Given below are two statements: Statement I: H2Se is more acidic than H2Te. Statement II: H2Se has higher bond enthalpy for dissociation than H2Te. In the light of the above statements, choose the correct answer from the options given below:

Let us analyze the periodic properties of chalcogen hydrides (Group 16): 1. **Bond Dissociation Enthalpy (DeltadisH)**: As we descend the group from Selenium to Tellurium, the size of the central atom increases significantly (rTe > rSe). This increase in size leads to poorer orbital overlap with the small 1s orbital of hydrogen, resulting in a longer and weaker M-H bond. Consequently, the bond dissociation enthalpy decreases: **Thus, Statement II is true.** 2. **Acidic Strength**: A weaker bond dissociates more easily in aqueous solution to release H^+ ions. Since the Te-H bond is weaker than the Se-H bond, H2Te releases protons much more readily than H2Se, making it a stronger acid: **Thus, Statement I is false.** For binary hydrides down any group (like Group 15, 16, or 17), atomic size increase weakens the covalent bond. A weaker bond releases protons more effectively, meaning that both **acidic strength and reducing character increase down the group**, while thermal stability decreases. Class 12 Chemistry: p-Block Elements

Given below are two statements: Statement I: H2Se is more acidic than H2Te. Statement II: H2Se has higher bond enthalpy for dissociation than H2Te. In the light of the above statements, choose the cor

Let us analyze the periodic properties of chalcogen hydrides (Group 16): 1. **Bond Dissociation Enthalpy (DeltadisH)**: As we descend the group from Selenium to Tellurium, the size of the central atom increases significantly (rTe > rSe). This increase in size leads to poorer orbital overlap with the small 1s orbital of hydrogen, resulting in a longer and weaker M-H bond. Consequently, the bond dissociation enthalpy decreases: **Thus, Statement II is true.** 2. **Acidic Strength**: A weaker bond

Given below are two statements: Statement I: H2Se is more acidic than H2Te. Statement II: H2Se has higher bond enthalpy for dissociation than H2Te. In the light of the above statements, choose the correct answer from the options given below:

Let us analyze the periodic properties of chalcogen hydrides (Group 16): 1. **Bond Dissociation Enthalpy (DeltadisH)**: As we descend the group from Selenium to Tellurium, the size of the central atom increases significantly (rTe > rSe). This increase in size leads to poorer orbital overlap with the small 1s orbital of hydrogen, resulting in a longer and weaker M-H bond. Consequently, the bond dissociation enthalpy decreases: **Thus, Statement II is true.** 2. **Acidic Strength**: A weaker bond dissociates more easily in aqueous solution to release H^+ ions. Since the Te-H bond is weaker than the Se-H bond, H2Te releases protons much more readily than H2Se, making it a stronger acid: **Thus, Statement I is false.** For binary hydrides down any group (like Group 15, 16, or 17), atomic size increase weakens the covalent bond. A weaker bond releases protons more effectively, meaning that both **acidic strength and reducing character increase down the group**, while thermal stability decreases. Class 12 Chemistry: p-Block Elements

Given below are two statements: Statement I: H2Se is more acidic than H2Te. Statement II: H2Se has higher bond enthalpy for dissociation than H2Te. In the light of the above statements, choose the cor

Let us analyze the periodic properties of chalcogen hydrides (Group 16): 1. **Bond Dissociation Enthalpy (DeltadisH)**: As we descend the group from Selenium to Tellurium, the size of the central atom increases significantly (rTe > rSe). This increase in size leads to poorer orbital overlap with the small 1s orbital of hydrogen, resulting in a longer and weaker M-H bond. Consequently, the bond dissociation enthalpy decreases: **Thus, Statement II is true.** 2. **Acidic Strength**: A weaker bond

Given below are two statements: Statement I: H2Se is more acidic than H2Te. Statement II: H2Se has higher bond enthalpy…

Year	2026	2025	2024	Total
Questions	16	13	13	42

Q jee_main_2024_29_january_evening Anomalous Behaviour of Oxygen

Anomalous behaviour of oxygen is due to its

A. Large size and high electronegativity
B. Small size and low electronegativity
C. Small size and high electronegativity
D. Large size and low electronegativity

Solution

Related Formula

Anomalous properties of second-period elements

Core Logic

The anomalous properties of oxygen compared to other chalcogens stem directly from its position in the second period of the periodic table. It is characterized by:

An exceptionally small atomic radius.
Highly pronounced electronegativity.
Complete absence of low-energy valence $d$ $d$-orbitals.

Step 1: Selection Verification

Therefore, the combination of small size and high electronegativity is the correct choice, matching option C.

Pattern Recognition

All first members of periodic blocks ( $N$ $\text{N}$, $O$ $\text{O}$, $F$ $\text{F}$) deviate significantly from their heavier group members due to their high charge density, high electronegativity, and lack of $d$ $d$-orbitals.

Chapter Mix

Class 11 Chemistry: p-Block Elements

Q73 jee_main_2024_27_jan_morning Properties of Boron

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Melting point of Boron (

2453 K

) is unusually high in group 13 elements. Reason (R) : Solid Boron has very strong crystalline lattice. In the light of the above statements, choose the most appropriate answer from the options given below;

A. Both (A) and (R) are correct but (R) is not the correct explanation of (A)
B. Both (A) and (R) are correct and (R) is the correct explanation of (A)
C. (A) is true but (R) is false
D. (A) is false but (R) is true

Solution

Core Logic

Boron forms a highly compact, robust icosahedral covalent polymeric three-dimensional framework structure ( $B₁₂$ $B_{12}$ units). This extremely solid, dense crystalline lattice organization requires immense thermal activation energy to rupture, explaining why its melting point ( $2453 K$ $2453\text{ K}$) is uniquely elevated among Group 13 elements. Both statements are true and (R) is the perfect explanation.

