Given below are two statements: Statement (I) : The first ionization energy of Pb is greater than that of Sn Statement (II) : The first ionization energy of Ge is greater than that of Si. In the light of the above statements, choose the correct answer from the options given below :

Q: Given below are two statements: Statement (I) : The first ionization energy of Pb is greater than that of Sn Statement (II) : The first ionization energy of Ge is greater than that of Si. In the light

Let's analyze the first ionization energy (IE1) values for Group 14 elements (C, Si, Ge, Sn, Pb): Generally, ionization energy decreases down a group as atomic size increases. However, heavy post-transition elements exhibit an anomaly: * Analysis of Statement I: Moving from Sn to Pb, the 4f orbital subshell becomes fully filled. Because 4f electrons provide very poor shielding, the outer valence electrons experience a significantly higher effective nuclear charge (Zeff). This inert pair effect c

A. \text{Statement I is true but Statement II is false}
B. \text{Both Statement I and Statement II are false}
C. \text{Statement I is false but Statement II is true}
D. \text{Both Statement I and Statement II are true}

Solution & Explanation

### Core Logic Let's analyze the first ionization energy (

textIE_1

) values for Group 14 elements (

mathrmC, Si, Ge, Sn, Pb

): Generally, ionization energy decreases down a group as atomic size increases. However, heavy post-transition elements exhibit an anomaly: * Analysis of Statement I: Moving from

mathrmSn

mathrmPb

, the

4f

orbital subshell becomes fully filled. Because

4f

electrons provide very poor shielding, the outer valence electrons experience a significantly higher effective nuclear charge (

Z_texteff

). This inert pair effect contractive behavior tightly binds the outer electrons, making the first ionization energy of Lead higher than that of Tin:

textIE_1(mathrmPb) = 715text kJ/mol > textIE_1(mathrmSn) = 708text kJ/mol

Thus, Statement I is true. * Analysis of Statement II: Following normal periodic trends down the group, the ionization energy decreases from Silicon to Germanium due to the increasing atomic radius:

textIE_1(mathrmSi) = 786text kJ/mol > textIE_1(mathrmGe) = 761text kJ/mol

Therefore, the claim that

mathrmGe > mathrmSi

is false. ### Pattern Recognition The overall first ionization energy trend for Group 14 is:

mathrmC > Si > Ge > Pb > Sn

. Notice that Lead breaks the downward trend and has a higher ionization energy than Tin due to poor shielding by

4f

electrons (lanthanide contraction). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 11 Chemistry: The p-Block Elements

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

Q34 2025 Periodic Trends 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 atom is always endothermic C. The

1^textst

ionization energy of the boron is less than that of the beryllium D. The electronegativity of C is 2.5 in

CH_4

and

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

### 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

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

More Classification of Elements and Periodicity in Properties Questions — jee_main_2025_24_jan_evening

Practice all Classification of Elements and Periodicity in Properties previous-year questions →

YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Under Development

PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics

Latest from the Blog

View all →

Loading articles...

Review & Tools

Mistakes To Revise

No conceptual mistakes detected.

🔥 Practice Session Live

Showing All Questions

Exam Questions

Test your concepts live. Choose options or enter numerical values, then verify your answer to reveal the double-box solution matrix.

Given below are two statements: Statement (I) : The first ionization energy of Pb is greater than that of Sn Statement (II) : The first ionization energy of Ge is greater than that of Si. In the light of the above statements, choose the correct answer from the options given below :

Solution & Explanation

Reference Study Guides

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

Solution

Solution

Solution

Solution

Solution

More Classification of Elements and Periodicity in Properties Questions — jee_main_2025_24_jan_evening

We Map Every Repeating Question in Competitive Exams.

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam