JEE Main · 2022 · Shift-IImediumATOM-121

The correct decreasing order of energy for orbitals having, following set of quantum numbers: (A) n=3,l=0 (B) n=4,l=0…

Structure of Atom · Class 11 · JEE Main Previous Year Question

Question

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

Options
  1. a

    (D) > (B) > (C) > (A)

    ✓
  2. b

    (B) > (D) > (C) > (A)

  3. c

    (C) > (B) > (D) > (A)

  4. d

    (B) > (C) > (D) > (A)

Correct Answera

(D) > (B) > (C) > (A)

Detailed Solution

🧠 The Aufbau Hierarchy Energy depends on the sum of nn and ll. This is why the 4s4s orbital fills before the 3d3d orbital in most atoms, but here we are comparing their relative heights.

🗺️ The Value Summation

  • (A): n=3,l=0  ⟹  n+l=3n=3, l=0 \implies n+l = 3 (3s)
  • (C): n=3,l=1  ⟹  n+l=4n=3, l=1 \implies n+l = 4 (3p)
  • (B): n=4,l=0  ⟹  n+l=4n=4, l=0 \implies n+l = 4 (4s)
  • (D): n=3,l=2  ⟹  n+l=5n=3, l=2 \implies n+l = 5 (3d)

Comparing (n+l)=4(n+l)=4 pairs: Orbital B (n=4n=4) is higher than C (n=3n=3). Final rank (Energy): D > B > C > A.

⚡ The "Shell vs shape" Hook Usually, 3d3d is the highest in this set because l=2l=2 adds more energy cost than shifting from n=3n=3 to n=4n=4 does for an s-orbital (l=0l=0).

⚠️ Common Traps Mixing up the ll values for subshells (s=0,p=1,d=2,f=3s=0, p=1, d=2, f=3).

Answer: (a)\boxed{\text{Answer: (a)}}

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