Question:medium

What is the value of spin only magnetic moment for $Cu^{2+}$ in BM?

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If $n=1, \mu \approx 1.73$; if $n=2, \mu \approx 2.84$; if $n=3, \mu \approx 3.87$.
Updated On: Jun 19, 2026
  • 2.84
  • 3.87
  • 1.73
  • 0.0
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
The spin-only magnetic moment ($\mu$) depends on the number of unpaired electrons ($n$) in the d-subshell of the metal ion.

Step 2: Key Formula or Approach:

1. Write the electronic configuration of the ion.
2. Identify the number of unpaired electrons ($n$).
3. Use the formula: \[ \mu = \sqrt{n(n+2)} \text{ BM} \]

Step 3: Detailed Explanation:

Atomic number of Copper (Cu) is 29.
Ground state electronic configuration: $[\text{Ar}] 3d^{10} 4s^1$
$\text{Cu}^{2+}$ configuration: $[\text{Ar}] 3d^9$ (electrons are lost from 4s then 3d).
In $3d^9$, there are 4 pairs and 1 unpaired electron.
So, $n = 1$.
Calculation: \[ \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \text{ BM} \]

Step 4: Final Answer:

The spin-only magnetic moment for $\text{Cu}^{2+}$ is 1.73 BM.
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