Question

Given below are the quantum numbers for 4 electrons.
A. n=3, l=2, ml=1,ms=+$\frac{1}{2}$
B. n=4, l=1, ml=0,ms=+$\frac{1}{2}$
C. n=4, l=2, ml=–2,ms=–$\frac{1}{2}$
D. n=3, l=1, ml=–1,ms=+$\frac{1}{2}$
The correct order of increasing energy is

• A. D < B < A < C
• B. D < A < B < C
• C. D < A < B < C
• D. B < D < A < C

Answer 1

Answer: (B)

D < A < B < C

Answer 2

Energy of the sub-shell is given by, (n + l) rule.

For	(n + l)
A	5
B	5
C	6
D	4

Hence, the correct order of increasing energy is D < A < B < C
The correct option is (B): D < A < B < C.