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JEE Main202227 Jul 2022Evening ShiftChemistryStructure of AtomActual

The correct decreasing order of energy, for the orbitals having, following set of quantum numbers: (A) n = 3 , 1 = 0 , m = 0 (B) n = 4 , l = 0 , m = 0 (C) n = 3 , l = 1 , m = 0 (D) n = 3 , 1 = 2 , m = 1

Options

  1. AD > B > C > A
  2. BB > D > C > A
  3. CC > B > D > A
  4. DB > C > D > A

Correct answer

A. D > B > C > A

Step-by-step solution

(A) n + l = 3 + 0 = 3 ⇒ 3 s (B) n + l = 4 + 0 = 4 ⇒ 4 s (C) n + l = 3 + 1 = 4 ⇒ 3 p (D) n + l = 3 + 2 = 5 ⇒ 3 d Higher the n + l value, higher the energy of the orbital and if two orbitals have same n + l value, then the orbital with higher n value has higher energy. Thus: D > B > C > A .

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