Q.Read the comprehension and answer the question that follows.
Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and anti bonding molecular orbital (ABMO). Energy of anti bonding orbital is raised above the parent atomic orbitals that have combined and the energy of the bonding orbital is lowered than the parent atomic orbitals. Energies of various molecular orbitals for elements hydrogen to nitrogen increase in the order:
σ1s < σ1s < σ2s < σ2s < (π2p_x ≈ π2p_y) < σ2p_z < (π2p_x ≈ π2p_y) < σ2p_z
and for oxygen and fluorine order of energy of molecular orbitals is given below:
σ1s < σ1s < σ2s < σ2s < σ2p_z < (π2p_x ≈ π2p_y) < (π2p_x ≈ π2p_y) < σ2p_z
Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which have comparable energies and proper orientation. Further, if the overlapping is head on, the molecular orbital is called 'Sigma', (σ) and if the overlap is lateral, the molecular orbital is called 'pi', (π). The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals. However, the order for filling is not the same for all molecules or their ions. Bond order is one of the most important parameters to compare the strength of bonds. Which of the following statements is correct?

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

The theory rests on three pillars:

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

The Three Common Hybridizations

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. …