Compare the energies of following sets of quantum numbers for multielectron system. (A) $$n = 4, l = 1$$ (B) $$n = 4, l = 2$$ (C) $$n = 3, l = 1$$ (D) $$n = 3, l = 2$$ (E) $$n = 4, l = 0$$. Choose the correct answer from the options given below :

We need to compare the energies of orbitals in a multi-electron system using the $$(n + l)$$ rule.

For multi-electron atoms, orbital energy increases with $$(n + l)$$. If two orbitals have the same $$(n + l)$$, the one with lower $$n$$ has lower energy.

Calculate $$(n + l)$$ for each:

(A) $$n = 4, l = 1$$: $$n + l = 5$$ (4p orbital)

(B) $$n = 4, l = 2$$: $$n + l = 6$$ (4d orbital)

(C) $$n = 3, l = 1$$: $$n + l = 4$$ (3p orbital)

(D) $$n = 3, l = 2$$: $$n + l = 5$$ (3d orbital)

(E) $$n = 4, l = 0$$: $$n + l = 4$$ (4s orbital)

Order by increasing energy:

$$n + l = 4$$: C (3p, $$n = 3$$) and E (4s, $$n = 4$$). Since both have $$n + l = 4$$, lower $$n$$ means lower energy: C < E

$$n + l = 5$$: D (3d, $$n = 3$$) and A (4p, $$n = 4$$). D < A

$$n + l = 6$$: B (4d)

So the increasing order of energy is: $$C < E < D < A < B$$

This matches Option D: (C) < (E) < (D) < (A) < (B).

The correct answer is Option D.