Which statements are NOT TRUE about XeO2F2? A. It has a see-saw shape. B. Xe has 5 electron pairs in its valence shell in XeO2F2. C. The O–Xe–O bond angle is close to 180. D. The F–Xe–F bond angle is close to 180. E. Xe has 16 valence electrons in XeO2F2. Choose the correct answer from the options given below.

Steric Number = Number of sigma-bonds + Number of lone pairs on central atom Evaluate the VSEPR structure of XeO2F2: Central atom Xe has 8 valence electrons. It forms 2 double bonds with Oxygen (2 x 2 = 4 electrons used) and 2 single bonds with Fluorine (2 x 1 = 2 electrons used). Total electrons used in bonding = 6. Remaining valence electrons on Xe = 8 - 6 = 2, which forms 1 lone pair. Steric Number = 4 (bond pairs) + 1 (lone pair) = 5 electron pairs. Hybridization is sp3d. Geometry is trigonal bipyramidal. According to Bent's rule, the highly electronegative F atoms occupy axial positions, while the double-bonded O atoms and the lone pair occupy equatorial positions. This results in a "see-saw" shape. A. True. The molecule has a see-saw shape. B. False. Xe is surrounded by 4 sigma-bonds, 2 pi-bonds, and 1 lone pair. In total, counting individual bonds, Xe has 7 electron pairs in its valence shell (4 sigma, 2 pi, 1 lp) = 14 valence electrons. C. False. The O atoms occupy equatorial positions, so the O-Xe-O bond angle is close to 120, not 180. D. True. The F atoms are axial, so the F-Xe-F bond angle is close to 180. E. False. The central Xe atom has 14 valence electrons involved (8 from octet, plus sharing from ligands), not 16. We need to find the INCORRECT statements, which are B, C, and E. In sp3d hybridization, bulky groups (double bonds, lone pairs) always occupy equatorial positions to minimize 90 repulsions. Highly electronegative groups (F) are forced into axial positions. Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: p-Block Elements

Which statements are NOT TRUE about XeO2F2? A. It has a see-saw shape. B. Xe has 5 electron pairs in its valence shell in XeO2F2. C. The O–Xe–O bond angle is close to 180. D. The F–Xe–F bond angle is

Steric Number = Number of sigma-bonds + Number of lone pairs on central atom Evaluate the VSEPR structure of XeO2F2: Central atom Xe has 8 valence electrons. It forms 2 double bonds with Oxygen (2 x 2 = 4 electrons used) and 2 single bonds with Fluorine (2 x 1 = 2 electrons used). Total electrons used in bonding = 6. Remaining valence electrons on Xe = 8 - 6 = 2, which forms 1 lone pair. Steric Number = 4 (bond pairs) + 1 (lone pair) = 5 electron pairs. Hybridization is sp3d. Geometry is trigona

Which statements are NOT TRUE about XeO2F2? A. It has a see-saw shape. B. Xe has 5 electron pairs in its valence shell in XeO2F2. C. The O–Xe–O bond angle is close to 180. D. The F–Xe–F bond angle is close to 180. E. Xe has 16 valence electrons in XeO2F2. Choose the correct answer from the options given below.

Steric Number = Number of sigma-bonds + Number of lone pairs on central atom Evaluate the VSEPR structure of XeO2F2: Central atom Xe has 8 valence electrons. It forms 2 double bonds with Oxygen (2 x 2 = 4 electrons used) and 2 single bonds with Fluorine (2 x 1 = 2 electrons used). Total electrons used in bonding = 6. Remaining valence electrons on Xe = 8 - 6 = 2, which forms 1 lone pair. Steric Number = 4 (bond pairs) + 1 (lone pair) = 5 electron pairs. Hybridization is sp3d. Geometry is trigonal bipyramidal. According to Bent's rule, the highly electronegative F atoms occupy axial positions, while the double-bonded O atoms and the lone pair occupy equatorial positions. This results in a "see-saw" shape. A. True. The molecule has a see-saw shape. B. False. Xe is surrounded by 4 sigma-bonds, 2 pi-bonds, and 1 lone pair. In total, counting individual bonds, Xe has 7 electron pairs in its valence shell (4 sigma, 2 pi, 1 lp) = 14 valence electrons. C. False. The O atoms occupy equatorial positions, so the O-Xe-O bond angle is close to 120, not 180. D. True. The F atoms are axial, so the F-Xe-F bond angle is close to 180. E. False. The central Xe atom has 14 valence electrons involved (8 from octet, plus sharing from ligands), not 16. We need to find the INCORRECT statements, which are B, C, and E. In sp3d hybridization, bulky groups (double bonds, lone pairs) always occupy equatorial positions to minimize 90 repulsions. Highly electronegative groups (F) are forced into axial positions. Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: p-Block Elements

Which statements are NOT TRUE about XeO2F2? A. It has a see-saw shape. B. Xe has 5 electron pairs in its valence shell in XeO2F2. C. The O–Xe–O bond angle is close to 180. D. The F–Xe–F bond angle is

Steric Number = Number of sigma-bonds + Number of lone pairs on central atom Evaluate the VSEPR structure of XeO2F2: Central atom Xe has 8 valence electrons. It forms 2 double bonds with Oxygen (2 x 2 = 4 electrons used) and 2 single bonds with Fluorine (2 x 1 = 2 electrons used). Total electrons used in bonding = 6. Remaining valence electrons on Xe = 8 - 6 = 2, which forms 1 lone pair. Steric Number = 4 (bond pairs) + 1 (lone pair) = 5 electron pairs. Hybridization is sp3d. Geometry is trigona

Which statements are NOT TRUE about XeO2F2? A. It has a see-saw shape. B. Xe has 5 electron pairs in its valence shell…

Year	2026	2025	2024	Total
Questions	16	14	13	43

Q68 jee_main_2024_01_february_morning Group 15 Elements

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

PH₃

has lower boiling point than

NH₃

. Reason (R): In liquid state

NH₃

molecules are associated through vander waal's forces, but

PH₃

molecules are associated through hydrogen bonding. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Solution

Core Logic

$NH₃$ $NH_3$ undergoes extensive intermolecular hydrogen bonding due to the high electronegativity and small size of Nitrogen. $PH₃$ $PH_3$ (Phosphine) molecules are only held together by weak van der Waals (dispersion) forces because Phosphorus is less electronegative and larger, unable to form strong hydrogen bonds.

Step 1: Evaluate Statements

Assertion (A) is correct: $PH₃$ $PH_3$ has a lower boiling point than $NH₃$ $NH_3$ because breaking H-bonds in $NH₃$ $NH_3$ requires more energy. Reason (R) is incorrect: It falsely claims $NH₃$ $NH_3$ has van der Waals association and $PH₃$ $PH_3$ has hydrogen bonding. It is exactly the opposite.

Pattern Recognition

N, O, and F are the only atoms electronegative enough to form stable hydrogen bonds in simple hydrides. Boiling point anomaly: $NH₃ > PH₃$ $NH_3 > PH_3$ purely due to H-bonding in $NH₃$ $NH_3$.

Chapter Mix

Class 12 Chemistry: The p-Block Elements Class 11 Chemistry: Chemical Bonding and Molecular Structure

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Check Statements

Pattern Recognition

Chapter Mix

Reference Study Guides

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

Solution

Core Logic

Step 1: Evaluate Statements

Pattern Recognition

Chapter Mix

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

More p-Block Elements Questions — jee_main_2026_23_january_evening

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