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 $1^textst$ $1^{\text{st}}$ ionization energy of the boron is less than that of the beryllium D. The electronegativity of C is 2.5 in $CH_4$ $CH_{4}$ and $CCl_4$ $CCl_{4}$ E. Li is the most electropositive among elements of group I Choose the correct answer from the options gives below

A. B and C only
B. A, C and D only
C. B and D only
D. B, C and E only

Solution & Explanation

### 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:**

textBe \ (1s^2 2s^2)

has a stable, fully-filled subshell configuration, making its first ionization energy higher than

textB \ (1s^2 2s^2 2p^1)

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 (

CCl_4 > CH_4

). * **E is incorrect:** Cesium (

textCs

) 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 (

textIE_1 text Be > textB

). Knowing these isolates option (1) immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Reference Study Guides

Classification Of Elements And Periodicity In Properties

More Classification of Elements and Periodicity in Properties Previous-Year Questions — Page 2

Q43 2025 Periodic Trends in Atomic Radii

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_2O_3

(2)

AO_2

(3) AO (4)

A_2O

A. $A_2O_3$ $A_{2}O_{3}$
B. $AO_2$ $AO_{2}$
C. AO
D. $A_2O$ $A_{2}O$

Solution

### 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_texteff

). * 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 Atomic Radii Trends

The incorrect decreasing order of atomic radii is :

A. $mathrmMg > mathrmAl > mathrmC > mathrmO$
B. $mathrmAl > mathrmB > mathrmN > mathrmF$
C. $mathrmBe > mathrmMg > mathrmAl > mathrmSi$
D. $mathrmSi > mathrmP > mathrmCl > mathrmF$

Solution

### Related Formula Atomic radius decreases across a period due to increase in effective nuclear charge (

Z_texteff

) and increases down a group due to addition of new electronic shells. ### Core Logic Let us analyze the elements in option (3): -

mathrmBe

and

mathrmMg

belong to Group 2. Down the group, atomic radius increases:

mathrmMg > mathrmBe

. -

mathrmMg

mathrmAl

, and

mathrmSi

belong to Period 3. Across the period, atomic radius decreases:

mathrmMg > mathrmAl > mathrmSi

. Combining these trends, the correct decreasing order is:

mathrmMg > mathrmBe > mathrmAl > mathrmSi

Therefore, the order given in option (3)

mathrmBe > mathrmMg > mathrmAl > mathrmSi

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;

mathrmMg

is larger than

mathrmBe

, making

mathrmBe > mathrmMg

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

Q40 2025 Atomic and Ionic Radii

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:

A. 31
B. 49
C. 13
D. 81

Solution

### 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:

textCorrect 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 null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 12 Chemistry: The p-Block Elements

Q31 2025 Atomic Radii Trends

Choose the incorrect trend in the atomic radii (

r

) of the elements:

A. $rtextBr < rtextK$ $r{\text{Br}} < r{\text{K}}$
B. $rtextMg < rtextAl$ $r{\text{Mg}} < r{\text{Al}}$
C. $r_textRb < r_textCs$ $r_{\text{Rb}} < r_{\text{Cs}}$
D. $r_textAt < r_textCs$ $r_{\text{At}} < r_{\text{Cs}}$

Solution

### Related Formula

textAtomic Radius (r) propto frac1Z_texteff quad text(Across a period)

textAtomic Radius (r) propto textNumber 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_texteff

) increases, drawing electrons closer to the nucleus. Hence, atomic radius decreases. -

textMg

and

textAl

sit in Period 3. Since

textAl

is further to the right (

Z=13

) than

textMg

(

Z=12

), the radius of

textAl

is smaller than that of

textMg

r_textMg > r_textAl

Thus, the stated trend

r_textMg < r_textAl

is incorrect. ### Step 1: Cross-Checking Valid Trends -

r_textBr < r_textK

: Correct, as potassium sits at the start of Period 4, bromine sits near the end. -

r_textRb < r_textCs

: Correct, atomic size increases down Group 1. -

r_textAt < r_textCs

: Correct, cesium sits far below and left compared to astatine. ### Pattern Recognition Period 3 size ranking baseline:

textNa > textMg > textAl > textSi > textP > textS > textCl

. 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

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