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The correct decreasing order of energy, for the orbitals having, following set of quantum numbers:(A) n = 3, l = 0, m = 0(B) n = 4, l = 0, m = 0(C) n = 3, l = 1, m = 0(D) n = 3, l = 2, m = 1(A) (D) > (B) > (C) > (A)(B) (B) > (D) > (C) > (A)(C) (C) > (B) > (D) > (A)(D) (B) > (C) > (D) > (A) |
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Answer» Correct option is (A) (D) > (B) > (C) > (A) (A) n + l = 3 + 0 = 3 (B) n + l = 4 + 0 = 4 (C) n + l = 3 + 1 = 4 (D) n + l = 3 + 2 = 5 Higher n + l value, higher the energy & if same n + l value, then higher n value, higher the energy. Thus: D > B > C > A. |
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