Identify the molecule(X) with maximum number of lone pairs of electrons (obtained using Lewis dot structure) among HNO3, H2SO4, NF3 and O3. Choose the correct bond angle made by the central atom of th

Draw Lewis structures for all given molecules and sum the total lone pairs on all atoms. HNO3: 7 lone pairs total. H2SO4: 8 lone pairs total. O3: 6 lone pairs total. NF3: Nitrogen has 1 lone pair, and each of the three fluorine atoms has 3 lone pairs. Total = 1 + (3 x 3) = 10 lone pairs. Molecule X is NF3 due to possessing the maximum number of lone pairs (10). In NF3, Nitrogen is sp3 hybridized. Due to the high electronegativity of Fluorine, the bond pair electron density shifts towards fluorin

Identify the molecule(X) with maximum number of lone pairs of electrons (obtained using Lewis dot structure) among HNO3, H2SO4, NF3 and O3. Choose the correct bond angle made by the central atom of th

Draw Lewis structures for all given molecules and sum the total lone pairs on all atoms. HNO3: 7 lone pairs total. H2SO4: 8 lone pairs total. O3: 6 lone pairs total. NF3: Nitrogen has 1 lone pair, and each of the three fluorine atoms has 3 lone pairs. Total = 1 + (3 x 3) = 10 lone pairs. Molecule X is NF3 due to possessing the maximum number of lone pairs (10). In NF3, Nitrogen is sp3 hybridized. Due to the high electronegativity of Fluorine, the bond pair electron density shifts towards fluorin

Identify the molecule(X) with maximum number of lone pairs of electrons (obtained using Lewis dot structure) among…

Year	2026	2025	2024	Total
Questions	12	14	16	42

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

Draw Lewis structures for all given molecules and sum the total lone pairs on all atoms. $HNO₃$ $HNO_3$: 7 lone pairs total. $H₂SO₄$ $H_2SO_4$: 8 lone pairs total. $O₃$ $O_3$: 6 lone pairs total. $NF₃$ $NF_3$: Nitrogen has 1 lone pair, and each of the three fluorine atoms has 3 lone pairs. Total = $1 + (3 \times 3) = 10$ $1 + (3 \times 3) = 10$ lone pairs.

Step 1: Identifying Molecule X

Molecule X is $NF₃$ $NF_3$ due to possessing the maximum number of lone pairs (10).

Lewis Structures and Lone Pairs diagram for Q65 - JEE Main 2026 Morning

Step 2: Bond Angle Analysis

In $NF₃$ $NF_3$, Nitrogen is $sp³$ $sp^3$ hybridized. Due to the high electronegativity of Fluorine, the bond pair electron density shifts towards fluorine. This reduces bond pair-bond pair repulsion around the central nitrogen compared to ammonia ( $NH₃$ $NH_3$). As a result, the lone pair compresses the F-N-F bond angle more severely than in $NH₃$ $NH_3$ ( $107^°$ $107^\circ$). The resulting F-N-F bond angle is approximately $102^°$ $102^\circ$.

Pattern Recognition

Electronegativity rules: If the surrounding atoms are more electronegative than the central atom, bond angles decrease because bonding electrons are pulled away from the central atom, allowing the lone pair to expand further and crush the angle.

Chapter Mix

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

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 7

Q jee_main_2024_27_jan_morning Dipole Moment

Choose the polar molecule from the following:

A. $CCl₄$ $CCl_4$
B. $CO₂$ $CO_2$
C. $CH₂=CH₂$ $CH_2=CH_2$
D. $CHCl₃$ $CHCl_3$

Solution

Core Logic

CCl₄ arrow μ = 0 (Symmetrical tetrahedral)

CO₂ arrow μ = 0 (Linear structure)

CH₂=CH₂ arrow μ = 0 (Planar symmetrical structure)

For $CHCl₃$ $CHCl_3$, the individual dipole vectors do not cancel due to differing electronegativities of H and Cl, leading to a permanent non-zero dipole moment ( $μ \neq 0$ $\mu \neq 0$).

Dipole vectors structural cancellation schema for Q69 - JEE Main 2024 Morning

Pattern Recognition

Symmetry yields vector cancellation $arrow μ=0$ $\rightarrow \mu=0$. Asymmetry in $CHCl₃$ $CHCl_3$ prevents cancellation $arrow$ $\rightarrow$ polar.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q89 jee_main_2024_27_jan_morning Molecular Orbital Theory

Sum of bond order of

CO

and

NO^+

Numerical Answer. Answer: 6 to 6

Solution

Step 1: Determine the bond order of $CO$ $\text{CO}$

Carbon monoxide ( $CO$ $\text{CO}$) contains $6 + 8 = 14$ $6 + 8 = 14$ total electrons. Its structural representation is $C$ $\text{C}\equiv\text{O}$, matching a bond order value of 3.

