ChemistryClass 11CBSE

Why a Water Molecule Is Bent While Carbon Dioxide Is Straight

Use VSEPR theory to predict shapes and bond angles with and without lone pairs, explain covalent bonds by orbital overlap and tell sigma from pi bonds, and apply sp, sp2, sp3, sp3d and sp3d2 hybridisation to molecules such as CH4, NH3, PCl5 and SF6.

What decides the three-dimensional shape of a molecule?

Water and carbon dioxide both have a central atom joined to two others, yet water is bent and CO

_{2}

is a straight line. That difference makes water a liquid that dissolves salt and CO

_{2}

a gas that does not.

Shape comes from how electron pairs around the central atom arrange themselves, and from which atomic orbitals overlap to form the bonds.

This part covers VSEPR theory, valence bond theory with sigma and pi bonds, and hybridisation.

How does VSEPR theory predict the shape and bond angles of molecules, with and without lone pairs?

Valence shell electron pairs around a central atom repel each other and spread as far apart as possible, with repulsion decreasing in the order lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, so lone pairs squeeze bond angles and change the shape.

Counting electron pairs. For a central atom with

V

valence electrons bonded to

M

monovalent atoms, the number of pairs is

\frac{1}{2} (V + M - cation charge + anion charge)

.

- **NH

_{3}

**:

\frac{1}{2} (5 + 3) = 4

pairs —

3

bond pairs,

1

lone pair
- **ClF

_{3}

**:

\frac{1}{2} (7 + 3) = 5

pairs —

3

bond pairs,

2

lone pairs
- **XeF

_{4}

**:

\frac{1}{2} (8 + 4) = 6

pairs —

4

bond pairs,

2

lone pairs

Shapes without lone pairs: BeCl

_{2}

linear (

180°

), BF

_{3}

trigonal planar (

120°

), CH

_{4}

tetrahedral (

109.5°

), PCl

_{5}

trigonal bipyramidal (

90°

and

120°

), SF

_{6}

octahedral (

90°

).

Shapes with lone pairs: NH

_{3}

trigonal pyramidal (

107°

), H

_{2}

O bent (

104.5°

), SF

_{4}

see-saw, ClF

_{3}

T-shaped, XeF

_{2}

linear, BrF

_{5}

square pyramidal, XeF

_{4}

square planar.

An everyday example. Tie four balloons together at their knots and they settle into a tetrahedron on their own — each balloon pushes the others as far away as it can.

The substance. The shape names only where the atoms are — water's four electron pairs are tetrahedral, but with two lone pairs the molecule is bent.

How does valence bond theory explain covalent bonds, and how do sigma and pi bonds differ?

A covalent bond forms when half-filled atomic orbitals of two atoms overlap and their electrons pair up; head-on overlap along the line joining the nuclei gives a stronger sigma bond, while sideways overlap of parallel p orbitals gives a weaker pi bond.

Hydrogen molecule. As two hydrogen atoms approach, attraction outweighs repulsion until the energy reaches a minimum at a distance of

74

pm; forming the bond releases

435.8

kJ/mol. Greater overlap means a stronger bond.

Sigma and pi bonds:

- **Sigma (

σ

) — s–s, s–p or end-on p–p overlap; electron density on the bond axis; allows rotation
- Pi (

π

) — sideways p–p overlap; electron density above and below the axis; prevents rotation
- A double bond** is

1 σ + 1 π

; a triple bond is

1 σ + 2 π

Worked example 1 — counting bonds.

- Ethene C

_{2}

_{4}

4

C–H +

1

C=C gives

5 σ

and

1 π

- Ethyne C

_{2}

_{2}

3 σ

and

2 π

- **N

_{2}

**:

1 σ

and

2 π

Worked example 2 — pi is weaker. C=C needs about

614

kJ/mol and C–C about

348

kJ/mol, so the pi part adds only about

614 - 348 = 266

kJ/mol, less than the sigma bond.

An everyday example. A firm head-on handshake grips more strongly than two people brushing hands side by side — like sigma versus pi overlap.

The substance. A pi bond never forms on its own — it is always added to an existing sigma bond.

What are sp, sp2, sp3, sp3d and sp3d2 hybridisation, and how do they explain molecular shapes?

**Hybridisation mixes atomic orbitals of similar energy on the central atom into an equal number of identical hybrid orbitals, whose arrangement matches the shape: sp linear, sp

^{2}

trigonal planar, sp

^{3}

tetrahedral, sp

^{3}

d trigonal bipyramidal and sp

^{3}

^{2}

octahedral.

The number of hybrid orbitals equals the number of electron pairs — bond pairs and lone pairs — around the central atom.

Worked examples.

