Match the shape of molecules in Column I with the type of hybridisation in Column II. Column I (i) Tetrahedral

NCERT Exemplar · Q60

Q. Match the shape of molecules in Column I with the type of hybridisation in Column II. Column I

(i) Tetrahedral

(ii) Trigonal

(iii) Linear
Column II

(a) sp^2

(b) sp

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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 shape is determined by the arrangement of electron domains around the central atom, which in turn is dictated by the type of hybridization. A tetrahedral shape corresponds to $\text{sp}^3$ hybridization, a trigonal planar shape to $\text{sp}^2$ hybridization, and a linear shape to $\text{sp}$ hybridization.

The shape of a molecule is fundamentally determined by the arrangement of electron pairs (both bonding and non-bonding) around the central atom. This arrangement is best explained by the concept of orbital hybridization.

Orbital Hybridization Theory

Atoms often mix their atomic orbitals (s, p, d) to form new, equivalent hybrid orbitals before forming bonds. This process, called hybridization, is crucial because:

It allows for the formation of stronger, more stable bonds by maximizing orbital overlap.
It minimizes electron-pair repulsion, leading to specific, predictable geometries around the central atom.

The number of hybrid orbitals formed is always equal to the number of atomic orbitals that mix. These hybrid orbitals then orient themselves in space to be as far apart as possible, defining the electron geometry of the molecule. The molecular shape is then derived from this electron geometry, considering only the positions of the atoms.

Here's how we match the given shapes to their corresponding hybridization types:

Determine the Number of Electron Domains (Steric Number)

The first step in predicting hybridization and shape is to count the total number of electron domains around the central atom. An electron domain can be:
- A single bond
- A double bond
- A triple bond
- A lone pair of electrons
Each of these counts as one electron domain for the purpose of determining hybridization. The total number of electron domains is often called the steric number.
Relate Steric Number to Hybridization

The steric number directly tells us how many hybrid orbitals are needed.
- If the steric number is 2, two hybrid orbitals are needed, formed by mixing one s and one p orbital ( $\text{sp}$ hybridization).
- If the steric number is 3, three hybrid orbitals are needed, formed by mixing one s and two p orbitals ( $\text{sp}^2$ hybridization).
- If the steric number is 4, four hybrid orbitals are needed, formed by mixing one s and three p orbitals ( $\text{sp}^3$ hybridization).
$\text{Steric Number} = (\text{Number of sigma bonds}) + (\text{Number of lone pairs})$
Match Shapes to Hybridization Types

(i) Tetrahedral Shape
- A tetrahedral shape arises when there are four electron domains around the central atom, and all four are bonding pairs (e.g., in $\text{CH}_4$ ).
- With four electron domains, the central atom requires four hybrid orbitals.
- This corresponds to the mixing of one s orbital and three p orbitals, resulting in $\text{sp}^3$ hybridization.
- The four $\text{sp}^3$ hybrid orbitals point towards the corners of a tetrahedron, with ideal bond angles of $109.5^\circ$ .
- Therefore, (i) Tetrahedral matches (c) $\text{sp}^3$ .
(ii) Trigonal Shape
- In the context of simple hybridization, "Trigonal" typically refers to trigonal planar geometry. This shape occurs when there are three electron domains around the central atom, and all three are bonding pairs (e.g., in $\text{BF}_3$ ).
- With three electron domains, the central atom requires three hybrid orbitals.
- This corresponds to the mixing of one s orbital and two p orbitals, resulting in $\text{sp}^2$ hybridization. …

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Q.Match the shape of molecules in Column I with the type of hybridisation in Column II. Column I

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Q.Match the shape of molecules in Column I with the type of hybridisation in Column II.
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