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A compound of Vanadium has a magnetic moment of 1.73 Bm. Work out the electric configuration of the Vanadium ion in the compound.

$(a)\;1s^2,2s^22p^6,3s^23p^63d^1\qquad(b)\;1s^2,2s^22p^6,3s^2\qquad(c)\;1s^2,2s^22p^6,3s^23p^5\qquad

(d)\;1s^2,2s^22p^6,3s^23p^63d^3,4s^2$

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Number of unpaired electron are given by
Magnetic moment = $\sqrt {n(n+2)}$
Where n is the No. of unpaired electrons
Or 1.73 = $\sqrt {n(n+2)}$
$1.73\times1.73 = n^2+2n$
$\therefore$ n = 1
Vanadium atom must have one unpaired electron and thus its configuration is
$_{23} V^{4+} : 1s^2,2s^22p^6,3s^23p^63d^1$
Hence answer is (a)
answered Jan 28, 2014 by sharmaaparna1
edited Mar 19, 2014 by sharmaaparna1
 

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