The trap you need to fix: When asked to arrange elements by atomic radius across a period, students include the noble gas and wonder why the "decreasing trend" breaks. It breaks because the noble gas radius reported in data tables is a van der Waals radius — measured from non-bonded contact distances — while every other element's radius is covalent (half the internuclear distance in a bonded pair). Comparing the two is like comparing shoe sizes measured in different systems.
The core concept (NCERT Class 11 Chemistry, Chapter 3, page 86):
Atomic radius is not a sharply defined quantity. Two operational definitions matter for NEET:
Noble gases (He, Ne, Ar…) don't form conventional covalent bonds under normal conditions. Their radii are van der Waals only.
Periodic trend — same radius type only:
Ionic radii — quick rules:
Watch-out for NEET: Any question asking you to compare atomic radii across a full period — check whether a noble gas is in the list. If yes, the comparison is invalid unless they specify the same radius type.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following correctly represents the trend in covalent radius across Period 3 (Na to Cl)?
Answer: D. Covalent radius decreases from left to right across a period due to increasing effective nuclear charge. Na has the largest covalent radius, Cl the smallest, in a monotonic decrease (NCERT Class 11 Chemistry, Chapter 3, page 86).
Why A is wrong: This is the reverse order (increasing). Covalent radius decreases across a period, so the largest atom is at the left (Na), not the right.
Why B is wrong: This reverses the trend. Increasing Z_eff across a period pulls electrons closer, so radius decreases left to right, not increases.
Why C is wrong: While the overall decreasing trend is correct for most, placing Cl > S at the end contradicts the steady decrease — S has a larger covalent radius than Cl.
Argon's atomic radius listed in data tables is larger than chlorine's. The correct explanation is:
Answer: B. Argon does not form a covalent bond under normal conditions, so its reported radius is van der Waals (non-bonded contact distance), which is always larger than the covalent radius used for Cl. The comparison is invalid because the measurement types differ (NCERT Class 11 Chemistry, Chapter 3, page 86).
Why A is wrong: Argon and chlorine are both in Period 3 — same number of shells (3). The size difference is not due to an extra shell.
Why C is wrong: Argon (Z = 18) has a higher nuclear charge than chlorine (Z = 17), not lower. This would predict a smaller radius, not larger.
Why D is wrong: There is no quantum-mechanical basis for 'expanded orbitals due to filled subshells.' The apparent larger size is simply a measurement-type artefact (van der Waals vs covalent).
Among the following isoelectronic species (10 electrons each), which has the smallest ionic radius?
Answer: C. In an isoelectronic series, all species have the same number of electrons (10 here). The one with the highest nuclear charge (Al³⁺, Z = 13) exerts the greatest pull on the electron cloud, giving the smallest radius. Order: O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺.
Why A is wrong: O²⁻ has Z = 8 pulling on 10 electrons — weakest pull among these, giving the largest radius, not the smallest.
Why B is wrong: F⁻ has Z = 9, larger than O²⁻ but much smaller nuclear charge than Al³⁺ (Z = 13). F⁻ is not the smallest in this series.
Why D is wrong: Na⁺ has Z = 11. While smaller than O²⁻ and F⁻, it still has a lower nuclear charge than Al³⁺, so its radius is larger than Al³⁺.
A cation is always smaller than its parent atom because:
Answer: C. When an atom loses electrons to form a cation, the nuclear charge (number of protons) stays constant but there are fewer electrons. Reduced electron-electron repulsion + same nuclear pull = electrons drawn closer = smaller radius.
Why A is wrong: Forming a cation removes electrons, not protons. The number of protons (nuclear charge) is unchanged.
Why B is wrong: A cation is still the same element in the same period. Losing an electron doesn't change its position in the periodic table.
Why D is wrong: Nuclear charge (proton count) never changes during ion formation. Only the electron count changes.
Which of the following statements about van der Waals radius is correct?
Answer: B. Van der Waals radius is defined as half the shortest distance between nuclei of two adjacent non-bonded atoms (e.g., in a molecular crystal). This is always larger than the covalent radius because bonded atoms are pulled closer than non-bonded ones (NCERT Class 11 Chemistry, Chapter 4, page 107).
Why A is wrong: Van der Waals radius is always larger than covalent radius for the same element, not smaller. Non-bonded contact distance exceeds bonded internuclear distance.
Why C is wrong: This describes the covalent radius (half the bond length in a homonuclear diatomic), not the van der Waals radius. Trap: confusing the two definitions.
Why D is wrong: Van der Waals radius applies to non-metallic and noble gas elements primarily. Metals typically use metallic radius. It is not restricted to metals.
In the isoelectronic pair N³⁻ and F⁻ (both with 10 electrons), which has the larger ionic radius and why?
Answer: A. N³⁻ has Z = 7 pulling on 10 electrons; F⁻ has Z = 9 pulling on 10 electrons. Lower nuclear charge means weaker pull and a more diffuse electron cloud, so N³⁻ is larger.
Why B is wrong: Electronegativity describes tendency to attract bonding electrons — it doesn't determine ionic radius in an isoelectronic comparison. The governing factor is nuclear charge vs electron count.
Why C is wrong: Same electron count does not mean same radius. The nuclear charge differs (Z = 7 vs Z = 9), and higher Z compresses the electron cloud more. They are not equal.
