Chapter 4 · Chemical Bonding and Molecular Structure

4.3 Valence Bond Theory and Hybridisation

Bonds form when half-filled atomic orbitals overlap. The energy curve of the hydrogen molecule, positive, negative and zero overlap, σ and π bonds, and hybridisation — sp, sp², sp³, sp³d and sp³d² — which explains the four equal bonds of methane, the double and triple bonds of ethene and ethyne, and the shapes of PCl₅ and SF₆.

VSEPR theory predicts shapes very well, but it does not say why two atoms bond, how strong the bond is, or why the bond angle in water is $104.5^{\circ}$ when the oxygen atom uses $2p$ orbitals that are only $90^{\circ}$ apart. To answer such questions we must look at the electrons as waves in atomic orbitals, as in Section 2.3. In 1927 Walter Heitler and Fritz London showed with quantum mechanics why two hydrogen atoms form a molecule; Linus Pauling and John Slater turned their idea into the valence bond (VB) theory, which describes a covalent bond as the overlap of two atomic orbitals each holding one electron.

This section follows that idea: first the energy of two hydrogen atoms as they approach, then the kinds of overlap and the two types of bond they produce — σ and π — and finally hybridisation, the mixing of atomic orbitals into new, equivalent orbitals pointing in the directions VSEPR predicts. Hybridisation is the everyday language of organic chemistry (Chapters 8 and 9).

Orbital Overlap and the Hydrogen Molecule

Imagine two hydrogen atoms, A and B, far apart: they do not interact and the energy of the pair is taken as zero. As they come closer, new forces appear. Each nucleus attracts its own electron and also the electron of the other atom; at the same time the two electrons repel each other and the two nuclei repel each other. At first the attractions win and the energy falls. At a distance of $74\ \mathrm{pm}$ attraction and repulsion balance and the energy is at a minimum, $435.8\ \mathrm{kJ\,mol^{-1}}$ below that of the separate atoms. If the atoms are pushed closer still, nuclear repulsion takes over and the energy rises steeply. The two atoms settle at $74\ \mathrm{pm}$, the bond length of $\mathrm{H_{2}}$, and $435.8\ \mathrm{kJ\,mol^{-1}}$ is its bond enthalpy (Section 4.1).

📘 Definition 4.9 — Orbital overlap and the valence bond picture
When two atoms approach, their atomic orbitals partly occupy the same region of space: they overlap. In valence bond theory a covalent bond forms when a half-filled orbital of one atom overlaps a half-filled orbital of the other and the two electrons, of opposite spin, pair up in the overlap region, where they are attracted by both nuclei. The greater the extent of overlap, the stronger the bond (the overlap criterion). An atom can form as many covalent bonds as it has half-filled orbitals available — which is why filled orbitals, as in two helium atoms, do not bond.

Orbitals are waves with phases, shown as signs $+$ and $-$ on their lobes (Section 2.3). How the lobes meet decides whether overlap helps:

✏️ Example 1 — Energetics of the hydrogen molecule
(a) How much energy is released when one molecule of $\mathrm{H_{2}}$ forms from two hydrogen atoms? Give the answer in joules and in electronvolts ($1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}$, $N_{A}=6.022\times10^{23}\ \mathrm{mol^{-1}}$). (b) What is the longest wavelength of light that could break the bond of one $\mathrm{H_{2}}$ molecule ($h=6.626\times10^{-34}\ \mathrm{J\,s}$, $c=3.0\times10^{8}\ \mathrm{m\,s^{-1}}$)? (c) Why does $\mathrm{He_{2}}$ not form, according to VB theory? (d) From the bond length, find the covalent radius of hydrogen.
(a) $E=\dfrac{435.8\times10^{3}}{6.022\times10^{23}}=7.237\times10^{-19}\ \mathrm{J}$; in eV: $7.237\times10^{-19}/1.602\times10^{-19}=4.52\ \mathrm{eV}$.
(b) $\lambda=\dfrac{hc}{E}=\dfrac{6.626\times10^{-34}\times3.0\times10^{8}}{7.237\times10^{-19}}=2.75\times10^{-7}\ \mathrm{m}=275\ \mathrm{nm}$ (ultraviolet).
(c) Each He atom has a completely filled $1s^{2}$ orbital and no half-filled orbital to share; there are no unpaired electrons to pair up, so no bond forms.
(d) $r_{\mathrm{H}}=\tfrac12(74)=37\ \mathrm{pm}$.
Interactive 1 — Two Hydrogen Atoms Approach: the Energy Curve

Drag the slider to change the distance $r$ between the two nuclei, or press ▶ Play to bring the atoms together from far away. The picture shows the two $1s$ clouds and their overlap; the graph shows the energy of the pair (a Morse curve fitted to $r_{e}=74\ \mathrm{pm}$ and a depth of $435.8\ \mathrm{kJ\,mol^{-1}}$). The readouts give the energy, the overlap of the two $1s$ orbitals (0 = none, 1 = complete), and whether the net force pulls the atoms together or pushes them apart.

150 pm
distance r—
energy—
1s–1s overlap—
net force—

Sigma and Pi Bonds

Orbitals can overlap in two essentially different ways: end to end along the line joining the two nuclei (the internuclear axis), or side by side, with their axes parallel.

📘 Definition 4.10 — σ and π bonds
  • A sigma (σ) bond is formed by head-on (axial) overlap of two orbitals along the internuclear axis: $s$–$s$ (as in $\mathrm{H_{2}}$), $s$–$p$ (as in $\mathrm{HCl}$), $p$–$p$ end on (as in $\mathrm{Cl_{2}}$), or hybrid orbitals. Its electron density is concentrated on the axis, and the bond is symmetrical about it.
  • A pi (π) bond is formed by sideways (lateral) overlap of two parallel $p$ orbitals, perpendicular to the internuclear axis. Its electron density lies in two regions, above and below the axis, with a nodal plane containing the axis.
📐 Rule 4.8 — σ and π bonds in single and multiple bonds
  1. A single bond is one σ bond. A double bond is one σ + one π; a triple bond is one σ + two π (the two π bonds lie in perpendicular planes). The first bond between two atoms is always σ.
  2. A σ bond is stronger than a π bond, because head-on overlap is more extensive than sideways overlap. (So $\mathrm{C{=}C}$, $614\ \mathrm{kJ\,mol^{-1}}$, is less than twice $\mathrm{C{-}C}$, $348$.)
  3. Rotation about a σ bond leaves the overlap unchanged, so groups joined by a single bond can rotate freely; rotating one end of a π bond destroys the sideways overlap, so a double bond is rigid (the basis of geometrical isomerism, Chapter 9).
✏️ Example 2 — Counting σ and π bonds
Count the σ and π bonds in (a) $\mathrm{N_{2}}$; (b) ethene, $\mathrm{CH_{2}{=}CH_{2}}$; (c) $\mathrm{CO_{2}}$; (d) propenenitrile (acrylonitrile), $\mathrm{CH_{2}{=}CH{-}C{\equiv}N}$; (e) benzene, $\mathrm{C_{6}H_{6}}$ (Kekulé structure with three $\mathrm{C{=}C}$).
(a), (b) $\mathrm{N{\equiv}N}$: 1σ, 2π. Ethene: four $\mathrm{C{-}H}$ σ + the $\mathrm{C{=}C}$ (1σ + 1π) → 5σ, 1π.
(c) $\mathrm{O{=}C{=}O}$: each double bond is 1σ + 1π → 2σ, 2π.
(d) σ: three $\mathrm{C{-}H}$, $\mathrm{C{=}C}$ (1), $\mathrm{C{-}C}$ (1), $\mathrm{C{\equiv}N}$ (1) → 6σ. π: $\mathrm{C{=}C}$ (1) + $\mathrm{C{\equiv}N}$ (2) → 3π.
(e) σ: six $\mathrm{C{-}H}$ + six $\mathrm{C{-}C}$ ring bonds → 12σ. π: three (one per $\mathrm{C{=}C}$) → 3π.
Check A quick rule: σ bonds = (number of atoms − 1) for any molecule without rings, + 1 for each ring. Acrylonitrile: 7 atoms → 6σ ✓; benzene: 12 atoms, 1 ring → 11 + 1 = 12σ ✓.

Hybridisation

Simple overlap of pure atomic orbitals runs into trouble at once. Carbon, $1s^{2}2s^{2}2p^{2}$, has only two unpaired electrons, yet it forms four bonds in methane. Even if one $2s$ electron is promoted to give $2s^{1}2p^{3}$ with four unpaired electrons, three bonds would use $p$ orbitals at $90^{\circ}$ to one another and the fourth an $s$ orbital — but all four $\mathrm{C{-}H}$ bonds in methane are identical and $109.5^{\circ}$ apart. Pauling's answer was that the atomic orbitals mix before bonding.

📘 Definition 4.11 — Hybridisation
Hybridisation is the intermixing of atomic orbitals of slightly different energies on the same atom to form an equal number of new orbitals, called hybrid orbitals, which are equivalent in energy and shape. Its main features are:
  1. the number of hybrid orbitals equals the number of atomic orbitals that mix;
  2. the hybrid orbitals are directed in space so that they are as far apart as possible — the geometries of VSEPR theory;
  3. a hybrid orbital has one large lobe and gives more effective (head-on) overlap than a pure $s$ or $p$ orbital, so hybrid orbitals form σ bonds and hold lone pairs; π bonds are formed by the unhybridised $p$ orbitals.
Conditions: only orbitals of the valence shell and of similar energy mix (e.g. $2s$ with $2p$; $3s$, $3p$ and $3d$); promotion of an electron is not essential; half-filled, filled and even empty orbitals can take part.
Hybridisationspsp²sp³sp³dsp³d²
Orbitals mixed$s+p$$s+2p$$s+3p$$s+3p+d$$s+3p+2d$
Hybrid orbitals23456
Geometrylinear, $180^{\circ}$trigonal planar, $120^{\circ}$tetrahedral, $109.5^{\circ}$trigonal bipyramidaloctahedral, $90^{\circ}$
s character50%33.3%25%20%16.7%
Example$\mathrm{BeCl_{2}}$, $\mathrm{C_{2}H_{2}}$$\mathrm{BCl_{3}}$, $\mathrm{C_{2}H_{4}}$$\mathrm{CH_{4}}$, $\mathrm{NH_{3}}$, $\mathrm{H_{2}O}$$\mathrm{PCl_{5}}$$\mathrm{SF_{6}}$

