Chemical Bonding JEE Main 2023 PYQ with Solution is explained here with a clear explanation. This question helps students understand Hybridization and VSEPR Theory for the JEE Main exam.

Question :

Match List I with List II

Choose the correct answer from the options given below :

(a) A – IV, B – III, C – II, D – I

(b) A – II, B – I, C – III, D – IV

(C) A – IV, B – I , C – II, D – III

(d) A – II, B – I, C – IV, D – III

Chemical Bonding JEE Main 2023 PYQ with Solution

Answer :

XeF₄ :

The Central atom is Xeon

The electronic configuration of Xeon is [Kr]4d¹⁰5s²5p_⁶

The Number of valence electrons present in Xeon is 8

• The Hybridization of Xeon in XeF₄

Hybridization : number of $\sigma$ bonds + lone pairs = 4$\sigma$ + 2 Lone pairs = 6$\sigma$[sp³d² hybridization ]

The shape of XeF₄ molecule is square planar

SF₄ :

The Central atom is Sulphur

The electronic configuration of Sulphur is [Ne]3s²3p⁴

The Number valence electrons present in Sulphur is 6

• The Hybridization of Sulphur in SF₄

Hybridization : Number of $\sigma$ bonds + lone pairs = 4$\sigma$ + 1 lone pair = 5$\sigma$[sp³d hybridization ]

The shape of SF₄ Molecule is see-saw

NH⁺₄ :

The central atom is Nitrogen

The electronic configuration of Nitrogen is 1s²2s²2p³

The valence electrons present in nitrogen is 5

• The hybridization of nitrogen in NH₄⁺

Hybridization : Number of $\sigma$ bonds + Lone pair = 4$\sigma$ + 0 lone pair = 4$\sigma$ [sp³ hybridization]

The shape of NH⁺₄ molecule is tetrahedral

BrF₃ :

The central atom is Bromine

The electronic configuration of Bromine is [Ar]3d¹⁰4s²4p⁵

The Number of valence electrons present in Bromine is 7

• The Hybridization of Bromine in BrF₃

Hybridization : Number of $\sigma$ bonds + Lone pair = 3$\sigma$ + 2lone pair = 5[sp³d hybridization ]

The shape of BrF₃ molecule is T- shape

Correct Answer : Option (d)

Concept Involved :

• VSEPR Theory
• Hybridization

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Practice Questions :

1. The hybridization of the central atom in ICl₂⁺ is :

Options :

(a) dsp²

(b) sp

(c) sp²

(d) sp³

Answer :

The central atom is iodine

The electronic configuration : [Kr]4d¹⁰5s²5p⁵

The number of valence electrons in iodine is 7

Hybridization : Number of $\sigma$ bonds + Lone pair = 2$\sigma$ + 2 lone pair = 4$\sigma$ [sp³ hybridization ]

Correct answer : option (d)

Concept involved :

• Hybridization

2. What is the state of hybridization of Xe in cationic part of solid XeF₆ ?

Options :

(a) sp³d³

(b) sp³d²

(c) sp³d

(d) sp³

Answer :

The central atom : Xenon

The electronic configuration : [Kr]4d¹⁰5s²5p⁶

The number of valence electrons in Xenon is 8

Hybridization : Number of $\sigma$ bonds + Lone pair = 6$\sigma$ + 1 lone pair = 7$\sigma$ [sp³d² hybridization]

Correct answer : Option (b)

Concept involved :

• Hybridization

3. Which one of the following is the correct set with respect to molecule, hybridization and shape ?

(a) BeCl₂, sp², linear

(b) BeCl₂ , sp², triangular planar

(c) BCl₃, sp², triangular planar

(d) BCl₃, sp³, tetrahedral

Answer :

Correct answer : Option (c)

The central atom is Boron

The electronic configuration of boron is [He]2s²2p¹

The number of valence electrons in boron is 3

Hybridization : Number of $\sigma$ bonds + Lone pair = 3$\sigma$ + 0 lone pair = 3$\sigma$ bonds [sp² hybridization ]

The shape of BCl₃ molecule is triangular planar

concept involved :

• Hybridization

Conclusion :

• Practicing Chemical Bonding JEE Main 2023 PYQ with Solution help student understand bonding concepts. and improves their performance in exam.