Question:medium

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.

Show Hint

Count unpaired electrons: Mn2+ and Fe3+ are both d5 with 5 unpaired electrons.
Updated On: Oct 1, 2026
  • (B) and (D) only
  • (A) and (E) only
  • (C) and (D) only
  • (A) and (D) only
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Plan:
Write the electronic configuration of each ion by removing the 4s electrons first and then the 3d electrons.

Step 2: Configurations:
Ti ($3d^24s^2$) loses 3 electrons and becomes $3d^1$. Cr ($3d^54s^1$) becomes $3d^3$. Mn ($3d^54s^2$) loses two and becomes $3d^5$. Fe ($3d^64s^2$) loses three and becomes $3d^5$. Sc ($3d^14s^2$) becomes $3d^0$.

Step 3: Compare:
The half-filled $d^5$ set has five unpaired electrons, and both Mn(II) and Fe(III) carry it. So both have $\mu = \sqrt{5\times7} = 5.92$ BM.

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