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Inorganic Chemistry is one of the highest scoring sections in JEE Main, but it is also the subject where tiny exceptions, special cases, and NCERT-based facts are frequently tested, you know. A lot of students lose easy marks because they just memorise the general patterns, but they overlook the key inorganic chemistry details, sometimes. In this article, the Inorganic Chemistry Exceptions for JEE Main 2027 guide, we have put together the most important inorganic chemistry exceptions from every chapter in one place, and practice questions related to these exceptions that are asked frequently in the JEE Main Exam.
Also Read: JEE Mains Top Inorganic Exceptions and Organic Named Reactions
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In this section, you will find the chapter-wise important inorganic chemistry exceptions notes.
Practice Quiz of Inorganic Chemistry Exceptions Download Free PDF
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Atomic Radius:
Metallic Character:
Diagonal Relationship:
Chemical Bonding is one of the most important chapters in JEE Main Chemistry. While a lot of questions are based on standard bonding concepts, NCERT exceptions and some special cases are asked again and again.
Topic | General Rule | Exception |
Octet Rule | Atoms complete their octet. | Incomplete octet: BeCl 2 , BF 3 , and AlCl 3 have less than 8 electrons around the central atom. |
Expanded Octet | Period 2 elements cannot expand their octet. | Period 3 and beyond can expand the octet, e.g., PCl 5 , SF 6 , ClF 3 , XeF 4 . |
Bond Angle | A lone pair always decreases the bond angle. | H₂O (104.5°) < NH₃ (107°) < CH₄ (109.5°) due to increasing lone pair repulsion. |
Bond Order | Bond order is always an integer. | Molecular orbital theory predicts fractional bond orders, e.g., O 2 + ( 2.5 ) , O 2 − ( 1.5 ) |
Paramagnetism | Molecules with all paired electrons are diamagnetic. | O₂ is paramagnetic because it contains two unpaired electrons in π* antibonding orbitals. |
The s-Block Elements chapter has a few high-yield exceptions that are repeatedly asked in JEE Main, kind of again and again, you know.
Topic | General Rule | Exception |
Reaction with Oxygen | Alkali metals form normal oxides (M₂O). | Li forms oxide (Li₂O), Na mainly forms peroxide ( Na 2 O 2 ) , while K, Rb, and Cs form superoxides ( ( KO 2 , RbO 2 , C s O 2 ) .. |
Nature of Hydroxides | All Group 2 hydroxides have similar solubility. | Mg ( OH ) 2 is sparingly soluble, whereas Ba ( OH ) 2 is highly soluble. Solubility increases down the group. |
Thermal Stability of Carbonates | Carbonates are generally stable on heating. | Li 2 CO 3 decomposes on heating, whereas other alkali metal carbonates are thermally stable. |
Thermal Stability of Nitrates | Alkali metal nitrates decompose similarly. | LiNO 3 decomposes to Li 2 O , while other alkali metal nitrates decompose to nitrites (MNO₂). |
Carbonates | All Group 2 carbonates decompose equally. | Thermal stability of Group 2 carbonates increases down the group ( BeCO 3 < MgCO 3 < CaCO 3 < SrCO 3 < BaCO 3 ) .. |
Hydroxides | Basic strength remains similar. | Basic strength of hydroxides increases down Group 2: Be ( OH ) 2 < Mg ( OH ) 2 < Ca ( OH ) 2 < Sr ( OH ) 2 < Ba ( OH ) 2 . |
Diagonal Relationship | Elements in different groups show different properties. | Lithium resembles Magnesium more than other alkali metals due to the diagonal relationship. |
In the p-Block Elements chapter, you’ll find the highest number of NCERT-based exceptions in Inorganic Chemistry. Here are some of the most important and usually asked ones for JEE Main 2027, these are exceptions you really should know.
Topic | General Rule | Exception |
Oxidation State | The group oxidation state remains constant. | Due to the inert pair effect, heavier p-block elements prefer lower oxidation states ( Tl + > Tl 3 + , Pb 2 + > Pb 4 + , Bi 3 + > Bi 5 + ) . |
Oxides of Nitrogen | All nitrogen oxides are coloured. | N 2 O and NO are colourless, while NO₂ is brown. |
Acid Strength | Hydracid strength follows electronegativity. | HF is the weakest acid among hydrogen halides despite fluorine being the most electronegative element because of its very strong H–F bond. |
Reducing Power of Hydrogen Halides | Higher electronegativity means a stronger reducing agent. | Reducing power increases as HI > HBr > HCl > HF. |
Oxyacids of Halogens | All halogens form all oxyacids. | Fluorine does not form oxyacids because it cannot exhibit positive oxidation states. |
Halogen Oxidation States | All halogens show positive oxidation states. | Fluorine shows only −1 oxidation state in its compounds. |
Maximum Covalency | All elements can expand their octet. | Second-period elements (C, N, O, F) cannot expand their octet, while heavier p-block elements can. |
Oxides | All oxides are acidic. | CO, NO and N 2 O are neutral oxides. |
Acidic Character | The acidity of oxides increases across a period. | Al 2 O 3 is amphoteric, not purely acidic. |
Noble Gases | Noble gases are completely inert. | Xe forms compounds such as XeF 2 , XeF 4 , XeF 6 , XeO 3 and XeOF 4 .. Krypton forms only a few compounds ( KrF 2 ) . |
The d- and f-Block Elements chapter includes a bunch of NCERT-based exceptions that are regularly asked in JEE Main. Here are a few of the most important ones you should go through again and revise properly. These are the exceptions that tend to pop up the most, so don’t skip them.
Topic | General Rule | Exception |
Electronic Configuration | Electrons fill orbitals according to the Aufbau principle. | C r : [ A r ] 3 d 5 4 s 1 and C u : [ A r ] 3 d 10 4 s 1 show exceptional electronic configurations due to the extra stability of half-filled and filled d-orbitals. |
Oxidation States | Elements show only one or two oxidation states. | Mn exhibits the maximum oxidation state of +7, while most transition elements show multiple oxidation states. |
Colour of Ions | All transition metal ions are coloured. | Sc 3 + , Ti 4 + , Zn 2 + , Cu + and Cd 2 + are colourless because they have d⁰ or d¹⁰ electronic configurations. |
Magnetic Properties | All transition metal ions are paramagnetic. | Zn 2 + , Cd 2 + , Hg 2 + , Sc 3 + and Ti 4 + are diamagnetic due to the absence of unpaired electrons. |
Lanthanoid Contraction | Atomic size increases down a group. | Lanthanoid contraction causes almost identical sizes of Zr and Hf, leading to very similar chemical properties. |
Oxidation States of Lanthanoids | All lanthanoids show a +3 oxidation state only. | Ce also shows +4, while Eu and Yb commonly show +2 oxidation states due to extra electronic stability. |
All d-block elements are transition elements. | Zn, Cd and Hg are d-block elements but not transition elements because their atoms and common ions have filled d-orbitals ( ( d 10 ) . | |
Ion Formation | 3d electrons are removed first during ionisation. | 4s electrons are removed before 3d electrons when transition metals form cations. |
The Coordination Compounds chapter has a bunch of NCERT-based questions that are asked over and over in JEE Main. Usually, the questions revolve around IUPAC naming, magnetic nature, oxidation state, hybridisation, and those special ligands. Below, I’m putting the most essential exceptions to revise again, because questions are asked about them again and again.
Topic | General Rule | Exception |
Ligand Name | Ligand names usually end with -o. | Neutral ligands have special names: H 2 O → aqua, NH 3 → ammine, CO → carbonyl, NO → nitrosyl. |
Anionic Complex | The metal name remains unchanged. | In anionic complexes, the metal name ends with -ate (Fe → ferrate, Cu → cuprate, Ag → argentate, Au → aurate). |
Oxidation Number | Ligands are neutral. | NO (nitrosyl) can behave as NO + (+1), unlike most neutral ligands. |
Chelating Ligands | All ligands are monodentate. | en, oxalate ( C 2 O 4 2 − ) ), and EDTA are multidentate (chelating) ligands. |
Magnetic Behaviour | All complexes of a metal have the same number of unpaired electrons. | The same metal ion can form high-spin or low-spin complexes depending on ligand strength (e.g., Fe 2 + , Co 3 + ). |
Hybridization | Hybridisation depends only on the coordination number. | Strong-field ligands ( CN , CO ) cause electron pairing, whereas weak-field ligands ( F − , Cl − , H 2 O ) usually do not. |
Coordination Number | Ligands contribute one donor atom. | Polydentate ligands contribute multiple donor atoms; e.g., EDTA has a coordination number of 6. |
Isomerism | All coordination compounds show isomerism. | Tetrahedral complexes generally do not show geometrical isomerism, whereas square planar and octahedral complexes often do. |
Also Read: JEE Main 2027 Chemistry Sample Paper
Question 1: Which element has a more negative electron gain enthalpy?
A) Fluorine
B) Chlorine
C) Bromine
D) Iodine
Solution: Answer (B)
Although fluorine is more electronegative, its very small size causes greater electron-electron repulsion. Hence, Cl has a more negative electron gain enthalpy than F.
Question 2: Which pair shows a diagonal relationship?
A) Na−Ca
B) Li−Mg
C) K−Ca
D) Be−Si
Solution: Answer (B)
Due to similar ionic size and charge density, Li resembles Mg.
Question 3: Which hydrogen halide is the weakest acid?
A) HCl
B) HBr
C) HI
D) HF
Solution: Answer (D)
HF has the strongest H–F bond, making it the weakest acid among hydrogen halides.
Question 4: Which halogen does not form oxyacids?
A) Chlorine
B) Bromine
C) Iodine
D) Fluorine
Solution: Answer (D)
Fluorine cannot exhibit positive oxidation states.
Question 5: Which oxide is neutral?
A) SO2
B) CO2
C) NO
D) P2O5
Solution: Answer (C)
CO, NO and N₂O are neutral oxides.
Question 6: Which element has the electronic configuration [Ar]3 d54 s1?
A) Iron
B) Chromium
C) Copper
D) Manganese
Solution: Answer (B)
Chromium has an exceptional electronic configuration due to the stability of a half-filled d-subshell.
Question 7: Which of the following is not a transition element?
A) Fe
B) Ni
C) Zn
D) Cr
Solution: Answer (C)
Zn and Zn2+ have a completely filled d10 configuration.
Question 8: Which ion is colourless?
A) Fe3+
B) Cu2+
C) Mn2+
D) Zn2+
Solution: Answer (D)
Zn2+ has a d10 configuration, so no d-d transition occurs.
Question 9: The IUPAC name of NH₃ as a ligand is
A) Amino
B) Amide
C) Ammine
D) Ammonia
Solution: Answer (C)
The neutral ligand NH3 is named ammine.
Question 10: Which is a strong-field ligand?
A) F−
B) Cl−
C) CN−
D) H2O
Solution: Answer (C)
CN⁻ causes the pairing of electrons and forms low-spin complexes.
Question 11: Which of the following pairs has an anomalous (exceptional) electronic configuration compared to what the Aufbau principle predicts?
1) Sc and Ti
2) Cr and Cu
3) Mn and Fe
4) Zn and Ga
Solution: Answer (2)
Cr is [Ar] 3 d 5 4 s 1 (not 3 d 4 4 s 2 ) and Cu is [Ar] 3 d 10 4 s 1 ( not 3 d 9 4 s 2 ) . This happens because half-filled ( d 5 ) and fully-filled ( d 10 ) subshells have extra stability due to symmetrical charge distribution and exchange energy.
Question 12: The first ionization enthalpy of Be is greater than that of B, even though B comes after Be in the same period. This exception occurs because:
1) B has a larger atomic radius than Be
(2) Removing an electron from B's 2p orbital is easier than removing from Be's fully-filled 2s orbital
3) B has a higher nuclear charge, making its electrons harder to remove
(4) Be has a smaller nuclear charge than B
Solution: Answer (2)
Be has configuration 2s2 (fully filled, extra stable), while B is 2s22p1. The 2p electron in B is at higher energy and better shielded from the nucleus by the 2 s2 pair, so it is removed more easily than an electron from Be's stable, fully-filled 2s orbital, reversing the expected trend.
Question 13: Similarly, N has a higher first ionization enthalpy than O. This exception is due to:
1) N has a larger size than O
2) O has a lower effective nuclear charge
3) N's 2p³ configuration is half-filled and extra stable, while removing an electron from O's 2p⁴ relieves electron-electron repulsion
4) N has a fully-filled 2s orbital, unlike O
Solution: Answer (3)
N(2p3) is half-filled, extra stable due to symmetry and exchange energy. O ( 2p4 ) has one orbital with a paired electron; removing that electron relieves inter-electronic repulsion, making ionisation comparatively easier. Hence, IE(N)>IE(O), against the general left-to-right increasing trend.
Question 14: Which element shows the diagonal relationship with Mg, giving it exceptional properties unlike other Group 1 elements?
1) Na
2) Li
3) K
4) Be
Solution: Answer: (2)
Li resembles Mg (diagonal relationship) due to a similar charge/size ratio (polarizing power). Both form covalent bonds, their carbonates decompose on heating, both form nitrides, and their chlorides are soluble in organic solvents, unlike typical alkali metal behaviour.
Question 15: Electron gain enthalpy of fluorine is less negative than that of chlorine. This exception (F should be more negative going by trend) is because:
1) F has a smaller nuclear charge than Cl
2) F's small atomic size causes high electron-electron repulsion in its compact 2p subshell when an extra electron is added
3) Cl has higher electronegativity than F
4) F has no d-orbitals
Solution: Answer (2)
Fluorine's very small size results in a high electron density in the 2p subshell. Adding another electron causes significant inter-electronic repulsion, reducing the energy released. Chlorine, being larger with a 3p subshell, accommodates the extra electron more easily, releasing more energy. Hence ΔegH: Cl > F (more negative for Cl).
Question 16: Noble gases have positive (unfavourable) electron gain enthalpy. This is an exception to the general periodic trend because:
(a) They have the largest atomic radius in their period
(b) Their fully-filled ns2np6 configuration is extremely stable, so an added electron must occupy a higher energy level, requiring energy input
(c) They have the highest ionisation enthalpy
(d) They are chemically inert
Solution: Answer (b)
Noble gases have completely filled, highly stable ns2np6 configurations. An incoming electron would have to enter the next higher shell/orbital, which is energetically unfavourable, giving a positive electron gain enthalpy, an exception to the trend of increasingly negative values across a period.
Question 17: Melting points generally decrease down Group 13, but which element breaks this trend by having a melting point lower than the element above and below it?
1) Al
2) Ga
3) In
4) Tl
Solution: Answer (2)
Gallium has an unusually low melting point (~30°C), lower than both Al (660°C) and In (157°C). This is due to its unique crystal structure, consisting of Ga₂ dimers held by weak metallic/covalent bonding rather than a simple close-packed metallic lattice.
Question 18: The stability of the +2 oxidation state increases down Group 14 (Pb2+ is more stable than Sn2+), which is an exception to the usual trend of higher oxidation states being more stable for lighter elements. This is best explained by:
(a) Lanthanide contraction
(b) Inert pair effect
(c) Diagonal relationship
(d) Screening effect of d-orbitals only
Solution: Answer (b)
Due to poor shielding by filled d and f orbitals and relativistic effects, the ns² electron pair in heavier p-block elements (like Pb, Bi, Tl) becomes reluctant to participate in bonding the "inert pair effect." Hence Pb2+ (retaining the 6s² pair) is more stable than Pb4+, reversing the normal group trend of highest oxidation state stability.
Question 19: HF has an abnormally high boiling point compared to HCl, HBr, and HI, breaking the expected trend of increasing boiling point with molecular mass down the group. This exception arises due to:
1) Higher polarizability of F
2) Strong intermolecular hydrogen bonding in HF due to high electronegativity and small size of F
3) HF being ionic in nature
4) Weaker van der Waals forces in HF
Solution: Answer (2)
Fluorine's small size and very high electronegativity allow HF molecules to form extensive, strong hydrogen bonds (even forming polymeric (HF)ₙ chains), giving it an anomalously high boiling point (~19.5°C) compared to HCl (−85°C), despite HF having the lowest molar mass among the hydrogen halides.
