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 2.
Q43 (2025)
The type of oxide formed by the element among Li, Na, Be, Mg, B and Al that has the least atomic radius is:
(1) $A_{2}O_{3}$
(2) $AO_{2}$
(3) AO
(4) $A_{2}O$
- $A_{2}O_{3}$
- $AO_{2}$
- AO
- $A_{2}O$
### Core Logic
Let's analyze the periodic trend among the listed elements: Li, Na, Be, Mg, B, Al.
* Atomic radius decreases across a period due to increasing effective nuclear charge ($Z_{\text{eff}}$).
* Atomic radius increases down a group due to addition of electron shells.
Comparing Period 2 elements (Li, Be, B): Boron ($B$) has the highest atomic number here and thus the smallest atomic radius.
Boron forms an oxide where its oxidation state is $+3$, which gives $B_2O_3$. This matches the structural template $A_2O_3$.
### Pattern Recognition
Smallest element in Period 2 (excluding noble gases) is on the far right. Boron belongs to Group 13, so it forms traditional trivalent acidic oxides ($A_2O_3$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q26 (2025)
The incorrect decreasing order of atomic radii is :
- $\mathrm{Mg} > \mathrm{Al} > \mathrm{C} > \mathrm{O}$
- $\mathrm{Al} > \mathrm{B} > \mathrm{N} > \mathrm{F}$
- $\mathrm{Be} > \mathrm{Mg} > \mathrm{Al} > \mathrm{Si}$
- $\mathrm{Si} > \mathrm{P} > \mathrm{Cl} > \mathrm{F}$
### Related Formula
Atomic radius decreases across a period due to increase in effective nuclear charge ($Z_{\text{eff}}$) and increases down a group due to addition of new electronic shells.
### Core Logic
Let us analyze the elements in option (3):
- $\mathrm{Be}$ and $\mathrm{Mg}$ belong to Group 2. Down the group, atomic radius increases: $\mathrm{Mg} > \mathrm{Be}$.
- $\mathrm{Mg}$, $\mathrm{Al}$, and $\mathrm{Si}$ belong to Period 3. Across the period, atomic radius decreases: $\mathrm{Mg} > \mathrm{Al} > \mathrm{Si}$.
Combining these trends, the correct decreasing order is:
$$\mathrm{Mg} > \mathrm{Be} > \mathrm{Al} > \mathrm{Si}$$
Therefore, the order given in option (3) $\mathrm{Be} > \mathrm{Mg} > \mathrm{Al} > \mathrm{Si}$ is incorrect.
### Pattern Recognition
Sees: Group 2 and Period 3 comparison.
Trap: Assuming atomic radius always increases down a group regardless of cross-period shifts; $\mathrm{Mg}$ is larger than $\mathrm{Be}$, making $\mathrm{Be} > \mathrm{Mg}$ fundamentally incorrect.
Shortcut: Check adjacent group/period boundaries to immediately isolate inversions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q34 (2025)
Which of the following statements are correct?
A. The process of the addition an electron to a neutral gaseous atom is always exothermic
B. The process of removing an electron from an isolated gaseous atom is always endothermic
C. The $1^{\mathrm{st}}$ ionization energy of the boron is less than that of the beryllium
D. The electronegativity of $\mathrm{C}$ is $2.5$ in $\mathrm{CH}_4$ and $\mathrm{CCl}_4$
E. $\mathrm{Li}$ is the most electropositive among elements of group I
Choose the correct answer from the options given below:
- B and C only
- A, C and D only
- B and D only
- B, C and E only
### Related Formula
$$\text{IE}_1(\text{Be}) = 1s^2 2s^2 > \text{IE}_1(\text{B}) = 1s^2 2s^2 2p^1$$
### Core Logic
Statement A: Electron gain enthalpy can be endothermic (e.g., noble gases, noble metals). (False)
Statement B: Removal of an electron requires energy to overcome nuclear attraction; ionization is strictly endothermic. (True)
Statement C: $\mathrm{Be}$ ($1s^2 2s^2$) has a fully filled stable $2s$ subshell, requiring higher ionization energy than $\mathrm{B}$ ($1s^2 2s^2 2p^1$). (True)
Statement D: Electronegativity of carbon varies with oxidation state / partial charge ($Z_{\text{eff}}$ in $\mathrm{CCl}_4 > \mathrm{CH}_4$). (False)
Statement E: $\mathrm{Cs}$ (or $\mathrm{Fr}$) is the most electropositive element in group I, not $\mathrm{Li}$. (False)
### Step 1: Final Conclusion
Statements B and C are correct.
