2s and 2p orbital energies of H and Li — JEE Advanced 2026 Paper 1 Q.5

QThe question — verbatim

JEE (Advanced) 2026, Paper 1, Chemistry, Section 2 — one or more options correct: +4 only if all the correct options are chosen, partial marks +3 / +2 / +1 for a correct subset, 0 if unanswered, −1 otherwise.

Q.5 The 2s and the 2p orbital energies of hydrogen atom are E_2s(H) and E_2p(H), respectively. The 2s and the 2p orbital energies of lithium atom are E_2s(Li) and E_2p(Li), respectively. The correct option(s) about the orbital energies is(are)

(A)E_2s(Li) < E_2p(Li)
(B)E_2s(H) = E_2p(H)
(C)E_2p(H) < E_2s(Li)
(D)E_2s(H) > E_2s(Li)

What is being asked. An orbital energy is the energy of an electron in that orbital — negative, because the electron is bound; more negative means more tightly held. In hydrogen the electron is alone with a +1 nucleus. In lithium the 2s or 2p electron shares the atom with two 1s core electrons that partly shield the +3 nucleus. How much shielding an electron feels depends on how often it gets inside the core — its penetration. The experiment solves for all four energies, measures the penetration of each orbital and then tests every option.

01Interactive experiment

An atom chamber and an energy ladder. Press START — for each of the four orbitals the experiment solves the Schrödinger equation for the electron in the field of the nucleus and the 1s core, shows where the electron spends its time, places its energy on the ladder, and finally tests options (A)–(D) against the ladder.

Ready

PrepareSolveElectron cloud Energy ladderTest optionsCalculateAnswer

Set your values or use the question defaults.

Atom chamber and energy ladder

no orbital yet

A hydrogen atom and a lithium atom wait in the chamber. The question's values are already loaded — press START EXPERIMENT.

✓ USING QUESTION VALUES

Optional — change anything, or nothing

Hydrogen is fixed: Z = 1, no core. For the second atom the question means lithium — Z = 3 with a filled 1s² core. Try other values to see what the core does.

Live data

orbital

—

nucleus / core

—

energy E

— eV

solver step

—

time inside core

— %

extra pull (penetration)

— eV

Z_eff = n√(E/E_H)

—

testing

—

true options

—

Where the electron is — radial probability P(r)

Shaded: the lithium 1s core. The inner bump of Li 2s reaches inside it; Li 2p barely does.

Effective nuclear charge of each orbital

Z_eff from the solved energy, E = −13.6 Z_eff²/n² eV. Higher Z_eff = lower energy.

Calculation — written as the experiment produces the numbers

Nothing measured yet.

Orbital log (0)

Nothing needs to be typed. Press START EXPERIMENT.

02Detailed solution

Run the experiment — the solution opens with the answer.

The solution is still locked

The experiment has not finished yet. Press START and let it run to the end — or use Reveal anyway in the header if you are revising.

Hydrogen — energy depends only on n

Hydrogen has one electron and nothing to shield it, so its energy is fixed by the principal quantum number alone:

E_n = −13.6 / n² eV ⇒ E_2s(H) = E_2p(H) = −13.6/4 = −3.40 eV

The 2s and 2p orbitals of hydrogen are degenerate. (B) is correct.

In the experiment the solver landed on exactly the same energy for H 2s and H 2p, and both had Z_eff = 1.00.
Lithium — energy depends on n and l (penetration)

In lithium the outer electron moves in the field of the +3 nucleus and the two 1s electrons. Outside the core it feels about +1; inside the core it feels up to +3. The 2s orbital has a small inner lobe that penetrates the core; the 2p orbital, with its angular node at the nucleus, stays mostly outside.

2s spends more time inside the core ⇒ higher Z_eff ⇒ lower energy: E_2s(Li) < E_2p(Li)

(A) is correct.

On the bench: the orange dots (electron found inside the core) were far more frequent for Li 2s than for Li 2p, and the P(r) graph showed the 2s inner bump sitting in the shaded core.
Hydrogen against lithium

The orbital energy follows E ≈ −13.6 Z_eff²/n² eV. For H, Z_eff = 1. For the 2s electron of Li, the nuclear charge is 3 and the 1s² core does not shield it completely, so Z_eff > 1 (about 1.26 from the measured energy; about 1.3 by Slater-type estimates).

Z_eff(Li, 2s) > Z_eff(H, 2s) ⇒ E_2s(Li) < E_2s(H) ⇒ E_2s(H) > E_2s(Li)

(D) is correct. And since E_2p(H) = E_2s(H), also E_2p(H) > E_2s(Li) — the reverse of what (C) says. (C) is wrong.
Numbers

—

Measured (spectroscopy): H 2s = 2p = −3.40 eV; Li 2s = −5.39 eV (minus the ionisation energy); Li 2p = −3.54 eV (the 2s → 2p red line of lithium, 671 nm, is 1.85 eV). The simple core model misses exchange with the core, so it under-binds Li 2s a little — but every inequality in the options comes out the same.

E_2s(Li) < E_2p(Li) < E_2s(H) = E_2p(H)

Answer: —

Independent verification

Numerical Schrödinger equation. The radial equation was solved by Numerov shooting on a logarithmic grid, separately from the page: H 2s = H 2p = −3.401 eV (the exact −13.6/4); Li 2s = −4.76 eV, Li 2p = −3.43 eV in the field of the nucleus plus a 1s² core (Slater exponent Z − 5/16).
Spectroscopic data. H n = 2 levels −3.40 eV; Li ionisation energy 5.39 eV; Li resonance line 670.8 nm (1.85 eV) → Li 2p −3.54 eV. Same order; (A), (B), (D) true, (C) false.
Slater's rules. Li 2s: σ = 2 × 0.85 = 1.70, Z_eff = 1.30 → E ≈ −13.6 × 1.30²/4 = −5.75 eV < −3.40 eV.
Robustness. With the core removed (Z = 3, no 1s electrons) the 2s and 2p become degenerate again and (A) turns false; for Be⁺, B²⁺ … the same three options stay true.
Solution websites. MathonGo's published solution (their Q37) gives (A), (B), (D) with the same reasoning — degeneracy in H, penetration in Li, Z_eff(Li, 2s) ≈ 1.3 > 1.
This page. The energies come from the page's own solver and the true options are found by comparing them at run time; the answer is not written into the page.
Answer key. Checked afterwards against the official final key printed with the paper: Q5 → (A), (B), (D).

03How to use this simulation

For a student opening it for the first time. The whole experiment runs by itself after one press.

STEP 1

Read the question

Four energies — 2s and 2p of hydrogen, 2s and 2p of lithium. Which of the four statements about them are true? More than one can be.

STEP 2

Check the values

The question's atoms are loaded: hydrogen (Z = 1) and lithium (Z = 3, two 1s core electrons). You do not need to enter anything.

STEP 3

Optional: change a value

Change the second atom's nuclear charge or the number of core electrons. An empty box means "use the question value". The badge switches to ⚙ USING CUSTOM VALUES; ↺ RESET TO QUESTION VALUES puts everything back.

STEP 4

Start

Press ▶ START EXPERIMENT. There is nothing else to click — the experiment runs from beginning to end on its own (about 40 seconds; 2× makes it faster).

STEP 5

Watch

For each orbital the solver homes in on the energy, the electron cloud appears — orange dots are the moments the electron is inside the 1s core — and the energy drops onto the ladder. Then each option is tested against the ladder: ✓ or ✗.

STEP 6

Answer reveal

When the experiment finishes it says ✓ EXPERIMENT COMPLETE and reveals the correct options. The header answer box unlocks at the same time.

STEP 7

Understand

Scroll to Detailed solution. Each step points back to what you saw in the experiment.

STEP 8

Experiment again

Press ▶ REPLAY EXPERIMENT, or set the core electrons to 0 and press ⚙ RUN WITH CUSTOM VALUES — what happens to the gap between 2s and 2p? CLASSROOM MODE enlarges everything for a projector.

Classification

Exam	JEE Advanced 2026, Paper 1, Chemistry, Section 2, Q.5
Chapter	Atomic Structure
Topic	Orbital energies - degeneracy in hydrogen, penetration and shielding in lithium
Difficulty	Easy-Moderate
Answer	— run the experiment