Question 20: Which of the following correctly explains why the atomic radius of Ga is nearly the same as (rather than larger than) Al, going against the usual trend of increasing size down a group?
1) Ga has a lower nuclear charge than Al
2) The d-block contraction: presence of filled 3d electrons in Ga poorly shields the nuclear charge, pulling electrons closer
3) Ga has fewer electrons than Al
4) Ga is more electronegative than Al
Solution: Answer (2)
Between Al and Ga lies the first transition series (d-block). The poor shielding effect of the filled 3d¹⁰ subshell in Ga means valence electrons experience a higher effective nuclear charge, causing the atomic radius of Ga (135 pm) to be almost equal to or even slightly less than Al (143 pm), an exception known as d-block contraction.
Treating "smooth trend" as a universal law: Students memorise "IE increases across a period, decreases down a group" and apply it blindly, forgetting that Be>B,Mg>Al,N>O, and P>S are the rule at those specific points, not the exception to memorize separately. The fix: learn the two triggers (ns2→np1 and np3→np4) as a pair, not four isolated facts.
Confusing electron gain enthalpy with electronegativity: A huge number of students assume fluorine has the most negative electron gain enthalpy because it's the most electronegative element. These are different properties — electronegativity is about attracting shared electron density in a bond; electron gain enthalpy is about a free gaseous atom accepting an extra electron. Chlorine wins on the second one because F's tiny 2p subshell repels the incoming electron.
Assuming every group behaves like row 3 onward : The single biggest error: applying "d orbitals are available for expanded covalency" to period-2 elements. Students write NF5 or OF6 as plausible species because they've seen PF5 and SF6, forgetting n=2 has no d orbitals at all.
Applying the inert pair effect in the wrong direction: A common slip is expecting the higher oxidation state to be more stable for heavy p-block elements (since that's true earlier in the group), when actually Tl+,Pb2+, and Bi3+ (two less than the group number) become more stable going down.
Not distinguishing d-block from transition elements: Zn, Cd, and Hg are frequently misclassified as ordinary transition metals in options, when by strict definition they're excluded (full d10 in all common states).
Assuming all transition metal ions are colored: Sc3+,Ti4+, and Cu+trip up a lot of students because they're d° or d 10 - no partially filled d subshell means no d-d transition, hence colourless. Students often forget to check the ion's configuration, not the element's.
The chelate effect confused with bond strength: Students think [Ni(en)3]2+ is more stable than [Ni(NH3)6]2+ because Ni-N bonds with en are individually stronger. It's actually an entropy effect from releasing more free particles, not bond enthalpy.
Missing that oxidation state can be zero: Metal carbonyls having the metal at 0 oxidation state feel "wrong" to students used to always assigning a positive charge to the central metal, so they misassign oxidation states in Ni(CO)4-type questions.
Choosing the right books is kinda essential if you want to master Inorganic Chemistry for JEE Main. Most of the JEE Main questions, in a very straightforward way, are linked to NCERT concepts, facts, and the odd exceptions too. So it makes sense that your preparation should start with NCERT first, then continue towards objective practice books, step by step.
Book | Author | Best For |
NCERT Chemistry Class 11 & 12 | NCERT | Theory, NCERT facts, exceptions, direct JEE Main questions |
J.D. Lee Concise Inorganic Chemistry for JEE (Main & Advanced) | J.D. Lee (Wiley Adaptation) | Conceptual understanding and advanced reference |
Problems in Inorganic Chemistry for JEE Main & Advanced | V.K. Jaiswal | Chapter-wise MCQ practice |
Inorganic Chemistry | O.P. Tandon | Theory with objective questions |
Arihant Visualise Inorganic Chemistry | Arihant Publications | Visual learning, concept revision, and illustrations |
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