Chapter Mix

Class 11 Chemistry: p-Block Elements

Q90 jee_main_2024_27_jan_morning Oxidation States of Sulphur

From the given list, the number of compounds with +4 oxidation state of Sulphur:

SO₃

,

H₂SO₃

,

SOCl₂

,

SF₄

,

BaSO₄

,

H₂S₂O₇

Numerical Answer. Answer: 3 to 3

Solution

Step 1: Audit oxidation numbers individually

Compound	Oxidation State of Sulphur Calculation
$SO₃$ $\text{SO}_3$	$x + 3(-2) = 0 x = +6$ $x + 3(-2) = 0 \implies x = +6$
$H₂SO₃$ $\text{H}_2\text{SO}_3$	$2(+1) + x + 3(-2) = 0 x = +4$ $2(+1) + x + 3(-2) = 0 \implies x = +4$
$SOCl₂$ $\text{SOCl}_2$	$x + (-2) + 2(-1) = 0 x = +4$ $x + (-2) + 2(-1) = 0 \implies x = +4$
$SF₄$ $\text{SF}_4$	$x + 4(-1) = 0 x = +4$ $x + 4(-1) = 0 \implies x = +4$
$BaSO₄$ $\text{BaSO}_4$	$+2 + x + 4(-2) = 0 x = +6$ $+2 + x + 4(-2) = 0 \implies x = +6$
$H₂S₂O₇$ $\text{H}_2\text{S}_2\text{O}_7$	$2(+1) + 2x + 7(-2) = 0 2x = 12 x = +6$

Step 2: Sum the targets

The compounds displaying an exact $+4$ $+4$ assignment are $H₂SO₃$ $\text{H}_2\text{SO}_3$, $SOCl₂$ $\text{SOCl}_2$, and $SF₄$ $\text{SF}_4$. The total number is 3.

Pattern Recognition

Sulfurous derivatives, thionyl groupings, and tetrafluoride configurations typically feature the $+4$ $+4$ oxidation level state.

Chapter Mix

Class 11 Chemistry: Redox Reactions Class 12 Chemistry: p-Block Elements

Q65 jee_main_2024_29_jan_morning Group 14 Elements Physical Properties

Given below are two statements : Statement I : The electronegativity of group 14 elements from Si to Pb gradually decreases. Statement II : Group 14 contains non-metallic, metallic, as well as metalloid elements. In the light of the above statements, choose the most appropriate from the options given below:

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

Solution

Core Logic

Analyzing Statement I: The electronegativity values for Group 14 elements according to the Pauling scale are approximately:

Carbon (C): $2.5$ $2.5$
Silicon (Si): $1.8$ $1.8$
Germanium (Ge): $1.8$ $1.8$
Tin (Sn): $1.8$ $1.8$
Lead (Pb): $1.9$ $1.9$

The electronegativity values from Si to Pb are almost identical, and it slightly increases at Pb due to the poor shielding effect of d and f-orbitals (inert pair effect). It does not "gradually decrease." Therefore, Statement I is false.

Analyzing Statement II: Group 14 consists of:

Carbon (C): Non-metal
Silicon (Si) & Germanium (Ge): Metalloids
Tin (Sn) & Lead (Pb): Metals

Therefore, the group contains non-metals, metalloids, and metals. Statement II is true.

Step 1: Final Conclusion

Statement I is false, but Statement II is true.

Pattern Recognition

Electronegativity in Group 13 and 14 does not follow a strict linear decrease due to d-block and f-block contraction (poor shielding by d and f electrons).

Chapter Mix

Class 11 Chemistry: The p Block Elements

Q65 jee_main_2024_30_january_evening Group 16 Elements

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A:

H₂Te

is more acidic than

H₂S

. Reason R: Bond dissociation enthalpy of

H₂Te

is lower than

H₂S

. In the light of the above statements, Choose the most appropriate from the options given below.

A. $Both A and R are true but R is NOT the correct explanation of A.$
B. $Both A and R are true and R is the correct explanation of A.$
C. $A is false but R is true.$ $\text{A is false but R is true.}$
D. $A is true but R is false.$ $\text{A is true but R is false.}$

Solution

Core Logic

As we move down Group 16, the atomic size of the central atom increases. The increased size of Tellurium compared to Sulphur leads to a longer and weaker Element-Hydrogen bond.

Consequently, the bond dissociation enthalpy of $H₂Te$ $H_2Te$ is lower than that of $H₂S$ $H_2S$. Because the $Te-H$ $Te-H$ bond is weaker and more easily broken, it ionizes to release $H^+$ $H^+$ ions more readily than $H₂S$ $H_2S$.

Thus, $H₂Te$ $H_2Te$ is more acidic than $H₂S$ $H_2S$, making both the assertion and reason true, with the reason correctly explaining the assertion.

Pattern Recognition

Down the group for p-block hydrides: Size increases → Bond length increases → Bond strength decreases → Acidity increases.

Chapter Mix

Class 12 Chemistry: The p Block Elements

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More p-Block Elements Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Selection Verification

Pattern Recognition

Chapter Mix

Solution

Core Logic

Chapter Mix

Solution

Step 1: Audit oxidation numbers individually

Step 2: Sum the targets

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

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