Match the species in Column I with the geometry/shape in Column II. Column I (i) H3O^+ (ii) HC≡CH (iii) ClO2^-

NCERT Exemplar · Q57

Q. Match the species in Column I with the geometry/shape in Column II. Column I

(i) H3O^+

(ii) HC≡CH

(iii) ClO2^-

(iv) NH4^+
Column II

(a) Linear

(b) Angular

(d) Trigonal bipyramidal

(e) Pyramidal

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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. …

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Molecular geometry is determined by the number of electron domains (bond pairs and lone pairs) around the central atom, following VSEPR theory.

H3O $^+$ is Pyramidal.
HC≡CH is Linear.
ClO $_2^-$ is Angular.
NH $_4^+$ is Tetrahedral.

The shape or geometry of a molecule is crucial for understanding its properties and reactivity. We can predict molecular geometry using the Valence Shell Electron Pair Repulsion (VSEPR) theory, often complemented by the concept of hybridization. VSEPR theory states that electron pairs (both bonding and non-bonding, i.e., lone pairs) around a central atom will arrange themselves as far apart as possible to minimize repulsion, thereby determining the electron geometry. The molecular geometry, or shape, is then determined by the arrangement of only the atoms, taking into account the influence of lone pairs.

Here's how we determine the geometry for each species:

Determine the central atom: This is usually the least electronegative atom (excluding hydrogen) or the unique atom in the formula.
Count total valence electrons: Sum the valence electrons of all atoms and adjust for any charge (add for negative charge, subtract for positive charge).
Draw the Lewis structure: Connect the central atom to the terminal atoms with single bonds. Distribute remaining electrons to satisfy octets, starting with terminal atoms, then the central atom.
Determine the steric number: This is the sum of the number of atoms bonded to the central atom and the number of lone pairs on the central atom. Each multiple bond (double or triple) counts as one "electron domain" for VSEPR purposes.
Predict electron geometry and hybridization: The steric number dictates the electron geometry and hybridization of the central atom.
- Steric number 2: Linear, $sp$
- Steric number 3: Trigonal planar, $sp^2$
- Steric number 4: Tetrahedral, $sp^3$
- Steric number 5: Trigonal bipyramidal, $sp^3d$
- Steric number 6: Octahedral, $sp^3d^2$
Predict molecular geometry: This is based on the electron geometry but considers only the positions of the atoms, not the lone pairs. Lone pairs occupy space and influence bond angles but are not part of the visible "shape" of the molecule.

Let's apply these steps to each species:

(i) H $_3$ O $^+$ (Hydronium ion)

Central atom: Oxygen (O).
Total valence electrons: $3 \times 1 (\text{H}) + 6 (\text{O}) - 1 (\text{charge}) = 3 + 6 - 1 = 8$ valence electrons.
Lewis structure: Oxygen is bonded to three hydrogen atoms. This uses $3 \times 2 = 6$ electrons. The remaining $8 - 6 = 2$ electrons form one lone pair on the oxygen atom.
Steric number: 3 bond pairs (O-H) + 1 lone pair = 4 electron domains.
Electron geometry and hybridization: With 4 electron domains, the electron geometry is tetrahedral, and the hybridization of oxygen is $sp^3$ .
Molecular geometry: Out of the 4 electron domains, 3 are bonding pairs and 1 is a lone pair. The lone pair repels the bonding pairs more strongly, distorting the tetrahedral arrangement of electron domains. This results in a pyramidal molecular geometry.

(ii) HC≡CH (Ethyne / Acetylene)

Central atoms: Both carbon atoms (C). The molecule is symmetrical.
Total valence electrons: $2 \times 1 (\text{H}) + 2 \times 4 (\text{C}) = 2 + 8 = 10$ valence electrons.
Lewis structure: H-C≡C-H. Each carbon atom forms a single bond with a hydrogen atom and a triple bond with the other carbon atom.
Steric number (for each carbon): Each carbon has 1 bond pair (C-H) and 1 bond pair (C≡C, which counts as one electron domain). So, $1 + 1 = 2$ electron domains.
Electron geometry and hybridization: With 2 electron domains, the electron geometry is linear, and the hybridization of each carbon is $sp$ .
Molecular geometry: Since there are no lone pairs on the central carbon atoms, the molecular geometry is the same as the electron geometry: linear.

(iii) ClO $_2^-$ (Chlorite ion)

Central atom: Chlorine (Cl).
Total valence electrons: $7 (\text{Cl}) + 2 \times 6 (\text{O}) + 1 (\text{charge}) = 7 + 12 + 1 = 20$ valence electrons. …

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