title: "Create a quiz on inorganic chemistry exceptions to help me revise for my jee main exam" description: "Struggling to remember inorganic chemistry exceptions? Read our comprehensive revision guide with an interactive practice quiz to ace your JEE Main exam." slug: "chemistry-create-a-quiz-on-inorganic-chemistry-exceptions-to-help-me-revise-for-my-jee-main-exam" date: "2026-06-29" author: "QuizPerCard" indexable: true
Create a Quiz on Inorganic Chemistry Exceptions to Help Me Revise for My JEE Main Exam
Ask any JEE aspirant about Inorganic Chemistry, and they'll tell you it feels like an endless list of rules followed by an even longer list of exceptions. But here is the secret: Inorganic exceptions are not random. They are governed by physical principles like shielding effects, electronic configurations, and atomic size.
JEE Main heavily tests these "exceptions" because they prove whether you understand the underlying concepts or are just memorizing trends. In this guide, we break down the top high-yield exceptions you must master, followed by an interactive revision quiz.
1. The d-Block and Lanthanoid Contractions (Group 13 & 14)
The Gallium vs. Aluminium Size Exception
- The Expected Rule: Atomic radius should increase down Group 13 ($\text{B} < \text{Al} < \text{Ga} < \text{In} < \text{Tl}$).
- The Exception: The atomic radius of Gallium (135 pm) is slightly smaller than Aluminium (143 pm).
- The Chemistry: Gallium is preceded by ten transition elements of the 3d-series. The 10 d-electrons offer very poor shielding of the nuclear charge. Consequently, the effective nuclear charge ($Z_{\text{eff}}$) increases significantly, pulling the outermost $4s$ and $4p$ electrons closer and shrinking the atom.
The Stability of Lower Oxidation States (Inert Pair Effect)
- The Expected Rule: Group oxidation state is the most stable.
- The Exception: In heavier p-block elements (Group 13, 14, 15), lower oxidation states become more stable down the group (e.g., $\text{Pb}^{2+}$ is more stable than $\text{Pb}^{4+}$; $\text{Bi}^{3+}$ is more stable than $\text{Bi}^{5+}$).
- The Chemistry: As you go down the group, the shielding of intervening $d$ and $f$ electrons is poor. The outer $s$-electrons ($ns^2$) experience a strong nuclear pull and are "reluctant" to participate in bond formation. Only the $p$-electrons participate, leading to a stable oxidation state that is $2$ less than the group state.
2. Anomalous Period 2 Trends
First Ionization Enthalpy: Nitrogen vs. Oxygen
- The Expected Rule: Ionization enthalpy increases across a period ($\text{C} < \text{N} < \text{O} < \text{F}$).
- The Exception: Nitrogen ($1402 \text{ kJ/mol}$) has a higher first ionization enthalpy than Oxygen ($1314 \text{ kJ/mol}$).
- The Chemistry: Nitrogen has a stable, half-filled $2p^3$ valence electronic configuration ($1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$). Removing an electron from this highly stable state requires significantly more energy than removing a paired electron from Oxygen's $2p^4$ shell, which actually experiences inter-electronic repulsion.
Electron Gain Enthalpy: Fluorine vs. Chlorine
- The Expected Rule: Electron gain enthalpy becomes more negative up a group ($\text{I} < \text{Br} < \text{Cl} < \text{F}$).
- The Exception: Chlorine ($-349 \text{ kJ/mol}$) has a more negative electron gain enthalpy than Fluorine ($-328 \text{ kJ/mol}$).
- The Chemistry: Fluorine is extremely compact. When an incoming electron enters Fluorine's small $2p$ subshell, it experiences severe inter-electronic repulsion from the existing $7$ valence electrons. In Chlorine, the larger $3p$ subshell distributes the charge density, making the addition of an electron much more energetically favorable.
🔥 High-Alert Exception: Fluorine's Weak F-F Bond Even though Fluorine is smaller than Chlorine, the bond dissociation energy of $\text{F}_2$ is lower than that of $\text{Cl}_2$ and $\text{Br}_2$ ($\text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2$). This is again because of the immense inter-electronic repulsion between lone pairs of the tiny, adjacent Fluorine atoms.
3. Group 16 & 17 Hydrides (Acidic Strength & Boiling Points)
Hydride Boiling Points: The Role of Hydrogen Bonding
- The Expected Rule: Boiling points increase with increasing molecular mass (due to stronger van der Waals forces).
- The Exception: $\text{H}_2\text{O}$ and $\text{HF}$ have anomalously high boiling points compared to their group counterparts ($\text{H}_2\text{O} > \text{H}_2\text{Te} > \text{H}_2\text{Se} > \text{H}_2\text{S}$).
- The Chemistry: Oxygen and Fluorine are highly electronegative and small, enabling the formation of strong intermolecular hydrogen bonds that hold the liquid state together. The other hydrides are held only by weak dipole-dipole or van der Waals interactions.
⚠️ Concept Trap: Acidic Strength vs. Boiling Point While $\text{H}_2\text{O}$ has the highest boiling point, it is the least acidic hydride in Group 16 ($\text{H}_2\text{O} < \text{H}_2\text{S} < \text{H}_2\text{Se} < \text{H}_2\text{T}\text{e}$). Acidic strength depends on bond length: as the central atom gets larger, the $E-H$ bond length increases and strength decreases, making it easier to release $\text{H}^+$.
Summary Table: Most Common JEE Inorganic Trends & Exceptions
| Property | Expected Trend | Actual Exception Trend | Primary Reason | | :--- | :--- | :--- | :--- | | Atomic Size (G13) | $\text{Al} < \text{Ga}$ | $\text{Ga} < \text{Al}$ | Poor shielding by 3d electrons (Transition Contraction) | | IE₁ (Period 2) | $\text{N} < \text{O}$ | $\text{O} < \text{N}$ | Stable half-filled $2p^3$ configuration in Nitrogen | | Electron Gain (G17) | $\text{Cl} < \text{F}$ (more negative) | $\text{F} < \text{Cl}$ (more negative) | Small size of fluorine causes high inter-electronic repulsion | | Bond Energy (G17) | $\text{F}_2 > \text{Cl}_2 > \text{Br}_2$ | $\text{Cl}_2 > \text{Br}_2 > \text{F}_2$ | Lone-pair repulsion on small Fluorine atoms | | Hydride Acidic Strength | $\text{H}_2\text{Te} < \text{H}_2\text{O}$ | $\text{H}_2\text{O} < \text{H}_2\text{S} < \text{H}_2\text{Se} < \text{H}_2\text{Te}$ | Lower $\text{H}-\text{Te}$ bond strength due to Ga/Te size mismatch |
Interactive Revision Quiz: Inorganic Exceptions (JEE Main Level)
Test your active recall. See if you can spot the trap in these actual JEE-style questions!
Q1. Which of the following has the highest hydration energy?
- A. $\text{Li}^+$
- B. $\text{Na}^+$
- C. $\text{K}^+$
- D. $\text{Cs}^+$
Correct Answer: A ($\text{Li}^+$)
Explanation: Hydration energy is inversely proportional to ionic size. Since Lithium is the smallest ion in Group 1, it has the highest charge density and attracts water molecules most aggressively, releasing the most energy.
Q2. Select the correct order of basic strength of Group 15 hydrides:
- A. $\text{NH}_3 < \text{PH}_3 < \text{AsH}_3 < \text{SbH}_3$
- B. $\text{NH}_3 > \text{PH}_3 > \text{AsH}_3 > \text{SbH}_3$
- C. $\text{PH}_3 > \text{NH}_3 > \text{AsH}_3 > \text{SbH}_3$
- D. $\text{SbH}_3 > \text{AsH}_3 > \text{PH}_3 > \text{NH}_3$
Correct Answer: B ($\text{NH}_3 > \text{PH}_3 > \text{AsH}_3 > \text{SbH}_3$)
Explanation: Basic strength depends on the availability of the lone pair of electrons. As the size of the central atom increases down the group, the lone pair occupies a larger volume of space (lower charge density), making it less available to coordinate with a proton ($\text{H}^+$).
Q3. The stability of $+1$ oxidation state in Group 13 elements increases in the order:
- A. $\text{B} < \text{Al} < \text{Ga} < \text{In} < \text{Tl}$
- B. $\text{Tl} < \text{In} < \text{Ga} < \text{Al} < \text{B}$
- C. $\text{B} < \text{Al} < \text{In} < \text{Ga} < \text{Tl}$
- D. $\text{Ga} < \text{In} < \text{Al} < \text{Tl} < \text{B}$
Correct Answer: A ($\text{B} < \text{Al} < \text{Ga} < \text{In} < \text{Tl}$)
Explanation: Due to the Inert Pair Effect, the stability of the lower oxidation state ($+1$) increases dramatically down Group 13, making $\text{Tl}^+$ extremely stable and $\text{Tl}^{3+}$ highly oxidizing.
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