Among H2S, H2O, NF3, NH3 and CHCl3, identify the molecule (X) with lowest dipole moment value. The number of lone pairs…

Q: Among H2S, H2O, NF3, NH3 and CHCl3, identify the molecule (X) with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule (X) is:

Step 1: Compare dipole moments: - H2O = 1.85 D - NH3 = 1.47 D - CHCl3 = 1.04 D - H2S = 0.95 D - NF3 = 0.23 D (lowest dipole moment) Step 2: Identify molecule X = NF3. Step 3: Central atom is N (2s2 2p3). It forms 3 single bonds with F atoms and retains 1 lone pair of electrons. Sees: Comparison between NH3 and NF3 dipoles. Shortcut: In NF3, fluorine's high electronegativity pulls electron density away from the lone pair direction, resulting in an exceptionally low dipole moment (0.23 D). Class 1

Year	2026	2025	2024	Total
Questions	12	14	16	42

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 2

Q65 jee_main_2026_23_january_morning Lewis Structures and Lone Pairs

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 the molecule (X).

A. $120°$ $120^{\circ}$
B. $107°$ $107^{\circ}$
C. $102°$ $102^{\circ}$
D. $116°$ $116^{\circ}$

Solution

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

Q54 jee_main_2026_24_january_morning Dipole Moment and Formal Charge

Given below are statements about some molecules/ions. Identify the CORRECT statements. A. The dipole moment value of

NF₃

is higher than that of

NH₃

. B. The dipole moment value of

BeH₂

is zero. C. The bond order of

O₂²⁻

and

F₂

is same. D. The formal charge on the central oxygen atom of ozone is -1. E. In

NO₂

, all the three atoms satisfy the octet rule, hence it is very stable. Choose the correct answer from the options given below :

A. $A, B, C, D & E$ $\text{A, B, C, D \& E}$
B. $B & C only$ $\text{B \& C only}$
C. $B, C & D only$ $\text{B, C \& D only}$
D. $A, C & D only$ $\text{A, C \& D only}$

Solution

Core Logic

(A) Dipole moment of $NF₃$ $\mathrm{NF}_3$ is lower than that of $NH₃$ $\mathrm{NH}_3$ because in $NF₃$ $\mathrm{NF}_3$, the orbital dipole due to the lone pair and the bond dipoles (N-F) are in opposite directions, whereas in $NH₃$ $\mathrm{NH}_3$ they reinforce each other. (Statement A is false).

(B) $BeH₂$ $\mathrm{BeH}_2$ is $sp$ $sp$ hybridized, a linear molecule. The two Be-H bond dipoles cancel out, resulting in a zero net dipole moment. (Statement B is true).

(C) Bond order of $O₂²⁻$ $\mathrm{O}_2^{2-}$ (peroxide ion, $18 e^-$ $18 e^-$) $= 1$ $= 1$. Bond order of $F₂$ $\mathrm{F}_2$ ( $18 e^-$ $18 e^-$) $= 1$ $= 1$. They are isoelectronic and have the same bond order. (Statement C is true).

(D) The formal charge on the central oxygen atom in ozone ( $O₃$ $\mathrm{O}_3$) is $+1$ $+1$, not $-1$ $-1$. (Statement D is false).

(E) In $NO₂$ $\mathrm{NO}_2$, nitrogen is an odd-electron species ( $7+16 = 23 e^-$ $7+16 = 23 e^-$), thus it does not follow the octet rule. (Statement E is false).

Step 1: Final Conclusion

Only statements B and C are correct.

Pattern Recognition

Ammonia vs Nitrogen trifluoride is a classic NCERT dipole exception. Central atoms with odd electrons (like N in $NO$ $\mathrm{NO}$, $NO₂$ $\mathrm{NO}_2$, $ClO₂$ $\mathrm{ClO}_2$) inherently violate the octet rule.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q56 jee_main_2026_24_january_morning VSEPR Theory and Shapes

Among the following, the CORRECT combinations are: A.

