Inorganic Chemistry forms a significant portion of Chemistry in JEE Main 2027; however, it also comes with various exceptions that may confuse students at times. Understanding these inorganic chemistry exceptions will not only make the subject easier for you to understand but also prevent you from making mistakes in the exam. This article will discuss some of the most essential inorganic chemistry exceptions for JEE Main along with their explanations. We have compiled the most important inorganic chemistry exceptions from every chapter in one place, along with practice questions related to these exceptions that are frequently asked in the JEE Main Exam.
Also Read: JEE Main 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. You can download Inorganic Chemistry Practice Quiz from the download button given below:
|
Topic |
General Trend |
Exception |
|
Orbitals fill according to the Aufbau principle. |
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. | |
|
First Ionisation Enthalpy |
Increases across a period. |
Be > B because the electron removed from B is a 2p electron, which is easier to remove than Be's 2s electron. |
|
First Ionisation Enthalpy |
Increases across a period. |
N > O because oxygen has paired 2p electrons, causing electron-electron repulsion and making electron removal easier. |
|
Becomes more negative across a period. |
Be, Mg, and Noble Gases have positive or nearly zero electron gain enthalpy due to stable electronic configurations. | |
|
Electron Gain Enthalpy |
More negative down a group. |
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 |
Increases down a group. |
Gallium (Ga) has a slightly smaller atomic radius than Aluminium (Al) due to poor shielding by d-electrons (d-block contraction). |
|
Generally decreases down a group. |
Noble gases usually have no standard electronegativity values because they rarely form compounds. | |
|
Metallic Character |
Increases down a group. |
Hydrogen is a unique element and cannot be classified strictly as a metal or non-metal. |
|
Oxidation States |
Group number often indicates the highest oxidation state. |
Fluorine always shows -1 oxidation state in its compounds because it is the most electronegative element. |
|
Position in the Periodic Table |
Elements show similar properties within a group. |
Hydrogen resembles both Group 1 and Group 17, but it is placed separately due to its unique properties. |
|
Diagonal Relationship |
Not common in the periodic table. |
Li–Mg, Be–Al, and B–Si show similar properties because of comparable ionic size, charge density, and electronegativity. |
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: $\mathrm{BeCl}_2, \mathrm{BF}_3$, and $\mathrm{AlCl}_3$ 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., $\mathrm{PCl}_5, \mathrm{SF}_6, \mathrm{ClF}_3, \mathrm{XeF}_4$. |
|
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., $\mathrm{O}_2^{+}(2.5), \mathrm{O}_2^{-}(1.5)$ |
|
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 $\left(\mathrm{Na}_2 \mathrm{O}_2\right)$, while K, Rb, and Cs form superoxides ($\left(\mathrm{KO}_2, \mathrm{RbO}_2, \mathrm{C} s \mathrm{O}_2\right)$.. |
|
Nature of Hydroxides |
All Group 2 hydroxides have similar solubility. |
$\mathrm{Mg}(\mathrm{OH})_2$ is sparingly soluble, whereas $\mathrm{Ba}(\mathrm{OH})_2$ is highly soluble. Solubility increases down the group. |
|
Thermal Stability of Carbonates |
Carbonates are generally stable on heating. |
$\mathrm{Li}_2 \mathrm{CO}_3$ decomposes on heating, whereas other alkali metal carbonates are thermally stable. |
|
Thermal Stability of Nitrates |
Alkali metal nitrates decompose similarly. |
$\mathrm{LiNO}_3$ decomposes to $\mathrm{Li}_2 \mathrm{O}$, 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 $\left(\mathrm{BeCO}_3<\mathrm{MgCO}_3<\mathrm{CaCO}_3<\mathrm{SrCO}_3<\mathrm{BaCO}_3\right)$.. |
|
Hydroxides |
Basic strength remains similar. |
Basic strength of hydroxides increases down Group 2: $\mathrm{Be}(\mathrm{OH})_2<\mathrm{Mg}(\mathrm{OH})_2<\mathrm{Ca}(\mathrm{OH})_2<\mathrm{Sr}(\mathrm{OH})_2<\mathrm{Ba}(\mathrm{OH})_2$. |
|
