JEE Main 2023 Chemical Bonding PYQ with Solution

This JEE Main 2023 Chemical Bonding PYQ with Solution helps students learn hybridization and VSEPR Theory for JEE Main exam. Practicing PYQs improves problem- solving skills and helps students prepare better for the JEE Main.

Question :

Match List – I with List – II

Options :

(a) A – II, B – III, C – IV, D – I

(b) A – IV, B – III, C – II, D – I

(c) A – II, B – I , C – IV, D – III

(d) A – IV, B – I, C – II, D – III

JEE Main 2023 Chemical Bonding PYQ with Solution

Answer :

IF₇ :

The central atom is Iodine

The electronic configuration of iodine is [Kr]4d¹⁰5s²5p⁵

The Number of valence electrons present in iodine is 7

• Hybridization of IF₇ molecule

Hybridization : Number of $\sigma$ bonds + Lone pair = 7$\sigma$ + 0 Lone pair

The Number of Lone pair present in IF₇ molecule is ‘zero’

ICl^–₄ :

The central atom is iodine

The electronic configuration is [Kr]4d¹⁰5s²5p⁵

The valence electrons present in iodine is 7

Hybridization : Number of $\sigma$ + Lone pair = 4$\sigma$ + 2 lone pair = 6$\sigma$ [sp³d²]

The Number of lone pair present in ICl₄^– molecule is ‘Two’

XeF₆ :

The central atom is Xenon

The electronic configuration of Xeon is [Kr]4d¹⁰5s²5p⁶

The number of valence electrons present in Xenon is 8

Hybridization : Number of $\sigma$ bonds + Lone pair = 6$\sigma$ + 1lone pair

The number of lone pair present in XeF₆ molecule is ‘one’

XeF₂ :

The central atom is Xenon

The electronic configuration of Xenon is [Kr]4d¹⁰5s²5p⁶

The number of valence electrons present in Xenon is 8

Hybridization : Number of $\sigma$ bonds + Lone pair = 2$\sigma$ + 3 lone pair

The number of lone pair present in XeF₂ molecule is ‘Three’

Correct answer : Option(b)

Concept Involved :

• Hybridization
• VSEPR Theory

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Practice Questions :

1. The hybridization of central iodine atom in IF₅, I₃^– and I₃⁺ respectively :

(a) sp³d², sp³d , sp³

(b) sp³d , sp³d, sp³

(c) sp³d², sp³d², sp³

(d) sp³d, sp³d², sp³

Answer :

IF₅ :

The central atom : Iodine

Electronic configuration : [Kr] 4d¹⁰ 5s² 5p⁵

The number of valence electrons in Iodine is 7

Hybridization : Number of $\sigma$ bonds + Lone pair = 5$\sigma$ + 1 lone pair = 6$\sigma$ [ sp³d² hybridization ]

I^–₃ :

The central atom : Iodine

Electronic configuration : [Kr] 4d¹⁰ 5s² 5p⁵

The number of valence electrons in iodine is 7

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

I₃⁺ :

The central atom : Iodine

The electronic configuration : [Kr] 4d¹⁰ 5s² 5p⁵

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 (a)

Concept involved :

• Hybridization

2. Which bonds are formed by a carbon atom with sp² hybridization ?

(a) 4$\pi$ bonds

(b) 2$\pi$ bonds and 2$\sigma$ bonds

(c) 1$\pi$ bonds and 3$\sigma$ bonds

(d) 4$\sigma$ bonds

Answer :

1$\pi$ bond and 3$\sigma$ bonds = sp² hybridization

Correct answer : option (c)

3. The H——O——H bond angles in H₃O⁺ are approximately 107^o The orbitals used by oxygen in these bonds are best described as :

(a) p orbitals

(b) sp hybrid orbitals

(c) sp² hybrid orbital

(d) sp³ hybrid orbital

Answer :

The central atom is Oxygen

The electronic configuration : [He] 2s² 2p⁴

The number of valence electrons present in oxygen is 6

Hybridization : Number of $\sigma$ bonds + Lone pair = 3$\sigma$ bonds + 1 lone pair = 4$\sigma$ [ sp³ hybridization ]

Correct answer : Option (d)