sp — beryllium chloride. Be is $2s^{2}$; one electron is promoted, $2s^{1}2p^{1}$, and the $2s$ and one $2p$ orbital mix into two sp hybrids at $180^{\circ}$. Each overlaps a $3p$ orbital of a chlorine atom: $\mathrm{Cl{-}Be{-}Cl}$ is linear. sp² — boron trichloride. B, $2s^{2}2p^{1}$, is promoted to $2s^{1}2p^{2}$; three sp² hybrids lie in a plane at $120^{\circ}$, and one empty $2p$ orbital is left perpendicular to the plane. sp³ — methane. C, promoted to $2s^{1}2p^{3}$, forms four sp³ hybrids pointing to the corners of a tetrahedron; each overlaps a hydrogen $1s$ orbital. In ammonia and water the central atom is also sp³ hybridised, but one or two of its hybrid orbitals hold lone pairs ($\mathrm{N}$: $2s^{2}2p^{3}$ gives sp³ hybrids holding $2,1,1,1$ electrons; $\mathrm{O}$: $2,2,1,1$), and lone-pair repulsion closes the angles to $107^{\circ}$ and $104.5^{\circ}$.

📐 Rule 4.9 — Finding the hybridisation of a central atom
The number of hybrid orbitals on a central atom equals the number of its electron pairs in the VSEPR sense — σ bonds plus lone pairs (π bonds are not counted). With the shortcut of Section 4.2, $\tfrac12(V+M-c+a)=2,3,4,5,6$ means sp, sp², sp³, sp³d, sp³d². The hybridisation fixes the electron-pair geometry; the shape then follows from the number of lone pairs.
✏️ Example 3 — Hybridisation of the central atom
State the hybridisation of the central atom and the shape of: (a) $\mathrm{BF_{4}^{-}}$; (b) $\mathrm{CO_{2}}$; (c) $\mathrm{SO_{2}}$; (d) $\mathrm{NH_{4}^{+}}$; (e) $\mathrm{SF_{4}}$; (f) $\mathrm{XeF_{4}}$; (g) $\mathrm{NO_{3}^{-}}$; (h) $\mathrm{H_{2}O}$.
(a), (b) $\mathrm{BF_{4}^{-}}$: $\tfrac12(3+4+1)=4$ → sp³, tetrahedral. $\mathrm{CO_{2}}$: C has 2 σ bonds, no lone pair → sp, linear (the two π bonds use the two unhybridised $2p$ orbitals).
(c), (d) $\mathrm{SO_{2}}$: 2 σ + 1 lone pair = 3 → sp², bent. $\mathrm{NH_{4}^{+}}$: $\tfrac12(5+4-1)=4$ → sp³, tetrahedral.
(e), (f) $\mathrm{SF_{4}}$: 5 → sp³d, see-saw. $\mathrm{XeF_{4}}$: 6 → sp³d², square planar.
(g), (h) $\mathrm{NO_{3}^{-}}$: 3 σ, no lone pair → sp², trigonal planar. $\mathrm{H_{2}O}$: 2 σ + 2 lp = 4 → sp³, bent.
Interactive 2 — Hybridisation Step by Step

Choose a molecule. The orbital boxes show the valence shell of the central atom: in the ground state, after promotion of an electron (if needed), after mixing into hybrid orbitals, and after bonding, when each half-filled hybrid receives an electron (red) from an outer atom to form a σ bond. Unhybridised half-filled $p$ orbitals form π bonds. Press ▶ Play to run through the steps; use the step menu to stop at one.

central atom—
hybridisation—
geometry · angle—
bonds from central atom—

Multiple Bonds in Carbon Compounds and Hybridisation Involving d Orbitals

Ethane, $\mathrm{C_{2}H_{6}}$. Both carbon atoms are sp³. One sp³ hybrid of each overlaps head-on to form the $\mathrm{C{-}C}$ σ bond ($154\ \mathrm{pm}$); the other three of each carbon overlap hydrogen $1s$ orbitals ($\mathrm{C{-}H}$ $109\ \mathrm{pm}$). All angles are close to $109.5^{\circ}$ and the two $\mathrm{CH_{3}}$ groups can rotate about the σ bond.

Ethene, $\mathrm{C_{2}H_{4}}$. Each carbon is sp²: three sp² hybrids in a plane at $120^{\circ}$, and one unhybridised $2p$ orbital, holding one electron, perpendicular to that plane. The sp² hybrids form the $\mathrm{C{-}C}$ σ bond and two $\mathrm{C{-}H}$ σ bonds on each carbon; the two parallel $2p$ orbitals overlap sideways to form the π bond. The whole molecule is planar, the double bond ($134\ \mathrm{pm}$) is shorter than the single bond, and rotation about it is blocked.

