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Q. Maximum value of spin multiplicity for electrons in $Cr$ atom is -

NTA AbhyasNTA Abhyas 2020

Solution:

$Cr\left(\right.24\left.\right)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^{1}$
Spin multiplicity $=2\left|\frac{6}{2}\right|+1=7$