The quantum numbers of four electrons are given below : I. n=4 ;...

(I) $n=4, I=2, m_I=-2, s=-\frac{1}{2}$; represents $4 d(n+I=6)$ (II) $n=3, I=2, m_l=1, s=+\frac{1}{2}$; represents $3 d(n+I=5)$ (III) $n=4, I=1, m_I=0, s=+\frac{1}{2}$; represents $4 p(n+I=5)$ (IV) $n=3, I=1, m_l=-1, s=+\frac{1}{2}$; represents $3 p(n+I=4)$ Order of energy depends on the $(n+I)$, greater is the $(n+I)$ value greater is the energy, if $(n+I)$ is same, then it depends on $n$; if ' $n$ ' is more, energy is more. Step-1: According to $(\mathrm{n}+\mathrm{I}$ ) Energy $=($\mathrm{I}$)\gt($\mathrm{II}$)=($ III $)\gt(\mathrm{IV})$ Step-2 : If $\mathrm{n} \uparrow$, then energy increases Energy $=($\mathrm{I}$)\gt($ III $)\gt($ II $)\gt($ IV)