Group 15 Electronegativity X & Y | JEE Main 2026 | Solved ✓

Element X and Y Belong to Group 15 – Electronegativity Difference JEE Main 2026 PYQ

Q: What are the electronegativity values of the Group 15 elements?

On the Pauling scale: N=3.04, P=2.19, As=2.18, Sb=2.05, Bi=2.02 (NCERT-rounded approximations: 3.0, 2.1, 2.0, 1.9, 1.9). Note the single big jump between N and P and the almost flat values from P to Bi.

Medium
⭐⭐⭐ (3/5)
· 31 Aug 2026
· JEE Main 2026 (23 Jan – Shift 2)

Nishant Kumar Gupta

Former Faculty · Allen · Aakash · Narayana

Chemistry mentor behind PadhoLikhoJEE and Quantum Chemistry Classes (Arrah, Bihar). Teaches the “why” behind every JEE trap — not just the trick.

Quick Summary

Question Type:: Trend application — comparing electronegativity differences
Chapter:: Classification of Elements – Periodic Trends (Group 15)
Difficulty:: ⭐⭐⭐ Medium
Time to Solve:: 2-3 minutes
Key Concept:: EN dives sharply from N to P, then plateaus (poor 3d/4f shielding)
Correct Answer:: (A) N and As
Why:: |EN(N) − EN(P)| = 0.85 is the only X-side gap that beats its P-side gap among the four official pairs

The Question

JEE Main 2026 (23 January – Shift 2) – Classification of Elements & Periodicity

Element X and Y belong to Group 15. The difference between the
electronegativity values of X and Phosphorus is higher than that of the difference
between Phosphorus and Y. X and Y are respectively:

(A) N and As ✓
(B) As and Bi
(C) Bi and N
(D) As and Sb

Quick Answer

Correct Option: (A) N and As

Reasoning: Electronegativity takes one massive dive between nitrogen and phosphorus, then barely moves down the rest of Group 15. Measuring every element’s gap from phosphorus settles it instantly:

Group 15 values (Pauling): N = 3.04, P = 2.19, As = 2.18, Sb = 2.05, Bi = 2.02
Gap above P: |EN(N) − EN(P)| = |3.04 − 2.19| = 0.85
Gaps below P: As = 0.01, Sb = 0.14, Bi = 0.17 — all tiny, because the values plateau
Only option (A) pairs a big X-side gap with a small P-side gap: 0.85 > 0.01 ✓ — (B), (C) and (D) all fail the inequality

Video Solution

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watch the video solution below:

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Understanding the Concept

Why Electronegativity Decreases Down a Group

Electronegativity is an atom’s power to pull the shared pair of a bond towards itself.
Two quantities decide that power: the effective nuclear charge (Z_eff) felt by the
bonding electrons, and their distance from the nucleus (the atomic radius). Moving down a group
adds a whole principal shell at every step, so radius grows and inner electrons shield the nucleus
more strongly. The net pull on bonding electrons drops, and electronegativity falls with it.

EN ∝ Z_eff / atomic radius

Z_eff = nuclear charge (Z) minus shielding by inner electrons — decides the pull strength
r = atomic radius — decides how far the bonding electrons sit from the nucleus
Down a group: r ↑ fast, shielding ↑ → EN ↓ (P = 2.19 is far below N = 3.04)

The Group 15 Plateau — Why the Trend “Flatlines” After Phosphorus

Group 15 hides a surprise. From arsenic onwards the expected decline almost stops:
the values read 2.19, 2.18, 2.05, 2.02 from P to Bi. The reason is poor shielding from the
inner d- and f-electrons. Arsenic (Period 4) is the first pnictogen with a filled
3d¹⁰ subshell, and bismuth (Period 6) carries a filled 4f¹⁴ subshell.
d- and f-electrons screen the nucleus badly, so Z_eff stays unusually high and cancels
most of the size increase. The result is an electronegativity plateau — exactly the
behaviour this PYQ tests. This is the “straight-line illusion” trap: students who picture a smooth
uniform slide down the group misjudge every gap below phosphorus.

