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Four electrons have the following quantum numbers: (P) n=4, l=1, m=0, s=+1/2 (a 4p electron) (Q) n=5, l=0, m=0, s=-1/2 (a 5s electron) (R) n=6, l=2, m=0, s=+1/2 (a 6d electron) (S) n=6, l=3, m=-1, s=+1/2 (a 6f electron) Arrange them in order of increasing energy (lowest to highest):

1. P < Q < R < S
2. Q < R < P < S
3. Q < R < S < P
4. S < Q < R < P

Correct answer: P < Q < R < S

Solution

(P) 4p: n+l = 4+1 = 5, n=4. (Q) 5s: n+l = 5+0 = 5, n=5. (R) 6d: n+l = 6+2 = 8. (S) 6f: n+l = 6+3 = 9. Using Aufbau rule: lower n+l means lower energy. R (n+l=8) < S (n+l=9). For P and Q (both n+l=5), lower n comes first: P (n=4) < Q (n=5). Overall: P < Q < R < S.

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