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

Q 2025 Electronegativity Trends

Electronic configuration of four elements A, B, C and D are given below : (A)

1mathrms^2 2mathrms^2 2mathrmp^3

(B)

1mathrms^2 2mathrms^2 2mathrmp^4

(C)

1mathrms^2 2mathrms^2 2mathrmp^5

(D)

1mathrms^2 2mathrms^2 2mathrmp^2

Which of the following is the correct order of increasing electronegativity (Pauling's scale)?

A. $mathrmA < D < B < C$ $\mathrm{A < D < B < C}$
B. $mathrmA < C < B < D$ $\mathrm{A < C < B < D}$
C. $mathrmA < B < C < D$ $\mathrm{A < B < C < D}$
D. $mathrmD < A < B < C$ $\mathrm{D < A < B < C}$

Solution

### Related Formula

chi_mathrmP propto Z_texteff propto frac1textAtomic Radius quad text(Across a period)

### Core Logic Let's first identify each element based on its electronic configuration: (A):

1mathrms^2 2mathrms^2 2mathrmp^3 implies textAtomic Number 7 implies textNitrogen (N)

(B):

1mathrms^2 2mathrms^2 2mathrmp^4 implies textAtomic Number 8 implies textOxygen (O)

(C):

1mathrms^2 2mathrms^2 2mathrmp^5 implies textAtomic Number 9 implies textFluorine (F)

(D):

1mathrms^2 2mathrms^2 2mathrmp^2 implies textAtomic Number 6 implies textCarbon (C)

### Step 1: Check Periodic Table Trends All four elements belong to the 2nd period. Electronegativity increases across a period from left to right because nuclear charge (

Z_texteff

) increases, and atomic radius decreases:

textCarbon (D) < textNitrogen (A) < textOxygen (B) < textFluorine (C)

On Pauling's scale, the precise values are: - Carbon (D) =

2.55

- Nitrogen (A) =

3.04

- Oxygen (B) =

3.44

- Fluorine (C) =

3.98

### Step 2: Conclusion The correct increasing order is

mathrmD < A < B < C

. ### Pattern Recognition Fluorine is the most electronegative element in the entire periodic table (Pauling electronegativity of

4.0

). Electronegativity always increases towards the top-right of the main-group elements. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Q28 2025 Periodic Trends in Halogens

The property/properties that show irregularity in first four elements of group-17 is/are : (A) Covalent radius (B) Electron affinity (C) Ionic radius (D) First ionization energy Choose the correct answer from the options given below:

A. $text(1) B and D only$ $\text{(1) B and D only}$
B. $text(2) A and C only$ $\text{(2) A and C only}$
C. $text(3) B only$ $\text{(3) B only}$
D. $text(4) A, B, C and D$ $\text{(4) A, B, C and D}$

Solution

### Related Formula Standard downward group variations typically follow predictable monotonic pathways:

textRadius propto textNumber of shells

textIonization Energy propto frac1textAtomic Size

### Core Logic Let's review the explicit values and trends across

mathrmF, Cl, Br, I

: * **Covalent radius**:

mathrmF < Cl < Br < I

(Perfect monotonic increase). * **Ionic radius**:

mathrmF^- < Cl^- < Br^- < I^-

(Perfect monotonic increase). * **First ionization energy**:

mathrmF > Cl > Br > I

(Perfect monotonic decrease). * **Electron affinity**:

mathrmCl > F > Br > I

(Irregular trend! Fluorine has an anomalously lower electron affinity than Chlorine due to high inter-electronic repulsions inside its exceptionally compact

2p

valence subshell). ### Step 1: Conclusion Therefore, only electron affinity (B) demonstrates an irregular trend line among the first four halogens. ### Pattern Recognition Fluorine anomalous properties are a classic JEE question archetype. Whenever a question asks about irregularities in halogens, immediately check Electron Gain Enthalpy (Electron Affinity) and Bond Dissociation Enthalpy, where Fluorine routinely breaks the monotonic descending order. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 12 Chemistry: The p-Block Elements

Q39 2025 Anomalous Atomic Radii in Boron Family

Given below are two statements : Statement (I): The metallic radius of Al is less than that of Ga. Statement (II): The ionic radius of

mathrmAl^3+

is less than that of

mathrmGa^3+

. In the light of the above statements, choose the most appropriate answer from the options given below:

