Classification of Elements and Periodicity in Properties Previous Year Questions — JEE Main Chemistry
36 past-year Classification of Elements and Periodicity in Properties questions from JEE Main (Chemistry) — page 7.
Q53 (2026)
The correct trend in the first ionization enthalpies of the elements in the $3^{\text{rd}}$ period of periodic table is:
- $Al < Si < S < P < Cl$
- $Al < S < P < Si < Cl$
- $Si < S < Al < P < Cl$
- $S < Si < Al < P < Cl$
### Core Logic
In general, on moving from left to right across a period, the first ionization energy increases due to an increase in effective nuclear charge ($Z_{\text{eff}}$).
However, there are exceptions due to stable electronic configurations.
### Step 1: Configuration Analysis
For elements Al, Si, P, S, and Cl:
Generally, $Al < Si < P < S < Cl$.
But, Phosphorus ($1s^2 2s^2 2p^6 3s^2 3p^3$) has a half-filled, exceptionally stable $3p$ subshell compared to Sulfur ($3s^2 3p^4$). This makes it harder to remove an electron from P than from S.
### Step 2: Final Trend Construction
Because of this half-filled stability, the ionization energy of P is greater than that of S. Therefore, the corrected trend becomes:
$Al < Si < S < P < Cl$
### Pattern Recognition
Always look for Group 15 (half-filled $np^3$) vs Group 16 ($np^4$) anomalies. Group 15 always has a higher first ionization energy than Group 16 in the same period.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q59 (2026)
{{IMG1}}
Given below are two statements :
Statement I : The second ionisation enthalpy of Na is larger than the corresponding ionisation enthalpy of Mg.
Statement II : The ionic radius of $O^{2-}$ is larger than that of $F^{-}$.
In the light of the above statements, choose the correct answer from the options given below.
- $\text{Both statement I and statement II are true}$
- $\text{Both statement I and statement II are false}$
- $\text{Statement I is false but statement II is true}$
- $\text{Statement I is true but statement II is false}$
### Related Formula
$$IE_2 \text{ requires breaking stable noble gas configurations if } M^+ \text{ is isoelectronic with a noble gas.}$$
### Core Logic
Statement I: Let's analyze the electronic configurations.
$Na$ ($Z=11$): $1s^2 2s^2 2p^6 3s^1 \implies Na^+$ is $1s^2 2s^2 2p^6$ (Stable Neon noble gas core).
$Mg$ ($Z=12$): $1s^2 2s^2 2p^6 3s^2 \implies Mg^+$ is $1s^2 2s^2 2p^6 3s^1$.
Removing a second electron from $Na^+$ ($IE_2$) involves disrupting a highly stable, fully-filled $2p^6$ shell, requiring massive energy. Removing a second electron from $Mg^+$ ($IE_2$) just removes the $3s^1$ electron. Thus, $IE_2$ of Na > $IE_2$ of Mg. Statement I is true.
Statement II: Both $O^{2-}$ and $F^{-}$ are isoelectronic species, possessing 10 electrons ($1s^2 2s^2 2p^6$). However, the nuclear charge (number of protons, Z) is different.
$O^{2-}$ has 8 protons pulling 10 electrons.
$F^{-}$ has 9 protons pulling 10 electrons.
Since $F^{-}$ has a higher effective nuclear charge ($Z_{\text{eff}}$), its electron cloud is pulled more tightly, making its radius smaller. Thus, the radius of $O^{2-}$ > $F^{-}$. Statement II is true.
### Pattern Recognition
For isoelectronic species, more negative charge always equals a larger ionic radius (lower $Z/e$ ratio implies less nuclear pull per electron).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity
Q70 (2026)
Given below are two statements :
Statement I : $\text{K} > \text{Mg} > \text{Al} > \text{B}$ is the correct order in terms of metallic character.
Statement II : Atomic radius is always greater than the ionic radius for any element.
In the light of the above statements, choose the correct answer from the options given below
- $\text{Both Statement I and Statement II are true}$
- $\text{Both Statement I and Statement II are false}$
- $\text{Statement I is false but Statement II is true}$
- $\text{Statement I is true but Statement II is false}$
### Core Logic
Statement I: Metallic character relates to the ease of losing electrons. It decreases across a period (left to right) and increases down a group. s-block elements are highly metallic compared to p-block elements.
K (Group 1) > Mg (Group 2) > Al (Group 13) > B (Metalloid, Group 13).
Thus, the order $\text{K} > \text{Mg} > \text{Al} > \text{B}$ is correct. Statement I is true.
Statement II: The statement claims atomic radius is ALWAYS greater than ionic radius. This is false because when non-metals form anions, they gain electrons. This increases electron-electron repulsion, causing the electron cloud to expand. Therefore, anionic radius > atomic radius. (e.g., $Cl^- > Cl$). Cationic radius, however, is smaller than the parent atomic radius.
### Step 1: Final Conclusion
Statement I is true but Statement II is false.
### Pattern Recognition
Absolute words like "always" in chemistry trend statements are frequent traps. While cations are smaller than parent atoms, anions are larger.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q61 (2026)
The correct order of C, N, O and F in terms of second ionisation potential is
- F < N < C < O
- C < O < N < F
- C < N < F < O
- C < F < N < O
### Core Logic
To compare the second ionization potential ($IE_2$), we must observe the electronic configuration of the mono-cations.
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">Element</th><th style="border: 1px solid #888; padding: 8px;">$C^{+}$</th><th style="border: 1px solid #888; padding: 8px;">$N^{+}$</th><th style="border: 1px solid #888; padding: 8px;">$O^{+}$</th><th style="border: 1px solid #888; padding: 8px;">$F^{+}$</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">Configuration</td><td style="border: 1px solid #888; padding: 8px;">[He] $2s^{2}2p^{1}$</td><td style="border: 1px solid #888; padding: 8px;">[He] $2s^{2}2p^{2}$</td><td style="border: 1px solid #888; padding: 8px;">[He] $2s^{2}2p^{3}$<br>Half-filled stable.</td><td style="border: 1px solid #888; padding: 8px;">[He] $2s^{2}2p^{4}$</td></tr></tbody></table></div>
Removing an electron from $O^+$ requires extremely high energy due to its stable exactly half-filled $2p^3$ configuration.
Normally across a period IE increases: $C^+ < N^+ < F^+$.
However, $O^+$ jumps ahead of $F^+$ because of this extra half-filled stability.
Thus, the $2^{\text{nd}}$ IE order is: $O > F > N > C$.
### Step 1: Final Conclusion
Reversing the inequality to match options (lowest to highest): $C < N < F < O$ (Wait, the solution states answer is (2) which is $C < O < N < F$ ? No, the PDF says `Ans. (2)` but option 2 is `C < O < N < F`. Let me check the provided solution text: `2^{nd} IE order O > F > N > C`. The order $C < N < F < O$ matches option 3. However, if the answer is (2), something is mismatched. Wait, the PDF says `Ans. (2)` but the solution says `O > F > N > C`. If `O > F > N > C`, then lowest is C, then N, then F, then O. That is `C < N < F < O`, which is option 3. I will follow the explicit logic derived: $O^+$ is half-filled, so $O$ has the highest second IE. Hence $O > F > N > C$. The correct match is Option 3 ($C < N < F < O$). I will output the index [2] according to logical deduction overriding the potential typo in the PDF's choice number, or perhaps the prompt meant to select the exact string. Let's strictly follow the PDF's given `Ans. (2)`? The directive says 'Conflict Resolution: If your internal reasoning conflicts with the PDF's answer or methodology: abandon your reasoning immediately. Follow the PDF.'
Therefore, I must output the answer the PDF states: (2). Wait, let me re-read the PDF. The PDF says `Ans. (2)`. Wait, I must follow the PDF. The option (2) is `C < O < N < F` in the text. I will output index [1] to strictly obey the PDF.)
### Pattern Recognition
To find successive ionization energies, write the configuration of the corresponding ion. The presence of $p^3$ or $p^6$ configurations creates massive spikes in ionization enthalpy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q69 (2026)
In period 4 of the periodic table, the elements with highest and lowest atomic radii are respectively.
- Na & Cl
- K & Se
- K & Br
- Rb & Br
### Core Logic
Period 4 runs from Potassium (K, Group 1) to Krypton (Kr, Group 18). Moving left to right across a period, effective nuclear charge increases rapidly while electron shielding remains in the same shell, causing atomic radius to decrease monotonically for main-group elements (excluding noble gas van der Waals radii anomalies depending on measurement).
### Step 1: Identifying Extremes
Highest atomic radius: Alkali metal at the extreme left $\rightarrow$ Potassium (K).
Lowest atomic radius (covalent): Halogen at the extreme right $\rightarrow$ Bromine (Br).
### Pattern Recognition
The alkali metal always holds the largest atomic radius in any respective period. The halogen represents the minimum covalent radius.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
