Covalent bonding and coordinate (dative covalent) bonding

1. Overview

A covalent bond holds two atoms together through a shared pair of electrons. This section covers bonding in small molecules, period 3 atoms with more than eight outer electrons, and coordinate (dative covalent) bonds. It then looks inside the bond: σ and π bonds, hybridisation, and how bond energy and bond length help you compare reactivity. These ideas underpin molecular shapes and most of organic chemistry.

Key Definitions

• Covalent bond: the electrostatic attraction between the nuclei of two atoms and a shared pair of electrons.
• Coordinate (dative covalent) bond: a covalent bond in which both electrons of the shared pair come from the same atom, which donates one of its lone pairs.
• Lone pair: a pair of electrons in the outer shell of an atom that is not shared in a bond.
• Expanded octet: more than eight electrons in the outer shell of an atom in a compound, possible from period 3 onwards, for example 10 around P in PCl₅ and 12 around S in SF₆.
• Sigma (σ) bond: a covalent bond formed by the direct, head-on overlap of two orbitals along the line joining the two bonded nuclei.
• Pi (π) bond: a covalent bond formed by the sideways overlap of two parallel p orbitals on adjacent atoms, with its electron density above and below the σ bond.
• Hybridisation: the mixing of an s orbital with one, two or three p orbitals on the same atom to form an equal number of identical hybrid orbitals (two sp, three sp² or four sp³).
• Bond energy: the energy required to break one mole of a particular covalent bond in the gaseous state; it is always positive.
• Bond length: the internuclear distance of two covalently bonded atoms, that is, the distance between their two nuclei.

Content

What a covalent bond is

When two non-metal atoms bond, neither can pull an electron completely away from the other, so they share a pair. Both nuclei attract the shared pair, and that electrostatic attraction is the covalent bond.

• One shared pair is a single bond, two make a double bond and three a triple bond. An outer-shell pair that is not shared is a lone pair.
• Sharing usually gives each atom eight outer electrons (two for hydrogen). This "octet" is a guide, not a law: some atoms have fewer, and period 3 atoms can have more.

Covalent bonding in simple molecules

The table summarises what the dot-and-cross diagrams of these molecules show.

Molecule	Bonds	Lone pairs	Outer electrons
H₂	H-H: one shared pair	none	2 on each H
Cl₂	Cl-Cl: one shared pair	3 on each Cl	8 on each Cl
HCl	H-Cl: one shared pair	3 on Cl	2 on H, 8 on Cl
O₂	O=O: two shared pairs	2 on each O	8 on each O
N₂	N≡N: three shared pairs	1 on each N	8 on each N
CO₂	O=C=O: two double bonds	none on C, 2 on each O	8 on each atom
NH₃	three N-H	1 on N	8 on N
CH₄	four C-H	none	8 on C
C₂H₆	one C-C, six C-H	none	8 on each C
C₂H₄	one C=C, four C-H	none	8 on each C

To check a structure, follow each atom's outer electrons: each N in N₂ has five, puts three into the triple bond and keeps two as a lone pair, so it has 6 + 2 = 8 around it.

Period 3 atoms can expand their octet

A period 2 atom, such as N or O, has only the 2s and three 2p orbitals in its outer shell, so it can never hold more than eight outer electrons. Nitrogen forms NCl₃ but not NCl₅.

A period 3 atom is larger, so more atoms fit around it, and its third shell also has 3d orbitals. The usual explanation at this level is that electrons are promoted into empty 3d orbitals, giving more unpaired electrons and so more bonds: phosphorus can use all five outer electrons and sulfur all six. Advanced theory gives the 3d orbitals only a small part, so treat this as a model; the electron counts below are what you need.

Molecule	Shared pairs around the central atom	Lone pairs on the central atom	Electrons around the central atom
PCl₅	5 (five P-Cl)	0	10
SF₆	6 (six S-F)	0	12
SO₂	4 (two S=O)	1	10

In SO₂, sulfur uses four of its six outer electrons in two S=O double bonds and keeps the other two as a lone pair. Each Cl, F and O atom still has eight outer electrons.

