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# Calculate the number of unpaired electrons in the following gaseous ions: Cr^3+

8.24 Calculate the number of unpaired electrons in the following gaseous ions:

$(ii)\; Cr^{3+}$

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Electronic configuration of chromium is $Cr = 3d^{5}4s^{1}$. The number of unpaired electron in $Cr^{3+}$ is 3
$Cr^{3+}(Z=24) =\left [ Ar \right ] 3d^{3}$
after losing 3 electron, Cr has 3 electron left d-orbital

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