- CH

_{4}

** —

4

pairs, **sp

^{3}

**, tetrahedral,

109.5°

- **NH

_{3}

** —

4

pairs including

1

lone pair, **sp

^{3}

**, pyramidal,

107°

- **H

_{2}

O** —

4

pairs including

2

lone pairs, **sp

^{3}

**, bent,

104.5°

- **BF

_{3}

and each carbon in C

_{2}

_{4}

— sp

^{2}

**,

120°

- **BeCl

_{2}

and the carbon in CO

_{2}

— sp**,

180°

- **PCl

_{5}

— sp

^{3}

d**, trigonal bipyramidal; its axial P–Cl bonds (

219

pm) are longer than the equatorial ones (

204

pm), because axial bonds face more repulsion
- **SF

_{6}

— sp

^{3}

^{2}

**, octahedral, all

90°

s-character. sp has

50%

, sp

^{2}

about

33%

and sp

^{3}

25%

s-character; more s-character holds electrons closer, giving shorter bonds.

An everyday example. Blending mango and milk into a mango shake turns two different ingredients into one uniform drink — just as s and p orbitals blend into identical hybrids.

The substance. Lone pairs occupy hybrid orbitals too, which is why ammonia and water are sp

^{3}

even though they are not tetrahedral in shape.

Exam tip

What earns full marks on shapes and hybridisation?

Count the electron pairs on the central atom first; that single number gives both the hybridisation and the arrangement of pairs.

- **

2

pairs: sp, linear
-

3

pairs**: sp

^{2}

, trigonal planar
- **

4

pairs**: sp

^{3}

, tetrahedral arrangement
- **

5

pairs**: sp

^{3}

d, trigonal bipyramidal arrangement
- **

6

pairs**: sp

^{3}

^{2}

, octahedral arrangement
- Repulsion: lp–lp > lp–bp > bp–bp

The trap. Naming the shape of water as tetrahedral. The electron-pair arrangement is tetrahedral, but the molecular shape is bent.

Did you know

Where does the tetrahedral angle of 109.5° actually come from?

Place four identical bonds as far apart as possible around a central atom, and they point to the corners of a regular tetrahedron. Geometry then fixes the angle

θ

between any two bonds through

cos θ = - \frac{1}{3} \Rightarrow θ \approx 109.47°

So the famous

109.5°

of methane is not measured and memorised by accident — it is the only angle that lets four equal electron pairs spread out evenly in three dimensions. Lone pairs, which repel more strongly, pull this down to

107°

in ammonia and

104.5°

in water.

Exam relevance

How are VSEPR shapes and hybridisation tested in JEE Main and NEET?

VSEPR theory and hybridisation are among the most heavily used ideas in Chemical Bonding for both JEE Main and NEET, and JEE Advanced extends them to interhalogen and xenon compounds and to organic molecules.

What gets asked. Shapes and hybridisation of molecules such as XeF

_{4}

, ClF

_{3}

, SF

_{4}

and PCl

_{5}

, ordering bond angles in molecules with lone pairs, counting sigma and pi bonds, and matching molecules with shapes. Hybridisation is reused throughout organic chemistry and in coordination compounds in Class 12.

Question types. Match-the-column, ordering and statement-based questions.

The trap that costs marks. Ignoring lone pairs when counting electron pairs, which gives the wrong hybridisation and shape.

Key takeaways

What must you be able to do from this part?

- VSEPR: NH

_{3}

has

4

pairs and is pyramidal; ClF

_{3}

has

5

pairs and is T-shaped; XeF

_{4}

has

6

pairs and is square planar
- Valence bond theory: H

_{2}

bonds at

74

pm; ethene has

5 σ

and

1 π

; the pi part of C=C is about

266

kJ/mol
- Hybridisation: CH

_{4}

, NH

_{3}

, H

_{2}

O sp

^{3}

; BF

_{3}

^{2}

; BeCl

_{2}

sp; PCl

_{5}

^{3}

d; SF

_{6}

^{3}

^{2}

Predict the hybridisation, shape and approximate bond angle of SF

_{4}

and of XeF

_{2}

, showing your count of electron pairs.

What decides the three-dimensional shape of a molecule?

How does VSEPR theory predict the shape and bond angles of molecules, with and without lone pairs?

How does valence bond theory explain covalent bonds, and how do sigma and pi bonds differ?

What are sp, sp2, sp3, sp3d and sp3d2 hybridisation, and how do they explain molecular shapes?

What earns full marks on shapes and hybridisation?

Where does the tetrahedral angle of 109.5° actually come from?

How are VSEPR shapes and hybridisation tested in JEE Main and NEET?

What must you be able to do from this part?

Related study notes

Ready to put this into practice?