KCET · Chemistry · d and f Block Elements
The calculated spin only magnetic moment of \(\text{Cr}^{2+}\) ion is
- A \(3.87\) BM
- B \(4.90\) BM
- C \(5.92\) BM
- D \(2.84\) BM
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
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
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