Chemistry - GENERAL CHEMISTRY Question with Solution | TestHub

In multi-electronic species, energy is decided on the basis of ($\mathrm{n}+\mathrm{l}$) rule.

$3\mathrm{s} = 3 + 0 + 0= 3$

$4\mathrm{s} = 4 + 0 +0 = 4$

$3\mathrm{p} = 3 + 1 +0 = 4$

$3\mathrm{d} = 3 + 2 + 1 = 6$

For 4s and 3p, an orbital with a higher value of n has higher energy.

So increasing the order of energy is $3\mathrm{d}>4\mathrm{s}>3\mathrm{p}>3\mathrm{s}$.