Valence Bond Theory and Hybridisation – Formation of Covalent Bonds (Senior 4 Chemistry) | Ganzaa

Formation of Covalent Bonds

Lessons

Senior 4 · Chemistry

Formation of Covalent Bonds

5 lessons

Lesson 3 of 5

Valence Bond Theory and Hybridisation

25 min

Carbon's puzzle

Write the configuration of carbon: $1s^2\,2s^2\,2p^2$ . Its outer shell has only two unpaired electrons, in two p orbitals at 90° to each other. If bonds formed only from unpaired electrons in these orbitals, carbon would make a compound $\mathrm{CH_2}$ with a 90° angle. In fact the simplest stable hydride is methane, $\mathrm{CH_4}$ : four identical C–H bonds, each the same length and strength, pointing to the corners of a tetrahedron at 109.5°. To explain this, Linus Pauling added two ideas to the theory of bonding by overlap: promotion and hybridisation.

Definition

Valence bond theory: a covalent bond forms when a half-filled orbital on one atom overlaps a half-filled orbital on another, and the two electrons pair with opposite spins. The number of bonds an atom forms equals its number of unpaired electrons.

Hybridisation: the mixing of atomic orbitals of similar energy on the same atom to form an equal number of new, identical hybrid orbitals, which point in definite directions.

Promotion, then mixing

First, one 2s electron of carbon is promoted into the empty 2p orbital. This needs energy, but it gives four unpaired electrons and so four bonds instead of two; the energy released by the two extra bonds more than repays the cost. Second, the 2s and the three 2p orbitals mix to form four identical $s p^{3}$ orbitals. They repel one another and point as far apart as possible: towards the corners of a tetrahedron, 109.5° apart. Each overlaps the 1s orbital of a hydrogen atom to form a σ bond, and the four C–H bonds are identical.

Carbon need not mix all three p orbitals. Mixing the s with two p orbitals gives three $s p^{2}$ orbitals, and mixing it with one p gives two $s p$ orbitals. The p orbitals left over are used for π bonds.

Table 3.1: Types of hybridisation

Hybridisation	Orbitals mixed	Hybrid orbitals	Arrangement and angle	Examples
$s p$	s + p	2	linear, 180°	$\mathrm{BeCl_2}$ ; C in $\mathrm{C_2H_2}$ and $\mathrm{CO_2}$
$s p^{2}$	s + 2p	3	trigonal planar, 120°	$\mathrm{BF_3}$ ; C in $\mathrm{C_2H_4}$
$s p^{3}$	s + 3p	4	tetrahedral, 109.5°	$\mathrm{CH_4}$ ; N in $\mathrm{NH_3}$ ; O in $\mathrm{H_2O}$
$s p^{3} d$	s + 3p + d	5	trigonal bipyramidal, 90° and 120°	$\mathrm{PCl_5}$
$s p^{3} d^{2}$	s + 3p + 2d	6	octahedral, 90°	$\mathrm{SF_6}$

A quick rule: count the σ bonds plus lone pairs on the atom (π bonds do not count). Two gives $s p$ , three $s p^{2}$ , four $s p^{3}$ , five $s p^{3} d$ , six $s p^{3} d^{2}$ .

Double and triple bonds: ethene and ethyne

In ethene, $\mathrm{C_2H_4}$ , each carbon is $s p^{2}$ hybridised. Its three $s p^{2}$ orbitals form σ bonds to the other carbon and to two hydrogen atoms, all in one plane at about 120°. The fourth electron is in the unhybridised 2p orbital, perpendicular to that plane. The two p orbitals overlap sideways to form the π bond. Because the π overlap would be broken by twisting, there is no free rotation about a C=C bond.

In ethyne, $\mathrm{C_2H_2}$ , each carbon is $s p$ hybridised: two σ bonds in a straight line (H–C and C–C) and two unhybridised p orbitals at right angles to each other and to the axis. They overlap with those of the other carbon to form two π bonds, which surround the axis like a cylinder. The molecule is linear.

Worked example 1: hybridisation of each atom

State the hybridisation of each carbon atom, and of the oxygen atom, in ethanal, $\mathrm{CH_3CHO}$ , and give the approximate bond angles around each carbon.

Solution. Count σ bonds + lone pairs on each atom.

$\mathrm{CH_3}$ carbon: four σ bonds (3 C–H, 1 C–C), no lone pairs → 4 → $s p^{3}$ , about 109.5°.
CHO carbon: three σ bonds (C–C, C–H, C–O; the C=O also has a π bond, not counted) → 3 → $s p^{2}$ , about 120°.
Oxygen: one σ bond (to C) + two lone pairs → 3 → $s p^{2}$ ; its unhybridised p orbital forms the π bond of C=O.

Worked example 2: phosphorus pentachloride

Explain how phosphorus ( $[\mathrm{Ne}]\,3s^2\,3p^3$ ) forms five equivalent-type bonds in $\mathrm{PCl_5}$ , and name the hybridisation.

