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The spin only magnetic moment of $Cr^{x+}$ is $4.9\;BM$ . Find $x$

$(a)\;x=4\\(b)\;x=2 \\(c)\;x=1 \\(d)\;x=3 $

1 Answer

$\mu=4.9\;BM=\sqrt {n(n+2)}$
unpaired $e^-=n=4$
Cont :- $3d^44s^0$
$Cr^{2+}$ ion poseses 4 uparide : x=2
Hence b is the correct answer.
answered Mar 12, 2014 by meena.p
 
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