JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Atomic Structure

The 2s and the 2p orbital energies of hydrogen atom are

E_{2 s} (H)

and

E_{2 p} (H)

, respectively. The 2s and the 2p orbital energies of lithium atom are

E_{2 s} (Li)

and

E_{2 p} (Li)

, respectively. The correct option (s) about the orbital energies is (are)

(A) $E_{2 s} (Li) < E_{2 p} (Li)$ ✓ Correct
(B) $E_{2 s} (H) = E_{2 p} (H)$ ✓ Correct
(C) $E_{2 p} (H) < E_{2 s} (Li)$
(D) $E_{2 s} (H) > E_{2 s} (Li)$ ✓ Correct

Answer: (A), (B), (D)

Solution

(A) For multi-electron systems like Li, energy depends on both n and l. Using the (n+l) rule:

For 2s: n+l = 2+0 = 2
For 2p: n+l = 2+1 = 3

Therefore,

E_{2 s} (Li) < E_{2 p} (Li)

. Correct.

(B) For single-electron systems like H, energy depends only on n. Since both 2s and 2p have n=2:

E_{2 s} (H) = E_{2 p} (H)

. Correct.

(C) Li has Z=3, so higher effective nuclear charge pulls electrons closer, making orbitals more stable (more negative energy). Thus

E_{2 s} (Li) < E_{2 s} (H)

, and similarly for 2p. This option is incorrect.

(D) Since Li has higher Z, its orbitals are more stable (lower/more negative energy):

E_{2 s} (H) > E_{2 s} (Li)

. Correct.

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JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Atomic Structure

Solution

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