Question

The electrons identified by quantum numbers $n$ and $l$:
(A) $n=4,l=1$
(B) $n=4,l=0$
(C) $n=3,l=2$
(D) $n=3,l=1$
can be placed in order of increasing energy as:

• A. (C) < (D) < (B) < (A)
• B. (D) < (B) < (C) < (A)
• C. (B) < (D) < (A) < (C)
• D. (A) < (C) < (B) < (D)

Answer 1

The correct option is B (D) < (B) < (C) < (A)

From the quantum numbers, the orbitals are
(a) 4p (b) 4s (c) 3d (d) 3p
According to Bohr Bury's $n+l$ rule, increasing order of energy (D) < (B) < (C) < (A)
Note: If the two orbitals have same value of $(n+l)$ then the orbital with lower value of $n$ will be filled first.