Solution. Ground state: three unpaired 3p electrons, enough for only $\mathrm{PCl_3}$ . Promote one 3s electron into an empty 3d orbital: $3s^1\,3p^3\,3d^1$ , five unpaired electrons. Mix the 3s, three 3p and one 3d orbital: five $s p^{3} d$ hybrid orbitals, pointing to the corners of a trigonal bipyramid. Each overlaps a 3p orbital of a chlorine atom to form a σ bond.

Answer: $s p^{3} d$ ; trigonal bipyramidal, with bond angles of 120° (in the triangle) and 90° (to the two axial chlorines).

Warning

Common mistakes. (1) Counting π bonds when deciding the hybridisation: only σ bonds and lone pairs count. (2) Saying a hybrid orbital is "between an s and a p": it is a new orbital, and the number of hybrids always equals the number of orbitals mixed. (3) Forgetting the leftover p orbitals in $s p^{2}$ and $s p$ atoms: they are where π bonds come from. (4) Hybridisation explains a shape that is already known; to predict a shape, use VSEPR (Lesson 4).

Tip

Learn the ladder: 2 → sp → 180°; 3 → sp² → 120°; 4 → sp³ → 109.5°; 5 → sp³d → 90° and 120°; 6 → sp³d² → 90°, where the number is σ bonds + lone pairs.

The Organic Chemistry Tutor: Hybridization of Atomic Orbitals - Sigma and Pi Bonds - sp, sp2, sp3 (11 min). Watch how the number of groups around an atom gives its hybridisation, and how σ and π bonds are counted in each example.

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Practice

Explain why carbon forms four bonds in methane, although its ground-state configuration has only two unpaired electrons.
Draw box diagrams to show the formation of $s p^{2}$ hybrid orbitals in boron, starting from $2s^2\,2p^1$ .
State the hybridisation of the central atom in: $\mathrm{BeCl_2}$ , $\mathrm{BF_3}$ , $\mathrm{NH_3}$ , $\mathrm{H_2O}$ , $\mathrm{SF_6}$ .
Describe the bonding in ethyne in terms of hybrid orbitals and σ and π bonds.
Give the hybridisation of each carbon atom in propene, $\mathrm{CH_2{=}CH{-}CH_3}$ .
Explain why there is no free rotation about the C=C bond in ethene.
In carbon dioxide, O=C=O, what is the hybridisation of carbon? How many π bonds are there, and how are their p orbitals arranged?

Show answersHide answers

One 2s electron is promoted to the empty 2p orbital, giving four unpaired electrons; the 2s and three 2p orbitals hybridise into four equal $s p^{3}$ orbitals, each forming a σ bond with H. The energy of the extra two bonds repays the promotion energy.
Ground: 2s ↑↓, 2p ↑ _ _. Promote: 2s ↑, 2p ↑ ↑ _. Mix 2s + two 2p: three $s p^{2}$ ↑ ↑ ↑, and one empty 2p.
$\mathrm{BeCl_2}$ sp; $\mathrm{BF_3}$ $s p^{2}$ ; $\mathrm{NH_3}$ $s p^{3}$ (3 σ + 1 lone pair); $\mathrm{H_2O}$ $s p^{3}$ (2 σ + 2 lone pairs); $\mathrm{SF_6}$ $s p^{3} d^{2}$ .
Each C is sp: one sp orbital forms a σ bond with H, the other a σ bond with the other C, in a straight line. Each C has two unhybridised p orbitals at 90° to each other; they overlap sideways with the other carbon's to give two π bonds. Total: 3 σ, 2 π; linear, 180°.
$\mathrm{CH_2}$ = carbon: $s p^{2}$ ; =CH– carbon: $s p^{2}$ ; $\mathrm{CH_3}$ carbon: $s p^{3}$ .
The π bond needs the two p orbitals to stay parallel. Rotating one carbon would twist them apart and break the π overlap, which needs a lot of energy.
sp (two σ bonds, no lone pairs). Two π bonds, one to each oxygen; the two p orbitals of carbon used for them are at right angles to each other.

Quick check

What is the hybridisation of each carbon atom in ethene, C₂H₄?

Key points

Valence bond theory: bonds form from overlapping half-filled orbitals; promotion can create more unpaired electrons.
Hybridisation mixes orbitals on one atom into equal hybrid orbitals: sp (2, 180°), $s p^{2}$ (3, 120°), $s p^{3}$ (4, 109.5°), $s p^{3} d$ (5), $s p^{3} d^{2}$ (6, 90°).
Hybridisation = σ bonds + lone pairs on the atom; leftover p orbitals form π bonds.
Ethene: $s p^{2}$ carbons, planar, one π bond, no free rotation. Ethyne: sp carbons, linear, two π bonds.

Sources and credits

Written by Ganzaa.
Video: Hybridization of Atomic Orbitals - Sigma and Pi Bonds - Sp Sp2 Sp3, The Organic Chemistry Tutor, embedded from YouTube.
The diagrams, table, worked examples and Practice items were written and drawn by Ganzaa.

1Carbon's puzzle

2Promotion, then mixing

3Double and triple bonds: ethene and ethyne

Carbon's puzzle

Promotion, then mixing

Double and triple bonds: ethene and ethyne