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# Magnetic moment of $Fe^{a+}(z=26)$ is $\sqrt{24}$BM.Hence number of unpaired electrons and value of a respectively are :

$(a)\;4,2\qquad(b)\;2,4\qquad(c)\;3,1\qquad(d)\;0,2$

Magnetic moment $\sqrt{n(n+2)}$BM
$\Rightarrow n(n+2)=24$
$\Rightarrow n=4$
Thus, $Fe^{2+}$ has four unpaired electrons,
i.e., It is $Fe^{2+}$ or $[Ar]3d^6$
Hence (a) is the correct answer.
edited Mar 19, 2014