Q
jee_main_2025_29_jan_morning
Periodic Trends in Properties
An element 'E' has the ionisation enthalpy value of 374 \, mathrmkJ \, mol^-1$374 \, \mathrm{kJ \, mol^{-1}}$ . 'E' reacts with elements A, B, C and D with electron gain enthalpy values of -328$-328$ , -349$-349$ , -325$-325$ and -295 \, mathrmkJ \, mol^-1$-295 \, \mathrm{kJ \, mol^{-1}}$ , respectively.
The correct order of the products EA, EB, EC and ED in terms of ionic character is :
- A. mathrmEB > mathrmEA > mathrmEC > mathrmED$\mathrm{EB} > \mathrm{EA} > \mathrm{EC} > \mathrm{ED}$
- B. mathrmED > mathrmEC > mathrmEA > mathrmEB$\mathrm{ED} > \mathrm{EC} > \mathrm{EA} > \mathrm{EB}$
- C. mathrmEA > mathrmEB > mathrmEC > mathrmED$\mathrm{EA} > \mathrm{EB} > \mathrm{EC} > \mathrm{ED}$
- D. mathrmED > mathrmEC > mathrmEB > mathrmEA$\mathrm{ED} > \mathrm{EC} > \mathrm{EB} > \mathrm{EA}$
Solution
### Related Formula
textIonic Character propto lvert Delta H_textIE - Delta H_textEGE rvert$$\text{Ionic Character} \propto \lvert \Delta H_{\text{IE}} - \Delta H_{\text{EGE}} \rvert$$
### Core Logic
The relative ionic quality of a standard binary bond rises as the gap scale between ionization enthalpy and negative electron gain enthalpy parameters widens. Comparing the values:
* For B: Delta H_textEGE = -349 mathrm~kJ/mol$\Delta H_{\text{EGE}} = -349 \mathrm{~kJ/mol}$ (largest energy release) rightarrow$\rightarrow$ Highest ionic character.
* For A: Delta H_textEGE = -328 mathrm~kJ/mol$\Delta H_{\text{EGE}} = -328 \mathrm{~kJ/mol}$.
* For C: Delta H_textEGE = -325 mathrm~kJ/mol$\Delta H_{\text{EGE}} = -325 \mathrm{~kJ/mol}$.
* For D: Delta H_textEGE = -295 mathrm~kJ/mol$\Delta H_{\text{EGE}} = -295 \mathrm{~kJ/mol}$ (smallest energy release) rightarrow$\rightarrow$ Lowest ionic character.
Arranging them in descending order of ionic character yields:
mathrmEB gt mathrmEA gt mathrmEC gt mathrmED$$\mathrm{EB} \gt \mathrm{EA} \gt \mathrm{EC} \gt \mathrm{ED}$$
### Pattern Recognition
A highly exothermic electron gain enthalpy value favors easier anion production, widening electronegativity variations to enhance ionic bond properties.
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q
jee_main_2025_29_jan_morning
Periodic Trends in Physical and Chemical Properties
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 -$\mathrm {M g} ^ {2 +} < \mathrm {N a} ^ {+} < \mathrm {F} ^ {-} < \mathrm {O} ^ {2 -}$
Statement (II) : The magnitude of electron gain enthalpy of halogen decreases in the order: mathrm C l > mathrm F > mathrm B r > mathrm I$\mathrm {C l} > \mathrm {F} > \mathrm {B r} > \mathrm {I}$
- A. Statement I is incorrect but Statement II is correct
- B. Both Statement I and Statement II are incorrect.
- C. Statement I is correct but Statement II is incorrect
- D. Both Statement I and Statement II are correct
Solution
### Related Formula
textIonic Radius propto frac1textNuclear Charge (Z) quad text(for Isoelectronic series)$$\text{Ionic Radius } \propto \frac{1}{\text{Nuclear Charge } (Z)} \quad \text{(for Isoelectronic series)}$$
### Core Logic
Evaluating each statement systematically :
* Statement (I) is correct: mathrmMg^2+, mathrmNa^+, mathrmF^-, mathrmO^2-$\mathrm{Mg}^{2+}, \mathrm{Na}^{+}, \mathrm{F}^{-}, \mathrm{O}^{2-}$ all possess exactly 10 electrons (isoelectronic). As the positive nuclear charge decreases (Z = 12$Z = 12$ for mathrmMg$\mathrm{Mg}$ down to Z = 8$Z = 8$ for mathrmO$\mathrm{O}$), the nucleus exerts less pull on the electron cloud, causing the ionic radius to increase :
mathrmMg^2+ < mathrmNa^+ < mathrmF^- < mathrmO^2-$$\mathrm{Mg}^{2+} < \mathrm{Na}^{+} < \mathrm{F}^{-} < \mathrm{O}^{2-}$$
* Statement (II) is correct: Chlorine has a higher electron gain enthalpy magnitude than fluorine due to lower electron-electron repulsion in its larger 3p$3p$ orbital. The standard halogen trend follows:
mathrmCl > mathrmF > mathrmBr > mathrmI$$\mathrm{Cl} > \mathrm{F} > \mathrm{Br} > \mathrm{I}$$
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
jee_main_2024_01_february_morning
Ionic Radii
In case of isoelectronic species the size of F^-$F^-$, Ne$Ne$ and Na^+$Na^+$ is affected by:
- A. textPrincipal quantum number (n)$\text{Principal quantum number (n)}$
- B. textNone of the factors because their size is the same$\text{None of the factors because their size is the same}$
- C. textElectron-electron interaction in the outer orbitals$\text{Electron-electron interaction in the outer orbitals}$
- D. textNuclear charge (z)$\text{Nuclear charge (z)}$
Solution
### Core Logic
F^-$F^-$, Ne$Ne$, Na^+$Na^+$ all have 1s^2, 2s^2, 2p^6$1s^2, 2s^2, 2p^6$ configuration (10 electrons).
