Arrange the following orbitals in decreasing order of energy.
A. n = 3, l = 0, m = 0
B. n = 4, l = 0, m = 0
C. n = 3, l = 1, m = 0
D. n = 3, l = 2, m = 1
The correct option for the order is:

Using the $$(n + l)$$ rule for ordering orbital energies (for multi-electron atoms):

A: $$n = 3, l = 0$$ → 3s orbital, $$(n+l) = 3$$

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

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

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

Higher $$(n+l)$$ means higher energy. For orbitals with same $$(n+l)$$, the one with higher $$n$$ has higher energy.

Ordering by decreasing energy:

D (n+l = 5) > B (n+l = 4, n = 4) > C (n+l = 4, n = 3) > A (n+l = 3)

The correct decreasing order is: D > B > C > A