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Q.
How many unpaired electrons would have been present in Fe atom, if the azimuthal quantum number (l) can have (n+1) values and all other laws are applicable?
NTA AbhyasNTA Abhyas 2022
Solution:
If l can take n+1 values,
$\therefore \, \, $ the following sublevels will be possible
$1s, \, 1p, \, 2s, \, 2p, \, 2d, \, 3s, \, 3p, \, 3d, \, 3f\ldots .etc.$
Order in which the sublevels will be filled (using n+1 rule)
$1s>1p>2s>2p>3s>2d>3p\ldots .etc$
$\therefore \, \, $ electronic configuration of $Fe^{26}$ is
$1s^{2}1p^{6}2s^{2}2p^{6}3s^{2}2d^{8}$
$\therefore \, \, $ it will have 2 unpaired electrons in 2d subshell.