Chemical Bonding JEE Main 2018 PYQ with solution

This Chemical Bonding JEE Main 2018 PYQ with solution helps student understand Hybridization and Molecular Geometry. Solving More PYQs helps student improve problem-solving skills.

Question :

Identify the pair in which the geometry of the species is T – shape and square pyramidal respectively

Options :

(a) ICI₂^– and ICI₅

(b) IO₃ and IO₂F₂^–

(c) ClF₃ and IO₄^–

(d) XeOF₂ and XeOF₄

Chemical Bonding JEE Main 2018 PYQ with solution

Answer :

Correct answer : Option (d)

XeOF₂ :

The central atom is Xenon

The electronic configuration of Xenon is [Kr] 4d¹⁰ 5s² 5p⁶

The number of valence electrons in Xenon is 8

• The hybridization of Xenon in XeOF₂

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

The shape of XeOF₂ molecule is T – shape

XeOF₄ :

The central atom : Xenon

The electronic configuration of Xenon is [Kr] 4d¹⁰ 5s² 5p⁶

The number of valence electrons in Xenon is 8

• The hybridization of Xenon in XeOF₄

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

The shape of XeOF₄ molecule is square pyramidal.

Concept Involved :

• Hybridization

Practice Questions :

1. The pair of species with similar shape :

(a) PCl₃, NH₃

(b) CF₄, SF₄

(c) PbCl₂, CO₂

(d) PF₅, IF₅

Answer :

Correct answer : Option (a)

The central atom : Phosphorus

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

The number of valence electrons in phosphorus is 5

• The hybridization of phosphorus in PCl₅ :

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

The shape of PCl₃ molecule is Trigonal pyramidal

NH₃ :

The central atom : Nitrogen

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

The number of valence electrons in nitrogen is 5

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

The shape of NH₃ molecule is Trigonal pyramidal

The PCl₃, NH₃ pair of species has same shape

2. The BCl₃ is a planar molecule where as NCl₃ is pyramidal because :

(a) B-Cl bond is more polar than N-Cl bond

(b) N – Cl bond is more covalent that B – Cl bond

(c) nitrogen atom is smaller than boron atom

(d) BCl₃ has no lone pair of electrons but NCl₃ has a lone pair of electrons

Answer :

Correct answer : Option (d)

BCl₃ :

The central atom : Boron

The electronic configuration of boron is [He] 2s² 2p¹

The number of valence electrons present in boron is 3

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

The Number of lone pair present in BCl₃ molecule is 0 lone pair

NCl₃ :

The central atom : Nitrogen

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

The number of valence electrons present in nitrogen is 5

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

The number of lone pair present in NCl₃ molecule is 1 lone pair

3. Correct statement regarding molecules SF₄, CF₄ and XeF₄ are :

Options :

(a) 2, 0, and 1 lone pairs of central atom respectively

(b) 1, 0 and 1 lone pair of central atom respectively

(c) 0, 0 and 2 lone pair of central atom respectively

(d) 1, 0 and 2 lone pair of central atom respectively

Answer :

SF₄ :

The central atom is Sulphur

The electronic configuration of Sulphur is [Ne]3s²3p⁴

The Number of valence electrons present in Sulphur is 6

• Hybridization of Sulphur in SF₄

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

The number of lone pair present in central atom of SF₄ molecule is 1 lone pair

CF₄ :

The central atom is Carbon

The electronic configuration of Carbon is [He] 2s² 2p²

The Number of valence electrons present in Carbon is 4

• Hybridization of carbon in CF₄ :

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

The number of lone pair present in central atom of CF₄ is 0 lone pair

XeF₄ :

The central atom is Xenon

The electronic configuration of Xenon is [Kr] 4d¹⁰ 5s² 5p⁶

The Number of valence electrons present in Carbon is 8

• Hybridization of Xenon in XeF₄

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

The number of lone pair present in central atom of XeF₄ is 2

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