Write the important conditions required for the linear combination of atomic orbitals to form molecular orbita

Exercises · 4.34

Q. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Jammu Kashmir JkboseTextbookSubjective· 2mImportance★★★★★est

34% · 38/112 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Orbital Hybridization Theory

Orbital Hybridization Theory – From Intuition to Precision

The Problem That Started It All

Imagine you are looking at a methane molecule, $\text{CH}_4$ . Carbon has four valence electrons: two in the $2s$ orbital and two in the $2p$ orbitals. If carbon used its pure atomic orbitals to bond, you would expect two bonds from the $2s$ (identical, but one direction) and two from the $2p$ (at $90^\circ$ to each other). That would give you three different bond types and bond angles of $90^\circ$ and something else.

But experiment says methane is perfectly tetrahedral: all four bonds are identical in length, strength, and energy, and the bond angle is $109.5^\circ$ , not $90^\circ$ . Something is fundamentally wrong with the "pure orbital" picture.

This is the puzzle that hybridization theory solves.

The Core Intuition

Think of atomic orbitals as shapes that an electron can occupy. The $s$ orbital is a sphere. The $p$ orbitals are dumbbells along the $x$ , $y$ , and $z$ axes. When an atom forms bonds, it wants to mix these shapes together to create new, hybrid shapes that point in directions that maximise bond strength and minimise repulsion.

It is like mixing primary colours to get new colours. You don't have to use red, blue, and yellow separately — you can blend them to get green, orange, or purple. Similarly, an atom can blend its $s$ and $p$ orbitals to get new hybrid orbitals that are better suited for bonding.

The key insight: hybridization is a mathematical mixing of atomic orbitals on the same atom to produce an equal number of new, equivalent hybrid orbitals. The number of hybrid orbitals formed always equals the number of atomic orbitals mixed.

The Precise Statement

s

The theory rests on three pillars:

Conservation of orbitals: Mixing $n$ atomic orbitals gives exactly $n$ hybrid orbitals. No orbitals are created or destroyed.
Energy averaging: The hybrid orbitals have energies that are intermediate between the original $s$ and $p$ energies.
Directionality: Hybrid orbitals point in specific directions to minimise electron pair repulsion, which directly determines molecular shape.

The Three Common Hybridizations

Hybridization	Orbitals Mixed	Number of Hybrids	Geometry	Bond Angle	Example
$sp$	one $s$ + one $p$	2	Linear	$180^\circ$	$\text{BeCl}_2$
$sp^2$	one $s$ + two $p$	3	Trigonal planar	$120^\circ$	$\text{BF}_3$
$sp^3$	one $s$ + three $p$	4	Tetrahedral	$109.5^\circ$	$\text{CH}_4$

Note

The superscript in $sp^2$ or $sp^3$ tells you how many $p$ orbitals were mixed. $sp^3$ means one $s$ and three $p$ orbitals were blended. It does not mean there are three $s$ orbitals — there is only one $s$ orbital per shell.

How It Works: The Methane Example

Carbon in its ground state has the configuration $1s^2 2s^2 2p_x^1 2p_y^1$ . Only two unpaired electrons — it should form only two bonds. But we know carbon forms four bonds.

Step 1: Promotion. One electron from the $2s$ orbital is promoted (excited) to the empty $2p_z$ orbital. This costs a small amount of energy, but it is more than compensated by the energy released when four strong bonds form instead of two.

Step 2: Hybridization. The one $2s$ orbital and three $2p$ orbitals mix to form four equivalent $sp^3$ hybrid orbitals. Each hybrid has 25% $s$ character and 75% $p$ character.

Step 3: Bonding. Each $sp^3$ hybrid overlaps with the $1s$ orbital of a hydrogen atom, forming four identical $\sigma$ bonds. The hybrids point to the corners of a tetrahedron, giving the $109.5^\circ$ angle. …

Start your 14-day free trial to unlock the full solution →

The formation of molecular orbitals via LCAO requires three conditions: the combining atomic orbitals must have comparable energies, they must have the same symmetry about the internuclear axis, and they must overlap significantly. These ensure effective constructive and destructive interference, yielding bonding and antibonding MOs.

The Core Idea: Why Conditions Matter

When two atomic orbitals combine to form molecular orbitals, we are essentially asking: can these wavefunctions interfere constructively and destructively in a stable, meaningful way? Just like two waves in a pond — if they have different frequencies or meet at an angle, they won't produce a clean standing wave. The same logic governs orbital mixing.

The Linear Combination of Atomic Orbitals (LCAO) method is built on the idea that the molecular orbital wavefunction $\psi_{MO}$ is a linear combination:

\psi_{MO} = c_1 \phi_A + c_2 \phi_B

where $\phi_A$ and $\phi_B$ are atomic orbitals on two atoms. For this combination to yield physically meaningful bonding ( $\psi_+$ ) and antibonding ( $\psi_-$ ) orbitals, three conditions must be satisfied.

The Three Essential Conditions

1. Comparable Energies of the Combining Orbitals

The atomic orbitals must have similar energies. Why? Because the extent of mixing depends on the energy difference $\Delta E = |E_A - E_B|$ . If the energies are too different, the resulting MO will be almost identical to the lower-energy AO, with negligible contribution from the higher-energy one — no real bonding occurs.

\frac{c_2}{c_1} \approx \frac{H_{AB}}{E_A - E_B}

Example: In HF, the 1s orbital of H ( $-13.6$ eV) and the 2p $_z$ orbital of F ( $-18.6$ eV) have a moderate energy difference, so they mix to form a polar bond. But the 1s of H and the 2s of F ( $-40.2$ eV) are too far apart — they do not combine.

Watch out

A common mistake is to think that any two orbitals with the same principal quantum number will combine. They won't if their energies differ significantly — e.g., 2s and 2p in the same atom have different energies (in multi-electron atoms), so they do not mix directly across atoms unless symmetry allows.

2. Same Symmetry About the Internuclear Axis

The orbitals must have the same symmetry with respect to rotation around the bond axis. This ensures that the overlap integral $\int \phi_A \phi_B \, d\tau$ is non-zero.

Sigma ( $\sigma$ ) symmetry: Orbitals that are cylindrically symmetric about the bond axis (e.g., s orbitals, p $_z$ orbitals if the axis is z). They overlap end-to-end.
Pi ( $\pi$ ) symmetry: Orbitals with one nodal plane containing the bond axis (e.g., p $_x$ , p $_y$ ). They overlap side-to-side.
Delta ( $\delta$ ) symmetry: Two nodal planes containing the axis (e.g., d $_{xy}$ ).

Only orbitals of the same symmetry type can combine. A p $_z$ (sigma) and a p $_x$ (pi) have zero net overlap — their positive and negative lobes cancel.

Tip

Think of it like matching puzzle pieces: a round peg (s orbital) fits into a round hole (end of a p $_z$ ), but not into a slot (side of a p $_x$ ). The symmetry must match for constructive interference to be possible.

3. Significant Overlap

Even if energies match and symmetry aligns, the orbitals must physically overlap to a sufficient degree. The overlap integral $S = \int \phi_A \phi_B \, d\tau$ must be appreciable (typically $S > 0.1$ for meaningful bonding).

Overlap depends on:

Distance between nuclei: Too far apart → negligible overlap. …

Start your 14-day free trial to unlock the full solution →

←Previous question · Exercises4.33 — Explain the formation of H₂ molecule on the basis of valence bond theo…→Next question · Exercises4.35 — Use molecular orbital theory to explain why the Be₂ molecule does not …

Q.Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

The Core Idea: Why Conditions Matter

The Three Essential Conditions

1. Comparable Energies of the Combining Orbitals

2. Same Symmetry About the Internuclear Axis

3. Significant Overlap

🎓Unlock everything free for 14 days

Unlock everything free for 14 days