ADV-2026-P2-CHE-Q07 · Periodic Properties Virtual Lab

01Original question — verbatim

The correct statement(s) regarding the periodic properties of elements is(are)

(A)

Second ionization enthalpy of carbon atom is less than that of boron atom.
(B)

Increasing order of ionic radii: Al³⁺ < Mg²⁺ < Na⁺
(C)

Under identical conditions, in solid state, the density of potassium metal is more than density of sodium metal.
(D)

The H–H bond is weaker than F–F bond.

Multiple options may be correct (+4 / partial / −1). Four separate claims about four different periodic trends — each one has its own instrument below.

02Ionisation bench — strip electrons one at a time

Pick an element, then pull electrons off it one by one and read the energy each pull costs. The orbital the electron leaves from is shown as you go — that is where the answer to (A) lives.

Choose an element

Period 2 is live. Boron and carbon are the pair the question asks about.

Electrons remaining

Successive ionisation energies

selected element comparison element

Compare against

IE₁

—

kJ mol⁻¹

IE₂

—

kJ mol⁻¹

IE₂ − IE₁

—

kJ mol⁻¹

2nd electron leaves

—

orbital

Pull two electrons off boron and off carbon, then compare.

03Isoelectronic bench — same electrons, different nucleus

These three ions all carry exactly ten electrons. Only the nuclear charge changes. Drag the proton count and watch what happens to the size of the cloud those ten electrons make.

Ten electrons, variable nucleus

Protons in the nucleus Z 11

O²⁻ (8) · F⁻ (9) · Ne (10) · Na⁺ (11) · Mg²⁺ (12) · Al³⁺ (13) — all with 10 electrons

Side by side

Move the slider across all three of Na⁺, Mg²⁺ and Al³⁺.

04Bulk bench — density and bond strength

The last two claims are about bulk matter, not single atoms. Weigh equal volumes of the alkali metals, then pull two diatomic molecules apart and read the work it took.

Weigh one cubic centimetre

Weigh sodium and potassium and compare.

Bond-breaking rig

Pull the atoms apart 0 %

Break H–H, then break F–F, and compare the work.

05Verdict on each statement

Each statement is marked from the instrument that tests it. Nothing here is looked up.

Lab notebook

Written as you work. Only measurements you actually made appear here.

Nothing recorded yet — work through the three benches above.

Predict, then check

06Step-by-step solution

The solution is still locked

Test all four statements on the bench above and the derivation opens by itself.
Statements tested: 0 of 4 — ionise B and C, sweep the isoelectronic series, weigh Na against K and break both bonds

(A) Second ionisation enthalpy — look at what is left behind

The second ionisation removes an electron from the cation, so write those configurations, not the neutral atoms'.

B⁺ : 1s² 2s² · C⁺ : 1s² 2s² 2p¹

B⁺ has a filled, strongly penetrating 2s subshell — a stable arrangement that resists losing another electron. C⁺ only has to give up a lone 2p electron, which is higher in energy and better shielded.

IE₂(C) = 2352.6 kJ mol⁻¹ < IE₂(B) = 2427.1 kJ mol⁻¹

Note the reversal: IE₁ rises across the period (B 800.6 < C 1086.5), but IE₂ falls. (A) is correct.
(B) Isoelectronic ions — only the nucleus changes

Na⁺, Mg²⁺ and Al³⁺ each have exactly 10 electrons. With the electron count fixed, the shielding is essentially fixed too, so the effective nuclear charge tracks the proton count.

Z: Na⁺ 11 < Mg²⁺ 12 < Al³⁺ 13 ⟹ radius: Al³⁺ < Mg²⁺ < Na⁺

102 pm (Na⁺) · 72 pm (Mg²⁺) · 53.5 pm (Al³⁺)

More protons pulling the same ten electrons means a smaller ion. The printed order is exactly that. (B) is correct.
(C) Density down group 1 — the one place the trend breaks

Density is mass per unit volume, and both rise going down the group. Usually mass wins and density rises. Between sodium and potassium it does not: the atomic volume jumps more than the atomic mass does.

Li 0.534 · Na 0.968 · K 0.856 · Rb 1.532 · Cs 1.879 g cm⁻³

Potassium is less dense than sodium — the only inversion in the group, and the reason both float on water while the heavier members do not. The statement claims the opposite. (C) is incorrect.
(D) H–H against F–F — not close

H–H = 436 kJ mol⁻¹ · F–F = 158.8 kJ mol⁻¹

H–H is one of the strongest single bonds there is: two 1s orbitals overlapping with nothing else in the way. F–F is anomalously weak — the two fluorines are small, so the three lone pairs on each sit close enough to repel strongly, and there are no d orbitals to relieve it. F–F is even weaker than Cl–Cl (242) and Br–Br (193).

The statement has it backwards by a factor of nearly three. (D) is incorrect.
Answer

(A) and (B)

Marking: +4 for choosing both; +2 for either one alone; −1 for including (C) or (D).

✓ Independent verification

Ionisation reversal is real, not a rounding artefact: the gap is 2427.1 − 2352.6 = 74.5 kJ mol⁻¹, far outside any measurement uncertainty, and it is explained by the configurations of the cations rather than the atoms.
Isoelectronic check: all three ions were confirmed to hold exactly 10 electrons before their radii were compared — the trend statement is only meaningful within an isoelectronic set.
Density is measured, not asserted: the bench weighs a fixed 1 cm³ of each metal, so the Na/K inversion appears as an actual reading rather than a remembered fact.
Bond enthalpies checked across the whole halogen series: Cl–Cl 242 > Br–Br 193 > F–F 158.8 > I–I 151 kJ mol⁻¹. Fluorine sits out of place — weaker than both chlorine and bromine despite being the smallest — which is the anomaly (D) trips over.
Distractor audit: (C) is believable because density does rise down almost every group; (D) is believable if you assume a small atom always makes a strong bond. Both are correct-sounding generalisations with a famous exception, which is what makes them good distractors.
Cross-checked against a published answer key for this paper after the derivation was complete — the key agrees on (A) and (B).

07Concepts and classification

Relations used

IE₂ removes an electron from the cation, not the atom

filled subshell in the cation ⟹ large IE₂

isoelectronic: radius falls as Z rises

ρ = mass / volume — both rise down group 1, volume faster at K

small atom + many lone pairs ⟹ weak single bond (F–F)

Z_eff = Z − σ

Classification

Simulation ID	ADV-2026-P2-CHE-Q07
Exam / Year / Paper	JEE Advanced · 2026 · Paper 2
Subject	Chemistry (Inorganic)
Chapter	Classification of Elements and Periodicity
Topic	Periodic trends in physical properties
Subtopics	Successive ionisation enthalpies · Isoelectronic ionic radii · Densities of the alkali metals · Bond dissociation enthalpy of the halogens · Effective nuclear charge
Question type	Multiple correct MCQ (+4 / partial / −1)
Difficulty	Moderate
Answer	— run the experiment

01Original question — verbatim

02Ionisation bench — strip electrons one at a time

Choose an element

Electrons remaining

Successive ionisation energies

03Isoelectronic bench — same electrons, different nucleus

Ten electrons, variable nucleus

Side by side

04Bulk bench — density and bond strength

Weigh one cubic centimetre

Bond-breaking rig

05Verdict on each statement

Lab notebook

Predict, then check

06Step-by-step solution

The solution is still locked

(A) Second ionisation enthalpy — look at what is left behind

(B) Isoelectronic ions — only the nucleus changes

(C) Density down group 1 — the one place the trend breaks

(D) H–H against F–F — not close

Answer

✓ Independent verification

07Concepts and classification

Relations used

Classification