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A vanadium chloride compound has a magnetic moment of 1.73 BM. What is the formula of this vanadium chloride?

1. VCl2
2. VCl3
3. VCl4
4. VCl5

Correct answer: VCl4

Solution

mu = sqrt(n(n+2)) = 1.73 BM -> n(n+2) = 3 -> n=1. V in VCl2 is V²+ with config [Ar]3d³ (3 unpaired). V in VCl3 is V³+ with [Ar]3d² (2 unpaired). V in VCl4 is V⁴+ with [Ar]3d¹ (1 unpaired). V in VCl5 is V⁵+ with [Ar]3d⁰ (0 unpaired). Only VCl4 gives 1 unpaired electron and mu = sqrt(3) = 1.732 BM.

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