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A compound of vanadium has a magnetic moment of `1.73BM`. Work out the electronic configuration of vanadium in the compoundA. `[Ar]3d^(1)`B. `[Ar]3d^(2)`C. `[Ar]3d^(3)`D. `[Ar]3d^(0)` |
Answer» Correct Answer - A `sqrt(n(n+2))=1.73` `n(n+2)=3` `n=1` So vandium has 1 unparied electron and its configuration is`[A]3d^(1)`. |
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