The calculated spin only magnetic moment of Cr^2+ ion is:

Correct answer: (D) (d) 4.90 BM. Answer: (d) 4.90 BM. Solution: Cr^2+ has the configuration [Ar] 3d^4, so it has n = 4 unpaired electrons. Using the spin-only formula: = √(n(n+2)) = √(4(4+2)) = √(24) = 4.90 BM. Hence option (d) is correct.

NEET 2026 (Phase 1) · ChemistryMagnetismPrevious Year Question

The calculated spin only magnetic moment of $Cr X^{2 +}$ ion is:

Answer: (D) (d) $4.90$ BM. \textbf{Answer:} (d) $4.90$ BM.

A.(a) $5.92$ BM
B.(b) $2.84$ BM
C.(c) $3.87$ BM
D.(d) $4.90$ BM✓

Correct Answer

(D) (d) $4.90$ BM

Solution & Explanation

\textbf{Answer:} (d) $4.90$ BM. \textbf{Solution:} $Cr X^{2 +}$ has the configuration $[Ar] 3 d X^{4}$ , so it has $n = 4$ unpaired electrons. Using the spin-only formula: $μ = n (n + 2) = 4 (4 + 2) = 24 = 4.90$ BM. Hence option (d) is correct.

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The calculated spin only magnetic moment of CrX2+ ion is:

Solution & Explanation

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The calculated spin only magnetic moment of $Cr X^{2 +}$ ion is: