Chemistry · Chemical Bonding and Molecular Structure

Valence Bond Theory

Valence Bond Theory is the first quantum-mechanical picture of the covalent bond, introduced by Heitler and London in 1927 and developed further by Pauling. Following NCERT §4.5, it explains bond formation as the overlap of half-filled atomic orbitals, traces the potential-energy curve of the $\ce{H2}$ molecule, and classifies bonds as sigma or pi. For NEET, the highest-yield skill from this section is counting $\sigma$ and $\pi$ bonds in molecules such as $\ce{N2}$, $\ce{C2H4}$ and $\ce{C2H2}$.

Why Valence Bond Theory Was Needed

The Lewis approach helps in writing the structure of molecules, but it fails to explain how a chemical bond actually forms. It offers no reason for the wide difference in bond dissociation enthalpies and bond lengths of molecules that, on paper, look identical. Both $\ce{H2}$ and $\ce{F2}$ contain a single covalent bond formed by sharing one electron pair, yet their properties are very different.

MoleculeBond dissociation enthalpyBond length
$\ce{H2}$435.8 kJ mol⁻¹74 pm
$\ce{F2}$155 kJ mol⁻¹144 pm

Lewis theory also gives no idea about the shapes of polyatomic molecules. The VSEPR theory does predict geometry, but only empirically; it does not explain why a particular geometry arises, and it has limited applications. To overcome these limitations, two theories rooted in quantum-mechanical principles were introduced: Valence Bond (VB) theory and molecular orbital theory.

Valence Bond Theory rests on the knowledge of atomic orbitals, electronic configurations of elements, the overlap criteria of atomic orbitals, the hybridisation of atomic orbitals, and the principles of variation and superposition. A rigorous mathematical treatment is beyond the NCERT scope, so the theory is presented qualitatively, beginning with the simplest molecule of all, $\ce{H2}$.

Formation of the H₂ Molecule

Consider two hydrogen atoms A and B approaching each other, with nuclei $N_A$ and $N_B$ and electrons $e_A$ and $e_B$. When the atoms are far apart there is no interaction. As they approach, new attractive and repulsive forces begin to operate simultaneously.

Attractive forces (pull atoms together)Repulsive forces (push atoms apart)
Nucleus and its own electron: $N_A\!-\!e_A$ and $N_B\!-\!e_B$Electron–electron: $e_A\!-\!e_B$
Nucleus and the other atom's electron: $N_A\!-\!e_B$, $N_B\!-\!e_A$Nucleus–nucleus: $N_A\!-\!N_B$

Experimentally, the magnitude of the new attractive forces exceeds that of the new repulsive forces. As a result, the two atoms move closer and the potential energy of the system decreases. A stage is finally reached where the net attraction exactly balances the net repulsion and the system acquires its minimum energy. At this point the two atoms are bonded together into a stable molecule with a bond length of 74 pm.

Because energy is released when the bond forms, the $\ce{H2}$ molecule is more stable than the two isolated atoms. The energy released is the bond enthalpy, equal in magnitude to the minimum of the energy curve. Conversely, the same quantity of energy must be supplied to break the bond:

$\ce{H2(g) + 435.8\,kJ\,mol^{-1} -> H(g) + H(g)}$

The H₂ Potential-Energy Curve

The competition between attraction and repulsion is captured by the potential-energy curve, plotted against the internuclear distance. At large separation the energy is taken as zero. As the atoms approach, attraction dominates and the curve descends into a well; at very short distances nuclear and electronic repulsion shoot up steeply. The minimum sits at 74 pm, the bond length, and its depth below zero is the bond enthalpy of 435.8 kJ mol⁻¹.

Figure 1 0 74 pm (bond length) 435.8 kJ mol⁻¹ Internuclear distance → Potential energy → most stable state of H₂
Figure 1. The potential-energy curve for $\ce{H2}$ formation as a function of internuclear distance. Energy falls as attraction dominates, reaches a minimum at the bond length of 74 pm, then rises sharply due to repulsion at shorter distances. The depth of the minimum equals the bond enthalpy, 435.8 kJ mol⁻¹.

The Orbital Overlap Concept

At the minimum-energy state, the two hydrogen atoms are so close that their atomic orbitals undergo partial interpenetration. This partial merging of atomic orbitals is called overlapping, and it results in the pairing of electrons. The covalent bond, in VB language, forms by the pairing of valence-shell electrons of opposite spin on the two atoms.

The single most important rule that follows is: the extent of overlap decides the strength of the covalent bond. In general, the greater the overlap, the stronger the bond. This one principle threads through the rest of the chapter, explaining the σ-versus-π hierarchy and, after hybridisation is introduced, the strengths of $sp$, $sp^2$ and $sp^3$ bonds.

Types of Overlap: s-s, s-p, p-p

When the orbitals of two atoms approach, the overlap may be positive, negative or zero, depending on the sign (phase) and the direction of the orbital wave-function amplitudes in space. For a bond to form, the overlapping lobes must have the same phase and be correctly oriented; this is positive overlap. Orbitals of opposite phase give negative overlap, and orthogonal orientations give zero net overlap. Three combinations build the axial (head-on) bonds.

Figure 2 s s s–s overlap s p s–p overlap p p p–p axial all overlap along the internuclear axis → σ bonds
Figure 2. The three axial (head-on) overlaps that produce sigma bonds. In s–s overlap two half-filled s-orbitals merge along the internuclear axis; in s–p overlap a half-filled s-orbital meets a half-filled p-orbital; in p–p axial overlap two half-filled p-orbitals meet end-to-end. Each maximises overlap along the line joining the nuclei.

Sigma and Pi Bonds

Depending on the mode of overlap, the covalent bond is classified into two types: the sigma ($\sigma$) bond and the pi ($\pi$) bond.

FeatureSigma (σ) bondPi (π) bond
Mode of overlapEnd-to-end (head-on / axial) overlap along the internuclear axisSidewise (lateral) overlap, orbital axes parallel and perpendicular to the internuclear axis
Orbitals involveds–s, s–p, or p–p (axial)p–p (parallel), sidewise
Electron cloudSymmetrical about the internuclear axisTwo saucer-shaped charge clouds above and below the plane of the atoms
Extent of overlapLargerSmaller
StrengthStrongerWeaker
OccurrencePresent in every covalent bond, single or multipleOnly in multiple bonds, always in addition to a σ bond
Figure 3 σ bond — axial overlap on the axis π bond — sidewise two clouds above & below
Figure 3. A sigma bond (left) arises from axial overlap concentrated along the line joining the nuclei. A pi bond (right) arises from sidewise overlap of parallel p-orbitals, giving two saucer-shaped charge clouds above and below the internuclear plane. The greater axial overlap makes the σ bond stronger.

Strength of Sigma vs Pi Bonds

The strength of a bond depends on the extent of overlapping. In a sigma bond the orbitals overlap to a larger extent, so it is stronger; in a pi bond the sidewise overlap is smaller in extent, so it is weaker. This is why, in a multiple bond, the σ component is the more robust skeleton and the π component is comparatively easy to break, a fact that underpins the reactivity of double and triple bonds.

Two consequences are worth fixing in memory before the bond-counting drills. First, a single bond is always a pure σ bond. Second, in any multiple bond the first bond is the σ bond and every additional bond is a π bond formed in addition to it.

Next Concept

Simple orbital overlap predicts the wrong bond angles for $\ce{CH4}$, $\ce{NH3}$ and $\ce{H2O}$. Pauling's fix is the next chapter section — see Hybridisation.

Counting Sigma and Pi Bonds

This is the highest-frequency skill NEET tests from this section. The recipe is mechanical once the rule is internalised:

  • Every single bond = 1 σ.
  • Every double bond = 1 σ + 1 π.
  • Every triple bond = 1 σ + 2 π.
  • Count every bond in the molecule, including all C–H and C–C bonds — not just the multiple bonds.
MoleculeBonds presentσ bondsπ bonds
$\ce{N2}$ (N≡N)one triple bond12
$\ce{C2H4}$ ethene4 C–H + 1 C=C4 + 1 = 51
$\ce{C2H2}$ ethyne2 C–H + 1 C≡C2 + 1 = 32
$\ce{C2H6}$ ethane6 C–H + 1 C–C70
$\ce{CO2}$ (O=C=O)two C=O double bonds22
Worked Example

Count the σ and π bonds in ethyne, $\ce{C2H2}$ (H–C≡C–H).

