JEE Advanced 2026 — Paper 1 — CHEM-II solution

JEE Advanced2026Paper 1CHEM-II

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

Solution

(A) Multi-electron rule for Li. In multi-electron atoms, orbital energies follow the $(n + ℓ)$ rule: 2s has $n + ℓ = 2$ and 2p has $n + ℓ = 3$ , so $E_{2 s} (Li) < E_{2 p} (Li)$ . Correct.

(B) Hydrogen is a one-electron atom. Its orbital energies depend only on $n$ , so $E_{2 s} (H) = E_{2 p} (H)$ . Correct.

(C) Compare Li 2s with H 2p. The 2s electron of Li experiences a larger effective nuclear charge ( $Z_{eff} \approx 1.3$ ) than the 2p electron of H (effective $Z = 1$ ). Greater $Z_{eff}$ means more negative (lower) energy, so $E_{2 s} (Li) < E_{2 p} (H)$ . The inequality stated in (C) reverses this. Incorrect.

(D) Compare H 2s with Li 2s. Following the same logic, the Li 2s is more tightly bound (lower energy) than the H 2s, so $E_{2 s} (H) > E_{2 s} (Li)$ . Correct.

Correct answers: (A), (B), (D).

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