Covalent character of ionic bond increases with: small cation, large anion, high charge on either ion, cation with d-electrons (pseudo-noble-gas configuration).
-- NCERT Class 11 Chemistry, Ch. 4, p. 112Electronegativity Fajans'
8 MCQs2 revision cards9-step worked example
Source: NCERT Chemical Bonding and Molecular StructurePYQ coverage: NEET 2021Official key: NTA-verifiedLast updated: 19 Sep 2026
Electronegativity Fajans', explained for NEET
Electronegativity is the tendency of a bonded atom to attract shared electron pairs toward itself. Pauling's scale (NCERT Class 11 Chemistry Chapter 4, page 107) assigns fluorine the highest value (4.0). Across a period, electronegativity increases (rising nuclear charge, shrinking radius); down a group, it decreases (more shielding, larger radius).
A common confusion: treating electronegativity and electron affinity as interchangeable. Electronegativity is a bond property — it describes pull on shared electrons within a molecule. Electron affinity is an isolated-atom property — the energy change when a gaseous atom gains an electron.
Fajan's rules predict when an ionic bond develops covalent character. The central idea: a small, highly charged cation polarises a large, highly charged anion, distorting its electron cloud toward the cation. This polarisation introduces covalent character into what would otherwise be a purely ionic bond.
Conditions favouring covalent character (Fajan's rules):
- Small cation with high charge (high charge density → strong polarising power).
- Large anion with high charge (easily polarisable electron cloud).
- Cation with pseudo-noble-gas configuration (e.g., Cu⁺, Ag⁺ with 18-electron core) polarises more than a noble-gas-configuration cation of similar size and charge.
The dipole moment formula μ = q × d (NCERT Class 11 Chemistry Chapter 4) connects here: the greater the electronegativity difference in a diatomic bond, the larger the partial charge q and hence the larger μ. For polyatomic molecules, individual bond dipoles are summed as vectors — symmetric molecules like CO₂ and BF₃ have zero net dipole despite having polar bonds.
Watch out: NEET questions frequently test whether you can distinguish polarising power (cation property) from polarisability (anion property), and whether you remember that pseudo-noble-gas cations polarise more than noble-gas cations of comparable size.
Can you answer these Electronegativity Fajans' MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which of the following elements has the highest electronegativity on the Pauling scale?
Show answer and why every option is right or wrong
Answer: C. Fluorine has the highest electronegativity (4.0) on the Pauling scale, as stated in NCERT Class 11 Chemistry Chapter 4, page 107. No other element exceeds this value.
Why A is wrong: A is wrong because oxygen (3.5) has a high electronegativity but is second to fluorine (4.0).
Why B is wrong: B is wrong because chlorine (3.0) is in the same group as fluorine but lower in the group, so its electronegativity is lower due to increased atomic size and shielding.
Why D is wrong: D is wrong because nitrogen (3.0) has lower electronegativity than fluorine; both are in period 2 but fluorine has higher nuclear charge with similar shielding.
MCQ 2Easy RecallPractice
Electronegativity of elements generally increases:
Show answer and why every option is right or wrong
Answer: C. Across a period (left to right), increasing nuclear charge pulls bonding electrons more strongly. Up a group (bottom to top), smaller atomic size and less shielding increase the pull on shared electrons. This is the standard periodic trend from NCERT Class 11 Chemistry Chapter 4.
Why A is wrong: A is wrong because electronegativity decreases down a group (increasing atomic radius and shielding reduce pull on shared electrons), not increases.
Why B is wrong: B is wrong because electronegativity increases from left to right across a period (not right to left), as nuclear charge increases across the period.
Why D is wrong: D is wrong on both counts: electronegativity increases left to right across a period and decreases (not increases) down a group.
MCQ 3Easy RecallPractice
Fajan's rules predict the degree of:
Show answer and why every option is right or wrong
Answer: B. Fajan's rules specifically address how much covalent character develops in a compound that is nominally ionic, based on the polarising power of the cation and polarisability of the anion (NCERT Class 11 Chemistry Chapter 4).
Why A is wrong: A is wrong because Fajan's rules address bond character in ionic compounds, not the metallic character of elements, which is a separate periodic property.
Why C is wrong: C is wrong because Fajan's rules start from an ionic compound and assess how much covalent character it gains — not the reverse direction. Ionic character in a covalent compound is assessed by electronegativity difference, not Fajan's rules.
