Question :

Number of lone pairs of electrons in the central atom of SCl₂ , O₃ , ClF₂ and SF₆ respectively are

Options :

(a) 0 ,1 ,2 and 2

(b) 2, 1 , 2 and 0

(c) 1, 2, 2 and 0

(d) None of these

Chemical Bonding JEE Main 2022 PYQ with Detailed Solution.

Answer :

SCl₂ :

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 SCl₂

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

The Number of Lone pairs present in SCl₂ Molecule : 2

O₃ :

The Central atom is Oxygen

The electronic Configuration of Oxygen is 1s²2s²2p⁴

The Number of valence electrons present in Oxygen is 6

• The Hybridization of Oxygen in O₃

Hybridization : Number of $\sigma$ bond + Lone pairs = 2$\sigma$ bonds + 1Lone pair

Number of Lone pairs present in oxygen : 1

ClF₃ :

The central atom is Cl

The Electronic configuration of chlorine is [Ne]3s²3p⁵

The Number of Valence electrons present in chlorine is 7

• The Hybridization of chlorine in ClF₃

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

Number of Lone pairs present in ClF₃ Molecule is 2

SF₆:

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 SF₆

Hybridization : Number of $\sigma$ bonds + Lone pair = 6$\sigma$ bonds + 0 lone pair

Number of Lone pairs present in SF₆ Molecule is 0

Correct Answer : Option (b)

Concept Involved :

• VSEPR Theory
• Hybridization

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Practice Questions :

1. The hybridization of atomic orbitals of N in NO₂⁺, NO₃^–, and NH₄⁺ are respectively :

(a) sp, sp², sp³

(b) sp, sp³, sp²

(c) sp², sp, sp³

(d) sp², sp³, sp

Answer :

NO₂⁺

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 NO₂⁺

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

NO₃^– :

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 NO₃^–

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

NH₄⁺ :

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]

Correct answer : Option (a)

2. Dipole moment of NF₃ is smaller than :

(a) NH₃

(b) CO₂

(c) BF₃

(d) CCl₄

Answer :

Correct answer is Option (a)

NH₃ :

The central atom is Nitrogen

The electronic configuration of Nitrogen is 1s²2s²2p³

The 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 shape of NH₃ molecule is pyramidal

Here, N—H dipole moment and N – orbital lone pair dipole moment in same direction so the dipole moment NF₃ is smaller than NH₃