Chemical Bonding JEE Main 2021 PYQ with solution

Question :

The hybridization of the atomic orbitals of nitrogen in NO₂^–, NO₂⁺ and NH₄⁺ respectively are :

Options :

(a) sp³, sp², and sp

(b) sp, sp², and sp³

(c) sp³, sp and sp²

(d) sp², sp and sp³

Chemical Bonding JEE Main 2021 PYQ with solution

Answer :

NO₂^– :

The central atom : Nitrogen

The electronic configuration of nitrogen is 1s²2s²2p³

The number of valence electrons present in nitrogen is 5

• Hybridization of nitrogen in NO₂^–

Hybridization : Number of $\sigma$ bonds + Lone pair = 2$\sigma$ bonds + 1 lone pair = 3$\sigma$ [ sp² hybridization ]

NO₂⁺ :

The central atom : Nitrogen

The electronic configuration of nitrogen is 1s²2s²2p³

The number of valence electrons present in nitrogen is 5

Hybridization of nitrogen in NO₂⁺ :

Hybridization : Number of $\sigma$ bonds + Lone pair = 2$\sigma$ bonds + 0 lone pair = 2$\sigma$ [ sp hybridization ]

NH₄⁺ :

The central atom : Nitrogen

The electronic configuration of nitrogen is 1s²2s²2p³

The number of valence electrons present in nitrogen is 5

• Hybridization of nitrogen in NH₄⁺

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

Correct answer : Option (d)

Concept involved :

• Hybridization

Practice Questions :

1. The shape of the noble gas compound XeF₄ is :

(a) square planar

(b) distorted tetrahedral

(c) tetrahedral

(d) octahedral

Answer :

XeF₄ :

The central atom : Xenon

The electronic configuration of xenon : [Kr] 4d¹⁰ 5s² 5p⁶

The number of valence electrons present in xenon is 8

• The hybridization of XeF₄ molecule :

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

The shape of XeF₄ molecule is square planar

correct answer : Option (a)

2. Which of the following statements is incorrect for PCl₅ ?

(a) Its three P – Cl bond length are equal

(b) It involves sp³d hybridization

(c) It has linear geometry

(d) Its shape is trigonal bipyramidal

Answer :

The central atom : phosphorus

The electronic configuration of phosphorus is [Ne] 3s² 3p³

The Number of valence electrons in Phosphorus is 5

• Hybridization of PCl₅ molecule :

Hybridization : Number of $\sigma$ bonds + lone pair = 5$\sigma$ bonds + 0 lone pair = 5$\sigma$ [ sp³d hybridization ]

Correct answer : Option (c)

The incorrect statement of PCl₅ molecule is linear geometry

3. The hybridization of atomic orbitals of central atom ” Xe” in XeO₄, XeO₂F₂ and XeOF₄ respectively.

(a) sp³, sp³d², sp³d²

(b) sp³d, sp³d, sp³d²

(c) sp³, sp³d², sp³d

(d) sp³, sp³d, sp³d²

Answer :

XeO₄ :

The central atom : Xenon

The electronic configuration of xenon : [Kr] 4d¹⁰ 5s² 5p⁶

The number of valence electrons present in xenon is 8

• The hybridization of XeO₄ molecule

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

XeO₂F₂ :

The central atom : Xenon

The electronic configuration of xenon : [Kr] 4d¹⁰ 5s² 5p⁶

The number of valence electrons present in xenon is 8

• The hybridization of XeO₂F₂ molecule

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

XeOF₄ :

The central atom : Xenon

The electronic configuration of xenon : [Kr] 4d¹⁰ 5s² 5p⁶

The number of valence electrons present in xenon is 8

• The hybridization of XeOF₄ molecule

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

Correct answer : Option (d)

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