CUET · CHEMISTRY · PYQ PAPER 2025
The calculated magnetic moment using the 'spin only' formula \(\mu=\sqrt{n(n+2)}\) for \(\mathrm{Cr}^{2+}\) and \(\mathrm{Fe}^{2+}\) respectively is
- A \(3.87,4.89\)
- B Both 4.89
- C Both 3.87
- D \(4.89,3.87\)
Answer & Solution
Correct Answer
(B) Both 4.89
Step-by-step Solution
Detailed explanation
For Cr²⁺: Cr: [Ar] 3d⁵ 4s¹ Cr²⁺: [Ar] 3d⁴ \(n = 4\) \(\mu = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24} = 4.89\) For Fe²⁺: Fe: [Ar] 3d⁶ 4s² Fe²⁺: [Ar] 3d⁶ \(n = 4\) \(\mu = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24} = 4.89\)
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