Question

Which of the following ions exhibit same value of spin only magnetic moment?
(A). \(\text{Ti}^{3+}\)
(B). \(\text{Cr}^{3+}\)
(C). \(\text{Mn}^{2+}\)
(D). \(\text{Fe}^{3+}\)
(E). \(\text{Sc}^{3+}\)
Choose the most appropriate answer from the options given below.

• (B) and (D) only
• (A) and (E) only
• (C) and (D) only
• (A) and (D) only

Hint:

Count unpaired electrons: Mn2+ and Fe3+ are both d5 with 5 unpaired electrons.

Answer 1

The Correct Option is C

Step 1: Key Formula:
Spin-only magnetic moment is \(\mu = \sqrt{n(n+2)}\) BM, where \(n\) is the number of unpaired electrons. Equal \(n\) means equal \(\mu\).

Step 2: Find the unpaired electrons:
(A) \(\text{Ti}^{3+}\): \(3d^1\), so \(n = 1\).
(B) \(\text{Cr}^{3+}\): \(3d^3\), so \(n = 3\).
(C) \(\text{Mn}^{2+}\): \(3d^5\), so \(n = 5\).
(D) \(\text{Fe}^{3+}\): \(3d^5\), so \(n = 5\).
(E) \(\text{Sc}^{3+}\): \(3d^0\), so \(n = 0\).

Step 3: Pick the pair:
Only (C) and (D) share the same \(n = 5\), and so \(\mu = \sqrt{35} = 5.92\) BM for both.
The other options fail: (B) and (D) have 3 and 5 unpaired electrons, (A) and (E) have 1 and 0, and (A) and (D) have 1 and 5.

Final Answer:
Mn(II) and Fe(III) have the same spin-only moment, option (C). \[ \boxed{\text{(C) and (D) only}} \]