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

The electron configurations for several metal ions in their +2 oxidation state are provided: \( 22\text{Ti}^{2+} \rightarrow [\text{Ar}]3d^2 \), \( 23\text{V}^{2+} \rightarrow [\text{Ar}]3d^3 \), \( 25\text{Mn}^{2+} \rightarrow [\text{Ar}]3d^5 \), and \( 26\text{Fe}^{2+} \rightarrow [\text{Ar}]3d^6 \). The spin-only magnetic moment (\(\mu_s\)) is calculated using the formula \( \mu_s = \sqrt{n(n+2)} \, \text{BM} \), where \(n\) represents the number of unpaired electrons. Given a spin-only magnetic moment of \( \mu_s = 3.86 \, \text{BM} \), we can determine the number of unpaired electrons: \( 3.86 = \sqrt{n(n+2)} \). Squaring both sides yields \( 3.86^2 = n(n+2) \), which simplifies to \( 14.9 \approx n(n+2) \). Solving this equation for \(n\) results in \( n = 3 \). To identify the element with \(n = 3\) unpaired electrons in its +2 oxidation state, we examine the configurations again: \( 22\text{Ti}^{2+} \rightarrow [\text{Ar}]3d^2 \, (n=2) \), \( 23\text{V}^{2+} \rightarrow [\text{Ar}]3d^3 \, (n=3) \), \( 25\text{Mn}^{2+} \rightarrow [\text{Ar}]3d^5 \, (n=5) \), and \( 26\text{Fe}^{2+} \rightarrow [\text{Ar}]3d^6 \, (n=4) \). Therefore, the element corresponding to \(n=3\) is \(V\) (Vanadium), which has an atomic number of 23.