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KCET · Chemistry · d and f Block Elements

The calculated spin only magnetic moment of \(\text{Cr}^{2+}\) ion is

  1. A \(3.87\) BM
  2. B \(4.90\) BM
  3. C \(5.92\) BM
  4. D \(2.84\) BM
Verified Solution

Answer & Solution

Correct Answer

(B) \(4.90\) BM

Step-by-step Solution

Detailed explanation

The atomic number of Chromium (Cr) is \(24\).

The electronic configuration of Cr atom is \([\text{Ar}] 3d^{5} 4s^{1}\).

For \(\text{Cr}^{2+}\) ion, two electrons are removed, giving the configuration \([\text{Ar}] 3d^{4}\).

The number of unpaired electrons (\(n\)) in \(\text{Cr}^{2+}\) is \(4\).

The spin only magnetic moment (\(\mu\)) is calculated using the formula:

\(\mu = \sqrt{n(n+2)} \text{ BM}\)

Substituting \(n = 4\):

\(\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}\)

Answer: \(4.90\) BM