Valence Bond Theory (VBT) & Orbital Overlap

Why We Need Valence Bond Theory

The Lewis and VSEPR pictures are wonderful for counting and shaping, but they treat electrons as static dots. They can't explain why a covalent bond actually forms, or how strong it is. Valence Bond Theory (VBT), developed by Heitler and London (1927) and extended by Pauling and Slater, brings in quantum mechanics and atomic orbitals.

Core idea: A covalent bond forms when an orbital of one atom containing an unpaired electron overlaps with an orbital of another atom containing an unpaired electron of opposite spin. The shared electron density between the two nuclei is the bond.

The energy picture for H$_2$

As two H atoms approach:

• At large distance — no interaction.
• As they get closer — the electron of each is attracted to both nuclei; potential energy falls.
• At the equilibrium distance ($74\,pm$ for H$_2$) — energy is at a minimum (most stable); $435.8\,kJ/mol$ is released.
• If pushed closer — nucleus-nucleus and electron-electron repulsions dominate; energy rises sharply.

Key Point: The bond length is the internuclear distance at the minimum of the potential energy curve. The depth of that minimum is the bond enthalpy.

[JEE Tip] Greater orbital overlap → stronger bond → more energy released → deeper potential-energy minimum.

Types of Orbital Overlap

The extent and direction of overlap decides bond strength and type. Orbitals can overlap along the internuclear axis (head-on) or sideways.

Overlap combinations

• s-s overlap: two s-orbitals overlap (e.g. H$_2$).
• s-p overlap: an s-orbital with a p-orbital (e.g. HCl, H–F).
• p-p overlap: two p-orbitals, which can be either axial (head-on) or lateral (sideways).

Strength of overlap

For the same type, the strength generally follows the directionality and concentration of the orbitals. Axial (head-on) overlap is more effective than sideways overlap because the electron density is concentrated directly between the nuclei.

Key Point: Overlap must be along directions that maximise electron density between the nuclei; this is why covalent bonds are directional and molecules have definite shapes.

[NEET Important] Only orbitals with electrons of opposite spin can pair up to form a bond — a direct consequence of the Pauli principle.

Sigma (σ) and Pi (π) Bonds

Based on the mode of overlap, covalent bonds are classified as sigma or pi.

Sigma (σ) bond

Formed by head-on (axial) overlap of orbitals along the internuclear axis. Possible by s-s, s-p, or p-p axial overlap. The electron density is cylindrically symmetric about the bond axis.

• A σ bond is strong (large, effective overlap).
• Allows free rotation about the bond axis.

Pi (π) bond

Formed by sideways (lateral) overlap of two parallel p-orbitals, perpendicular to the internuclear axis. The electron density lies above and below the axis.

• A π bond is weaker than a σ bond (less effective overlap).
• Restricts rotation (which is why cis-trans isomers exist).
• A π bond never exists alone — it always accompanies a σ bond.

Counting σ and π in multiple bonds

• Single bond: 1 σ.
• Double bond: 1 σ + 1 π.
• Triple bond: 1 σ + 2 π.

Key Point: σ bonds form first and are stronger; π bonds form from the leftover sideways overlap and are weaker but crucial for shape and reactivity.

[JEE Tip] σ allows rotation, π forbids it — this single fact explains geometric (cis-trans) isomerism.

Strengths and Limitations of VBT

What VBT explains well

• The formation and directional nature of covalent bonds.
• Bond strength in terms of degree of overlap.
• The need for unpaired electrons of opposite spin.
• With hybridisation (next section), the shapes of many molecules.

Limitations of VBT

1. It cannot explain the paramagnetism of $O_2$ (predicts it to be diamagnetic) — only MOT can.
2. It does not satisfactorily explain bonding in electron-deficient species (e.g. diborane $B_2H_6$) or in many coordination compounds.
3. It treats bonds as fully localised between two atoms, so it struggles with delocalised systems unless resonance is invoked.
4. It gives no direct information about bond energies of excited states.

These gaps are exactly what Molecular Orbital Theory (Section 11) was designed to fill.

Key Point: VBT's biggest single failure is $O_2$'s paramagnetism — remember this, it links Sections 9, 4 and 11.

[NEET Important] VBT + hybridisation explains shapes; MOT explains magnetic behaviour and bond order more generally.

Solved Examples

Example 1: Bond formation in H₂

Describe the bond in $H_2$ in terms of orbital overlap.

Solution:

1. Each H has one electron in a $1s$ orbital.
2. Overlap: the two $1s$ orbitals overlap head-on (s-s overlap), pairing the two opposite-spin electrons.
3. Result: a σ bond with electron density concentrated between the nuclei; bond length $74\,pm$, bond enthalpy $435.8\,kJ/mol$.

Takeaway: $H_2$ is the simplest σ bond, formed by s-s overlap of two singly occupied orbitals.

Example 2: σ and π in N₂

How many σ and π bonds are in $N_2$, and from which overlaps?

Solution:

1. N≡N is a triple bond: 1 σ + 2 π.
2. σ bond: axial (head-on) p-p overlap along the bond axis.
3. Two π bonds: lateral overlap of the remaining two pairs of parallel p-orbitals.

Takeaway: A triple bond = one strong axial σ + two weaker sideways π bonds.

Example 3: Why is a σ bond stronger than a π bond?

Explain in terms of overlap.

Solution:

1. σ bond: head-on overlap concentrates electron density directly between the nuclei → large, effective overlap.
2. π bond: sideways overlap places density above and below the axis → smaller, less effective overlap.
3. Conclusion: greater overlap in σ → stronger bond.

Takeaway: Effectiveness of overlap (axial > lateral) determines bond strength.

Example 4: Restricted rotation about a double bond

Why can cis and trans isomers of but-2-ene exist?

Solution:

1. The C=C double bond has a π component from sideways p-orbital overlap.
2. Rotation about the bond would break this sideways overlap (cost energy), so rotation is restricted.
3. Result: the groups are locked in place → distinct cis and trans isomers exist.

Takeaway: π bonds prevent free rotation, allowing geometric (cis-trans) isomerism.

Example 5: Identifying overlap type in HF

What type of overlap forms the H–F bond?

Solution:

1. H has a half-filled $1s$ orbital; F has a half-filled $2p$ orbital.
2. Overlap: the $1s$ of H overlaps head-on with the $2p_z$ of F (s-p axial overlap).
3. Result: a σ bond.

Takeaway: s-p axial overlap gives a σ bond (as in all hydrogen halides).

Example 6: Potential-energy curve interpretation

On the PE vs internuclear-distance curve for $H_2$, what do the minimum and its depth represent?

Solution:

1. The minimum occurs at the equilibrium internuclear distance — this is the bond length ($74\,pm$).
2. The depth of the minimum (energy released on bond formation) is the bond dissociation enthalpy ($435.8\,kJ/mol$).
3. Beyond the minimum (closer), repulsions raise the energy steeply.

Takeaway: Minimum position = bond length; minimum depth = bond strength.

Example 7: Spin requirement for bonding

Why must the two electrons forming a covalent bond have opposite spins?

Solution:

1. Pauli exclusion principle: two electrons in the same region (overlapping orbitals) must have opposite spins.
2. Opposite spins allow the electrons to occupy the shared bonding region simultaneously, lowering energy.
3. Same spins would be forbidden in the same space → no bond.

Takeaway: Covalent bonding requires pairing of opposite-spin electrons (Pauli principle).

Example 8: σ and π count in ethene

How many σ and π bonds are in ethene ($C_2H_4$)?

Solution:

1. Bonds present: four C–H single bonds + one C=C double bond.
2. σ bonds: 4 (C–H) + 1 (C–C) = 5.
3. π bonds: 1 (the second component of the C=C double bond).
4. Total: 5 σ + 1 π.

Takeaway: Count one σ per bonded pair of atoms; extra bonds in multiple bonds are π.

Example 9: VBT failure for O₂

What property of $O_2$ does VBT fail to explain?

Solution:

1. VBT/Lewis structure $O=O$ shows all electrons paired → predicts diamagnetic.
2. Reality: $O_2$ is paramagnetic (2 unpaired electrons), attracted by a magnet.
3. Conclusion: VBT cannot account for this; only Molecular Orbital Theory can.

Takeaway: $O_2$'s paramagnetism is the famous failure of VBT — resolved by MOT.

Example 10: Axial vs lateral overlap

Classify the bonds formed by (a) axial p-p overlap and (b) lateral p-p overlap.

Solution:

1. Axial (head-on) p-p overlap: electron density along the bond axis → σ bond.
2. Lateral (sideways) p-p overlap: electron density above and below the axis → π bond.
3. Conclusion: axial → σ; lateral → π.

Takeaway: Direction of p-orbital overlap decides whether the bond is σ or π.

Example 11: Number of σ and π bonds in CO₂

Using VBT, count the σ and π bonds in $CO_2$ ($O=C=O$).

Solution:

1. Two C=O double bonds, each = 1 σ + 1 π.
2. σ bonds: 2 (one per C=O).
3. π bonds: 2.
4. Total: 2 σ + 2 π.

Takeaway: Each double bond contributes exactly one σ and one π bond.

Example 12: Bond strength and overlap in halogens

Why does the F–F bond (in $F_2$) involve only a σ bond, while $N_2$ has σ + 2π?

Solution:

1. $F_2$: each F needs one more electron; a single head-on p-p (axial) overlap gives one σ bond — octet complete.
2. $N_2$: each N needs three more electrons; one axial σ + two lateral π bonds complete the triple bond.
3. Conclusion: the number of unpaired electrons available for overlap sets the number of σ and π bonds.

Takeaway: More unpaired p-electrons → more π bonds in addition to the single σ bond.