The Key Principle

To solve any “difference from the anchor element” question of this family:

Write the actual scale values for the group — never argue from the trend arrow alone
Compute the absolute gap |ΔEN| of every candidate from the anchor (here, phosphorus)
Apply the given inequality option-wise, respecting the order “X and Y respectively”

Detailed Step-by-Step Solution

Step 1: List the Actual Electronegativity Values

N = 3.04 (Period 2 — the small, greedy outlier)
P = 2.19 (Period 3 — our anchor element)
As = 2.18, Sb = 2.05, Bi = 2.02 (Periods 4–6 — the plateau)

Step 2: Measure Each Element’s Gap from Phosphorus

Above P: |EN(N) − EN(P)| = |3.04 − 2.19| = 0.85
Below P: |EN(As) − EN(P)| = 0.01; |EN(Sb) − EN(P)| = 0.14; |EN(Bi) − EN(P)| = 0.17
Picture the shape: one cliff between N and P, then a flat plain — 0.85 ≫ 0.17

Step 3: Test the Condition on Every Option

The demand: |EN(X) − EN(P)| > |EN(P) − EN(Y)|, with X first and Y second:

(A) X = N, Y = As: 0.85 > 0.01 → ✓ satisfied
(B) X = As, Y = Bi: 0.01 > 0.17 → ✗ fails
(C) X = Bi, Y = N: 0.17 > 0.85 → ✗ fails (the reversed order — the “respectively” trap)
(D) X = As, Y = Sb: 0.01 > 0.14 → ✗ fails

Step 4: Cross-Check with the Trend Story

X must sit far from P on the electronegativity scale — only nitrogen does (the 0.85 cliff).
Y must sit close to P on the lower side — arsenic is the closest of all (0.01).
Both readings agree, and no other option survives Step 3.

Final Answer

Option (A): N and As ✓

Nitrogen lies 0.85 electronegativity units above phosphorus — the largest gap in the group —
while arsenic sits a mere 0.01 below it, so (A) is the only pair where the X-side difference
genuinely exceeds the P-side difference.

Essential Facts for This Topic

The Working Relationship

Electronegativity trend:
- Increases across a period (size ↓, Z_eff ↑); decreases down a group (size ↑, shielding ↑)
- Comparing two elements: |ΔEN| = |EN₁ − EN₂| — the number, not the arrow, answers JEE questions
Group 15 plateau logic:
- Filled 3d¹⁰ (As onwards) and 4f¹⁴ (Bi) subshells shield poorly
- Poor shielding keeps Z_eff high → EN nearly constant from P to Bi

Values You Must Memorise (Pauling Scale)

Group 15: N = 3.04, P = 2.19, As = 2.18, Sb = 2.05, Bi = 2.02
Gaps from phosphorus: N − P = 0.85 (cliff); P − As = 0.01, P − Sb = 0.14, P − Bi = 0.17 (plateau)
NCERT-rounded approximations: 3.0, 2.1, 2.0, 1.9, 1.9 — same story, same answer
Reference point: F = 3.98 is the most electronegative element; Cs ≈ 0.79 the least

Common Mistakes to Avoid

❌ Mistake 1: Trusting the Straight-Line Illusion

Wrong Thinking: “Electronegativity falls by roughly equal steps down every group,
so the P→Y gap must be about half of the N→P gap for arsenic.”

Correct Approach: Trend arrows show direction, not magnitude. Here the fall is one
0.85 cliff (N→P) followed by a 0.01–0.17 crawl (P→Bi). Always pull up the actual numbers.

❌ Mistake 2: Ignoring d- and f-Orbital Shielding

Wrong Thinking: “Size keeps increasing, so electronegativity must keep dropping
significantly from As to Bi.”

Correct Approach: The filled 3d¹⁰ subshell (from arsenic) and 4f¹⁴
subshell (at bismuth) shield the nucleus poorly, keeping Z_eff high and the values pinned
between 2.18 and 2.02.

❌ Mistake 3: Comparing Arrows Instead of Computing Differences

Wrong Thinking: “Nitrogen is above P and bismuth is far below, so X = N and Y = Bi
gives the biggest differences.”

Correct Approach: The question compares two differences. Subtract:
|N − P| = 0.85 versus |P − Bi| = 0.17. A pair qualifies only when the first number genuinely
beats the second — and the order “X first, Y second” must match the option.

❌ Mistake 4: Misreading “Respectively” in Reversed Pairs

Wrong Thinking: “The pair {Bi, N} contains both elements the question talks about,
so it must be right.”

Correct Approach: “X and Y respectively” fixes the roles: X is measured against P
first, Y second. Option (C) “Bi and N” assigns X = Bi (gap 0.17) and Y = N (gap 0.85) — the
inequality flips and fails. Order matters as much as values.

Key Concept Summary

What You Must Remember

The Group 15 value chain: N 3.04 → P 2.19 → As 2.18 → Sb 2.05 → Bi 2.02 — one cliff, then a plateau
N→P is the only big jump: a 0.85 crash caused by nitrogen’s uniquely small Period 2 atom
The plateau has a cause: filled 3d¹⁰ and 4f¹⁴ subshells shield poorly, so Z_eff stays high and EN barely falls from P to Bi
Differences decide, not directions: compute |ΔEN| for both sides of the anchor before judging any pair
“Respectively” locks the order: X is the element whose P-gap must be the larger one — check reversed pairs separately

The Golden Rule for Electronegativity Trends

“Down Group 15, electronegativity makes one big dive from N to P — then it crawls. Compute the gap; never trust the straight-line trend.”

