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 4.
Q1320 (2025)
Given below are two statements :
Statement (I) : The radii of isoelectronic species increases in the order: $\mathrm {M g} ^ {2 +} < \mathrm {N a} ^ {+} < \mathrm {F} ^ {-} < \mathrm {O} ^ {2 -}$
Statement (II) : The magnitude of electron gain enthalpy of halogen decreases in the order: $\mathrm {C l} > \mathrm {F} > \mathrm {B r} > \mathrm {I}$
- Statement I is incorrect but Statement II is correct
- Both Statement I and Statement II are incorrect.
- Statement I is correct but Statement II is incorrect
- Both Statement I and Statement II are correct
### Related Formula
$$\text{Ionic Radius } \propto \frac{1}{\text{Nuclear Charge } (Z)} \quad \text{(for Isoelectronic series)}$$
### Core Logic
Evaluating each statement systematically :
* Statement (I) is correct: $\mathrm{Mg}^{2+}, \mathrm{Na}^{+}, \mathrm{F}^{-}, \mathrm{O}^{2-}$ all possess exactly 10 electrons (isoelectronic). As the positive nuclear charge decreases ($Z = 12$ for $\mathrm{Mg}$ down to $Z = 8$ for $\mathrm{O}$), the nucleus exerts less pull on the electron cloud, causing the ionic radius to increase :
$$\mathrm{Mg}^{2+} < \mathrm{Na}^{+} < \mathrm{F}^{-} < \mathrm{O}^{2-}$$
* Statement (II) is correct: Chlorine has a higher electron gain enthalpy magnitude than fluorine due to lower electron-electron repulsion in its larger $3p$ orbital. The standard halogen trend follows:
$$\mathrm{Cl} > \mathrm{F} > \mathrm{Br} > \mathrm{I}$$
Thus, both statements are correct.
### Pattern Recognition
For species with the same number of electrons, a higher negative charge always leads to a larger electron cloud radius due to reduced nuclear traction.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q65 (2024)
In case of isoelectronic species the size of $F^-$, $Ne$ and $Na^+$ is affected by:
- $\text{Principal quantum number (n)}$
- $\text{None of the factors because their size is the same}$
- $\text{Electron-electron interaction in the outer orbitals}$
- $\text{Nuclear charge (z)}$
### Core Logic
$F^-$, $Ne$, $Na^+$ all have $1s^2, 2s^2, 2p^6$ configuration (10 electrons).
However, their atomic numbers (nuclear charge, $Z$) are different:
$F$: $Z = 9$
$Ne$: $Z = 10$
$Na$: $Z = 11$
Because they have the same number of electrons but different nuclear charges, the attraction between the nucleus and the valence shell electrons will differ.
### Step 1: Final Conclusion
Higher nuclear charge strongly attracts the isoelectronic electron cloud, decreasing the ionic radius. Hence, their size is primarily affected by the nuclear charge ($z$).
### Pattern Recognition
For isoelectronic species, size is inversely proportional to atomic number $Z$. The greater the $Z$, the smaller the size.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q88 (2024)
Among the following oxide of p-block elements, number of oxides having amphoteric nature is
$Cl_2O_7$, $CO$, $PbO_2$, $N_2O$, $NO$, $Al_2O_3$, $SiO_2$, $N_2O_5$, $SnO_2$
### Core Logic
Let's classify the nature of each given oxide:
- $Cl_2O_7$: Non-metal oxide in highest oxidation state $\rightarrow$ Strongly Acidic.
- $CO$: Neutral oxide.
- $PbO_2$: Heavy metal oxide near metalloid line $\rightarrow$ Amphoteric.
- $N_2O$: Neutral oxide.
- $NO$: Neutral oxide.
- $Al_2O_3$: Classic amphoteric oxide.
- $SiO_2$: Weakly acidic oxide.
- $N_2O_5$: Non-metal oxide $\rightarrow$ Acidic.
- $SnO_2$: Heavy metal oxide near metalloid line $\rightarrow$ Amphoteric.
### Step 1: Count Amphoteric Oxides
The amphoteric oxides in the list are $Al_2O_3$, $SnO_2$, and $PbO_2$.
Total count = 3.
### Pattern Recognition
Memorize the main neutral oxides ($N_2O, NO, CO$) and the classic amphoteric oxides ($Al_2O_3, ZnO, PbO, PbO_2, SnO, SnO_2, BeO, As_2O_3, Sb_2O_3$). High oxidation state non-metals are always acidic.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Class 12 Chemistry: The p-Block Elements
Q1486 (2024)
The element having the highest first ionization enthalpy is
- Si
- Al
- N
- C
### Related Formula
$$\text{Ionization Enthalpy } (IE_1) \propto \frac{1}{\text{Atomic Size}} \quad \text{and} \quad \text{Stable configuration enhancements.}$$
### Core Logic
Analyzing periodic trends:
1. Ionization energy increases across a period from left to right and decreases down a group.
2. Nitrogen ($N$) and Carbon ($C$) belong to Period 2, while Aluminum ($Al$) and Silicon ($Si$) belong to Period 3. Consequently, Period 2 elements have smaller atomic radii and higher ionization energies.
3. Comparing Nitrogen and Carbon, Nitrogen ($1s^2 2s^2 2p^3$) has a highly stable, half-filled $p$-subshell configuration, giving it a much higher ionization energy than Carbon.
### Step 1: Trend Layout
The overall first ionization enthalpy trend follows the sequence:
$$\mathrm{Al} < \mathrm{Si} < \mathrm{C} < \mathrm{N}$$
### Pattern Recognition
Nitrogen exhibits an exceptionally high first ionization energy due to its small size combined with a stable, half-filled $2p^3$ valence subshell.
### Evaluation Rubric / Model Answer
None
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity
Q79 (2024)
Given below are two statements:
Statement I: Fluorine has most negative electron gain enthalpy in its group.
Statement II: Oxygen has least negative electron gain enthalpy in its group.
In the light of the above statements, choose the most appropriate from the options given below.
- Both Statement I and Statement II are true.
- Statement I is true but Statement II is false
- Both Statement I and Statement II are false.
- Statement I is false but Statement II is true.
### Related Formula
Electron Gain Enthalpy ($\Delta_{eg}H$) trends in Group 16 and Group 17 elements.
### Core Logic
Evaluating both statements:
* **Statement I is false**: Due to its small size, strong inter-electronic repulsions in the compact $2p$ subshell of Fluorine limit incoming electrons compared to Chlorine. As a result, Chlorine ($\text{Cl}$) has the *most negative* electron gain enthalpy in Group 17.
* **Statement II is true**: Similarly, the exceptionally small size of the Oxygen atom creates intense electron-electron repulsions. Consequently, it has the *least negative* electron gain enthalpy among all elements in Group 16.
### Step 1: Final Assessment
Therefore, Statement I is false while Statement II is true, matching choice (4).
### Pattern Recognition
Third-period elements ($\text{Cl}$, $\text{S}$) exhibit more negative electron gain enthalpies than their second-period counterparts ($\text{F}$, $\text{O}$) due to lower inter-electronic repulsion in their larger valence shells.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity