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 $\left(\mathrm{Tl}^{+}>\mathrm{Tl}^{3+}, \mathrm{Pb}^{2+}>\mathrm{Pb}^{4+}, \mathrm{Bi}^{3+}>\mathrm{Bi}^{5+}\right)$. |
|
Oxides of Nitrogen |
All nitrogen oxides are coloured. |
$\mathrm{N}_2 \mathrm{O}$ 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 $\mathrm{N}_2 \mathrm{O}$ are neutral oxides. |
|
Acidic Character |
The acidity of oxides increases across a period. |
$\mathrm{Al}_2 \mathrm{O}_3$ is amphoteric, not purely acidic. |
|
Noble gases are completely inert. |
Xe forms compounds such as $\mathrm{XeF}_2, \mathrm{XeF}_4$, $\mathrm{XeF}_6, \mathrm{XeO}_3$ and $\mathrm{XeOF}_4$.. Krypton forms only a few compounds $\left(\mathrm{KrF}_2\right)$. |
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. |
$\mathrm{Sc}^{3+}, \mathrm{Ti}^{4+}, \mathrm{Zn}^{2+}, \mathrm{Cu}^{+}$and $\mathrm{Cd}^{2+}$ are colourless because they have d⁰ or d¹⁰ electronic configurations. |
|
Magnetic Properties |
All transition metal ions are paramagnetic. |
$\mathrm{Zn}^{2+}, \mathrm{Cd}^{2+}, \mathrm{Hg}^{2+}, \mathrm{Sc}^{3+}$ and $\mathrm{Ti}^{4+}$ 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 ($\left(d^{10}\right)$. | |
|
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: $\mathrm{H}_2 \mathrm{O} \rightarrow$ aqua, $\mathrm{NH}_3 \rightarrow$ ammine, $\mathrm{CO} \rightarrow$ 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 $\mathrm{NO}^{+}$ (+1), unlike most neutral ligands. |
|
Chelating Ligands |
All ligands are monodentate. |
en, oxalate $\left(\mathrm{C}_2 \mathrm{O}_4^{2-}\right)$), 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., $\mathrm{Fe}^{2+}, \mathrm{Co}^{3+}$ ). |
|
Hybridization |
Hybridisation depends only on the coordination number. |
Strong-field ligands $(\mathrm{CN}, \mathrm{CO})$ cause electron pairing, whereas weak-field ligands $\left(\mathrm{F}^{-}, \mathrm{Cl}^{-}, \mathrm{H}_2 \mathrm{O}\right)$ 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
Get expert advice on college selection, admission chances, and career path in a personalized counselling session.
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 11: Which of the following elements has a less negative electron gain enthalpy despite having a stable electronic configuration?
A) Sodium
B) Magnesium
C) Chlorine
D) Sulfur
Solution: Answer (B)
Magnesium has the electronic configuration [Ne] 3s². Its 3s subshell is completely filled, so addition of another electron requires entering the higher-energy 3p orbital. Hence, Mg has a positive or nearly zero electron gain enthalpy compared with elements having a greater tendency to accept an electron.
Question 12: Which of the following correctly represents the atomic radius relationship?
A) Ga > Al
B) Ga < Al
C) Ga = Al
D) Ga is much larger than Al
Solution: Answer (B)
Although atomic radius generally increases down a group, gallium has a slightly smaller atomic radius than aluminium. This is due to poor shielding by the intervening 3d electrons, which increases the effective nuclear charge experienced by the outer electrons.
Therefore, Ga < Al
Question 13: Which of the following pairs exhibits a diagonal relationship?
A) Na and Mg
B) Li and Mg
C) K and Mg
D) Na and Ca
Solution: Answer (B)
Lithium and magnesium show a diagonal relationship because they have comparable ionic size, charge density and electronegativity.
Some similarities include the formation of nitrides and relatively covalent compounds.
Therefore, the correct pair is: Li and Mg
Question 14 :Which of the following species has a fractional bond order?
A) N₂
B) O₂
C) O₂⁺
D) F₂
Solution: Answer (C)
According to molecular orbital theory, the bond order of O₂ is 2.
When one electron is removed from O₂ to form $\mathrm{O}_2^{+}$, the electron is removed from an antibonding molecular orbital. Therefore, the bond order increases by 0.5.
Bond order of $\mathrm{O}_2^{+}$ = 2 + 0.5 = 2.5
Hence, $\mathrm{O}_2^{+}$ has a fractional bond order.
Question 15: Which of the following compounds is amphoteric?
A) Na₂O
B) MgO
C) Al₂O₃
D) SO₃
Solution: Answer (C)
Al₂O₃ is an important exception to the general trend of oxide character.
It reacts with both acids and bases. Therefore, it shows amphoteric behaviour.
For example:
$\begin{aligned} & \mathrm{Al}_2 \mathrm{O}_3+6 \mathrm{HCl} \rightarrow 2 \mathrm{AlCl}_3+3 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{NaOH} \rightarrow 2 \mathrm{NaAlO}_2+\mathrm{H}_2 \mathrm{O}\end{aligned}$
Question 16: Which of the following pairs of elements have nearly identical atomic sizes due to lanthanoid contraction?
A) Na and K
B) Zr and Hf
C) Sc and Y
D) Ti and Zr
Solution: Answer (B)
Lanthanoid contraction causes a gradual decrease in the atomic and ionic radii of lanthanoids due to ineffective shielding by 4f electrons.
As a result, the atomic radii of Zr and Hf become nearly identical, despite Hf being present in a lower period.
Therefore, Zr and Hf show very similar chemical properties.
Question 17: Which lanthanoid commonly exhibits the +4 oxidation state due to additional electronic stability?
A) La
B) Ce
C) Eu
D) Yb
Solution: Answer (B)
Cerium commonly exhibits the +3 oxidation state but can also show the +4 oxidation state.
Ce⁴⁺ has an electronic configuration close to a stable xenon configuration, which provides additional stability.
Therefore, among the given options, Ce is the lanthanoid known for the +4 oxidation state.
Question 18: During the formation of Fe²⁺ from Fe, electrons are removed first from which orbital?
A) 3d
B) 4s
C) 4p
D) 3p
Solution: Answer (B)
The electronic configuration of Fe is:
$\mathrm{Fe}=[\mathrm{Ar}] 3 \mathrm{~d}^6 4 \mathrm{~s}^2$
Although the 4s orbital is filled before the 3d orbital during the Aufbau process, the 4s electrons are removed first during ionisation.
Therefore:
$\begin{aligned} & \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \\ & \mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^6\end{aligned}$
Hence, electrons are removed from the 4s orbital first.
Question 19: What is the correct name of the metal in the anionic complex ion [Fe(CN)₆]⁴⁻?
A) Iron
B) Ferric
C) Ferrate
D) Ferrous
Solution: Answer (C)
When the coordination complex is anionic, the name of the metal is modified by adding the suffix “-ate”.
Therefore:
Iron → ferrate
Thus, $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^4$ contains the metal named ferrate in its IUPAC name.
The complete name is hexacyanidoferrate(II).
Question 20: Which type of complex generally does not show geometrical isomerism?
A) Square planar
B) Octahedral
C) Tetrahedral
D) Both square planar and octahedral
Solution: Answer (C)
Square planar and octahedral complexes can show geometrical isomerism because different spatial arrangements of ligands are possible.
Tetrahedral complexes, however, generally do not show geometrical isomerism because all four positions around the central metal are equivalent.
Therefore, the correct answer is: Tetrahedral complex
Question 21: Which of the following statements is correct?
A. Fluorine has the most negative electron gain enthalpy among all halogens.
B. Chlorine has a more negative electron gain enthalpy than fluorine.
C. Bromine has more negative electron gain enthalpy than chlorine.
D. Iodine has more negative electron gain enthalpy than chlorine.
Solution: Answer (B)
Fluorine has a very small atomic size, so the added electron experiences greater electron–electron repulsion in its compact 2p orbital. Therefore, its electron gain enthalpy is less negative than that of chlorine.
Hence:
$\mathrm{Cl}>\mathrm{F}$
in terms of the magnitude of negative electron gain enthalpy.
Question 22: Which of the following elements does not show variable oxidation states due to its filled d-subshell?
A. Fe
B. Cu
C. Zn
D. Mn
Solution: Answer (C)
Zinc has the electronic configuration:
$\mathrm{Zn}=[A r] 3 d^{10} 4 s^2$
After losing two electrons:
$\mathrm{Zn}^{2+}=[A r] 3 d^{10}$
The d-subshell remains filled. Therefore, Zn predominantly shows the +2 oxidation state and is not considered a transition element.
$\mathrm{Zn}^{2+}: 3 \mathrm{~d}^{10}$
Question 23: Which of the following comparisons of first ionisation enthalpy is correct?
