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: Electronic configurations:
Cr is $3d^5 4s^1$, so $\text{Cr}^{2+}$ is $3d^4$. Fe is $3d^6 4s^2$, so $\text{Fe}^{2+}$ is $3d^6$ (one paired orbital, four singly filled).

Step 2: Compare:
Both have 4 unpaired electrons, so both have the same spin only value, about 4.90 BM. In the other pairs the unpaired counts are (1, 0), (3, 1) and (0, 1). Answer (B).

Final Answer:
$\text{Cr}^{2+}$ and $\text{Fe}^{2+}$. \[ \boxed{\text{(B)}} \]