However, their atomic numbers (nuclear charge, Z$Z$) are different:
F$F$: Z = 9$Z = 9$
Ne$Ne$: Z = 10$Z = 10$
Na$Na$: Z = 11$Z = 11$
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$z$).
### Pattern Recognition
For isoelectronic species, size is inversely proportional to atomic number Z$Z$. The greater the Z$Z$, the smaller the size.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q88
jee_main_2024_01_february_morning
Periodic Trends in Chemical Properties
Among the following oxide of p-block elements, number of oxides having amphoteric nature is
Cl_2O_7$Cl_2O_7$, CO$CO$, PbO_2$PbO_2$, N_2O$N_2O$, NO$NO$, Al_2O_3$Al_2O_3$, SiO_2$SiO_2$, N_2O_5$N_2O_5$, SnO_2$SnO_2$
Numerical Answer. Answer: 3 to 3
Solution
### Core Logic
Let's classify the nature of each given oxide:
- Cl_2O_7$Cl_2O_7$: Non-metal oxide in highest oxidation state rightarrow$\rightarrow$ Strongly Acidic.
- CO$CO$: Neutral oxide.
- PbO_2$PbO_2$: Heavy metal oxide near metalloid line rightarrow$\rightarrow$ Amphoteric.
- N_2O$N_2O$: Neutral oxide.
- NO$NO$: Neutral oxide.
- Al_2O_3$Al_2O_3$: Classic amphoteric oxide.
- SiO_2$SiO_2$: Weakly acidic oxide.
- N_2O_5$N_2O_5$: Non-metal oxide rightarrow$\rightarrow$ Acidic.
- SnO_2$SnO_2$: Heavy metal oxide near metalloid line rightarrow$\rightarrow$ Amphoteric.
### Step 1: Count Amphoteric Oxides
The amphoteric oxides in the list are Al_2O_3$Al_2O_3$, SnO_2$SnO_2$, and PbO_2$PbO_2$.
Total count = 3.
### Pattern Recognition
Memorize the main neutral oxides (N_2O, NO, CO$N_2O, NO, CO$) and the classic amphoteric oxides (Al_2O_3, ZnO, PbO, PbO_2, SnO, SnO_2, BeO, As_2O_3, Sb_2O_3$Al_2O_3, ZnO, PbO, PbO_2, SnO, SnO_2, BeO, As_2O_3, Sb_2O_3$). 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
Q76
jee_main_2024_29_january_evening
Ionization Enthalpy Trends
The element having the highest first ionization enthalpy is
Solution
### Related Formula
textIonization Enthalpy (IE_1) propto frac1,
textAtomic Size quad textand quad textStable configuration enhancements.$$\text{Ionization Enthalpy } (IE_1) \propto \frac{1},
{\text{Atomic Size}} \quad \text{and} \quad \text{Stable configuration enhancements.}$$
### Core Logic
Analyzing periodic trends:
1. Ionization energy increases across a period from left to right and decreases down a group.
2. Nitrogen (N$N$) and Carbon (C$C$) belong to Period 2, while Aluminum (Al$Al$) and Silicon (Si$Si$) 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$1s^2 2s^2 2p^3$) has a highly stable, half-filled p$p$-subshell configuration, giving it a much higher ionization energy than Carbon.
### Step 1: Trend Layout
The overall first ionization enthalpy trend follows the sequence:
mathrmAl < mathrmSi < mathrmC < mathrmN$$\mathrm{Al} < \mathrm{Si} < \mathrm{C} < \mathrm{N}$$
### Pattern Recognition
Nitrogen exhibits an exceptionally high first ionization energy due to its small size combined with a stable, half-filled 2p^3$2p^3$ valence subshell.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity
