Exams › JEE Main › Chemistry
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 : (1) (B) > (A) > (C) > (E) > (D) (2) (E) > (C) > (D) < (A) < (B) (3) (E) > (C) > (A) > (D) > (B) (4) (C) < (E) < (D) < (A) < (B)
- (1) (B) > (A) > (C) > (E) > (D)
- (2) (E) > (C) > (D) < (A) < (B)
- (3) (E) > (C) > (A) > (D) > (B)
- (4) (C) < (E) < (D) < (A) < (B)
Correct answer: (4) (C) < (E) < (D) < (A) < (B)
Solution
The correct option reflects the increasing energy levels of the orbitals based on the principal quantum number (n) and azimuthal quantum number (l). As n increases, the energy generally increases, and for the same n, higher l values correspond to higher energy due to increased angular momentum.
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