Ethyne, $\mathrm{C_{2}H_{2}}$. Each carbon is sp: two sp hybrids at $180^{\circ}$ and two unhybridised $2p$ orbitals at right angles to each other and to the axis. The sp hybrids form the $\mathrm{C{-}C}$ σ bond and one $\mathrm{C{-}H}$ σ bond each; the two pairs of $2p$ orbitals overlap sideways to give two π bonds in perpendicular planes, whose electron clouds merge into a cylinder around the axis. The molecule is linear and the triple bond ($120\ \mathrm{pm}$) is the shortest of the three.

sp³d — phosphorus pentachloride. P is $[\mathrm{Ne}]3s^{2}3p^{3}$. Promotion of one $3s$ electron into an empty $3d$ orbital gives $3s^{1}3p^{3}3d^{1}$: five half-filled orbitals, which mix into five sp³d hybrids pointing to the corners of a trigonal bipyramid. Three equatorial $\mathrm{P{-}Cl}$ bonds ($202\ \mathrm{pm}$) lie in a plane at $120^{\circ}$; the two axial bonds ($240\ \mathrm{pm}$), at $90^{\circ}$ to that plane, suffer more repulsion, are longer and weaker — which is why $\mathrm{PCl_{5}}$ readily loses $\mathrm{Cl_{2}}$ and is so reactive. sp³d² — sulphur hexafluoride. S is $[\mathrm{Ne}]3s^{2}3p^{4}$; promotion of a $3s$ and a $3p$ electron gives $3s^{1}3p^{3}3d^{2}$, six half-filled orbitals and six sp³d² hybrids directed to the corners of a regular octahedron: six identical $\mathrm{S{-}F}$ bonds at $90^{\circ}$. Period-2 atoms have no $2d$ orbitals, so nitrogen and oxygen can never form $\mathrm{NCl_{5}}$ or $\mathrm{OF_{6}}$.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — when the central atom hardly hybridises

Not required by the CBSE syllabus. It builds on hybridisation and on the hydrides of groups 15 and 16, which JEE Main, NEET and JEE Advanced all list; the rule itself is not named in any syllabus.

The bond angles of the hydrides fall sharply down groups 15 and 16: $\mathrm{NH_{3}}\ 107^{\circ}$, $\mathrm{PH_{3}}\ 93.6^{\circ}$, $\mathrm{AsH_{3}}\ 92^{\circ}$; $\mathrm{H_{2}O}\ 104.5^{\circ}$, $\mathrm{H_{2}S}\ 92^{\circ}$. Angles so close to $90^{\circ}$ suggest that in $\mathrm{PH_{3}}$, $\mathrm{AsH_{3}}$ and $\mathrm{H_{2}S}$ the large central atom bonds through almost pure $p$ orbitals and keeps its lone pair in an orbital of mostly $s$ character: the $s$ and $p$ orbitals of these big atoms differ more in energy and size, and hybridisation would not pay. (This empirical generalisation is often called Drago's rule.) JEE questions on the order of bond angles in these hydrides rely on it.
✏️ Example 4 — Hybridisation along a carbon chain
For pent-3-en-1-yne, $\mathrm{\overset{1}{C}H{\equiv}\overset{2}{C}{-}\overset{3}{C}H{=}\overset{4}{C}H{-}\overset{5}{C}H_{3}}$: (a) give the hybridisation of each carbon; (b) count the σ and π bonds; (c) which $\mathrm{C{-}C}$ bond is the shortest? (d) Which $\mathrm{C{-}H}$ bond has the greatest $s$ character?
(a) C1, C2 (triple bond, 2 σ each): sp. C3, C4 (double bond, 3 σ each): sp². C5 (4 σ): sp³.
(b) Formula $\mathrm{C_{5}H_{6}}$: 11 atoms, no ring → $10$ σ (six $\mathrm{C{-}H}$ + four $\mathrm{C{-}C}$). π: 2 (triple) + 1 (double) = $3$.
(c) The $\mathrm{C1{\equiv}C2}$ triple bond (bond order 3).
(d) The $\mathrm{C1{-}H}$ bond: C1 uses an sp hybrid (50% $s$). More $s$ character holds electrons closer to carbon, so this $\mathrm{C{-}H}$ bond is the shortest ($\approx106\ \mathrm{pm}$) and its hydrogen the most acidic — the reason terminal alkynes are weakly acidic (Chapter 9).
✏️ Example 5 — Using d orbitals
(a) Write the valence-shell configurations of phosphorus in its ground and excited states in $\mathrm{PCl_{5}}$, and of sulphur in $\mathrm{SF_{6}}$. (b) Solid $\mathrm{PCl_{5}}$ consists of the ions $[\mathrm{PCl_{4}}]^{+}$ and $[\mathrm{PCl_{6}}]^{-}$. Give the hybridisation and shape of phosphorus in each. (c) Why can nitrogen form $\mathrm{NCl_{3}}$ but not $\mathrm{NCl_{5}}$?
(a) P: ground $3s^{2}3p^{3}$ (3 unpaired) → excited $3s^{1}3p^{3}3d^{1}$ (5 unpaired) → sp³d. S: ground $3s^{2}3p^{4}$ (2 unpaired) → excited $3s^{1}3p^{3}3d^{2}$ (6 unpaired) → sp³d².
(b) $[\mathrm{PCl_{4}}]^{+}$: $\tfrac12(5+4-1)=4$ → sp³, tetrahedral. $[\mathrm{PCl_{6}}]^{-}$: $\tfrac12(5+6+1)=6$ → sp³d², octahedral.
(c) N ($2s^{2}2p^{3}$) has three unpaired electrons and no $d$ orbitals in its valence shell ($n=2$), so it cannot promote electrons to make five half-filled orbitals.
✏️ Example 6 — Hybridisation can change in a reaction
Give the hybridisation of the named atom before and after each change: (a) B in $\mathrm{BF_{3}+NH_{3}\rightarrow H_{3}N{\rightarrow}BF_{3}}$ (N gives its lone pair to B); (b) C in $\mathrm{CH_{4}}$, the methyl cation $\mathrm{CH_{3}^{+}}$ and the methyl anion $\mathrm{CH_{3}^{-}}$; (c) Al in $\mathrm{AlCl_{3}+Cl^{-}\rightarrow[AlCl_{4}]^{-}}$.
(a) B in $\mathrm{BF_{3}}$: 3 σ, no lone pair → sp² (planar). In the adduct B has 4 σ bonds → sp³ (tetrahedral). N stays sp³ (3 σ + 1 lp before, 4 σ after).
(b) $\mathrm{CH_{4}}$: $\tfrac12(4+4)=4$, sp³. $\mathrm{CH_{3}^{+}}$: $\tfrac12(4+3-1)=3$, sp² — planar, with an empty $2p$ orbital. $\mathrm{CH_{3}^{-}}$: $\tfrac12(4+3+1)=4$, sp³ — pyramidal, with a lone pair. (These ions reappear in Chapter 8.)
(c) $\mathrm{AlCl_{3}}$: 3 σ → sp²; $[\mathrm{AlCl_{4}}]^{-}$: 4 σ → sp³ (tetrahedral).

