Question

Which cation from following exhibits no magnetic moment?

• A. $\text{Cr}^{3+}$
• B. $\text{Sc}^{3+}$
• C. $\text{Cu}^{2+}$
• D. $\text{V}^{3+}$

Hint:

The magnetic moment formula is $\mu = \sqrt{n(n+2)}$ Bohr Magnetons (BM), where $n$ is the number of unpaired electrons. If $n=0$, then $\mu=0$! Ions with $d^0$ or $d^{10}$ configurations always have zero magnetic moment.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The magnetic moment ($\mu$) of an ion is determined by the number of unpaired electrons it contains. A magnetic moment of zero means the ion is completely diamagnetic (zero unpaired electrons).

Step 2: Detailed Explanation:
Let's write the electronic configuration for each transition metal ion by removing electrons first from the 4s orbital and then the 3d orbital.
(a) $\text{Cr^{3+}$:} Cr (Z=24) is $[\text{Ar}] 4s^1 3d^5$. Removing 3 electrons gives $[\text{Ar}] 3d^3$. It has 3 unpaired electrons ($\mu \approx 3.87$ BM).
(b) $\text{Sc^{3+}$:} Sc (Z=21) is $[\text{Ar}] 4s^2 3d^1$. Removing all 3 valence electrons gives $[\text{Ar}] 3d^0$. It has exactly 0 unpaired electrons, making it diamagnetic with a magnetic moment of 0.
(c) $\text{Cu^{2+}$:} Cu (Z=29) is $[\text{Ar}] 4s^1 3d^{10}$. Removing 2 electrons gives $[\text{Ar}] 3d^9$. It has 1 unpaired electron ($\mu \approx 1.73$ BM).
(d) $\text{V^{3+}$:} V (Z=23) is $[\text{Ar}] 4s^2 3d^3$. Removing 3 electrons gives $[\text{Ar}] 3d^2$. It has 2 unpaired electrons ($\mu \approx 2.83$ BM).

Step 3: Final Answer:
The cation with no magnetic moment is $\text{Sc}^{3+}$, matching option (b).