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Question

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)

Solution

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