JEE MAIN - Chemistry (2022 - 24th June Morning Shift - No. 2)
Consider the following pairs of electrons
(A) (a) n = 3, $$l$$ = 1, m1 = 1, ms = + $${1 \over 2}$$
(b) n = 3, 1 = 2, m1 = 1, ms = + $${1 \over 2}$$
(B) (a) n = 3, $$l$$ = 2, m1 = $$-$$2, ms = $$-$$$${1 \over 2}$$
(b) n = 3, $$l$$ = 2, m1 = $$-$$1, ms = $$-$$$${1 \over 2}$$
(C) (a) n = 4, $$l$$ = 2, m1 = 2, ms = + $${1 \over 2}$$
(b) n = 3, $$l$$ = 2, m1 = 2, ms = + $${1 \over 2}$$
The pairs of electrons present in degenerate orbitals is/are :
Only (A)
Only (B)
Only (C)
(B) and (C)
Explanation
For degenerate orbitals, only the value of m must be different. The value of (n + l) must be the same.
Hence, the pair of electrons with quantum numbers given in (B) are degenerate.
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