Question

Some species are given: \[ \mathrm{Ni^{2+},\ Fe^{2+},\ Co^{2+},\ V^{3+}\ \text{and}\ Ti^{2+}} \] How many species have magnetic moment (spin-only) less than \(3\) BM?

Hint:

For quick checks:

\(n = 1,2 \Rightarrow \mu<3\) BM
\(n \ge 3 \Rightarrow \mu>3\) BM
First find unpaired electrons, then apply the formula

Answer 1

Correct Answer: 3

Concept: The spin-only magnetic moment is given by: \[ \mu = \sqrt{n(n+2)} \ \text{BM} \] where \(n\) is the number of unpaired electrons. A species will have magnetic moment less than 3 BM if: \[ \sqrt{n(n+2)}<3 \] Testing values: \[ n = 1 \Rightarrow \mu = \sqrt{3} \approx 1.73 \] \[ n = 2 \Rightarrow \mu = \sqrt{8} \approx 2.83 \] \[ n = 3 \Rightarrow \mu = \sqrt{15} \approx 3.87 \ (>3) \] Hence, species with \(n \le 2\) satisfy the condition.
Step 1: Determine electronic configuration and unpaired electrons

\(\mathrm{Ni^{2+}}\): Ni = [Ar] \(3d^{8}4s^{2}\) \( \Rightarrow \mathrm{Ni^{2+}} = 3d^{8} \Rightarrow n = 2 \)
\(\mathrm{Fe^{2+}}\): Fe = [Ar] \(3d^{6}4s^{2}\) \( \Rightarrow \mathrm{Fe^{2+}} = 3d^{6} \Rightarrow n = 4 \)
\(\mathrm{Co^{2+}}\): Co = [Ar] \(3d^{7}4s^{2}\) \( \Rightarrow \mathrm{Co^{2+}} = 3d^{7} \Rightarrow n = 3 \)
\(\mathrm{V^{3+}}\): V = [Ar] \(3d^{3}4s^{2}\) \( \Rightarrow \mathrm{V^{3+}} = 3d^{2} \Rightarrow n = 2 \)
\(\mathrm{Ti^{2+}}\): Ti = [Ar] \(3d^{2}4s^{2}\) \( \Rightarrow \mathrm{Ti^{2+}} = 3d^{2} \Rightarrow n = 2 \)
 
Step 2: Apply the condition Species with \(n \le 2\): \[ \mathrm{Ni^{2+},\ V^{3+},\ Ti^{2+}} \] Total number of such species: \[ \boxed{3} \] Final Answer: \[ \boxed{3} \]