Step 2: Determine the bond order of $NO^+$ $\text{NO}^+$

The nitrosonium ion ( $NO^+$ $\text{NO}^+$) contains $7 + 8 - 1 = 14$ $7 + 8 - 1 = 14$ total electrons. Since it is isoelectronic with $N₂$ $\text{N}_2$ and $CO$ $\text{CO}$ ( $14 electrons$ $14\text{ electrons}$), its corresponding bond order value is also 3.

Step 3: Sum the results

Sum = 3 + 3 = 6

Pattern Recognition

Isoelectronic species possessing 14 total electrons consistently demonstrate a bond order value of 3.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q81 jee_main_2024_29_jan_morning VSEPR Theory

Number of compounds with one lone pair of electrons on central atom amongst following is

O₃

H₂O

SF₄

ClF₃

NH₃

BrF₅

XeF₄

Numerical Answer. Answer: 4 to 4

Solution

Core Logic

Let us determine the steric number ( $Z$ $Z$) and number of lone pairs ( $LP$ $LP$) for the central atom in each given molecule. Formula: $Z = (1)/(2) (V + M - C + A)$ $Z = \frac{1}{2} (V + M - C + A)$ Where $V$ $V$ = valence electrons on central atom, $M$ $M$ = number of monovalent atoms, $C$ $C$ = cationic charge, $A$ $A$ = anionic charge. $LP = Z - Bond Pairs (B.P.)$ $LP = Z - \text{Bond Pairs (B.P.)}$

$O₃$ $O_3$: Central atom O ( $V=6$ $V=6$). It forms one double bond and one dative bond. It has 1 lone pair remaining.
$H₂O$ $H_2O$: Central atom O ( $V=6$ $V=6$). $Z = (1)/(2)(6 + 2) = 4$ $Z = \frac{1}{2}(6 + 2) = 4$. $LP = 4 - 2 = 2$ $LP = 4 - 2 = 2$.
$SF₄$ $SF_4$: Central atom S ( $V=6$ $V=6$). $Z = (1)/(2)(6 + 4) = 5$ $Z = \frac{1}{2}(6 + 4) = 5$. $LP = 5 - 4 = 1$ $LP = 5 - 4 = 1$ (See-saw shape).
$ClF₃$ $ClF_3$: Central atom Cl ( $V=7$ $V=7$). $Z = (1)/(2)(7 + 3) = 5$ $Z = \frac{1}{2}(7 + 3) = 5$. $LP = 5 - 3 = 2$ $LP = 5 - 3 = 2$ (T-shape).
$NH₃$ $NH_3$: Central atom N ( $V=5$ $V=5$). $Z = (1)/(2)(5 + 3) = 4$ $Z = \frac{1}{2}(5 + 3) = 4$. $LP = 4 - 3 = 1$ $LP = 4 - 3 = 1$ (Pyramidal).
$BrF₅$ $BrF_5$: Central atom Br ( $V=7$ $V=7$). $Z = (1)/(2)(7 + 5) = 6$ $Z = \frac{1}{2}(7 + 5) = 6$. $LP = 6 - 5 = 1$ $LP = 6 - 5 = 1$ (Square Pyramidal).
$XeF₄$ $XeF_4$: Central atom Xe ( $V=8$ $V=8$). $Z = (1)/(2)(8 + 4) = 6$ $Z = \frac{1}{2}(8 + 4) = 6$. $LP = 6 - 4 = 2$ $LP = 6 - 4 = 2$ (Square Planar).

Step 1: Final Counting

VSEPR Theory diagram for Q81 - JEE Main 2024 Morning

The compounds containing exactly ONE lone pair on the central atom are $O₃$ $O_3$, $SF₄$ $SF_4$, $NH₃$ $NH_3$, and $BrF₅$ $BrF_5$.

