Inorganic chemistry in JEE Main isn’t really about memorizing rules — it’s about knowing exactly where those rules break. Chromium refuses to follow the Aufbau principle. Fluorine acts weaker than it should. Beryllium behaves more like aluminium than like its own family. This quiz puts 15 of the most commonly tested exceptions in front of you, one at a time, so you can find your weak spots before exam day does.
⚡ Quick Answer
The exceptions JEE Main tests most often fall into four buckets: anomalous electronic configurations (Cr, Cu, and the d10 metals Zn/Cd/Hg), diagonal relationships (Li–Mg, Be–Al, B–Si), first-member anomalies in each p-block group (N, O, F behaving unlike their heavier siblings), and acid-base/oxidation-state irregularities (like HF being the weakest hydrohalic acid despite fluorine’s high electronegativity). Master these four categories and most “exception” questions become predictable rather than surprising.
📋 TL;DR
- Cr and Cu adopt [Ar]3d⁵4s¹ and [Ar]3d¹⁰4s¹ instead of the “expected” configuration, for extra stability from half-filled/fully-filled d-orbitals.
- Li and Be behave more like Mg and Al (diagonal relationship) than like the rest of their own groups, due to similar charge/size ratios.
- Fluorine and oxygen are anomalous within their own groups because they lack accessible d-orbitals and have unusually small atomic size.
- HF is the weakest acid among the hydrohalic acids (HF, HCl, HBr, HI) despite fluorine’s extreme electronegativity, because of very strong H–F bond enthalpy.
- Zn, Cd, and Hg are technically not transition metals, since their d-orbitals are completely filled (d¹⁰) in both the element and common ions.
- This quiz uses collapsible answers — try each question honestly before revealing the explanation.
Why These Exceptions Trip Students Up
- Most periodic trends are taught as clean, predictable rules — exceptions feel like the “unfair” part of the syllabus
- JEE Main deliberately tests exceptions because they separate students who memorized trends from students who understand the underlying reasons (electron-electron repulsion, orbital stability, size effects)
- The same handful of exceptions reappear year after year in slightly different question formats
- Knowing the reason behind an exception lets you answer variations you haven’t seen before, instead of just recalling a memorized fact
Section 1: s-Block Exceptions and Diagonal Relationships
Q1. Lithium shows a diagonal relationship with which element, rather than behaving like the rest of Group 1?
A) Sodium B) Magnesium C) Beryllium D) Aluminium
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Answer: B) Magnesium. Lithium’s small size and high charge density give it a charge/radius ratio close to magnesium’s, so Li shows Mg-like behavior — forming a covalent, thermally unstable carbonate/nitride and reacting directly with nitrogen, unlike other alkali metals.
Q2. Beryllium differs sharply from the rest of the alkaline earth metals mainly because:
A) It has a fully filled d-subshell B) Its ionization energy is the lowest in the group C) Its small size and high polarizing power give its compounds significant covalent character D) It doesn’t form oxides
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Answer: C. Beryllium’s tiny ionic radius gives it very high polarizing power (Fajans’ rule), making BeCl₂ covalent and even soluble in organic solvents — a hallmark of its diagonal relationship with aluminium rather than typical alkaline-earth behavior.
Q3. Which oxide is amphoteric due to the diagonal relationship between boron and silicon?
A) B₂O₃ B) SiO₂ C) Both behave similarly in showing weakly acidic/network covalent character D) Neither forms a covalent oxide
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Answer: C. Boron and silicon both form acidic, network-covalent oxides (B₂O₃ and SiO₂) rather than the ionic oxides typical of their broader periodic neighbors — a classic diagonal-relationship signature.
Section 2: p-Block First-Member Anomalies
Q4. Nitrogen, unlike phosphorus, does not form a stable N₅⁺-type extended chain or N₂O₅-analogous catenated network mainly because:
A) Nitrogen has no valence electrons available B) Nitrogen’s small size causes strong lone-pair repulsion, weakening N–N single bonds relative to its very strong N≡N triple bond C) Nitrogen cannot expand its octet D) Nitrogen is more electronegative than phosphorus
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Answer: B. N–N single bonds are comparatively weak due to lone-pair repulsion between small nitrogen atoms, so nitrogen strongly prefers existing as N₂ with a robust triple bond rather than catenating like phosphorus (which forms P₄ and longer chains).
Q5. Why is the bond angle in NH₃ (~107°) larger than in PH₃ (~93°)?
A) Nitrogen is more electronegative, pulling bonding pairs closer and increasing repulsion between them B) Phosphorus has more lone pairs C) NH₃ is planar while PH₃ is pyramidal D) There is no real difference
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Answer: A. Nitrogen’s higher electronegativity pulls bonding electron density closer to itself, increasing bond-pair/bond-pair repulsion and widening the angle compared to the larger, less electronegative phosphorus in PH₃.
Q6. Oxygen is considered anomalous within Group 16 mainly because it:
A) Cannot expand its octet due to the absence of accessible d-orbitals B) Is the least electronegative element in the group C) Doesn’t form double bonds D) Has the largest atomic radius in the group
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Answer: A. Unlike sulfur and its heavier congeners, oxygen has no accessible d-orbitals in its valence shell, so it cannot show variable covalency beyond two, and forms strong pπ–pπ multiple bonds (as in O=O) rather than the extended single-bond networks seen with sulfur.
Q7. Among the hydrohalic acids (HF, HCl, HBr, HI), which is the weakest acid in water, despite fluorine being the most electronegative halogen?
A) HCl B) HBr C) HI D) HF
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Answer: D) HF. The H–F bond is exceptionally strong (short bond length, high bond enthalpy) and HF also shows extensive hydrogen bonding, both of which oppose dissociation in water — making HF the weakest acid of the four despite fluorine’s extreme electronegativity.
Q8. The F–F bond in F₂ is unexpectedly weak compared to Cl–Cl. What’s the main reason?
A) Fluorine’s small size causes strong lone-pair/lone-pair repulsion between the two atoms B) Fluorine has fewer electrons C) Fluorine is less electronegative D) Fluorine has a larger atomic radius than chlorine
Show Answer
Answer: A. Fluorine’s very small atomic size forces the non-bonding lone pairs on each atom close together, creating strong repulsion that weakens the F–F bond relative to the larger Cl–Cl bond, where lone pairs are further apart.
Section 3: d-Block Configuration and Property Exceptions
Q9. Chromium’s actual ground-state electronic configuration is [Ar]3d⁵4s¹ instead of the “expected” [Ar]3d⁴4s². Why?
A) A half-filled d-subshell provides extra exchange energy and symmetry, making it more stable B) Chromium has no 4s electrons C) It’s simply an experimental error corrected later D) Chromium follows Hund’s rule differently from other elements
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Answer: A. A half-filled 3d⁵ configuration has extra stability from exchange energy and spherical symmetry, so one 4s electron shifts into the 3d subshell to reach this more stable arrangement.
Q10. Copper’s ground-state configuration is [Ar]3d¹⁰4s¹ rather than [Ar]3d⁹4s². This is explained by:
A) A completely filled 3d¹⁰ subshell is exceptionally stable B) Copper has 11 valence electrons only in excited states C) Copper never loses 4s electrons D) This configuration was only observed in Cu²⁺, not Cu
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Answer: A. A fully filled 3d¹⁰ subshell offers extra stability (symmetry and exchange energy), so copper “borrows” one 4s electron to complete its d-subshell, just as chromium does to half-fill it.
Q11. Zinc, cadmium, and mercury are usually excluded from the “true” transition metals. Why?
A) They don’t form colored compounds B) Their d-subshell is completely filled (d¹⁰) in both the element and their common ions, so they can’t use partially filled d-orbitals the way transition metals do C) They aren’t metals at all D) They only exist in the +1 oxidation state
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Answer: B. By the standard IUPAC definition, a transition element must have a partially filled d-subshell in at least one common oxidation state. Zn, Cd, and Hg have a full d¹⁰ configuration in the element and in their typical ions (Zn²⁺, Cd²⁺, Hg²⁺), so they’re classified separately, even though they sit within the d-block.
Q12. Manganese and zinc show unusually low melting points compared to their neighbors in the 3d transition series. What’s the key reason for manganese specifically?
A) Its half-filled 3d⁵4s² configuration provides no unpaired d-electrons available for strong metallic bonding B) It has the smallest atomic radius in the series C) It is not a solid at room temperature D) It has the highest nuclear charge in the series
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Answer: A. Manganese’s stable, symmetric 3d⁵4s² arrangement leaves fewer unpaired d-electrons to participate in metallic bonding compared to its neighbors, weakening the metallic lattice and lowering its melting point relative to the overall trend.
Section 4: Acid-Base and Oxidation State Exceptions
Q13. Hypophosphorous acid (H₃PO₂) is a strong reducing agent even though phosphorus is in the +1 oxidation state. What structural feature explains this?
