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Q. What is the magnetic moment for $Mn^{2+}$ ion?

JIPMERJIPMER 2015The d-and f-Block Elements

Solution:

$Mn^{2+} : 3d^5,$ no. of unpaired electrons $(n) = 5$
Magnetic moment $(\mu) = \sqrt{n(n+2)} \, B.M.$
$= \sqrt{5(5+2)} \, B.M. = \sqrt{35}$
$= 5.92 \,B.M.$