A. $(1)\ textBoth Statement I and Statement II are incorrect$
B. $(2)\ textStatement I is incorrect but Statement II is correct$
C. $(3)\ textStatement I is correct but Statement II is incorrect$
D. $(4)\ textBoth Statement I and Statement II are correct$

Solution

### Related Formula Effective nuclear charge scaling expression layout:

Z_texteff = Z - sigma

### Core Logic Let's analyze both properties systematically: * **Statement I**: The metallic radius of Gallium (

mathrmGa

) is anomalously *smaller* than Aluminum (

mathrmAl

) due to the poor shielding effect of the

10

d-electrons inserted before it. This increases

Z_texteff

, pulling the outer shell inward tightly. Thus,

textRadius of Al > textRadius of Ga

, making Statement I incorrect. * **Statement II**: When looking at completely stripped ionic configurations (

mathrmAl^3+

and

mathrmGa^3+

), the extra shell layer in Gallium resumes its dominant role, meaning standard periodic increase down a group holds true:

mathrmAl^3+ < Ga^3+

. (Statement II is correct). ### Step 1: Evaluation Thus, Statement I is incorrect but Statement II is correct. ### Pattern Recognition This is a classic trap: d-block contraction reverses the atomic/metallic radius trend line between Al and Ga, but does NOT reverse the ionic radius sequence line where Ga3+ is larger than Al3+. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 11 Chemistry: The p-Block Elements

Q 2025 Modern Periodic Table and Elements

Match the LIST-I with LIST-II.

LIST-I (Family)	LIST-II (Symbol of Element)
A. Pnicogen (group 15)	I. Ts
B. Chalcogen	II. Og
C. Halogen	III. Lv
D. Noble gas	IV. Mc

Choose the correct answer from the options given below :

A. A-IV, B-I, C-II, D-III
B. A-IV, B-III, C-I, D-II
C. A-III, B-I, C-IV, D-II
D. A-II, B-III, C-IV, D-I

Solution

### Related Formula The groups of the modern periodic table correspond to standard IUPAC group families: - Group 15 (Pnictogens): Nitrogen family - Group 16 (Chalcogens): Oxygen family - Group 17 (Halogens): Fluorine family - Group 18 (Noble Gases): Helium family ### Core Logic Map the heavy transactinide elements (Period 7) to their respective periodic groups using their atomic numbers: - Moscovium (Mc,

Z=115

): Group 15 (Pnicogen) - Livermorium (Lv,

Z=116

): Group 16 (Chalcogen) - Tennessine (Ts,

Z=117

): Group 17 (Halogen) - Oganesson (Og,

Z=118

): Group 18 (Noble Gas) ### Step 1: Match the symbols - A (Pnicogen)

rightarrow

IV (Mc) - B (Chalcogen)

rightarrow

III (Lv) - C (Halogen)

rightarrow

I (Ts) - D (Noble Gas)

rightarrow

II (Og) This maps to A-IV, B-III, C-I, D-II, matching Option (2). ### Pattern Recognition Modern IUPAC nomenclature adds heavy synthetic elements to complete Period 7. They correspond to group properties:

115

is below bismuth (Pnicogen),

116

is below polonium (Chalcogen),

117

is below astatine (Halogen), and

118

is below radon (Noble Gas). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Q32 2025 Ionization Enthalpy

The atomic number of the element from the following with the lowest

1^textst

ionisation enthalpy is:

A. 32
B. 35
C. 87
D. 19

Solution

### Core Logic Let's list the identity of the elements given by their atomic numbers (

Z

): *

Z = 19 implies textPotassium (K)

, an alkali metal in Period 4. *

Z = 32 implies textGermanium (Ge)

, a metalloid in Group 14, Period 4. *

Z = 35 implies textBromine (Br)

, a halogen in Group 17, Period 4. *

Z = 87 implies textFrancium (Fr)

, an alkali metal in Period 7. **Ionization Enthalpy Trends**: 1. Ionization enthalpy increases across a period from left to right due to an increase in effective nuclear charge. 2. Ionization enthalpy decreases down a group due to increasing atomic radius and increasing screening effects. Comparing alkali metals (

K

and

Fr

), Francium (

Fr, [Rn]7s^1

) is located far lower in Group 1, possessing an immense atomic size and the highest shielding. This makes its outermost valence electron exceptionally loose and effortless to remove. ### Pattern Recognition Alkali metals always define the absolute minimum first ionization energy in their respective horizontal rows. Among alkali metals, values monotonically decrease downwards, validating Francium (

Z=87

) immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

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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

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