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The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is

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Mn²⁺

TI²⁺

V²⁺

Cr²⁺

answer is C.

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Detailed Solution

M.M = $\sqrt{n (n + 2)}$ (n = no.of unpaired e^- s)
Ti²⁺ → [Ar] 3d² n = 2 M.M = 2.44
Mn²⁺ → [Ar] 3d⁵ n = 5 M.M = 5.91
V²⁺ → [Ar] 3d³ n = 3 M.M = 3.87
Cu²⁺ → [Ar] 3d⁹ n = 1 M.M = 1.73

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