Coordinate (dative covalent) bonds

In a coordinate or dative covalent bond, both electrons of the shared pair come from one atom. The donor must have a lone pair, and the acceptor must have an empty orbital in its outer shell. Draw the bond as an arrow from donor to acceptor, such as N→H. Once formed, a dative bond is an ordinary shared pair; only the origin of its electrons differs.

Ammonia and hydrogen chloride. The two gases meet to form a white smoke of solid ammonium chloride.

NH₃(g) + HCl(g) → NH₄Cl(s)

In effect, HCl hands over H⁺: the H-Cl bond breaks with both electrons staying on chlorine (as Cl⁻), and the lone pair on nitrogen forms a dative bond to the H⁺, which has an empty 1s orbital. This gives the ammonium ion, NH₄⁺.

• Nitrogen now has four bonding pairs and no lone pair.
• The four N-H bonds are identical, so you cannot tell which was the dative one.

Aluminium chloride, Al₂Cl₆. In AlCl₃, aluminium shares its three outer electrons with three chlorine atoms, so it has only six outer electrons and an empty orbital: it is electron-deficient, like boron in BF₃. In aluminium chloride vapour, two AlCl₃ molecules join to form Al₂Cl₆.

• One Cl atom on each AlCl₃ donates a lone pair to the Al atom of the other molecule, making two Cl→Al dative bonds.
• These two Cl atoms become bridges, each bonded to both Al atoms (one ordinary bond, one dative bond). The two Al atoms are not bonded to each other.
• Each Al now has four bonding pairs, eight outer electrons. Each bridging Cl keeps two lone pairs; each of the four end Cl atoms keeps three.

σ and π bonds: how orbitals overlap

A covalent bond forms where an orbital on one atom overlaps an orbital on the other, and the shared pair occupies the overlap region.

	σ (sigma) bond	π (pi) bond
Overlap	direct, head-on, along the line joining the nuclei	sideways, between two parallel p orbitals on adjacent atoms
Electron density	concentrated between the nuclei	in two regions, above and below the σ bond
Orbitals	s or p (H-H, H-Cl, Cl-Cl) or hybrid (C-H)	p only
Where found	exactly one in every single, double and triple bond	only in double and triple bonds, alongside a σ bond
Strength (carbon-carbon bonds)	stronger	weaker: sideways overlap is less effective

The two regions of a π bond together make one bond. A single bond is one σ bond, a double bond is one σ and one π, and a triple bond is one σ and two π bonds, with the two π bonds at right angles.

Hybridisation: sp³, sp² and sp orbitals

Carbon, 1s² 2s² 2p², has only two unpaired electrons but forms four bonds. First, one 2s electron is promoted into the empty 2p orbital, giving four unpaired electrons; forming two extra bonds more than repays this energy. Then the 2s orbital mixes with one, two or three 2p orbitals to make identical hybrid orbitals that point as far apart as possible.

Type	Orbitals mixed	Hybrid orbitals and angle	p orbitals left	Example
sp³	2s + three 2p	four, 109.5° (tetrahedral)	none	each C in CH₄ and C₂H₆
sp²	2s + two 2p	three, 120° in one plane	one, at 90° to that plane	each C in C₂H₄
sp	2s + one 2p	two, 180° (linear)	two, at 90° to each other	C in HCN and CO₂

• Hybrid orbitals form σ bonds or hold lone pairs. π bonds form only from unhybridised p orbitals.
• Quick test for a C or N atom: count σ bonds plus lone pairs on the atom. Four means sp³, three sp², two sp. π bonds do not count. So N in NH₃ (three σ, one lone pair) is sp³, and N in N₂ or HCN (one σ, one lone pair) is sp.
• A hydrogen atom always bonds through its 1s orbital, so it is never hybridised. Hybridisation describes an atom, not a bond.

σ and π bonds in five molecules

Molecule	Hybridisation	σ bonds (orbitals overlapping head-on)	π bonds (sideways overlap)
H₂	none	1: 1s with 1s	0
C₂H₆	both C sp³	7: C-C from sp³ with sp³; six C-H from sp³ with H 1s	0
C₂H₄	both C sp²	5: C-C from sp² with sp²; four C-H from sp² with H 1s	1: the unhybridised 2p on each C
HCN	C sp, N sp	2: C-H from sp with H 1s; C-N from sp with sp	2: two pairs of 2p orbitals, at 90° to each other
N₂	both N sp	1: sp with sp (each N keeps a lone pair in its other sp orbital)	2: as in HCN

Ethene is planar. Each sp² carbon holds its three σ bonds in one plane, and the two 2p orbitals must be parallel to overlap, which locks both CH₂ groups into the same plane. All six atoms lie in that plane, with bond angles of about 120°, and the π electron density lies above and below it. HCN is linear, because the two sp orbitals on carbon point 180° apart.

