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A compound of vanadium has a magnetic moment of $$1.73$$ $$BM$$. The electronic configuration of vanadium ion in the compound is:


A
[Ar]3d2
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B
[Ar]3d14s0
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C
[Ar]3d3
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D
[Ar]3d04s1
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Solution

The correct option is B $$\left[ Ar \right] 3{ d }^{ 1 }4{ s }^{ 0 }$$
The electronic configuration of $$V(Z=23) = 1s^22s^22p^63s^23p^64s^23d^3$$

If the number of unpaired electrons be n, then the Magnetic moment = $$ \sqrt{n(n+2)}$$ BM

 If n=1, then the magnetic momentum = $$ \sqrt{1(1+2)} = \sqrt3 = 1.73$$ BM

Therefore vanadium has one unpaired electron and the electronic configuration of the ion is $$[Ar]3d^14s^0$$. since the electrons are first removed from the outermost shell.

Chemistry

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