Question

Which of the following pairs of ions will have same spin only magnetic moment values within the pair?
• [A.] \( \text{Zn}^{2+}, \text{Ti}^{2+} \)
• [B.] \( \text{Cr}^{2+}, \text{Fe}^{2+} \)
• [C.] \( \text{Ti}^{3+}, \text{Cu}^{2+} \)
• [D.] \( \text{V}^{2+}, \text{Cu}^{+} \) Choose the correct answer from the options given below:

• A. C and D only
• B. A and D only
• C. A and B only
• D. B and C only

Hint:

To quickly find the number of unpaired electrons for a \( 3\text{d}^m \) configuration where \( m > 5 \), simply subtract the number of electrons from 10 (\( n = 10 - m \)). For example, for \( \text{Fe}^{2+} \) (\( 3\text{d}^6 \)), \( n = 10 - 6 = 4 \). This allows an instant matching shortcut with \( \text{Cr}^{2+} \) (\( 3\text{d}^4 \)) without drawing orbital diagrams!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The spin-only magnetic moment \(\mu\) is given by \(\sqrt{n(n+2)}\) Bohr Magnetons (BM), where \(n\) is the number of unpaired electrons.
Therefore, two ions will have the same magnetic moment if and only if they have the same number of unpaired electrons in their \(d\)-orbitals.
Step 2: Detailed Explanation:
Let's check each pair for the number of unpaired electrons (\(n\)):
A: \(Zn^{2+}\) ([Ar]\(3d^{10}\), \(n=0\)) and \(Ti^{2+}\) ([Ar]\(3d^2\), \(n=2\)). Different.
B: \(Cr^{2+}\) ([Ar]\(3d^4\), \(n=4\)) and \(Fe^{2+}\) ([Ar]\(3d^6\)). In \(3d^6\), there is one pair and 4 unpaired electrons \(\implies n=4\). Same.
C: \(Ti^{3+}\) ([Ar]\(3d^1\), \(n=1\)) and \(Cu^{2+}\) ([Ar]\(3d^9\)). In \(3d^9\), there are four pairs and 1 unpaired electron \(\implies n=1\). Same.
D: \(V^{2+}\) ([Ar]\(3d^3\), \(n=3\)) and \(Cu^{+}\) ([Ar]\(3d^{10}\), \(n=0\)). Different.
Since B and C have matching \(n\) values, they have same magnetic moments.
Step 3: Final Answer:
The statement corresponds to option (D) B and C only.