Exercises

I
Orbital overlap, σ and π bonds.
  • 1 What is meant by orbital overlap? State the overlap criterion of bond strength.
  • 2 Describe the changes in potential energy as two hydrogen atoms approach from a large distance to well inside the bond length.
  • 3 Distinguish positive, negative and zero overlap, with one example of each.
  • 4 Taking $z$ as the internuclear axis, say which combinations give a σ bond, which a π bond, and which no bond: $s$ + $s$; $s$ + $p_{z}$; $p_{z}$ + $p_{z}$; $p_{x}$ + $p_{x}$; $s$ + $p_{x}$.
  • 5 Why is a σ bond stronger than a π bond?
  • 6 Count the σ and π bonds in $\mathrm{HCN}$, $\mathrm{C_{2}H_{2}}$, $\mathrm{H_{2}C{=}O}$ and propadiene, $\mathrm{CH_{2}{=}C{=}CH_{2}}$.
  • 7 Can two atoms be joined by a π bond alone? Why is rotation about a double bond restricted?
  • 8 Using mean bond enthalpies $\mathrm{C{-}C}$ $348$ and $\mathrm{C{=}C}$ $614\ \mathrm{kJ\,mol^{-1}}$, estimate the strength of the π bond in $\mathrm{C{=}C}$ and compare it with the σ bond.
1 Partial merging of orbitals of two atoms in the same region of space; electrons of opposite spin pair in the overlap region. The greater the overlap, the stronger the bond.
2 Far apart: energy zero. Approaching: nucleus–electron attractions exceed electron–electron and nucleus–nucleus repulsions — energy falls. At $74\ \mathrm{pm}$: minimum ($-435.8\ \mathrm{kJ\,mol^{-1}}$), forces balance. Closer: nuclear repulsion dominates and the energy rises steeply.
3 Positive: lobes of the same phase meet ($1s$–$1s$ in $\mathrm{H_{2}}$) — bonding. Negative: lobes of opposite phase meet — antibonding, repulsive. Zero: contributions cancel, e.g. $s$ approaching $p_{x}$ along $z$.
4 σ: $s+s$, $s+p_{z}$, $p_{z}+p_{z}$. π: $p_{x}+p_{x}$ (sideways). No bond (zero overlap): $s+p_{x}$.
5 Head-on overlap along the axis is much more extensive than sideways overlap, and the electron density lies directly between the nuclei.
6 $\mathrm{H{-}C{\equiv}N}$: 2σ, 2π. $\mathrm{H{-}C{\equiv}C{-}H}$: 3σ, 2π. $\mathrm{H_{2}C{=}O}$: 3σ, 1π. Propadiene: 4 $\mathrm{C{-}H}$ + 2 $\mathrm{C{-}C}$ = 6σ, 2π.
7 No: a σ bond always forms first, and π bonds are added to it. Twisting one end of a double bond turns the two $p$ orbitals out of parallel and destroys the sideways overlap, which costs the energy of the π bond.
8 π $\approx614-348=266\ \mathrm{kJ\,mol^{-1}}$, weaker than the σ bond ($348$) — as Rule 4.8 says.
II
Hybridisation: sp, sp², sp³.
  • 1 Define hybridisation and state three of its features.
  • 2 Give the hybridisation of the central atom in $\mathrm{BeF_{2}}$, $\mathrm{BH_{3}}$, $\mathrm{SiH_{4}}$ and $\mathrm{H_{3}O^{+}}$.
  • 3 What is the percentage $s$ character of sp, sp² and sp³ hybrid orbitals?
  • 4 Can the $1s$ and $2p$ orbitals of an atom hybridise? Explain.
  • 5 Explain with orbital box diagrams how carbon forms four equivalent bonds in methane.
  • 6 Nitrogen in $\mathrm{NH_{3}}$ is sp³ hybridised although no electron is promoted. How many of its hybrid orbitals are used in bonding, and what does the remaining one hold?
  • 7 Why is the bond angle in $\mathrm{BCl_{3}}$ $120^{\circ}$ but in $\mathrm{NCl_{3}}$ about $107^{\circ}$?
  • 8 Give the hybridisation of carbon in $\mathrm{CO_{2}}$, $\mathrm{CO_{3}^{2-}}$ and $\mathrm{CH_{3}OH}$.
1 Mixing of atomic orbitals of similar energy on one atom to give the same number of equivalent hybrid orbitals. Features: number of hybrids = number mixed; equivalent energy and shape; directed as far apart as possible; they form σ bonds and hold lone pairs; better overlap than pure orbitals.
2 $\mathrm{BeF_{2}}$ sp; $\mathrm{BH_{3}}$ sp² ($\tfrac12(3+3)=3$); $\mathrm{SiH_{4}}$ sp³; $\mathrm{H_{3}O^{+}}$ sp³ ($\tfrac12(6+3-1)=4$).
3 sp $\tfrac12=50\%$; sp² $\tfrac13=33.3\%$; sp³ $\tfrac14=25\%$.
4 No: they belong to different shells and differ greatly in energy; only valence-shell orbitals of similar energy mix.
5 Ground: $2s$ [↑↓], $2p$ [↑][↑][ ]. Excited: $2s$ [↑], $2p$ [↑][↑][↑]. Mixing: four sp³ [↑][↑][↑][↑] at $109.5^{\circ}$; each overlaps an H $1s$ orbital to form four identical σ bonds.
6 $2s^{2}2p^{3}$ mix into four sp³ hybrids holding $2,1,1,1$ electrons: three half-filled hybrids bond to H; the fourth holds the lone pair.
7 B in $\mathrm{BCl_{3}}$ is sp² with no lone pair: planar, $120^{\circ}$. N in $\mathrm{NCl_{3}}$ is sp³ with one lone pair, whose repulsion closes the angle below $109.5^{\circ}$.
8 $\mathrm{CO_{2}}$: sp (2 σ). $\mathrm{CO_{3}^{2-}}$: sp² (3 σ). $\mathrm{CH_{3}OH}$: sp³ (4 σ).
III
Multiple bonds in carbon compounds.
  • 1 Describe the bonding in ethene in terms of hybridisation and σ and π bonds. Why is the molecule planar?
