Q99 / 240 · KCET 2026 · Chemistry
The calculated spin only magnetic moment of $\text{Cr}^{2+}$ ion is
- A$3.87$ BM
- Correct answerB$4.90$ BM
- C$5.92$ BM
- D$2.84$ BM
Solution
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
Hence, the correct answer is option: 'B'.