Compare the energies of following sets of quantum numbers for multielectron system. (A) n = 4, l = 1 (B) n = 4, l = 2 (C) n = 3, l = 1 (D) n = 3, l = 2 (E) n = 4, l = 0 Choose the correct answer from the options given below :
- (E) > (C) > (A) > (D) > (B)
- (B) > (A) > (C) > (E) > (D)
- (C) < (E) < (D) < (A) < (B)
- (E) < (C) < (D) < (A) < (B)
Answer
(3) (C) < (E) < (D) < (A) < (B)
Solution
In a multielectron atom energy follows the (n + l) rule: lower (n + l), lower energy; for equal (n + l), the orbital with lower n has lower energy.
(A) 4p: n + l = 5
(B) 4d: n + l = 6
(C) 3p: n + l = 4
(D) 3d: n + l = 5
(E) 4s: n + l = 4
∵ (C) and (E) both have n + l = 4 (3p is lower than 4s) and (D) and (A) both have n + l = 5 (3d is lower than 4p), the order is 3p < 4s < 3d < 4p < 4d.
Hence (C) < (E) < (D) < (A) < (B).
Answer: (3)