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Chemical Bonding and Molecular Structure – Class 11 Chemistry notes, important questions and MCQs

NCERT Class 11 Chemistry, chapter 4. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Atoms bond to reach stable configurations. Ionic and covalent bonds, VSEPR shapes, hybridisation, molecular orbital theory and hydrogen bonding explain structure and properties.

Bonds and shapes

Octet rule; ionic bond by electron transfer, covalent by sharing; formal charge helps choose Lewis structures.
VSEPR: BeCl₂ linear, BF₃ trigonal planar, CH₄ tetrahedral (109.5°), NH₃ pyramidal (107°), H₂O bent (104.5°), PCl₅ trigonal bipyramidal, SF₆ octahedral.
Hybridisation: sp (180°), sp² (120°), sp³ (109.5°), sp³d, sp³d².

MOT and H-bonding

Bond order = ½(Nb − Na); higher bond order → shorter, stronger bond; zero bond order → molecule does not exist (He₂).
O₂ is paramagnetic (two unpaired electrons); N₂ bond order 3.
Hydrogen bonding raises boiling points (H₂O, HF); dipole moment of CO₂ is zero due to symmetry.

Key terms

Hybridisation: mixing of atomic orbitals to form equivalent hybrid orbitals
Bond order: half the difference between bonding and antibonding electrons
Dipole moment: µ = charge × distance; a measure of bond or molecule polarity (unit debye)

Formulas

Bond order = ½(Nb − Na)

Important questions with answers

Why is NH₃ pyramidal while BF₃ is planar?2 marks

N in NH₃ has 3 bond pairs and 1 lone pair (sp³), and the lone pair pushes the bonds down, giving a pyramid. B in BF₃ has 3 bond pairs and no lone pair (sp²), so it is trigonal planar.

Why does H₂O have a higher boiling point than H₂S?3 marks

Oxygen is small and highly electronegative, so water forms strong intermolecular hydrogen bonds. Sulphur is larger and less electronegative and H₂S does not form significant hydrogen bonds, so less energy is needed to boil it.

Draw (describe) the molecular orbital energy diagram of O₂, find its bond order and explain its magnetic nature. Compare the bond orders of O₂⁺, O₂ and O₂⁻.5 marks

O₂ has 16 electrons: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x¹ π*2p_y¹. Bonding electrons Nb = 10, antibonding Na = 6, bond order = ½(10 − 6) = 2. The two unpaired electrons in π* orbitals make O₂ paramagnetic. O₂⁺ (one less π* electron): ½(10 − 5) = 2.5; O₂⁻ (one more): ½(10 − 7) = 1.5. Order of bond strength: O₂⁺ > O₂ > O₂⁻, and bond length the reverse.

Quick revision

O₂ is paramagnetic.
CO₂ is non-polar.
H-bonding raises boiling point.

Exam tips

Lone pairs repel more than bond pairs: lp–lp > lp–bp > bp–bp.
Draw the Lewis structure first; count bond pairs and lone pairs on the central atom to predict the VSEPR shape.

Practice MCQs

Shape of CH₄ is
1. square planar
2. pyramidal
3. linear
4. tetrahedral
Show answer
Answer: (d) tetrahedral. sp³.
A bond formed by sharing electrons is
1. ionic
2. hydrogen
3. metallic
4. covalent
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Answer: (d) covalent. Covalent.
Bond angle in water is about
1. 109.5°
2. 120°
3. 107°
4. 104.5°
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Answer: (d) 104.5°. Two lone pairs.
Hybridisation of carbon in ethene is
1. sp²
2. sp
3. sp³
4. sp³d
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Answer: (a) sp². Double bond.
Which molecule has zero dipole moment
1. H₂O
2. NH₃
3. HCl
4. BF₃
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Answer: (d) BF₃. Symmetrical trigonal planar.
Bond order of N₂ is
1. 3
2. 2
3. 1
4. 2.5
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Answer: (a) 3. (10 − 4)/2.
Which molecule is paramagnetic
1. N₂
2. H₂
3. F₂
4. O₂
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Answer: (d) O₂. Two unpaired π* electrons.
Hybridisation of S in SF₆ is
1. sp³
2. sp³d
3. dsp²
4. sp³d²
Show answer
Answer: (d) sp³d². Six bond pairs, octahedral.

All chapters of Class 11 Chemistry

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.