AIIMS2018Chemistry-Atomic Structure

AIIMS 2018 Chemistry Electronic Configuration MCQ Question

Type: MCQ-numerical-Medium-Class 11

A metal having +2 oxidation state has approximately 23e⁻. M²⁺ → 23e⁻ Therefore, M → 25 e⁻ Electronic configuration of M is equal to 3d⁵ 4s² The spin magnetic moment is calculated as: μ=n(n+2)\mu = \sqrt{n(n+2)}

A

5.9

B

4.9

C

3.9

D

2.9

Correct Answer

Option C

Detailed Explanation

To determine the spin magnetic moment (μ\mu) of the metal M in the +2 oxidation state, we first identify the number of unpaired electrons in its electronic configuration, which is 3d⁵ 4s². In this configuration, the 3d subshell has 5 electrons, all of which are unpaired, leading to n=5n = 5. Using the formula μ=n(n+2)\mu = \sqrt{n(n+2)}, we calculate μ=5(5+2)=355.9\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9, indicating that option C (3.9) is incorrect. The other options (A, B, D) do not match the calculated value, as they either overestimate or underestimate the magnetic moment based on the number of unpaired electrons.

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