(I) $$n=4, I=2, m I=-2, s=-\frac{1}{2}$$; represents $4 d(n+I=6)$

(II) $$n=3, I=2, m_I=1, s=+\frac{1}{2}$$; 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) $$n=3, I=1, m_1=-1, s=+\frac{1}{2}$$; 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 $$(\mathrm{n}+\mathrm{I})$$

$$\text { Energy }=(\mathrm{I})>(\mathrm{II})=(\mathrm{III})>(\text { IV) }$$

Step-2 : If $n \uparrow$, then energy increases

$$\text { Energy }=\text { (I) }>\text { (III) }>\text { (II) }>\text { (IV) }$$