Why D is wrong: D is wrong because while covalent character may indirectly affect conductivity, Fajan's rules predict bond character (covalent character in ionic compounds), not electrical conductivity directly.
MCQ 4Direct ApplicationPractice
Among LiF, LiCl, LiBr, and LiI, which lithium halide has the most covalent character according to Fajan's rules?
Show answer and why every option is right or wrong
Answer: D. The cation (Li⁺) is the same in all four. By Fajan's rules, the larger the anion, the more easily it is polarised. I⁻ is the largest halide anion, so LiI has the most covalent character.
Why A is wrong: A is wrong because F⁻ is the smallest halide ion, making it the hardest to polarise. LiF has the least covalent character in this series.
Why B is wrong: B is wrong because Cl⁻ is smaller than both Br⁻ and I⁻, so LiCl has less covalent character than LiBr or LiI.
Why C is wrong: C is wrong because although Br⁻ is larger than Cl⁻ and F⁻, it is still smaller than I⁻. LiBr has less covalent character than LiI.
MCQ 5Direct ApplicationPractice
Which of the following cations has the greatest polarising power?
Show answer and why every option is right or wrong
Answer: C. Polarising power increases with higher cation charge and smaller ionic radius. Al³⁺ has the highest charge (+3) and the smallest radius among these isoelectronic-period cations, giving it the greatest charge density and hence polarising power.
Why A is wrong: A is wrong because Na⁺ has a charge of only +1 and a larger radius than Mg²⁺ or Al³⁺, resulting in low charge density and weak polarising power.
Why B is wrong: B is wrong because although Mg²⁺ has a higher charge than Na⁺ and K⁺, Al³⁺ has both a higher charge (+3 vs +2) and a smaller radius, making Al³⁺ a stronger polariser.
Why D is wrong: D is wrong because K⁺ has the largest ionic radius in this set and only a +1 charge, giving it the lowest charge density and weakest polarising power.
MCQ 6Direct ApplicationPractice
The dipole moment of CO₂ is zero despite C=O bonds being polar. This is because:
Show answer and why every option is right or wrong
Answer: B. CO₂ is a linear molecule (O=C=O, 180°). The two equal C=O bond dipoles point in opposite directions and cancel vectorially, resulting in a net dipole moment of zero. This is the standard vector-sum argument from NCERT Class 11 Chemistry Chapter 4.
Why A is wrong: A is wrong because carbon does have a non-zero electronegativity (~2.5 on the Pauling scale). Each C=O bond is indeed polar; the net cancellation arises from geometry, not from carbon having zero electronegativity.
Why C is wrong: C is wrong because oxygen is more electronegative than carbon and attracts electron density away from carbon, not toward it. The zero net dipole is due to symmetric vector cancellation, not electron donation by oxygen.
Why D is wrong: D is wrong because double bonds absolutely can produce dipole moments (e.g., C=O in formaldehyde has a significant dipole). The zero net dipole in CO₂ is due to the symmetric linear geometry, not the bond order.
MCQ 7Concept TrapPractice
Cu⁺ (ionic radius 77 pm) and Na⁺ (ionic radius 102 pm) both have a +1 charge. According to Fajan's rules, CuCl is expected to have more covalent character than NaCl. What is the additional reason beyond size?
Show answer and why every option is right or wrong
Answer: B. Cu⁺ has an 18-electron configuration ([Ar] 3d¹⁰), known as a pseudo-noble-gas configuration. The d-electrons provide poor shielding of the nuclear charge, so the effective nuclear charge experienced by the anion's electron cloud is higher than for a noble-gas-configuration cation of similar size. This makes Cu⁺ a stronger polariser than Na⁺, giving CuCl more covalent character.
Why A is wrong: A is wrong because Cu⁺ does NOT have a noble gas configuration. It has [Ar] 3d¹⁰, which is a pseudo-noble-gas (18-electron) configuration. The distinction matters: the filled d-shell provides poor shielding, which is precisely why Cu⁺ polarises more strongly.
Why C is wrong: C is wrong because Na⁺ has the configuration [He] 2s² 2p⁶ (noble gas type) with no d-orbitals filled. It is Cu⁺ that has filled d-orbitals, and those d-electrons shield poorly, which increases (not decreases) polarising power.
