Question

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?

Answer 1

Answer: 2

If l can take n+1 values,
$\therefore$ the following sublevels will be possible
$1s,1p,2s,2p,2d,3s,3p,3d,3f\dots .etc.$
Order in which the sublevels will be filled (using n+1 rule)
$1s>1p>2s>2p>3s>2d>3p\dots .etc$
$\therefore$ electronic configuration of $F{e}^{26}$ is
$1s^{2}1p^{6}2s^{2}2p^{6}3s^{2}2d^8$
$\therefore$ it will have 2 unpaired electrons in 2d subshell.