A. IE₁(B) > IE₁(Be)
B. IE₁(Be) > IE₁(B)
C. IE₁(O) > IE₁(N)
D. IE₁(Be) = IE₁(B)
Solution: Answer (B)
The electronic configurations are:
$\begin{gathered}
\mathrm{Be}: 1 s^2 2 s^2 \\
\mathrm{~B}: 1 s^2 2 s^2 2 p^1
\end{gathered}$
In boron, the electron is removed from the higher-energy 2p orbital, whereas Be has a stable completely filled 2s orbital.
Therefore:
$I E_1(B e)>I E_1(B)$
Therefore, option B is correct.
Question 24: Which of the following ions is expected to be colourless?
A. Ti³⁺
B. Fe³⁺
C. Cu²⁺
D. Zn²⁺
Solution: Answer (D)
The electronic configuration of Zn is:
$\mathrm{Zn}=[A r] 3 d^{10} 4 s^2$
Therefore:
$\mathrm{Zn}^{2+}=[A r] 3 d^{10}$
A filled d-subshell has no possibility of a $\mathbf{d}-\mathbf{d}$ transition. Hence, $\mathrm{Zn}^{2+}$ compounds are generally colourless.
$\mathrm{Zn}^{2+}\left(3 \mathrm{~d}^{10}\right)$
Question 25: Which of the following statements about the inert pair effect is correct?
A. It becomes weaker down the group.
B. It favours higher oxidation states of heavier p-block elements.
C. It favours lower oxidation states of heavier p-block elements.
D. It occurs only in transition elements.
Solution: Answer (C)
The inert pair effect refers to the tendency of the outermost s-electrons to remain less involved in bonding. It becomes more significant as we move down the p-block groups.
As a result, lower oxidation states become more stable in heavier elements.
For example:
$\mathrm{Tl}^{+}>\mathrm{Tl}^{3+}$
in terms of stability.
Thus, the inert pair effect favours the lower oxidation state of heavier p-block elements.
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.
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.
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.
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.
Watch this video to know more
Inorganic chemistry is full of exceptions that can easily confuse students during JEE Main preparation. Use the table below to quickly identify the most common traps and the correct concepts to avoid mistakes in the exam.
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Use these 30 minutes to quickly revise the most important inorganic chemistry exceptions, unusual trends, and commonly confused concepts. Focus on high-yield facts rather than trying to learn new topics at the last moment.
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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 |
Frequently Asked Questions (FAQs)
No, skipping Inorganic Chemistry is not recommended for JEE Main; I mean, it’s generally a bad idea. Inorganic Chemistry brings a pretty big share to the Chemistry part, and a lot of the questions are quite direct, NCERT-based, and also more time-friendly than the heavy numerical kind of stuff.
Inorganic chemistry isn’t really conceptually hard, but it’s pretty factual, and honestly, it asks for a lot of memorisation too, which can feel kinda like a chore.
Yeah, honestly, NCERT is pretty enough for most of the Inorganic Chemistry questions that show up in JEE Main. Almost every year, a big chunk of those questions is either lifted straight from NCERT ,or kind of built around the same ideas ,like tables, reactions,and the little exceptions that they explain in the book.
Inorganic chemistry exceptions for JEE Main matter because many questions test unusual properties, electronic configurations, oxidation states, and trends. Knowing them can help students avoid common mistakes.
Important inorganic chemistry exceptions are commonly found in topics such as periodic trends, electronic configuration, chemical bonding, coordination compounds, p-block elements, d-block elements, and oxidation states.
On Question asked by student community
Dear Student,
You can access JEE Main Mathematics previous-year questions on Careers360. Practising previous-year questions will help you understand the exam pattern and question types.
Hello Student,
You can check the resources provided below:
Hope this helps!
Hello, since you are in Intermediate 2nd year with the MPC group and preparing for JEE Main, focus on completing the syllabus, revising important concepts and solving previous years’ JEE Main question papers. Take regular mock tests and analyse your mistakes to improve your score. Keep a realistic study schedule
Dear Student
Please specify whether you want Physics, Chemistry, or Mathematics, and which JEE Main/Advanced exam.
Hi Ansh Chauhan
Check the link below for the high-weightage Physics chapters for JEE Main 2027 based on recent exam trends. Students can use these chapters to prioritise their preparation and focus more on topics that have appeared frequently in previous papers.
https://engineering.careers360.com/articles/most-weightage-chapters-for-jee-main-physics-2027
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