Why D is wrong: The number of electrons gained doesn't directly determine relative size in an isoelectronic pair. What matters is the resulting nuclear charge vs total electron count. Both have 10 electrons; the one with fewer protons (N³⁻) is larger.
Consider the species: Ne, Na⁺, and F⁻. All have 10 electrons. The correct order of their radii is:
Answer: A. F⁻ (Z = 9, 10 e⁻) has the lowest nuclear charge among the ions/atom, so largest radius. Na⁺ (Z = 11, 10 e⁻) has the highest nuclear charge, so smallest. Ne (Z = 10, 10 e⁻) falls between. However, Ne's reported radius is van der Waals (not ionic), which complicates direct comparison. In a strict isoelectronic ionic comparison, F⁻ > Na⁺. When Ne's van der Waals radius (~154 pm) is included, it falls between F⁻ (~133 pm ionic) and Na⁺ (~95 pm ionic) — but note this mixes radius types. The standard NCERT-level answer for the isoelectronic-including-noble-gas ordering is F⁻ > Ne > Na⁺ based on decreasing Z pulling on the same electron cloud.
Why B is wrong: This places Ne as the largest, but Ne with Z = 10 has more nuclear pull than F⁻ with Z = 9 on the same 10 electrons. F⁻ should be largest.
Why C is wrong: This places Na⁺ > Ne, but Na⁺ has Z = 11 (highest pull) making it the smallest, not intermediate.
Why D is wrong: This reverses the correct trend entirely. Na⁺ with Z = 11 has the strongest pull on 10 electrons and is the smallest, not the largest.
An element X in Period 2 has a covalent radius of 77 pm. It forms an anion X⁴⁻. Compared to the neutral atom, the anion's radius will be:
Answer: D. Gaining electrons increases electron-electron repulsion. The nuclear charge (6 protons for carbon, which has ~77 pm covalent radius) remains the same but must now hold 10 electrons. The electron cloud expands, making the anion larger than the neutral atom.
Why A is wrong: Gaining electrons does not increase effective nuclear charge. The nuclear charge stays constant; it's the electron count that rises, reducing Z_eff per electron and expanding the cloud.
Why B is wrong: Noble gas configuration does not mean 'compact.' The stability is electronic (full shell), not geometric. An anion with noble gas configuration is still larger than its parent atom because extra electrons expand the cloud.
Why C is wrong: While the proton count is unchanged, the electron count has increased from 6 to 10. More electrons with the same nuclear pull means greater repulsion and a larger radius — not the same size.
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Given
Five isoelectronic species, each with 10 electrons:• N³⁻ (Z = 7)• O²⁻ (Z = 8)• F⁻ (Z = 9)• Na⁺ (Z = 11)• Mg²⁺ (Z = 12)
Required
Arrange in order of increasing ionic radius.
Concept
In an isoelectronic series, all species have the same number of electrons. The species with the highest nuclear charge (Z) pulls the electron cloud most tightly → smallest radius. Radius decreases as Z increases.
Formula
No formula needed — this is a direct application of the isoelectronic radius rule: for constant electron count, radius ∝ 1/Z (qualitative inverse relationship).
Substitution
Order by increasing Z: N³⁻ (7) < O²⁻ (8) < F⁻ (9) < Na⁺ (11) < Mg²⁺ (12).
Since radius decreases with increasing Z, the radius order is the reverse.
Calculation
Increasing radius = decreasing Z order:
Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻
Published values (pm): Mg²⁺ (72) < Na⁺ (95) < F⁻ (136) < O²⁻ (140) < N³⁻ (146).
Final answer
Order of increasing ionic radius: Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻
Note: Z values are exact integers (counting numbers) and do not affect any precision consideration.
Common trap
Students sometimes include Ne (Z = 10) in this series. But Ne's reported radius is van der Waals, not ionic — mixing radius types gives a misleading comparison (Trap: trap: atomic radius inert gas applied to the isoelectronic context).
Similar NEET-style question
"Arrange the following isoelectronic species in order of decreasing radius: Al³⁺, Na⁺, F⁻, O²⁻, N³⁻."
(Same principle — order by increasing Z to get decreasing radius.)
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Across period: atomic radius decreases (effective nuclear charge increases). Down group: atomic radius increases (additional shells). Cation < neutral atom < anion (for same element).
-- NCERT Class 11 Chemistry, Ch. 3, p. 86These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Student includes inert-gas radius in atomic-radius trends. But inert gases use van der Waals radius (much larger than covalent), making 'monotonic decrease across period' look broken.
Atomic radius comparison includes a noble gas or trends across period 2/3.
Compare like with like: covalent radii for non-noble gases. Noble gas radii are van der Waals (no covalent bond). Don't compare noble-gas radius directly to halogen.
Root cause: concept gap
Don't compare different radius types. Noble gases use vdW radius (much larger); halogens use covalent radius. Compare like-with-like.
More in Classification of Elements and Periodicity in Properties: 2 exam traps and mistakes · 1 formula from its other lessons.
2 questions from NEET 2023, 2025. Answers verified against NTA official keys.
The element expected to form largest ion to achieve the nearest noble gas configuration is
All 10 past-paper questions from Classification of Elements and Periodicity in Properties →
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
swapped classes
Tempts surface-level recall.
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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