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).
Q410 (2025)
Electronic configuration of four elements A, B, C and D are given below :
(A) $1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^3$
(B) $1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^4$
(C) $1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^5$
(D) $1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^2$
Which of the following is the correct order of increasing electronegativity (Pauling's scale)?
- $\mathrm{A < D < B < C}$
- $\mathrm{A < C < B < D}$
- $\mathrm{A < B < C < D}$
- $\mathrm{D < A < B < C}$
### Related Formula
$$\chi_{\mathrm{P}} \propto Z_{\text{eff}} \propto \frac{1}{\text{Atomic Radius}} \quad \text{(Across a period)}$$
### Core Logic
Let's first identify each element based on its electronic configuration:
<b>(A):</b> $1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^3 \implies \text{Atomic Number } 7 \implies \text{Nitrogen (N)}$<br><br>
<b>(B):</b> $1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^4 \implies \text{Atomic Number } 8 \implies \text{Oxygen (O)}$<br><br>
<b>(C):</b> $1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^5 \implies \text{Atomic Number } 9 \implies \text{Fluorine (F)}$<br><br>
<b>(D):</b> $1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^2 \implies \text{Atomic Number } 6 \implies \text{Carbon (C)}$
### Step 1: Check Periodic Table Trends
All four elements belong to the 2nd period. Electronegativity increases across a period from left to right because nuclear charge ($Z_{\text{eff}}$) increases, and atomic radius decreases:
$$\text{Carbon (D)} < \text{Nitrogen (A)} < \text{Oxygen (B)} < \text{Fluorine (C)}$$
On Pauling's scale, the precise values are:
- Carbon (D) = $2.55$
- Nitrogen (A) = $3.04$
- Oxygen (B) = $3.44$
- Fluorine (C) = $3.98$
### Step 2: Conclusion
The correct increasing order is $\mathrm{D < A < B < C}$.
### Pattern Recognition
Fluorine is the most electronegative element in the entire periodic table (Pauling electronegativity of $4.0$). Electronegativity always increases towards the top-right of the main-group elements.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q28 (2025)
The property/properties that show irregularity in first four elements of group-17 is/are :
(A) Covalent radius
(B) Electron affinity
(C) Ionic radius
(D) First ionization energy
Choose the correct answer from the options given below:
- $\text{(1) B and D only}$
- $\text{(2) A and C only}$
- $\text{(3) B only}$
- $\text{(4) A, B, C and D}$
### Related Formula
Standard downward group variations typically follow predictable monotonic pathways:
$$\text{Radius} \propto \text{Number of shells}$$
$$\text{Ionization Energy} \propto \frac{1}{\text{Atomic Size}}$$
### Core Logic
Let's review the explicit values and trends across $\mathrm{F, Cl, Br, I}$:
* **Covalent radius**: $\mathrm{F < Cl < Br < I}$ (Perfect monotonic increase).
* **Ionic radius**: $\mathrm{F^- < Cl^- < Br^- < I^-}$ (Perfect monotonic increase).
* **First ionization energy**: $\mathrm{F > Cl > Br > I}$ (Perfect monotonic decrease).
* **Electron affinity**: $\mathrm{Cl > F > Br > I}$ (Irregular trend! Fluorine has an anomalously lower electron affinity than Chlorine due to high inter-electronic repulsions inside its exceptionally compact $2p$ valence subshell).
### Step 1: Conclusion
Therefore, only electron affinity (B) demonstrates an irregular trend line among the first four halogens.
### Pattern Recognition
Fluorine anomalous properties are a classic JEE question archetype. Whenever a question asks about irregularities in halogens, immediately check Electron Gain Enthalpy (Electron Affinity) and Bond Dissociation Enthalpy, where Fluorine routinely breaks the monotonic descending order.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Class 12 Chemistry: The p-Block Elements
Q39 (2025)
Given below are two statements :
Statement (I): The metallic radius of Al is less than that of Ga.
Statement (II): The ionic radius of $\mathrm{Al}^{3+}$ is less than that of $\mathrm{Ga}^{3+}$.
In the light of the above statements, choose the most appropriate answer from the options given below:
- $(1)\ \text{Both Statement I and Statement II are incorrect}$
- $(2)\ \text{Statement I is incorrect but Statement II is correct}$
- $(3)\ \text{Statement I is correct but Statement II is incorrect}$
- $(4)\ \text{Both Statement I and Statement II are correct}$
### Related Formula
Effective nuclear charge scaling expression layout:
$$Z_{\text{eff}} = Z - \sigma$$
### Core Logic
Let's analyze both properties systematically:
* **Statement I**: The metallic radius of Gallium ($\mathrm{Ga}$) is anomalously *smaller* than Aluminum ($\mathrm{Al}$) due to the poor shielding effect of the $10$ d-electrons inserted before it. This increases $Z_{\text{eff}}$, pulling the outer shell inward tightly. Thus, $\text{Radius of Al} > \text{Radius of Ga}$, making Statement I incorrect.
