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 1st ionization energy of the boron is less than that of the beryllium D. The electronegativity of C is 2.5 in CH4 and CCl4 E. Li is the most electropositive among elements of group I Choose the correct answer from the options gives below

### Core Logic Let us check each criteria statement: * **A is incorrect:** Electron gain can be endothermic for stable configurations like noble gases or alkaline earth metals. * **B is correct:** Removing an electron from a stable atomic nucleus always requires input energy, hence Delta H > 0 (endothermic). * **C is correct:** Be (1s2 2s2) has a stable, fully-filled subshell configuration, making its first ionization energy higher than B (1s2 2s2 2p1) where the electron is removed from a higher energy p-orbital. * **D is incorrect:** Due to inductive withdrawal and shifting effective charge distribution, electronegativity alters slightly contextually across different molecular systems (CCl4 > CH4). * **E is incorrect:** Cesium (Cs) is the most electropositive Group 1 element. ### Step 1: Match with Choices Statements B and C are definitively evaluated to be correct, corresponding to option (1). ### Pattern Recognition Shortcut: Ionization energy is strictly endothermic (+ Delta H). Beryllium versus Boron is a classic fully-filled subshell anomaly (IE1 Be > B). Knowing these isolates option (1) immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

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

### Core Logic Let us check each criteria statement: * **A is incorrect:** Electron gain can be endothermic for stable configurations like noble gases or alkaline earth metals. * **B is correct:** Removing an electron from a stable atomic nucleus always requires input energy, hence Delta H > 0 (endothermic). * **C is correct:** Be (1s2 2s2) has a stable, fully-filled subshell configuration, making its first ionization energy higher than B (1s2 2s2 2p1) where the electron is removed from a higher

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 1st ionization energy of the boron is less than that of the beryllium D. The electronegativity of C is 2.5 in CH4 and CCl4 E. Li is the most electropositive among elements of group I Choose the correct answer from the options gives below

### Core Logic Let us check each criteria statement: * **A is incorrect:** Electron gain can be endothermic for stable configurations like noble gases or alkaline earth metals. * **B is correct:** Removing an electron from a stable atomic nucleus always requires input energy, hence Delta H > 0 (endothermic). * **C is correct:** Be (1s2 2s2) has a stable, fully-filled subshell configuration, making its first ionization energy higher than B (1s2 2s2 2p1) where the electron is removed from a higher energy p-orbital. * **D is incorrect:** Due to inductive withdrawal and shifting effective charge distribution, electronegativity alters slightly contextually across different molecular systems (CCl4 > CH4). * **E is incorrect:** Cesium (Cs) is the most electropositive Group 1 element. ### Step 1: Match with Choices Statements B and C are definitively evaluated to be correct, corresponding to option (1). ### Pattern Recognition Shortcut: Ionization energy is strictly endothermic (+ Delta H). Beryllium versus Boron is a classic fully-filled subshell anomaly (IE1 Be > B). Knowing these isolates option (1) immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

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

### Core Logic Let us check each criteria statement: * **A is incorrect:** Electron gain can be endothermic for stable configurations like noble gases or alkaline earth metals. * **B is correct:** Removing an electron from a stable atomic nucleus always requires input energy, hence Delta H > 0 (endothermic). * **C is correct:** Be (1s2 2s2) has a stable, fully-filled subshell configuration, making its first ionization energy higher than B (1s2 2s2 2p1) where the electron is removed from a higher

Which of the following statements are correct? A. The process of the addition an electron to a neutral gaseous atom is…

Year	2026	2025	2024	Total
Questions	9	16	11	36

Q65 jee_main_2024_01_february_morning Ionic Radii

In case of isoelectronic species the size of

F^-

,

Ne

and

Na^+

is affected by:

A. $Principal quantum number (n)$ $\text{Principal quantum number (n)}$
B. $None of the factors because their size is the same$
C. $Electron-electron interaction in the outer orbitals$
D. $Nuclear charge (z)$ $\text{Nuclear charge (z)}$

Solution

Core Logic

$F^-$ $F^-$, $Ne$ $Ne$, $Na^+$ $Na^+$ all have $1s², 2s², 2p⁶$ $1s^2, 2s^2, 2p^6$ 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.