A) It contains two P–H bonds, which are easily oxidized, in addition to one P–OH bond B) It has no oxygen atoms C) It is unstable and decomposes instantly D) Phosphorus shows its highest possible oxidation state here
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Answer: A. H₃PO₂ is dibasic (only one ionizable H, via the P–OH group) despite having three hydrogens, because two of those hydrogens are directly bonded to phosphorus (P–H bonds) rather than to oxygen. These P–H bonds are easily oxidized, which is exactly why the compound behaves as a strong reducing agent.
Q14. Thallium shows a stable +1 oxidation state alongside its group-expected +3 state, and Tl⁺ is actually more stable than Tl³⁺. This is an example of:
A) Diagonal relationship B) The inert pair effect C) Lanthanide contraction D) Back bonding
Show Answer
Answer: B) The inert pair effect. The heavier p-block elements’ outer ns² electron pair becomes increasingly reluctant to participate in bonding due to poor shielding by intervening d/f electrons, so the lower oxidation state (Tl⁺) becomes more stable than the group-typical higher one (Tl³⁺).
Q15. BF₃ acts as a weaker Lewis acid than BCl₃, even though fluorine is far more electronegative than chlorine. Why?
A) Stronger pπ–pπ back bonding from fluorine’s lone pairs into boron’s empty p-orbital in BF₃ partially fills that orbital, reducing boron’s electron-accepting ability B) BF₃ is not a real compound C) Chlorine is a better electron donor than fluorine in all cases D) BCl₃ has no empty orbital on boron
Show Answer
Answer: A. Fluorine’s small size allows better orbital overlap for pπ–pπ back bonding into boron’s empty 2p orbital, partially satisfying boron’s electron deficiency in BF₃. This back bonding is weaker in BCl₃ (poorer orbital overlap due to size mismatch), leaving boron more electron-deficient and making BCl₃ the stronger Lewis acid — the reverse of what electronegativity alone would predict.
Quick Reference: Top Exceptions at a Glance
| Exception | Element(s) | Root Cause |
|---|---|---|
| Anomalous ground-state configuration | Cr, Cu | Extra stability of half-filled/fully-filled d-subshell |
| Diagonal relationship | Li–Mg, Be–Al, B–Si | Similar charge/size ratio across the diagonal |
| No octet expansion | N, O, F | No accessible d-orbitals in the valence shell |
| Weak X–X bond | F–F, N–N, O–O | Strong lone-pair repulsion from small atomic size |
| Not “true” transition metals | Zn, Cd, Hg | Completely filled d¹⁰ in element and common ions |
| Inert pair effect | Tl, Pb, Bi | Poor shielding of ns² pair by inner d/f electrons |
| Reversed Lewis acidity | BF₃ vs BCl₃ | pπ–pπ back bonding strength differs by halogen size |
How to Use This Quiz Effectively
- Attempt every question before expanding the answer — guessing and checking builds better recall than reading passively
- For each exception, try explaining the “why” out loud in your own words, not just the “what”
- Revisit this list a day before your exam as a rapid-fire check rather than a first-time study session
- Group questions you got wrong by category (s-block, p-block, d-block) to spot your weakest area
Frequently Asked Questions
Are inorganic chemistry exceptions a big scoring area in JEE Main?
Yes. Because exceptions test conceptual understanding rather than pure recall, they appear frequently and reliably across JEE Main papers, often as single-concept questions that are quick to answer correctly if you know the underlying reason.
What’s the fastest way to memorize these exceptions?
Don’t memorize them as isolated facts — group them by cause (size effects, d-orbital availability, exchange energy, inert pair effect). Once you know the four or five underlying reasons, most individual exceptions become logical rather than something to memorize separately.
Do these exceptions also apply to JEE Advanced and NEET?
Yes, these are core NCERT-based concepts, so they’re equally relevant for JEE Advanced and NEET, though JEE Advanced tends to combine them with more complex multi-step reasoning.
For more revision material across science topics, browse our science and technology resources section. For the official JEE Main syllabus and exam pattern, refer to the National Testing Agency’s official JEE Main portal.
Written by Freddy John, Exam Prep & Science Desk, SeminarsOnly News. Freddy develops revision material for competitive exams including JEE Main and NEET, cross-checking every concept against NCERT chemistry references before publishing.
Last reviewed for accuracy: July 2026. Concepts covered are drawn from the standard NCERT Class 11/12 inorganic chemistry curriculum used as the basis for JEE Main.