IF₃ arrow T-shaped (sp³d)

IF₅ arrow Square pyramidal (sp³d²)

IF₇ arrow Pentagonal bipyramidal (sp³d³)

ClO₄^- arrow Square planar (sp²d)

Choose the correct answer from the options given below :

A. $A, B and C only$ $\text{A, B and C only}$
B. $A and B only$ $\text{A and B only}$
C. $A, B, C and D$ $\text{A, B, C and D}$
D. $B, C and D Only$ $\text{B, C and D Only}$

Solution

Core Logic

(A) $IF₃$ $\mathrm{IF}_3$: Iodine has 7 valence electrons. $3$ $3$ are shared with F, leaving $2$ $2$ lone pairs. Total electron pairs = $3 (bond) + 2 (lone) = 5$ . Hybridization is $sp³d$ $sp^3d$. Geometry is trigonal bipyramidal, shape is T-shaped. (Correct)

(B) $IF₅$ $\mathrm{IF}_5$: Iodine has 7 valence electrons. $5$ $5$ are shared with F, leaving $1$ $1$ lone pair. Total electron pairs = $5 (bond) + 1 (lone) = 6$ . Hybridization is $sp³d²$ $sp^3d^2$. Shape is Square pyramidal. (Correct)

(C) $IF₇$ $\mathrm{IF}_7$: Iodine has 7 valence electrons, all $7$ $7$ shared with F. Zero lone pairs. Total pairs = 7. Hybridization is $sp³d³$ $sp^3d^3$. Shape is Pentagonal bipyramidal. (Correct)

(D) $ClO₄^-$ $\mathrm{ClO}_4^-$: Chlorine forms 4 bonds (mostly double) with Oxygen, plus one extra electron overall. $4 bond regions + 0 lone pairs = 4$ . Hybridization is $sp³$ $sp^3$. Shape is Tetrahedral. (Incorrect)

Step 1: Final Conclusion

Only combinations A, B, and C are correct.

Pattern Recognition

Steric Number (SN) = (Valence $e^-$ $e^-$ on central atom + Monovalent atoms - Cation charge + Anion charge) / 2. Use SN to quickly deduce hybridization: 4 $arrow sp³$ $\rightarrow sp^3$, 5 $arrow sp³d$ $\rightarrow sp^3d$, 6 $arrow sp³d²$ $\rightarrow sp^3d^2$, 7 $arrow sp³d³$ $\rightarrow sp^3d^3$.

Chapter Mix

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

Q56 jee_main_2026_24_january_evening Molecular Orbital Theory

Pair of species among the following having same bond order as well as paramagnetic character will be-

A. $O₂⁺, N₂²⁻$ $O_{2}^{+}, N_{2}^{2-}$
B. $O₂⁻, N₂⁺$ $O_{2}^{-}, N_{2}^{+}$
C. $O₂⁺, N₂⁻$ $O_{2}^{+}, N_{2}^{-}$
D. $O₂⁻, N₂⁻$ $O_{2}^{-}, N_{2}^{-}$

Solution

Core Logic

We need to determine the bond order and magnetic nature of the given diatomic species using Molecular Orbital Theory.

Species	Bond order	Magnetic Nature
$O₂⁺$ $O_{2}^{+}$	2.5	Paramagnetic
$O₂⁻$ $O_{2}^{-}$	1.5	Paramagnetic
$N₂⁺$ $N_{2}^{+}$	2.5	Paramagnetic
$N₂⁻$ $N_{2}^{-}$	2.5	Paramagnetic
$N₂²⁻$ $N_{2}^{2-}$	2	Paramagnetic

Molecular Orbital Theory diagram for Q56 - JEE Main 2026 Evening

As seen, both $O₂^+$ $O_2^+$ and $N₂^-$ $N_2^-$ have a bond order of 2.5 and contain unpaired electrons making them paramagnetic.

Pattern Recognition

Isoelectronic species or species with equivalent valence electrons (e.g., both having 15 electrons) will exhibit identical bond orders (2.5) and have 1 unpaired electron, ensuring paramagnetism.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Among $H₂S$ $\text{H}_2\text{S}$, $H₂O$ $\text{H}_2\text{O}$, $NF₃$ $\text{NF}_3$, $NH₃$ $\text{NH}_3$ and $CHCl₃$ $\text{CHCl}_3$, identify the molecule (X) with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule (X) is:

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 2

Solution

Core Logic

Step 1: Identifying Molecule X

Step 2: Bond Angle Analysis

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

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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Among H₂S$\text{H}_2\text{S}$, H₂O$\text{H}_2\text{O}$, NF₃$\text{NF}_3$, NH₃$\text{NH}_3$ and CHCl₃$\text{CHCl}_3$, identify the molecule (X) with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule (X) is:

Solution & Explanation

Related Formula

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More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 2

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Step 1: Identifying Molecule X

Step 2: Bond Angle Analysis

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Step 1: Final Conclusion

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Among $H₂S$ $\text{H}_2\text{S}$, $H₂O$ $\text{H}_2\text{O}$, $NF₃$ $\text{NF}_3$, $NH₃$ $\text{NH}_3$ and $CHCl₃$ $\text{CHCl}_3$, identify the molecule (X) with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule (X) is: