Match List-I with List-II. List-I (Molecule) List-II (Shape) (A) BrF5 (I) T-shape (B) H2O (II) See saw (C) ClF3 (III)…

Q: Match List-I with List-II. List-I (Molecule) List-II (Shape) (A) BrF5 (I) T-shape (B) H2O (II) See saw (C) ClF3 (III) Bent (D) SF4 (IV) Square pyramidal

### Core Logic Using VSEPR theory: (A) BrF5: Br has 7 valence electrons. 5 form bonds with F, leaving 1 lone pair. (5 bp + 1 lp) arrow sp3d2 hybridization arrow Square pyramidal shape. (B) H2O: O has 6 valence electrons. 2 form bonds with H, leaving 2 lone pairs. (2 bp + 2 lp) arrow sp3 hybridization arrow Bent shape. (C) ClF3: Cl has 7 valence electrons. 3 form bonds with F, leaving 2 lone pairs. (3 bp + 2 lp) arrow sp3d hybridization arrow T-shape. (D) SF4: S has 6 valence electrons. 4 form bo

Year	2026	2025	2024	Total
Questions	12	14	16	42

Year

2026

2025

2024

Total

Questions

List-I (Molecule)	List-II (Shape)
(A) $BrF₅$ $BrF_5$	(I) T-shape
(B) $H₂O$ $H_2O$	(II) See saw
(C) $ClF₃$ $ClF_3$	(III) Bent
(D) $SF₄$ $SF_4$	(IV) Square pyramidal

List-I (Molecule)

List-II (Shape)

(A)

BrF₅

(I) T-shape

(B)

H₂O

(II) See saw

(C)

ClF₃

(III) Bent

(D)

SF₄

(IV) Square pyramidal

Solution & Explanation

Core Logic

Using VSEPR theory: (A) $BrF₅$ $BrF_5$: Br has 7 valence electrons. 5 form bonds with F, leaving 1 lone pair. (5 bp + 1 lp) $arrow$ $\rightarrow$ $sp³d²$ $sp^3d^2$ hybridization $arrow$ $\rightarrow$ Square pyramidal shape. (B) $H₂O$ $H_2O$: O has 6 valence electrons. 2 form bonds with H, leaving 2 lone pairs. (2 bp + 2 lp) $arrow$ $\rightarrow$ $sp³$ $sp^3$ hybridization $arrow$ $\rightarrow$ Bent shape. (C) $ClF₃$ $ClF_3$: Cl has 7 valence electrons. 3 form bonds with F, leaving 2 lone pairs. (3 bp + 2 lp) $arrow$ $\rightarrow$ $sp³d$ $sp^3d$ hybridization $arrow$ $\rightarrow$ T-shape. (D) $SF₄$ $SF_4$: S has 6 valence electrons. 4 form bonds with F, leaving 1 lone pair. (4 bp + 1 lp) $arrow$ $\rightarrow$ $sp³d$ $sp^3d$ hybridization $arrow$ $\rightarrow$ See-saw shape.

Step 1: Matching

(A) - (IV) (B) - (III) (C) - (I) (D) - (II)

VSEPR Theory solution diagram for Q71 - JEE Main 2024 Morning

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Class 11 Chemistry: Chemical Bonding and Molecular Structure

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 9

Q77 jee_main_2024_31_jan_morning Molecular Orbital Theory

The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals A. have the same energy B. have the minimum overlap C. have same symmetry about the molecular axis D. have different symmetry about the molecular axis Choose the most appropriate from the options given below:

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

Solution

Core Logic

Conditions for the linear combination of atomic orbitals (LCAO) to form molecular orbitals:

The combining atomic orbitals must have the same or nearly the same energy.
The combining atomic orbitals must have the same symmetry about the molecular axis.
The combining atomic orbitals must overlap to the maximum extent (not minimum).

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Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q85 jee_main_2024_31_jan_morning Hybridization

The number of species from the following in which the central atom uses

sp³

hybrid orbitals in its bonding is

NH₃, SO₂, SiO₂, BeCl₂, CO₂, H₂O, CH₄, BF₃

Numerical Answer. Answer: 4 to 4

Solution

Core Logic

Analyzing the hybridization of the central atom in each species:

$NH₃$ $NH_3$: 3 bp + 1 lp = 4 electron domains $arrow sp³$ $\rightarrow sp^3$
$SO₂$ $SO_2$: 2 bp + 1 lp = 3 electron domains $arrow sp²$ $\rightarrow sp^2$
$SiO₂$ $SiO_2$: A giant covalent network where each Si is bonded to 4 oxygens tetrahedrally $arrow sp³$ $\rightarrow sp^3$
$BeCl₂$ $BeCl_2$: 2 bp + 0 lp = 2 electron domains $arrow sp$ $\rightarrow sp$
$CO₂$ $CO_2$: 2 bp + 0 lp = 2 electron domains $arrow sp$ $\rightarrow sp$
$H₂O$ $H_2O$: 2 bp + 2 lp = 4 electron domains $arrow sp³$ $\rightarrow sp^3$
$CH₄$ $CH_4$: 4 bp + 0 lp = 4 electron domains $arrow sp³$ $\rightarrow sp^3$
$BF₃$ $BF_3$: 3 bp + 0 lp = 3 electron domains $arrow sp²$ $\rightarrow sp^2$

Total species with $sp³$ $sp^3$ hybridization: $NH₃, SiO₂, H₂O, CH₄$ $NH_3, SiO_2, H_2O, CH_4$. Total count = 4.

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Class 11 Chemistry: Chemical Bonding and Molecular Structure

Match List-I with List-II.
List-I (Molecule) List-II (Shape)
(A) $BrF₅$ $BrF_5$ (I) T-shape
(B) $H₂O$ $H_2O$ (II) See saw
(C) $ClF₃$ $ClF_3$ (III) Bent
(D) $SF₄$ $SF_4$ (IV) Square pyramidal

Solution & Explanation

Core Logic

Step 1: Matching

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Solution

Core Logic

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Core Logic

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Match List-I with List-II. List-I (Molecule)List-II (Shape)(A) BrF₅$BrF_5$(I) T-shape(B) H₂O$H_2O$(II) See saw(C) ClF₃$ClF_3$(III) Bent(D) SF₄$SF_4$(IV) Square pyramidal

Solution & Explanation

Core Logic

Step 1: Matching

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More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 9

Solution

Core Logic

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Solution

Core Logic

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Match List-I with List-II.
List-I (Molecule) List-II (Shape)
(A) $BrF₅$ $BrF_5$ (I) T-shape
(B) $H₂O$ $H_2O$ (II) See saw
(C) $ClF₃$ $ClF_3$ (III) Bent
(D) $SF₄$ $SF_4$ (IV) Square pyramidal