Question

Which from following pair of elements in their respective oxidation states develops same value of calculated spin only magnetic moment ?

• A. \(\text{Ti}^{3+}\) and \(\text{Zn}^{2+}\)
• B. \(\text{Cr}^{2+}\) and \(\text{Fe}^{2+}\)
• C. \(\text{Cr}^{3+}\) and \(\text{Ti}^{3+}\)
• D. \(\text{Zn}^{2+}\) and \(\text{Cu}^{2+}\)

Hint:

Count unpaired electrons: equal n gives equal spin only moment.

Answer 1

The Correct Option is B

Step 1: Key Formula or Approach:
\(\mu = \sqrt{n(n+2)}\) BM, where n is the number of unpaired electrons. Same n gives same \(\mu\).

Step 2: Find n for each ion:
\(\text{Ti}^{3+}\): \(3d^1\), n = 1. \(\text{Zn}^{2+}\): \(3d^{10}\), n = 0.
\(\text{Cr}^{2+}\): \(3d^4\), n = 4. \(\text{Fe}^{2+}\): \(3d^6\), n = 4.
\(\text{Cr}^{3+}\): \(3d^3\), n = 3. \(\text{Cu}^{2+}\): \(3d^9\), n = 1.

Step 3: Match the pairs:
(A) 1 and 0, different. (B) 4 and 4, same. (C) 3 and 1, different. (D) 0 and 1, different. So (B) is correct, with \(\mu = \sqrt{24} = 4.9\) BM.

Final Answer:
\(\text{Cr}^{2+}\) and \(\text{Fe}^{2+}\) both have four unpaired electrons, option (B). \[ \boxed{\text{Cr}^{2+}\text{ and Fe}^{2+}} \]