JEE Main Chemical Bonding 2016 PYQ with solution

Question :

The group of molecules having identical shape is

Options :

(a) PCI₅, IF₅, XeO₂F₂

(b) BF₃ , PCI₃, XeO₃

(c) SF₄,XeF₄,CCI₄

(d) CIF₃, XeOF_2, XeF₃⁺

JEE Main Chemical Bonding 2016 PYQ with solution

Answer :

Correct answer : option (d)

CIF₃ :

The central atom : Chlorine

The electronic configuration of chlorine is [Ne] 3s² 3p⁵

The number of valence electrons present in chlorine is 7

Hybridization : Number of $\sigma$ bonds + lone pairs = 3$\sigma$ bonds + 2 lone pairs = 5$\sigma$ { Sp³d hybridization }

The shape of CIF₃ molecule is T shape

XeOF₂ :

The central atom : Xenon

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

The number of valence electrons in Xenon is 8

Hybridization : Number of $\sigma$ bonds + lone pairs = 3$\sigma$ bonds + 2 lone pairs = 5$\sigma$ { Sp³d hybridization }

The shape of XeOF₂ molecules is T shape

XeF₃⁺ :

The central atom : Xenon

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

The number of valence electrons in Xenon is 8

Hybridization : Number of $\sigma$ bonds + lone pairs = 3$\sigma$ bonds +2 lone pairs = 5$\sigma$ { Sp³d hybridization }

The shape of XeF₃⁺ molecule is T shape

Practice question :

Which of the following pairs of species have identical shapes ?

(a) NO₂ and NO₂^–

(b) PCl₅ and BrF₅

(c) XeF₄ and ICI₄^–

(d) TeCI₄ and XeO₄

Answer :

Correct answer : option (C)

XeF₄ :

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 XeF₄

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

The shape of XeF₄ molecule is T shape

ICl₄^– :

The central atom : Iodine

The electronic configuration of Iodine is [Kr] 4d¹⁰ 5s² 5p⁵

The number of valence electrons in iodine is 7

• The hybridization of ICl₄^–

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

The shape of ICl₄^– molecule is T shape

2. Which is not correctly matched ?

(a) XeO₃ – Trigonal bipyramidal

(b) CIF₃ – bent T – shape

(c) XeOF₄ – Square pyramidal

(d) XeF₂ – Linear shape

Answer :

Correct answer : Option (a)

XeO₃ :

The central atom is Xenon

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

The number of valence electrons in Xenon is 8

• The hybridization of XeO₃ molecule

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

The shape XeO₃ molecule is Trigonal pyramidal not the trigonal bipyramidal

3. In which of the following pairs, both the species have the same hybridization ?

(I) SF₄ , XeF₄

(II) I₃^– , XeF₂

(III) ICl₄⁺, SiCl₄

(IV) ClO₃^–, PO₄^3-

(a) I, II

(b) II, III

(c) II, IV

(d) I, II, III

Answer :

I₃^– :

The central atom : Iodine

The electronic configuration of iodine is [Kr] 4d¹⁰ 5s² 5p⁵

The number of valence electrons in iodine is 7

• The hybridization of I₃^–

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

XeF₂ :

The central atom is Xenon

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

The number of valence electrons in Xenon is 8

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

CIO₃^– :

The central atom : Chlorine

The electronic configuration of chlorine [Ne] 3s² 3p⁵

The number of valence electrons in chlorine is 7

• The hybridization of CIO₃^– molecule

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

PO₄^3- :

The central atom : Phosphorus

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

The number of valence electrons in phosphorus is 5

• The hybridization of PO₄^3- molecule

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

Correct Option : (c)

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