quad 23\text{V}^{2+} \rightarrow [\text{Ar}]3d^3 \quad 25\text{Mn}^{2+} \rightarrow [\text{Ar}]3d^5 \quad 26\text{Fe}^{2+} \rightarrow [\text{Ar}]3d^6 \]
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.