  • 2 Describe the bonding in ethyne. Why is it linear?
  • 3 Give the hybridisation of each carbon atom in propene, $\mathrm{CH_{3}{-}CH{=}CH_{2}}$.
  • 4 Give the hybridisation of each carbon atom in propadiene, $\mathrm{CH_{2}{=}C{=}CH_{2}}$.
  • 5 Give the hybridisation of each carbon and of nitrogen in ethanenitrile, $\mathrm{CH_{3}{-}C{\equiv}N}$.
  • 6 Arrange the carbon–carbon bonds of ethane, ethene, ethyne and benzene ($139\ \mathrm{pm}$) in order of increasing length, and explain the position of benzene.
  • 7 How many sp² carbon atoms are there in benzene and in buta-1,3-diene, $\mathrm{CH_{2}{=}CH{-}CH{=}CH_{2}}$?
  • 8 Arrange sp, sp² and sp³ carbon in order of increasing electronegativity and explain.
1 Each C is sp²: one sp²–sp² σ bond between the carbons, two sp²–$1s$ σ bonds to H on each; the unhybridised $2p$ orbitals overlap sideways to form a π bond. Sideways overlap needs the two $p$ orbitals parallel, which locks all six atoms in one plane.
2 Each C is sp: sp–sp σ bond and sp–$1s$ σ bonds to H; two pairs of unhybridised $2p$ orbitals form two π bonds in perpendicular planes. The sp hybrids are $180^{\circ}$ apart, so $\mathrm{H{-}C{\equiv}C{-}H}$ is linear.
3 $\mathrm{CH_{3}}$: sp³; $\mathrm{CH}$: sp²; $\mathrm{CH_{2}}$: sp².
4 End carbons sp² (3 σ each); central carbon sp (2 σ, two π bonds).
5 $\mathrm{CH_{3}}$ carbon sp³; nitrile carbon sp (2 σ); N sp (1 σ + 1 lone pair).
6 Ethyne $120\lt$ ethene $134\lt$ benzene $139\lt$ ethane $154\ \mathrm{pm}$. Benzene's bonds are a resonance hybrid of single and double bonds (bond order 1.5), so their length is in between.
7 Benzene: 6 (every carbon has 3 σ bonds). Buta-1,3-diene: 4.
8 sp³ $\lt$ sp² $\lt$ sp. More $s$ character means the hybrid orbital is closer to the nucleus, so electrons in it are held more tightly.
IV
Hybridisation involving d orbitals.
  • 1 Describe the hybridisation and shape of $\mathrm{PCl_{5}}$. Why are its axial bonds longer than its equatorial bonds?
  • 2 Describe the hybridisation of sulphur in $\mathrm{SF_{6}}$ and give the $\mathrm{F{-}S{-}F}$ bond angles.
  • 3 Give the hybridisation of the central atom in $\mathrm{XeF_{2}}$, $\mathrm{XeF_{4}}$, $\mathrm{ClF_{3}}$ and $\mathrm{BrF_{5}}$.
  • 4 Why does $\mathrm{SF_{6}}$ exist but $\mathrm{OF_{6}}$ does not?
  • 5 Give the hybridisation of iodine in $\mathrm{ICl_{4}^{-}}$ and in $\mathrm{I_{3}^{-}}$.
  • 6 When $\mathrm{NH_{3}}$ accepts a proton, does the hybridisation of N change? What happens to the bond angle?
  • 7 Which of the following have an sp³d-hybridised central atom: $\mathrm{PCl_{5}}$, $\mathrm{SF_{4}}$, $\mathrm{ClF_{3}}$, $\mathrm{XeF_{4}}$, $\mathrm{I_{3}^{-}}$, $\mathrm{SF_{6}}$?
  • 8 Applied. $\mathrm{SF_{6}}$ is used as an insulating gas in high-voltage switchgear because it is chemically very inert, while $\mathrm{PCl_{5}}$ fumes and reacts with moisture. Use hybridisation and shape to suggest reasons for the difference.
1 P: $3s^{1}3p^{3}3d^{1}$ → sp³d, trigonal bipyramidal. Each axial bond pair has three bond pairs at $90^{\circ}$ (equatorial ones only two), so axial bonds are repelled more and lengthen ($240$ vs $202\ \mathrm{pm}$).
2 S: $3s^{1}3p^{3}3d^{2}$ → sp³d², octahedral; adjacent $\mathrm{F{-}S{-}F}$ angles $90^{\circ}$ (opposite ones $180^{\circ}$).
3 $\mathrm{XeF_{2}}$: 5 pairs, sp³d. $\mathrm{XeF_{4}}$: 6, sp³d². $\mathrm{ClF_{3}}$: 5, sp³d. $\mathrm{BrF_{5}}$: 6, sp³d².
4 S ($n=3$) has vacant $3d$ orbitals and can make six half-filled orbitals by promotion; O ($n=2$) has no $d$ orbitals and at most two unpaired electrons.
5 $\mathrm{ICl_{4}^{-}}$: 6 pairs → sp³d². $\mathrm{I_{3}^{-}}$: 5 pairs → sp³d.
6 No — N stays sp³ (4 pairs before and after). The lone pair becomes a fourth bond pair, so the angle opens from $107^{\circ}$ to the regular $109.5^{\circ}$ of $\mathrm{NH_{4}^{+}}$.
7 $\mathrm{PCl_{5}}$, $\mathrm{SF_{4}}$, $\mathrm{ClF_{3}}$ and $\mathrm{I_{3}^{-}}$ (5 pairs each). $\mathrm{XeF_{4}}$ and $\mathrm{SF_{6}}$ are sp³d².
8 In $\mathrm{SF_{6}}$ six identical strong S–F bonds point to the corners of an octahedron; the small F atoms surround and shield the sulphur atom completely, and S has no lone pair or vacant low-energy orbital for an attacking water molecule. In $\mathrm{PCl_{5}}$ the two axial bonds are long and weak, the trigonal bipyramid leaves room for attack at P, and Cl is easily replaced.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET The hybridisation of the three carbon atoms in propadiene, $\mathrm{CH_{2}{=}C{=}CH_{2}}$, from left to right is (A) sp², sp², sp² (B) sp², sp, sp² (C) sp, sp², sp (D) sp³, sp, sp³

Idea End carbons: 3 σ bonds (2 H + 1 C) → sp². Middle carbon: 2 σ bonds and two π bonds → sp.
Answer (B).

E2. JEE Main The numbers of σ and π bonds in tetracyanoethene, $\mathrm{(NC)_{2}C{=}C(CN)_{2}}$, are (A) 9σ, 9π (B) 5σ, 9π (C) 9σ, 7π (D) 8σ, 8π

σ $\mathrm{C{=}C}$: 1; four $\mathrm{C{-}CN}$ single bonds: 4; four $\mathrm{C{\equiv}N}$: 4 → $9$ (check: 10 atoms, no ring → 9).
π $\mathrm{C{=}C}$: 1; four $\mathrm{C{\equiv}N}$: $4\times2=8$ → $9$.
Answer (A).

E3. Board · 3 marks Explain the formation of the ethyne molecule on the basis of hybridisation, naming the orbitals that form each σ and π bond.

1 mark Each C ($2s^{1}2p^{3}$ after promotion) mixes $2s$ and one $2p$ into two sp hybrids at $180^{\circ}$; two $2p$ orbitals remain unhybridised.
1 mark σ bonds: sp(C)–sp(C) between the carbons, sp(C)–$1s$(H) to each hydrogen; the molecule is linear.
1 mark π bonds: sideways overlap of $2p_{x}$–$2p_{x}$ and $2p_{y}$–$2p_{y}$ gives two π bonds in perpendicular planes. Triple bond = 1σ + 2π.

E4. JEE Main Numerical answer. How many of the following have an sp³d-hybridised central atom? $\mathrm{PCl_{5}}$, $\mathrm{SF_{4}}$, $\mathrm{ClF_{3}}$, $\mathrm{XeF_{2}}$, $\mathrm{BrF_{5}}$, $\mathrm{SF_{6}}$, $\mathrm{I_{3}^{-}}$, $\mathrm{XeF_{4}}$

Count pairs 5 pairs (sp³d): $\mathrm{PCl_{5}}$, $\mathrm{SF_{4}}$, $\mathrm{ClF_{3}}$, $\mathrm{XeF_{2}}$, $\mathrm{I_{3}^{-}}$. 6 pairs (sp³d²): $\mathrm{BrF_{5}}$, $\mathrm{SF_{6}}$, $\mathrm{XeF_{4}}$.
Answer 5.

E5. JEE Advanced One or more options correct. (A) In $\mathrm{PCl_{5}}$ the axial $\mathrm{P{-}Cl}$ bonds are longer than the equatorial ones. (B) π bonds are formed by overlap of hybrid orbitals. (C) The two π bonds of ethyne lie in mutually perpendicular planes. (D) The carbon atom in $\mathrm{CH_{3}^{-}}$ is sp² hybridised.

(A) True ($240$ vs $202\ \mathrm{pm}$).
(B) False: π bonds use unhybridised $p$ orbitals.
(C) True: they come from $p_{x}$ and $p_{y}$.
(D) False: $\tfrac12(4+3+1)=4$ pairs → sp³ (pyramidal); $\mathrm{CH_{3}^{+}}$ is the sp² one.
Answer (A), (C).

E6. Board · Case study When two hydrogen atoms approach, the energy of the system falls to a minimum of $-435.8\ \mathrm{kJ\,mol^{-1}}$ at an internuclear distance of $74\ \mathrm{pm}$ and then rises steeply at shorter distances. (i) Explain the fall and the rise in energy in terms of attractive and repulsive forces. (2 marks) (ii) Calculate the energy released per molecule of $\mathrm{H_{2}}$ formed ($N_{A}=6.022\times10^{23}\ \mathrm{mol^{-1}}$). (1 mark) (iii) What do $74\ \mathrm{pm}$ and $435.8\ \mathrm{kJ\,mol^{-1}}$ represent? (1 mark)

(i) As the atoms approach, each nucleus attracts both electrons; these attractions outweigh the electron–electron and nucleus–nucleus repulsions, so energy falls. Inside $74\ \mathrm{pm}$ the nuclei repel strongly and the energy rises.
(ii) $435.8\times10^{3}/6.022\times10^{23}=7.24\times10^{-19}\ \mathrm{J}$.
(iii) The bond length and the bond (dissociation) enthalpy of $\mathrm{H_{2}}$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Diamond, graphite and graphene are all pure carbon. Why is diamond the hardest natural material and an electrical insulator, graphite a soft lubricant that conducts electricity, and graphene — a single sheet of graphite — one of the strongest and most conductive materials known?

Why it matters. Graphene was first isolated in 2004 at the University of Manchester by peeling graphite with sticky tape, and won the 2010 Nobel Prize in Physics. It is being developed for flexible electronics, water-filtration membranes, batteries and sensors; industrial diamond is essential for cutting and drilling tools worldwide.

What to do.

  1. In diamond each carbon is bonded to four others; in graphite and graphene to three others in a flat hexagonal sheet. Give the hybridisation and bond angle of carbon in each.
  2. In a graphite sheet, each carbon has one unhybridised $2p$ electron. Explain how these electrons form π bonding spread over the whole sheet, and why this lets graphite conduct along the sheets but not between them.
  3. Each carbon in graphite forms three σ bonds and shares one π bond among its three neighbours. Find the average carbon–carbon bond order, and predict whether the bond length ($142\ \mathrm{pm}$) should be closer to that of ethane ($154\ \mathrm{pm}$) or ethene ($134\ \mathrm{pm}$).
  4. Graphite sheets are $335\ \mathrm{pm}$ apart. Compare this with twice the covalent radius of carbon ($77\ \mathrm{pm}$) and explain why the sheets slide easily over each other.

Tools: Definition 4.11, Rule 4.8, the resonance idea of Section 4.1, and a reliable materials-science source on graphene.

Hand in: A labelled sketch of the bonding in diamond and in a graphite sheet, your bond-order and distance calculations, and a paragraph linking hybridisation to the properties of the three materials.

Going further. Carbon nanotubes are graphene sheets rolled into cylinders, and fullerene $\mathrm{C_{60}}$ is a closed cage of 12 pentagons and 20 hexagons. Is each carbon in $\mathrm{C_{60}}$ still close to sp²? What makes the cage curve?

Instructor note: diamond sp³, $109.5^{\circ}$, all four electrons locked in σ bonds → hard, insulating; graphite/graphene sp², $120^{\circ}$, π electrons delocalised over each sheet → conduction in-plane. Bond order: $1+\tfrac13=\tfrac43\approx1.33$, length between single and double, nearer ethene's than a simple average would suggest (benzene, order 1.5, is $139\ \mathrm{pm}$). Inter-sheet distance $335\ \mathrm{pm}\gg2\times77=154\ \mathrm{pm}$: no covalent bonds between sheets, only weak van der Waals forces, so they slide. In $\mathrm{C_{60}}$ carbons are slightly pyramidalised sp²; the pentagons introduce curvature.

🧪 Self-check quiz

0/5 correct

Q1. The hybridisation of each carbon atom in ethyne is

Q2. The hybridisation of sulphur in $\mathrm{SF_{6}}$ is

Q3. How many π bonds are there in cyanogen, $\mathrm{N{\equiv}C{-}C{\equiv}N}$?

Q4. A π bond is formed by

Q5. The percentage s character of an sp² hybrid orbital is

← 4.2 VSEPR Theory, Molecular Shapes and Polarity4.4 Molecular Orbital Theory and Hydrogen Bonding →