Frequently Asked Questions

Q1: Why does electronegativity decrease down a group?

A: As we move down a group, a new principal shell is added at every element, so atomic size increases and the outer electrons sit farther from the nucleus. The greater distance plus stronger shielding by the inner electrons lowers the atom’s pull on shared bonding electrons, so electronegativity falls.

Q2: Why is the drop from nitrogen to phosphorus so much bigger than the drops after phosphorus?

A: Nitrogen is a tiny Period 2 element (EN 3.04), so moving to Period 3 phosphorus (2.19) adds a whole new shell and the value crashes by 0.85. After phosphorus, the poorly shielding 3d10 subshell (filled at arsenic) and 4f14 subshell (filled at bismuth) keep the effective nuclear charge high, so the values flatline between 2.18 and 2.02.

Q3: What are the electronegativity values of the Group 15 elements?

A: On the Pauling scale: N = 3.04, P = 2.19, As = 2.18, Sb = 2.05, Bi = 2.02 (NCERT-rounded approximations: 3.0, 2.1, 2.0, 1.9, 1.9). Note the single big jump between N and P and the almost flat values from P to Bi.

Q4: Which Group 15 element has the highest electronegativity and why?

A: Nitrogen (3.04) — the highest in the whole group and far above phosphorus. Its Period 2 atom is the smallest of the family, so its bonding electrons experience the strongest effective nuclear attraction.

Q5: From phosphorus to bismuth, how much does electronegativity actually change?

A: Very little — from 2.19 (P) to 2.02 (Bi), a total drop of only 0.17 across three periods. The individual steps are 0.01 (P→As), 0.13 (As→Sb) and 0.03 (Sb→Bi), which is why this region is called the electronegativity plateau of Group 15.

Prerequisites to Solve This Question

Before attempting this problem, you should understand:

Electronegativity basics: what the Pauling scale measures and why F (3.98) sits at the top
Periodic trend machinery: how atomic radius, shielding and Z_eff jointly decide the value
The Group 15 family: N, P, As, Sb, Bi sitting in Periods 2–6, with d- and f-block filling in between
Absolute-difference arithmetic: reading “the difference is higher than” as |ΔEN| comparisons

After Solving This, You Can:

Predict which Group 15 bonds are most polar using real EN gaps, not vague arrows
Explain the electronegativity plateau with d/f-block shielding in one line
Test any “difference is higher/lower than” pair inequality in under 30 seconds
Handle twin PYQs on ionisation enthalpy and covalency down Group 15 with the same plateau logic

Study Tips for This Topic

For JEE Main:

Memorise the chain as data, not a slope: 3.04 → 2.19 → 2.18 → 2.05 → 2.02. Ten seconds of recall kills this whole question family.
Tag the plateau as a keyword trigger: whenever a question contrasts an element with phosphorus, arsenic or bismuth, expect the flat region to be the trap.
Budget 60–90 seconds: write the five values, subtract from P, test the inequality option-wise — no derivation needed beyond subtraction.

Common JEE Variants:

“Correct order of electronegativity in Group 15” — answer N > P ≈ As > Sb ≈ Bi, quoting the values
Same plateau question on ionisation enthalpy or oxidising power down Group 15/16 — identical values-first method
Bond-polarity ranking (e.g. N–H vs P–H vs Bi–H) built from the same gap table
“Which pair shows anomalous behaviour and why?” — N/P and the d-block entry at As

Difficulty Rating & Exam Frequency

Difficulty Level: ⭐⭐⭐ (3/5) – Medium (the subtraction is easy; the trap is trusting the straight-line trend instead of the values — that is what earns the negative mark)

JEE Main Frequency: High — periodic trends give about 1 question per session, and Group 15/16 data-trend versions repeat almost every year

JEE Advanced Frequency: Medium — Advanced embeds the same plateau logic inside p-block and bond-analysis questions rather than asking it directly

Topic Importance: Very High — electronegativity drives bond polarity, acidity, oxidising power and half the p-block inorganic chemistry you will face

Written by Nishant Kumar Gupta
Former Faculty · Allen · Aakash · Narayana — Quantum Chemistry Classes, Arrah (Bihar)
Last Updated: 31 Aug 2026 ·
Question Source: JEE Main 2026 (23 January – Shift 2) Previous Year Question ·
Topic: Classification of Elements and Periodicity – Electronegativity Trends in Group 15