Walk through every bond. The two C–H bonds are single bonds: 2 σ. The carbon–carbon link is a triple bond: it contributes 1 σ + 2 π. Adding the σ contributions: $2 + 1 = 3$ σ bonds. The π count comes only from the triple bond: 2 π bonds. Result: 3 σ and 2 π.

NEET Trap

Don't forget the C–H sigma bonds

The most common error is counting only the σ bond hidden inside a double or triple bond and ignoring every single bond around it. In $\ce{C2H4}$, students often write "1 σ" because they see one C=C; the correct answer is 5 σ (four C–H plus one C–C σ) and 1 π. Every line in the structural formula is at least one σ bond.

Rule: total σ = total number of bonds (single + multiple counted once each); total π = (number of double bonds) + 2 × (number of triple bonds).

Directional Properties and Limitations

Valence Bond Theory accounts for the directional nature of bonds: $\ce{H2}$ forms by overlap of two 1s orbitals, and in polyatomic molecules the geometry matters as much as bond formation. But simple atomic-orbital overlap runs into trouble with shape. Carbon's ground-state configuration $\text{[He]}2s^2 2p^2$ becomes $\text{[He]}2s^1 2p_x^1 2p_y^1 2p_z^1$ on excitation, giving four half-filled orbitals to overlap with four hydrogen 1s orbitals.

The catch is direction. The three 2p orbitals lie at 90° to one another, so pure p-overlap would force three of the H–C–H angles to 90°, while the spherical 2s orbital can overlap in any direction, leaving the fourth bond undefined. This does not match the observed tetrahedral angle of 109.5° in $\ce{CH4}$. The same flaw predicts 90° angles for $\ce{NH3}$ and $\ce{H2O}$, against the real values of 107° and 104.5°.

MoleculeVBT (pure overlap) predictsObserved angle
$\ce{CH4}$90°109.5°
$\ce{NH3}$90°107°
$\ce{H2O}$90°104.5°

To repair this directional failure, Pauling introduced the concept of hybridisation, in which atomic orbitals of slightly different energies intermix to form an equivalent set of hybrid orbitals that point in the correct directions. Valence Bond Theory also cannot explain the paramagnetism of $\ce{O2}$, a gap filled by molecular orbital theory. Despite these limits, VBT remains the working language of the covalent bond and the foundation for everything that follows.

Quick Recap

Valence Bond Theory in one screen

  • Introduced by Heitler and London (1927); developed by Pauling. A covalent bond forms when half-filled valence orbitals of opposite-spin electrons overlap.
  • The extent of overlap decides bond strength: greater overlap → stronger bond.
  • The $\ce{H2}$ energy curve minimum lies at 74 pm (bond length); its depth is the bond enthalpy 435.8 kJ mol⁻¹.
  • σ bond = axial (head-on) overlap (s–s, s–p, p–p), stronger; π bond = sidewise overlap of parallel p-orbitals, weaker.
  • Single bond = 1 σ; double = 1 σ + 1 π; triple = 1 σ + 2 π. Count every bond. ($\ce{N2}$: 1 σ, 2 π · $\ce{C2H4}$: 5 σ, 1 π · $\ce{C2H2}$: 3 σ, 2 π.)
  • Limitation: pure overlap predicts 90° angles for $\ce{CH4}$, $\ce{NH3}$, $\ce{H2O}$ — solved by hybridisation; paramagnetism of $\ce{O2}$ needs MO theory.

NEET PYQ Snapshot — Valence Bond Theory

Counting σ and π bonds and identifying bond types between two carbon atoms are the recurring NEET themes from this section.

NEET 2024

Match List I (Molecule) with List II (Number and types of bond/s between two carbon atoms). A. ethane B. ethene C. carbon molecule $\ce{C2}$ D. ethyne, against: I. one σ-bond and two π-bonds; II. two π-bonds; III. one σ-bond; IV. one σ-bond and one π-bond.

  1. A-I, B-IV, C-II, D-III
  2. A-IV, B-III, C-II, D-I
  3. A-III, B-IV, C-II, D-I
  4. A-III, B-IV, C-I, D-II
Answer: (3)

Between the two carbons: ethane is a single bond → one σ-bond (III); ethene is a double bond → one σ + one π (IV); $\ce{C2}$ has two π-bonds and no net σ between the carbons (II); ethyne is a triple bond → one σ + two π (I). Hence A-III, B-IV, C-II, D-I.

NEET 2023

The number of σ bonds, π bonds and lone pairs of electrons in pyridine, respectively, are:

  1. 12, 2, 1
  2. 11, 2, 0
  3. 12, 3, 0
  4. 11, 3, 1
Answer: (4)

Pyridine ($\ce{C5H5N}$) has 11 σ bonds (5 C–H + 5 ring + counting the C–N σ framework), 3 π bonds from the three formal double bonds of the aromatic ring, and 1 lone pair on the nitrogen. So 11 σ, 3 π, 1 lone pair.

Concept

According to Valence Bond Theory, which statement about the strength of a covalent bond is correct?

  1. A π bond is stronger than a σ bond because of sidewise overlap
  2. Bond strength is independent of the extent of orbital overlap
  3. The greater the extent of orbital overlap, the stronger the bond
  4. σ and π bonds are always equal in strength
Answer: (3)

VBT states that bond strength depends on the extent of overlap; greater overlap gives a stronger bond. Axial (σ) overlap is larger in extent than sidewise (π) overlap, so the σ bond is stronger.

Concept

The total number of σ and π bonds in ethyne ($\ce{C2H2}$), respectively, are:

  1. 2 σ and 2 π
  2. 3 σ and 2 π
  3. 3 σ and 1 π
  4. 5 σ and 1 π
Answer: (2)

Two C–H single bonds give 2 σ; the C≡C triple bond gives 1 σ + 2 π. Total = 3 σ and 2 π.

FAQs — Valence Bond Theory

The doubts that surface most often when revising VBT for NEET.

What is Valence Bond Theory in simple terms?
Valence Bond Theory, introduced by Heitler and London in 1927 and developed further by Pauling, explains that a covalent bond forms when two atoms come close enough for their half-filled valence atomic orbitals to overlap. The overlap pairs up electrons of opposite spin and releases energy, so the bonded molecule is more stable than the isolated atoms. The greater the overlap, the stronger the bond.
Why is a sigma bond stronger than a pi bond?
A sigma bond is formed by end-to-end (axial) overlap of orbitals along the internuclear axis, which allows orbitals to overlap to a large extent. A pi bond is formed by sidewise overlap of parallel p-orbitals perpendicular to the internuclear axis, where the extent of overlap is smaller. Because bond strength depends on the extent of overlap, the sigma bond, with greater overlap, is stronger than the pi bond.
How do you count sigma and pi bonds in a molecule?
Every single bond is one sigma bond. In a multiple bond, the first bond is always a sigma bond and the remaining bonds are pi bonds. So a double bond is one sigma plus one pi, and a triple bond is one sigma plus two pi. Count every bond between any pair of bonded atoms, including all C-H and C-C bonds, to get the total number of sigma and pi bonds.
What does the H2 potential-energy curve show?
As two hydrogen atoms approach, attractive forces dominate over repulsive forces, so the potential energy of the system falls. The curve reaches a minimum at an internuclear distance of 74 pm, which is the bond length of H2. This minimum corresponds to the most stable state of the molecule, and the depth of the well equals the bond enthalpy of 435.8 kJ per mole.
What are the limitations of Valence Bond Theory?
Simple atomic orbital overlap fails to predict the correct bond angles of polyatomic molecules; for example it predicts 90 degree angles for CH4, NH3 and H2O instead of the observed 109.5, 107 and 104.5 degrees. To overcome this, Pauling introduced the concept of hybridisation. Valence Bond Theory also does not explain the paramagnetism of O2, which Molecular Orbital Theory accounts for.