Match the species in Column I with the type of hybrid orbitals in Column II. Column I Column II (i) SF 4 (a) sp 3 d 2 (ii) IF 5 (b) d 2 sp 3 (iii) NO 2 + (c) sp 3 d (iv) NH 4 + (d) sp 3 (e) sp

(i) SF 4 -----(c) sp 3 d S is surrounded by 4 bond pairs and 1 lone pair ( sp 3 d hybridisation). (ii) IF 5 -----(a) sp 3 d 2 I is surrounded by 5 bond pairs and 1 lone pair ( sp 3 d 2 hybridisation) (iii) NO 2 + -----(e) sp N has 2 bond pairs and no lone pair ( sp hybridisation) (iv) NH 4 + -----(d) sp 3 N has 4 bond pairs and no lone pair ( sp 3 hybridisation). (i) → (c) (ii) → (a) (iii) → (e) (iv) → (d) Column I Column II (i) SF 4 (c) sp 3 d (ii) IF 5 (a) sp 3 d 2 (iii) NO 2 + (e) sp (iv) NH 4 + (d) sp 3

Match the species in Column I with the geometry/shape in Column II. Column I Column II (i) H 3 O + (a) Linear (ii) HC ≡ CH (b) Angular (iii) ClO 2 - (c) Tetrahedral (iv) NH 4 + (d) Trigonal bipyramidal (e) Pyramidal

(i) H 3 O + -----(e)Pyramidal Oxygen has 3 bond pairs and 1 lone pair (pyramidal shape). (ii) HC ≡ CH -----(a)Linear In the formation of ethyne molecule, both the carbon atoms undergo sp -hybridisation having two unhybridised orbital. It forms linear molecules with an angle of 180 ° . (iii) ClO 2 - -----(b)Angular Chlorine has 2 bond pairs and 2 lone pairs (Angular shape). (iv) NH 4 + -----(c)Tetrahedral Nitrogen has 4 bond pairs and no lone pair (Tetrahedral). Column I Column II (i) H 3 O + (e) Pyramidal (ii) HC ≡ CH (a) Linear (iii) ClO 2 - (b) Angular (iv) NH 4 + (c) Tetrahedral

Match the species in Column I with the bond order in Column II. Column I Column II (i) NO (a) 1 . 5 (ii) CO (b) 2 . 0 (iii) O 2 - (c) 2 . 5 (iv) O 2 (d) 3 . 0

(i) NO -----(c) 2 . 5 Total number of electrons in given molecule is 7 + 8 = 15 electrons . Molecular electronic configuration is σ 1 s 2 < σ * 1 s 2 < σ 2 s 2 < σ * 2 s 2 < σ 2 p z 2 < π 2 p x 2 = π 2 p y 2 < π ∗ 2 p x 1 = π ∗ 2 p y < σ ∗ 2 p z Bond order = Number of bonding electrons - Number of antibonding electrons 2 Bond order = 10 - 5 2 = 2 . 5 (ii) CO -----(d) 3 Total number of electrons in given molecule is 6 + 8 = 14 electrons . Molecular electronic configuration is σ 1 s 2 < σ ∗ 1 s 2 < σ 2 s 2 < σ ∗ 2 s 2 < π 2 p x 2 = π 2 p y 2 < σ 2 p z 2 Bond order = N b − N a / 2 = ( 10 − 4 ) / 2 = 3 (iii) O 2 − -----(a) 1 . 5 Total number of electrons in given molecule is 8 + 8 + 1 = 17 electrons . Molecular electronic configuration is σ 1 s 2 < σ * 1 s 2 < σ 2 s 2 < σ * 2 s 2 < σ 2 p z 2 < π 2 p x 2 = π 2 p y 2 < π ∗ 2 p x 2 = π ∗ 2 p y 1 < σ ∗ 2 p z Bond order= N b − N a / 2 = ( 10 − 7 ) / 2 = 1 .5 (iv) O 2 -----(b) 2 Total number of electrons in given molecule is 8 + 8 = 16 electrons . Molecular electronic configuration is σ 1 s 2 < σ ∗ 1 s 2 < σ 2 s 2 < σ ∗ 2 s 2 < σ 2 p z 2 < π 2 p x 2 = π 2 p y 2 < [ π * 2 p x 1 = π * 2 p y 1 ] Bond order = N b − N a / 2 = ( 10 − 6 ) / 2 = 2 Column I Column II (i) NO (c) 2 . 5 (ii) CO (d) 3 . 0 (iii) O 2 - (a) 1 . 5 (iv) O 2 (b) 2 . 0

Match the items given in Column I with examples given in Column II. Column I Column II (i) Hydrogen bond (a) C (ii) Resonance (b) LiF (iii) Ionic solid (c) H 2 (iv) Covalent solid (d) HF (e) O 3

(i)Hydrogen bond-----(d) HF Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom ( F , O or N ) of another molecule. (ii)Resonance-----(e) O 3 According to the concept of resonance, whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, positions of nuclei, bonding and non-bonding pairs of electrons are taken as the canonical structures of the hybrid which describes the molecule accurately. (iii)Ionic solid-----(b) LiF Ionic solids are solids held together by ionic bonds. They are lattices composed of oppositely charged ions. Ionic solids tend to have high melting points. (iv)Covalent solid-----(a) C A network solid or covalent network solid is a chemical compound in which the atoms are bonded by covalent bonds in a continuous network extending throughout the material. Example: Diamond. (i) → (d) (ii) → (e) (iii) → (b) (iv) → (a) Column I Column II (i) Hydrogen bond (d) HF (ii) Resonance (e) O 3 (iii) Ionic solid (b) LiF (iv) Covalent solid (a) C

Match the shape of molecules in Column I with the type of hybridisation in Column II. Column I Column II (i) Tetrahedral (a) sp 2 (ii) Trigonal (b) sp (iii) Linear (c) sp 3

(i)Tetrahedral-----(c) sp 3 When one ‘ s ’ orbital and 3 ‘ p ’ orbitals belonging to the same shell of an atom mix together to form four new equivalent orbitals. The new orbitals formed are called sp 3 hybridised orbitals. It forms tetrahedral molecules with an angle of 109 . 5 ° . (ii)Trigonal-----(a) sp 2 sp 2 hybridisation is observed when one s and two p orbitals of the same shell of an atom mix to form 3 equivalent orbitals. The new orbitals formed are called sp 2 hybridised orbitals. It forms trigonal planar molecules with an angle of 120 ∘ . (iii)Linear-----(b) sp sp hybridisation is observed when one s and one p orbital in the same main shell of an atom mix to form two new equivalent orbitals. The new orbitals formed are called sp hybridised orbitals. It forms linear molecules with an angle of 180 ° . (i) → (c); (ii) → (a); (iii) → (b) Column I Column II (i) Tetrahedral (c) sp 3 (ii) Trigonal (a) sp 2 (iii) Linear (b) sp

Draw the resonating structure of Ozone molecule

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 4 - Chemical Bonding and Molecular Structure>Short Answer Type>Q 49

Predict the shapes of the following molecules on the basis of hybridisation.

$BC l_{3}, {CH}_{4}, {CO}_{2}, {NH}_{3}$

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 4 - Chemical Bonding and Molecular Structure>Short Answer Type>Q 50

All the

C — O

bonds in carbonate ion

({CO}_{3}^{2 -})

are equal in length. Explain.

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 4 - Chemical Bonding and Molecular Structure>Short Answer Type>Q 51

What is meant by the term average bond enthalpy? Why is there a difference in bond enthalpy of

O — H

bond in ethanol

(C_{2} H_{5} OH)

and water?

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 4 - Chemical Bonding and Molecular Structure>Matching Type>Q 52

Match the species in Column I with the type of hybrid orbitals in Column II.

	Column I		Column II
(i)	${SF}_{4}$	(a)	${sp}^{3} d^{2}$
(ii)	${IF}_{5}$	(b)	$d^{2} {sp}^{3}$
(iii)	${NO}_{2}^{+}$	(c)	${sp}^{3} d$
(iv)	${NH}_{4}^{+}$	(d)	${sp}^{3}$
		(e)	$sp$

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 4 - Chemical Bonding and Molecular Structure>Matching Type>Q 53

Match the species in Column I with the geometry/shape in Column II.

	Column I		Column II
(i)	$H_{3} O^{+}$	(a)	Linear
(ii)	$HC \equiv CH$	(b)	Angular
(iii)	${ClO}_{2}^{-}$	(c)	Tetrahedral
(iv)	${NH}_{4}^{+}$	(d)	Trigonal bipyramidal
		(e)	Pyramidal

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 4 - Chemical Bonding and Molecular Structure>Matching Type>Q 54

Match the species in Column I with the bond order in Column II.

	Column I		Column II
(i)	$NO$	(a)	$1.5$
(ii)	$CO$	(b)	$2.0$
(iii)	$O_{2}^{-}$	(c)	$2.5$
(iv)	$O_{2}$	(d)	$3.0$

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 4 - Chemical Bonding and Molecular Structure>Matching Type>Q 55

Match the items given in Column I with examples given in Column II.

	Column I		Column II
(i)	Hydrogen bond	(a)	$C$
(ii)	Resonance	(b)	$LiF$
(iii)	Ionic solid	(c)	$H_{2}$
(iv)	Covalent solid	(d)	$HF$
		(e)	$O_{3}$

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 4 - Chemical Bonding and Molecular Structure>Matching Type>Q 56

Match the shape of molecules in Column I with the type of hybridisation in Column II.

	Column I		Column II
(i)	Tetrahedral	(a)	${sp}^{2}$
(ii)	Trigonal	(b)	$sp$
(iii)	Linear	(c)	${sp}^{3}$

Draw the resonating structure of

Ozone molecule

Important Questions on Chemical Bonding and Molecular Structure

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