JEE Advanced 2026 · Paper 1 · Chemistry Q5

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ChemistryMultiple correct+4 / -1Question 37 of 48

The 2s and the 2p orbital energies of hydrogen atom are $E_{2s}(\mathrm H)$ and $E_{2p}(\mathrm H)$ , respectively. The 2s and the 2p orbital energies of lithium atom are $E_{2s}(\mathrm{Li})$ and $E_{2p}(\mathrm{Li})$ , respectively. The correct option(s) about the orbital energies is(are)

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

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

E_{2p}(\mathrm H)<E_{2s}(\mathrm{Li})

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

Answer and worked solution

Official answer: A, B, D

(B) In hydrogen (one electron) the energy depends only on $n$ : $E_{2s}(\mathrm H)=E_{2p}(\mathrm H)=-3.4$ eV. True.

(A) In lithium the 2s electron penetrates the $1s^2$ core more than a 2p electron, so it feels a larger effective nuclear charge and lies lower. True.

(C), (D) The ionization energy of Li is 5.4 eV, so $E_{2s}(\mathrm{Li})\approx-5.4$ eV, lower than $-3.4$ eV. Hence $E_{2s}(\mathrm H)>E_{2s}(\mathrm{Li})$ : (D) true. And $E_{2p}(\mathrm H)$ is not below $E_{2s}(\mathrm{Li})$ : (C) false.

Topics:Electronic configuration Ionisation enthalpy

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JEE Advanced 2026 · Paper 1, Chemistry question 5