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

Cl₂O₇

,

CO

,

PbO₂

,

N₂O

,

NO

,

Al₂O₃

,

SiO₂

,

N₂O₅

,

SnO₂

Numerical Answer. Answer: 3 to 3

Solution

Core Logic

Let's classify the nature of each given oxide:

$Cl₂O₇$ $Cl_2O_7$: Non-metal oxide in highest oxidation state $arrow$ $\rightarrow$ Strongly Acidic.
$CO$ $CO$: Neutral oxide.
$PbO₂$ $PbO_2$: Heavy metal oxide near metalloid line $arrow$ $\rightarrow$ Amphoteric.
$N₂O$ $N_2O$: Neutral oxide.
$NO$ $NO$: Neutral oxide.
$Al₂O₃$ $Al_2O_3$: Classic amphoteric oxide.
$SiO₂$ $SiO_2$: Weakly acidic oxide.
$N₂O₅$ $N_2O_5$: Non-metal oxide $arrow$ $\rightarrow$ Acidic.
$SnO₂$ $SnO_2$: Heavy metal oxide near metalloid line $arrow$ $\rightarrow$ Amphoteric.

Step 1: Count Amphoteric Oxides

The amphoteric oxides in the list are $Al₂O₃$ $Al_2O_3$, $SnO₂$ $SnO_2$, and $PbO₂$ $PbO_2$.

Total count = 3.

Pattern Recognition

Memorize the main neutral oxides ( $N₂O, NO, CO$ $N_2O, NO, CO$) and the classic amphoteric oxides ( $Al₂O₃, ZnO, PbO, PbO₂, SnO, SnO₂, BeO, As₂O₃, Sb₂O₃$ ). High oxidation state non-metals are always acidic.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 12 Chemistry: The p-Block Elements

Q jee_main_2024_29_january_evening Ionization Enthalpy Trends

The element having the highest first ionization enthalpy is

A. Si
B. Al
C. N
D. C

Solution

Related Formula

Ionization Enthalpy (IE₁) \propto 1Atomic Size and Stable configuration enhancements.

Core Logic

Analyzing periodic trends:

Ionization energy increases across a period from left to right and decreases down a group.
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.
Comparing Nitrogen and Carbon, Nitrogen ( $1s² 2s² 2p³$ $1s^2 2s^2 2p^3$) 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:

Al < Si < C < N

Pattern Recognition

Nitrogen exhibits an exceptionally high first ionization energy due to its small size combined with a stable, half-filled $2p³$ $2p^3$ valence subshell.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity

Q79 jee_main_2024_29_january_evening Electron Gain Enthalpy Trends

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.

A. Both Statement I and Statement II are true.
B. Statement I is true but Statement II is false
C. Both Statement I and Statement II are false.
D. Statement I is false but Statement II is true.

Solution

Related Formula

Electron Gain Enthalpy ( $ΔegH$ $\Delta_{eg}H$) 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 $2p$ $2p$ subshell of Fluorine limit incoming electrons compared to Chlorine. As a result, Chlorine ( $Cl$ $\text{Cl}$) 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 ( $Cl$ $\text{Cl}$, $S$ $\text{S}$) exhibit more negative electron gain enthalpies than their second-period counterparts ( $F$ $\text{F}$, $O$ $\text{O}$) due to lower inter-electronic repulsion in their larger valence shells.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity

Q63 jee_main_2024_27_jan_morning Oxidation States

Element not showing variable oxidation state is:

A. Bromine
B. Iodine
C. Chlorine
D. Fluorine

Solution

Core Logic

Fluorine is the most electronegative element and lacks vacant d-orbitals in its valence shell. Consequently, it exhibits only a $-1$ $-1$ oxidation state (and $0$ $0$ in elemental form) and cannot show variable positive oxidation states unlike other halogens.

Pattern Recognition

First element of a group lacks d-orbitals $arrow$ $\rightarrow$ anomalous properties $arrow$ $\rightarrow$ Fluorine only exhibits $-1$ $-1$.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Solution & Explanation

Core Logic

Step 1: Match with Choices

Pattern Recognition

Chapter Mix

More Classification of Elements and Periodicity in Properties Previous-Year Questions — Page 6

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Count Amphoteric Oxides

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Trend Layout

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Assessment

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

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