### Pattern Recognition
Ionization energy = ALWAYS endothermic.
Be vs B IE1 = Be higher due to stable full $2s$ subshell.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q40 (2025)
Given below are the pairs of group 13 elements showing their relation in terms of atomic radius:
$(B < Al)$, $(Al < Ga)$, $(Ga < In)$ and $(In < Tl)$
Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius $(M^{3+})$ than the other one. The atomic number of the element (X) is:
- 31
- 49
- 13
- 81
### Core Logic
Let's evaluate the anomalies within Group 13 trends:
* **Atomic Radius Order:** Due to the poor shielding effect of the filled $3d$ electron subshell in Gallium (transition contraction), its outer valence shell experiences a stronger nuclear pull. Consequently, the atomic radius sequence exhibits an inversion:
$$\text{Correct Atomic Size: } B < Ga < Al < In < Tl$$
Therefore, the pair $(Al < Ga)$ provided in the problem statement is incorrect.
* **Ionic Radius Order ($M^{3+}$):** In the fully ionized $+3$ configuration state, the transition contraction anomaly is overridden by standard shell count physics. The ionic radius follows the uniform down-the-group trend:
$$Al^{3+} < Ga^{3+}$$
Hence, Gallium ($Ga$) has the larger ionic radius between the two elements. Its atomic number is 31.
### Pattern Recognition
Transition contraction heavily distorts the *atomic* radius profile of Gallium, but standard descending size progression rules are fully restored when evaluating the *ionic* $M^{3+}$ radius profile.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Class 12 Chemistry: The p-Block Elements
Q31 (2025)
Choose the incorrect trend in the atomic radii ($r$) of the elements:
- $r{\text{Br}} < r{\text{K}}$
- $r{\text{Mg}} < r{\text{Al}}$
- $r_{\text{Rb}} < r_{\text{Cs}}$
- $r_{\text{At}} < r_{\text{Cs}}$
### Related Formula
$$\text{Atomic Radius } (r) \propto \frac{1}{Z_{\text{eff}}} \quad \text{(Across a period)}$$
$$\text{Atomic Radius } (r) \propto \text{Number of Shells } (n) \quad \text{(Down a group)}$$
### Core Logic
Let's review the periodic table positioning and trends:
- Across a period from left to right, effective nuclear charge ($Z_{\text{eff}}$) increases, drawing electrons closer to the nucleus. Hence, atomic radius decreases.
- $\text{Mg}$ and $\text{Al}$ sit in Period 3. Since $\text{Al}$ is further to the right ($Z=13$) than $\text{Mg}$ ($Z=12$), the radius of $\text{Al}$ is smaller than that of $\text{Mg}$:
$$r_{\text{Mg}} > r_{\text{Al}}$$
Thus, the stated trend $r_{\text{Mg}} < r_{\text{Al}}$ is incorrect.
### Step 1: Cross-Checking Valid Trends
- $r_{\text{Br}} < r_{\text{K}}$: Correct, as potassium sits at the start of Period 4, bromine sits near the end.
- $r_{\text{Rb}} < r_{\text{Cs}}$: Correct, atomic size increases down Group 1.
- $r_{\text{At}} < r_{\text{Cs}}$: Correct, cesium sits far below and left compared to astatine.
### Pattern Recognition
Period 3 size ranking baseline: $\text{Na} > \text{Mg} > \text{Al} > \text{Si} > \text{P} > \text{S} > \text{Cl}$. Moving rightward always drops the radius size unless noble gas van der Waals constraints interfere.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
