Inorganic Chemistry Exceptions JEE Main Quiz: Complete Revision Guide

Important Exceptions In Inorganic Chemistry for JEE Main

In this section, you will find the chapter-wise important inorganic chemistry exceptions notes.

Practice Quiz of Inorganic Chemistry Exceptions Download Free PDF

Exceptions of Periodic Classification

Electronic Configuration:

• General Trend: Orbitals fill according to the Aufbau principle.
• Exception: Chromium (Cr): $[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^1$ instead of $3 \mathrm{~d}^4 4 \mathrm{~s}^2$; Copper (Cu): $[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^1$ instead of $3 \mathrm{~d}^9 4 \mathrm{~s}^2$ due to extra stability of half-filled and completely filled d-orbitals.

Electron Gain Enthalpy:

• General Trend of period: Becomes more negative across a period.
• Exception: Be, Mg, and Noble Gases have positive or nearly zero electron gain enthalpy due to stable electronic configurations.
• General Trend for group: More negative down a group.
• Exception: Chlorine has a more negative electron gain enthalpy than Fluorine, although fluorine is more electronegative. The very small size of fluorine causes strong electron-electron repulsion.

Atomic Radius:

• General Trend: Increases down a group.
• Exception: Gallium (Ga) has a slightly smaller atomic radius than Aluminium (Al) due to poor shielding by d-electrons (d-block contraction).

Electronegativity:

• General Trend: Generally decreases down a group.
• Exception: Noble gases usually have no standard electronegativity values because they rarely form compounds.

Metallic Character:

• General Trend: Increases down a group.
• Exception: Hydrogen is a unique element and cannot be classified strictly as a metal or non-metal.

Diagonal Relationship:

• General Trend: Not common in the periodic table.
• Exception: Li–Mg, Be–Al, and B–Si show similar properties because of comparable ionic size, charge density, and electronegativity.

Exceptions of Chemical Bonding

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.

Octet Rule:

• General Trend: Atoms complete their octet.
• Exception: Incomplete octet: $\mathrm{BeCl}_2, \mathrm{BF}_3$, and $\mathrm{AlCl}_3$ have less than 8 electrons around the central atom.

Expanded Octet

• General Trend: Period 2 elements cannot expand their octet.
• Exception: Period 3 and beyond can expand the octet, e.g., $\mathrm{PCl}_5, \mathrm{SF}_6, \mathrm{ClF}_3, \mathrm{XeF}_4$.

Bond Angle

• General Trend: A lone pair always decreases the bond angle.
• Exception: $\mathrm{H}_2 \mathrm{O}\left(104.5^{\circ}\right)<\mathrm{NH}_3\left(107^{\circ}\right)<\mathrm{CH}_4\left(109.5^{\circ}\right)$) due to increasing lone pair repulsion.

Bond Order

• General Trend: Bond order is always an integer.
• Exception: Molecular orbital theory predicts fractional bond orders, e.g., $\mathrm{O}_2^{+}(2.5), \mathrm{O}_2^{-}(1.5)$

Paramagnetism

• General Trend: Molecules with all paired electrons are diamagnetic.
• Exception: O₂ is paramagnetic because it contains two unpaired electrons in π* antibonding orbitals.

Exceptions of s-Block Elements

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.

Exceptions of p-Block Elements

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.

Exceptions of d- and f-Block Elements

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.

Exceptions of Coordination Compounds

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.

Also Read: JEE Main 2027 Chemistry Sample Paper

Inorganic Chemistry Exceptions Quiz for JEE Main

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) $\mathrm{Na}-\mathrm{Ca}$

B) $\mathrm{Li}-\mathrm{Mg}$

C) $\mathrm{K}-\mathrm{Ca}$

D) $\mathrm{Be}-\mathrm{Si}$

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) $\mathrm{SO}_2$

B) $\mathrm{CO}_2$

C) NO

D) $\mathrm{P}_2 \mathrm{O}_5$

Solution: Answer (C)

CO, NO and N₂O are neutral oxides.

Question 6: Which element has the electronic configuration $[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^1$?

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 $\mathrm{Zn}^{2+}$ have a completely filled $\mathrm{d}^{10}$ configuration.

Question 8: Which ion is colourless?

A) $\mathrm{Fe}^{3+}$

B) $\mathrm{Cu}^{2+}$

C) $\mathrm{Mn}^{2+}$

D) $\mathrm{Zn}^{2+}$

Solution: Answer (D)

$\mathrm{Zn}^{2+}$ has a $\mathrm{d}^{10}$ 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 $\mathrm{NH}_3$ is named ammine.

Question 10: Which is a strong-field ligand?

A) $\mathrm{F}^{-}$

B) $\mathrm{Cl}^{-}$

C) $\mathrm{CN}^{-}$

D) $\mathrm{H}_2 \mathrm{O}$

Solution: Answer (C)

CN⁻ causes the pairing of electrons and forms low-spin complexes.

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 $2 s^2$ (fully filled, extra stable), while $B$ is $2 s^2 2 p^1$. The $2 p$ electron in $B$ is at higher energy and better shielded from the nucleus by the $2 \mathrm{~s}^2$ 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)

$\mathrm{N}\left(2 \mathrm{p}^3\right)$ is half-filled, extra stable due to symmetry and exchange energy. O ( $2 \mathrm{p}^4$ ) has one orbital with a paired electron; removing that electron relieves inter-electronic repulsion, making ionisation comparatively easier. Hence, $\operatorname{IE}(\mathrm{N})>\operatorname{IE}(\mathrm{O})$, 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 (polarising 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 $n s^2 n p^6$ 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 (2)

Noble gases have completely filled, highly stable $n s^2 n p^6$ 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 ($\mathrm{Pb}^{2+}$ is more stable than $\mathrm{Sn}^{2+}$), 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 (2)

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 $\mathrm{Pb}^{2+}$ (retaining the 6s² pair) is more stable than $\mathrm{Pb}^{4+}$, 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.

Common Mistakes Students Make While Learning Inorganic Chemistry Exceptions

Classification of Elements & Periodicity

Treating "smooth trend" as a universal law: Students memorise "IE increases across a period, decreases down a group" and apply it blindly, forgetting that $\mathrm{Be}>\mathrm{B}, \mathrm{Mg}>\mathrm{Al}, \mathrm{N}>\mathrm{O}$, and $\mathrm{P}>\mathrm{S}$ are the rule at those specific points, not the exception to memorize separately. The fix: learn the two triggers $\left(n s^2 \rightarrow n p^1\right.$ and $\left.n p^3 \rightarrow n p^4\right)$ 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.

p-Block Elements

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 $\mathrm{NF}_5$ or $\mathrm{OF}_6$ as plausible species because they've seen $\mathrm{PF}_5$ and $\mathrm{SF}_6$, forgetting $\mathrm{n}=2$ 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 $\mathrm{Tl}^{+}, \mathrm{Pb}^{2+}$, and $\mathrm{Bi}^{3+}$ (two less than the group number) become more stable going down.

d- and f-Block Elements

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 $d^{10}$ in all common states).

Assuming all transition metal ions are colored: $\mathrm{Sc}^{3+}, \mathrm{Ti}^{4+}$, and $\mathrm{Cu}^{+}$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.

Coordination Compounds

The chelate effect confused with bond strength: Students think $\left[\mathrm{Ni}(\mathrm{en})_3\right]^{2+}$ is more stable than $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$ 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 $\mathrm{Ni}(\mathrm{CO})_4$-type questions.

Best Books of JEE Mains for Inorganic Chemistry

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.