Total count = 4.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q88 jee_main_2024_29_jan_morning Molecular Orbital Theory

The number of species from the following which are paramagnetic and with bond order equal to one is

H₂, He₂^+, O₂^+, N₂²⁻, O₂²⁻, F₂, Ne₂^+, B₂

Numerical Answer. Answer: 1 to 1

Solution

Core Logic

Using Molecular Orbital (MO) Theory, we evaluate the bond order ( $BO = (Nb - Nₐ)/(2)$ $BO = \frac{N_b - N_a}{2}$) and magnetic nature (unpaired electrons = paramagnetic, all paired = diamagnetic) for each species:

Species	Magnetic behaviour	Bond order
$H₂$ $H_2$	Diamagnetic	1
$He₂^+$ $He_2^+$	Paramagnetic	0.5
$O₂^+$ $O_2^+$	Paramagnetic	2.5
$N₂²⁻$ $N_2^{2-}$	Paramagnetic	2
$O₂²⁻$ $O_2^{2-}$	Diamagnetic	1
$F₂$ $F_2$	Diamagnetic	1
$Ne₂^+$ $Ne_2^+$	Paramagnetic	0.5
$B₂$ $B_2$	Paramagnetic	1

Step 1: Final Selection

We need the species that satisfies BOTH conditions:

Paramagnetic
Bond Order = 1

Looking at the table, $B₂$ $B_2$ is the only molecule that is paramagnetic (it has 2 unpaired electrons in degenerate $π₂ₚ$ $\pi_{2p}$ orbitals) and has a bond order of 1.

Total number of such species = 1.

Pattern Recognition

$B₂$ $B_2$ (10 electrons) and $O₂$ $O_2$ (16 electrons) are the classic exceptions in MO theory that are paramagnetic despite having an even number of electrons. $B₂$ $B_2$ has BO = 1, and $O₂$ $O_2$ has BO = 2.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q jee_main_2024_30_january_evening VSEPR Theory and Molecular Shapes

The molecule/ion with square pyramidal shape is:

A. $[Ni(CN)₄]²⁻$ $\text{[Ni(CN)}_4]^{2-}$
B. $PCl₅$ $\text{PCl}_5$
C. $BrF₅$ $\text{BrF}_5$
D. $PF₅$ $\text{PF}_5$

Solution

Core Logic

According to VSEPR theory:

$[Ni(CN)₄]²⁻$ $[Ni(CN)_4]^{2-}$: $dsp²$ $dsp^2$ hybridization $arrow$ $\rightarrow$ Square Planar.
$PCl₅$ $PCl_5$: $sp³d$ $sp^3d$ hybridization with 0 lone pairs $arrow$ $\rightarrow$ Trigonal Bipyramidal.
$BrF₅$ $BrF_5$: Bromine has 7 valence electrons. It forms 5 single bonds with Fluorine, leaving 1 lone pair. $sp³d²$ $sp^3d^2$ hybridization $arrow$ $\rightarrow$ geometry is octahedral, but shape is Square Pyramidal.
$PF₅$ $PF_5$: $sp³d$ $sp^3d$ hybridization with 0 lone pairs $arrow$ $\rightarrow$ Trigonal Bipyramidal.

Square Pyramidal structure of BrF5 diagram for Q69 - JEE Main 2024 Evening

Pattern Recognition

$AX₅E₁$ $AX_5E_1$ configuration (5 bond pairs + 1 lone pair) typically yields a Square Pyramidal shape.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

Identify the molecule(X) with maximum number of lone pairs of electrons (obtained using Lewis dot structure) among HNO₃$HNO_{3}$, H₂SO₄$H_{2}SO_{4}$, NF₃$NF_{3}$ and O₃$O_{3}$. Choose the correct bond angle made by the central atom of the molecule (X).

Solution & Explanation

Core Logic

Step 1: Identifying Molecule X

Step 2: Bond Angle Analysis

Pattern Recognition

Chapter Mix

Reference Study Guides

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 7

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Step 1: Determine the bond order of CO$\text{CO}$

Step 2: Determine the bond order of NO^+$\text{NO}^+$

Step 3: Sum the results

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Counting

Chapter Mix

Solution

Core Logic

Step 1: Final Selection

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

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Identify the molecule(X) with maximum number of lone pairs of electrons (obtained using Lewis dot structure) among $HNO₃$ $HNO_{3}$, $H₂SO₄$ $H_{2}SO_{4}$, $NF₃$ $NF_{3}$ and $O₃$ $O_{3}$. Choose the correct bond angle made by the central atom of the molecule (X).

Step 1: Determine the bond order of $CO$ $\text{CO}$

Step 2: Determine the bond order of $NO^+$ $\text{NO}^+$