Why D is wrong: D is wrong because metallic bonding is a property of bulk metals, not of ionic compounds like CuCl. The additional covalent character in CuCl arises from the polarising power of the Cu⁺ cation, not from metallic bonding.
MCQ 8CalculationPractice
Arrange the following in order of increasing covalent character: NaCl, MgCl₂, AlCl₃.
Show answer and why every option is right or wrong
Answer: B. All three have Cl⁻ as the anion (same polarisability). Covalent character depends on the cation's polarising power: higher charge and smaller radius → greater charge density → more polarisation. Na⁺ (+1, largest) < Mg²⁺ (+2, smaller) < Al³⁺ (+3, smallest). Therefore covalent character: NaCl < MgCl₂ < AlCl₃. This is a two-step reasoning: first rank cation charge density, then apply Fajan's rule.
Why A is wrong: A is wrong because this is the reverse order. Al³⁺ has the highest charge and smallest radius among these cations, giving it the greatest polarising power and AlCl₃ the most covalent character — not the least.
Why C is wrong: C is wrong because NaCl must have the least covalent character (Na⁺ has the lowest charge and largest radius). Placing MgCl₂ below NaCl contradicts the trend of increasing cation charge density.
Why D is wrong: D is wrong because Al³⁺ has a higher charge (+3) and smaller radius than Mg²⁺ (+2). AlCl₃ therefore has more covalent character than MgCl₂, not less.
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Electronegativity Fajans': quick recall before you leave
How do you solve a Electronegativity Fajans' question? A worked example
- 1
Given
HF has a bond length of 0.92 Å (9.2 × 10⁻¹¹ m) and an observed dipole moment of 1.91 D.
HCl has a bond length of 1.27 Å (1.27 × 10⁻¹⁰ m) and an observed dipole moment of 1.03 D. - 2
Required
Explain why HF has a larger dipole moment than HCl despite having a shorter bond length d.
- 3
Concept
Dipole moment μ = q × d. A larger d alone would increase μ, yet HF (shorter d) has a larger μ than HCl. This means the partial charge q on HF must be significantly larger, overriding the shorter distance.
- 4
Formula
μ = q × d, therefore q = μ / d.
- 5
Substitution
For HF: q = 1.91 D / 0.92 Å (in relative terms)
For HCl: q = 1.03 D / 1.27 Å (in relative terms) - 6
Calculation
Relative charge density comparison:• HF: 1.91 / 0.92 ≈ 2.08 D/Å• HCl: 1.03 / 1.27 ≈ 0.81 D/Å
The ratio q(HF)/q(HCl) ≈ 2.08/0.81 ≈ 2.6. Fluorine's much higher electronegativity (4.0 vs chlorine's 3.0) pulls electron density far more strongly, creating a partial charge roughly 2.6 times larger in HF.
Note: The bond lengths and dipole moments are given data — they are treated as exact for this comparison. The ratio calculation uses them directly. - 7
Final answer
HF has a larger dipole moment than HCl because fluorine's greater electronegativity (4.0 vs 3.0) creates a much larger partial charge q, which more than compensates for the shorter bond length. The electronegativity difference dominates the μ = q × d product.
- 8
Common trap
Students sometimes assume that longer bond length automatically means larger dipole moment (since d appears in μ = qd). This ignores the charge term. When electronegativity difference changes significantly between two molecules, q can dominate over d.
- 9
Similar NEET-style question
"Among HF, HCl, HBr, and HI, which has the highest dipole moment? Justify using electronegativity and the dipole moment formula."
Answer: HF — highest electronegativity difference (F = 4.0, H = 2.1) produces the largest partial charge, outweighing the effect of bond length.
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What to remember before solving Electronegativity Fajans' questions
More in Chemical Bonding and Molecular Structure: 4 exam traps and mistakes · 2 formulas · 3 question patterns from its other lessons.
Electronegativity Fajans' questions from past NEET papers
1 question from NEET 2021. Answers verified against NTA official keys.
NEET 2021
1Beryllium chloride
2Calcium chloride
3Strontium chloride
4Magnesium chloride
NTA Answer: Option 1(final)
All 18 past-paper questions from Chemical Bonding and Molecular Structure →
Sources
NCERT refs: Class 11 Chemistry Chapter 4, p.107
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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