Question

Magnetic moment 2.84 B.M. is given by (At. nos, Ni=28, Ti=22, Cr=24, Co=27)

• A. ${ Cr^{2+} } $
• B. ${Co^{2+} } $
• C. ${Ni^{2+} } $
• D. ${Ti^{3+} } $

Answer 1

The Correct Option is C

This question involves determining the magnetic moment of a given ion and matching it to the options provided.

The magnetic moment of a transition metal ion can be calculated using the formula:

\(\mu = \sqrt{n(n+2)}\)

where \(\mu\) is the magnetic moment in Bohr Magnetons (B.M.) and \(n\) is the number of unpaired electrons.

We need to find which of the given ions has a magnetic moment of 2.84 B.M.

1. Calculate the number of unpaired electrons for each ion:
   • { Cr^{2+} }: The electron configuration will be [Ar] 3d⁴. Cr²⁺ has 4 unpaired electrons.
   • { Co^{2+} }: The electron configuration will be [Ar] 3d⁷. Co²⁺ has 3 unpaired electrons.
   • { Ni^{2+} }: The electron configuration will be [Ar] 3d⁸. Ni²⁺ has 2 unpaired electrons.
   • { Ti^{3+} }: The electron configuration will be [Ar] 3d¹. Ti³⁺ has 1 unpaired electron.
2. Calculate magnetic moment for each ion using the formula:
   • \(\mu_{Cr^{2+}} = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.90\)\, B.M.
   • \(\mu_{Co^{2+}} = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87\)\, B.M.
   • \(\mu_{Ni^{2+}} = \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.83\)\, B.M.
   • \(\mu_{Ti^{3+}} = \sqrt{1(1 + 2)} = \sqrt{3} \approx 1.73\)\, B.M.

From the calculations, the ion with a magnetic moment closest to 2.84 B.M. is { Ni^{2+} } (2.83 B.M.).

Conclusion: Therefore, the correct answer is { Ni^{2+} }.