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A compound of vanadium has a magnetic moment of `1.73BM` . Work out the electronic configuration of vanadium ion in the compound. |
Answer» No. of unpaird electron are given by Magnetic moment `= sqrt ( [n ( n + 20 )]` ( where (n) is no , of unpaired electrons ) or ` 1. 7 3 = sqrt ( [n (n + 2)])` or ` 1. 73 xx 1. 73 = n^2 + 2 n :. N=1` Now vanadium atom must have one unpaired electron and thus its configuration is ` ._(23) V^(4+) : 1s^2 , 2s^2 , 2s^2 . 2p^6 . 3 s^2 3p^6 3d^1`. |
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