Question

What is the value of spin only magnetic moment for \(\text{Cu}^{2+}\) in BM?

• A. \(2.84\)
• B. \(3.87\)
• C. \(1.73\)
• D. \(0.0\)

Hint:

Find unpaired electrons in d9 and use sqrt(n(n+2)).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The spin-only magnetic moment depends on the number of unpaired electrons \(n\): \(\mu = \sqrt{n(n+2)}\) BM.

Step 2: Electronic configuration:
Cu has atomic number 29: \([Ar]3d^{10}4s^1\). Cu\(^{2+}\) loses the \(4s\) electron and one \(3d\) electron, giving \([Ar]3d^9\).

Step 3: Unpaired electrons:
In \(3d^9\), four orbitals are full and one orbital holds a single electron. So \(n = 1\).

Step 4: Calculate:
\[ \mu = \sqrt{1\times(1+2)} = \sqrt{3} = 1.73\ \text{BM} \]

Step 5: Check the options:
2.84 is for \(n=2\), 3.87 is for \(n=3\), and 0 is for \(n=0\). So the answer is (C).

Final Answer:
Cu2+ has one unpaired electron, so mu is 1.73 BM. \[ \boxed{1.73\ \text{BM}} \]