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Chemical bonding underlies almost every later chemistry chapter, since molecular shape and bond type explain reactivity, polarity and physical properties. NEET tests this chapter heavily through VSEPR-predicted shapes and hybridisation, so building a reliable step-by-step method matters more than memorising individual molecules.
NEET weightage: High · NCERT Chemistry Class 11, Part 1, Chapter 4
An ionic bond forms through electrostatic attraction after electron transfer between atoms of very different electronegativity (e.g. Na⁺Cl⁻). A covalent bond forms through the sharing of electron pairs, typically between atoms of similar electronegativity, and can be single, double or triple depending on how many electron pairs are shared.
Valence Shell Electron Pair Repulsion theory predicts molecular shape from the idea that electron pairs (bonding and lone) around a central atom arrange themselves to minimise repulsion. Lone pairs repel more strongly than bonding pairs, which is why lone pairs compress bond angles below the 'ideal' geometric value (e.g. NH₃'s H-N-H angle is 107°, less than the tetrahedral 109.5°, due to one lone pair).
Atomic orbitals on the central atom mix to form new, equivalent hybrid orbitals that explain observed bond angles: sp (linear, 180°), sp² (trigonal planar, 120°), sp³ (tetrahedral, 109.5°). Hybridisation and VSEPR-predicted geometry should always agree with each other — use one to check the other.
Bond length is the average distance between two bonded nuclei; it decreases as bond order increases (triple bonds are shorter than double, which are shorter than single). Bond enthalpy is the energy needed to break one mole of a particular bond in the gas phase; it increases as bond order increases.
Atomic orbitals combine to form bonding (lower energy) and antibonding (higher energy) molecular orbitals. Bond order = ½(electrons in bonding MOs − electrons in antibonding MOs). A positive bond order predicts a stable molecule; a bond order of zero (as calculated for He₂) predicts the molecule doesn't exist.
sp hybridisation
Linear, 180° (e.g. BeCl₂)
sp² hybridisation
Trigonal planar, 120° (e.g. BF₃)
sp³ hybridisation (no lone pairs)
Tetrahedral, 109.5° (e.g. CH₄)
sp³ with 1 lone pair
Pyramidal, ~107° (e.g. NH₃)
sp³ with 2 lone pairs
Bent/angular, ~104.5° (e.g. H₂O)
Bond order formula
½ × (bonding e⁻ − antibonding e⁻)
Predict the shape and approximate bond angle of an NH₃ molecule using VSEPR theory.
Nitrogen has 5 valence electrons: 3 are used in N-H bonds, leaving 1 lone pair. Total electron pairs around N = 4 (3 bonding + 1 lone), giving a tetrahedral electron-pair arrangement. Since one position is a lone pair, the observed molecular shape is pyramidal (trigonal pyramidal), and because lone pair–bond pair repulsion is stronger than bond pair–bond pair repulsion, the H-N-H angle compresses slightly from 109.5° to about 107°.
Using molecular orbital theory, is the O₂⁺ ion expected to be more or less stable (higher or lower bond order) than neutral O₂?
O₂ has a bond order of 2 (it has 2 unpaired electrons in antibonding π* orbitals contributing to the standard MO diagram result of bond order 2). Removing one electron to form O₂⁺ removes an electron from an antibonding orbital, which increases the bond order to 2.5. A higher bond order means a shorter, stronger bond, so O₂⁺ is more stable (in terms of bond strength) than O₂.
What is the hybridisation and shape of the carbon atom in CH₄?
sp³ hybridisation, tetrahedral shape, with H-C-H bond angles of 109.5°. Carbon has 4 bonding pairs and no lone pairs, so the electron-pair geometry and the molecular shape are the same (tetrahedral).
Why is the bond angle in H₂O (about 104.5°) smaller than in NH₃ (about 107°)?
Both are sp³ hybridised with a tetrahedral electron-pair arrangement, but water has 2 lone pairs on oxygen versus ammonia's 1 lone pair on nitrogen. Since lone pair–lone pair repulsion is the strongest type of repulsion in VSEPR theory, water's two lone pairs compress its bond angle more than ammonia's single lone pair does.
Arrange the following in order of increasing bond length: N≡N, N=N, N–N (all nitrogen-nitrogen bonds).
N≡N < N=N < N–N. Bond length decreases as bond order increases, so the triple bond (bond order 3) is shortest and the single bond (bond order 1) is longest.
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