* **Statement II**: When looking at completely stripped ionic configurations ($\mathrm{Al^{3+}}$ and $\mathrm{Ga^{3+}}$), the extra shell layer in Gallium resumes its dominant role, meaning standard periodic increase down a group holds true: $\mathrm{Al^{3+} < Ga^{3+}}$. (Statement II is correct).
### Step 1: Evaluation
Thus, Statement I is incorrect but Statement II is correct.
### Pattern Recognition
This is a classic trap: d-block contraction reverses the atomic/metallic radius trend line between Al and Ga, but does NOT reverse the ionic radius sequence line where Ga3+ is larger than Al3+.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Class 11 Chemistry: The p-Block Elements
Q179 (2025)
Match the LIST-I with LIST-II.
<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;">LIST-I (Family)</th><th style="border: 1px solid #888; padding: 8px;">LIST-II (Symbol of Element)</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A. Pnicogen (group 15)</td><td style="border: 1px solid #888; padding: 8px;">I. Ts</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. Chalcogen</td><td style="border: 1px solid #888; padding: 8px;">II. Og</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. Halogen</td><td style="border: 1px solid #888; padding: 8px;">III. Lv</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. Noble gas</td><td style="border: 1px solid #888; padding: 8px;">IV. Mc</td></tr></tbody></table></div>
Choose the correct answer from the options given below :
- A-IV, B-I, C-II, D-III
- A-IV, B-III, C-I, D-II
- A-III, B-I, C-IV, D-II
- A-II, B-III, C-IV, D-I
### Related Formula
The groups of the modern periodic table correspond to standard IUPAC group families:
- Group 15 (Pnictogens): Nitrogen family
- Group 16 (Chalcogens): Oxygen family
- Group 17 (Halogens): Fluorine family
- Group 18 (Noble Gases): Helium family
### Core Logic
Map the heavy transactinide elements (Period 7) to their respective periodic groups using their atomic numbers:
- Moscovium (Mc, $Z=115$): Group 15 (Pnicogen)
- Livermorium (Lv, $Z=116$): Group 16 (Chalcogen)
- Tennessine (Ts, $Z=117$): Group 17 (Halogen)
- Oganesson (Og, $Z=118$): Group 18 (Noble Gas)
### Step 1: Match the symbols
- A (Pnicogen) $\rightarrow$ IV (Mc)
- B (Chalcogen) $\rightarrow$ III (Lv)
- C (Halogen) $\rightarrow$ I (Ts)
- D (Noble Gas) $\rightarrow$ II (Og)
This maps to A-IV, B-III, C-I, D-II, matching Option (2).
### Pattern Recognition
Modern IUPAC nomenclature adds heavy synthetic elements to complete Period 7. They correspond to group properties: $115$ is below bismuth (Pnicogen), $116$ is below polonium (Chalcogen), $117$ is below astatine (Halogen), and $118$ is below radon (Noble Gas).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q32 (2025)
The atomic number of the element from the following with the lowest $1^{\text{st}}$ ionisation enthalpy is:
- 32
- 35
- 87
- 19
### Core Logic
Let's list the identity of the elements given by their atomic numbers ($Z$):
* $Z = 19 \implies \text{Potassium (K)}$, an alkali metal in Period 4.
* $Z = 32 \implies \text{Germanium (Ge)}$, a metalloid in Group 14, Period 4.
* $Z = 35 \implies \text{Bromine (Br)}$, a halogen in Group 17, Period 4.
* $Z = 87 \implies \text{Francium (Fr)}$, an alkali metal in Period 7.
**Ionization Enthalpy Trends**:
1. Ionization enthalpy increases across a period from left to right due to an increase in effective nuclear charge.
2. Ionization enthalpy decreases down a group due to increasing atomic radius and increasing screening effects.
Comparing alkali metals ($K$ and $Fr$), Francium ($Fr, [Rn]7s^1$) is located far lower in Group 1, possessing an immense atomic size and the highest shielding. This makes its outermost valence electron exceptionally loose and effortless to remove.
### Pattern Recognition
Alkali metals always define the absolute minimum first ionization energy in their respective horizontal rows. Among alkali metals, values monotonically decrease downwards, validating Francium ($Z=87$) immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