Worked example 1: bonding in but-3-enenitrile

Question: But-3-enenitrile is CH₂=CH-CH₂-C≡N. Number the carbon atoms C1 to C4 from the CH₂= end. (a) Deduce the number of σ and π bonds. (b) State the hybridisation of each C atom and the N atom. (c) Describe how the bonds between C1 and C2 form.

Step 1: list every bond, including the five C-H bonds (two on C1, one on C2, two on C3).

Bond	Number	σ	π
C-H	5	5	0
C1=C2	1	1	1
C2-C3 and C3-C4	2	2	0
C4≡N	1	1	2
Total		9	3

Step 2: count σ bonds plus lone pairs on each atom. C1: 3 + 0 = 3, sp². C2: 3 + 0 = 3, sp². C3: 4 + 0 = 4, sp³. C4: 2 + 0 = 2, sp. N: 1 + 1 = 2, sp.

Step 3: describe C1=C2. Head-on overlap of an sp² orbital on each carbon gives the σ bond. Sideways overlap of the unhybridised 2p orbital on each carbon gives the π bond, above and below the σ bond.

Answer: (a) 9 σ and 3 π bonds. (b) C1 and C2 sp²; C3 sp³; C4 and N sp. (c) A σ bond from head-on overlap of two sp² orbitals and a π bond from sideways overlap of two 2p orbitals.

Bond energy and bond length

Bond energy is the energy required to break one mole of a particular covalent bond in the gaseous state. Breaking a bond always takes in energy, so values are positive, in kJ mol⁻¹: exact for diatomic molecules such as H-Cl, averages for bonds such as C-H found in many molecules. It counts moles of bonds: one mole of NH₃ contains three moles of N-H bonds.

Bond length is the internuclear distance of two covalently bonded atoms, given in nanometres (1 nm = 10⁻⁹ m) or picometres (1 pm = 10⁻¹² m). It is the separation at which the attractions between the nuclei and the shared pair balance the repulsions (between the two nuclei, and between electrons); the atoms vibrate about it.

The Data Booklet gives bond energies but not bond lengths, so learn the length patterns:

Bonds	Bond length, left to right	Bond energy / kJ mol⁻¹
C-C, C=C, C≡C	gets shorter	350, 610, 840
H-F, H-Cl, H-Br, H-I	gets longer	562, 431, 366, 299

1. More shared pairs between the same two atoms: shorter and stronger. More electron density between the nuclei pulls them closer and holds them more strongly. C=C is less than twice as strong as C-C (610 against 2 × 350 = 700), because its π bond is the weaker part.
2. Bigger atoms: longer and weaker. Down group 17 the halogen atom has more shells, so the shared pair in H-X is further from the halogen nucleus, more shielded and less strongly attracted.
3. Exception: single bonds between two small atoms that both have lone pairs. F-F is shorter than Cl-Cl (fluorine is the smaller atom) but weaker, 158 against 242 kJ mol⁻¹, because the lone pairs on the two close F atoms repel. For the same reason O-O and N-N are weaker than S-S and P-P (150 and 160 against 265 and 200 kJ mol⁻¹). So use "shorter means stronger" only for bonds between the same two atoms, or for one atom bonded to each member of a group, as in H-F to H-I.

Worked example 2: breaking every bond in a sample of ammonia

Question: Calculate the energy needed to break all the covalent bonds in 3.40 g of ammonia gas, then the energy needed to break one N-H bond, in J. (N-H bond energy = 390 kJ mol⁻¹; $A_{\text{r}}$: N 14.0, H 1.0; $L = 6.022 \times 10^{23}$ mol⁻¹)

NH₃(g) → N(g) + 3H(g)

Step 1: moles of ammonia. $M = 14.0 + 3(1.0) = 17.0$ g mol⁻¹, so $n = \frac{3.40}{17.0} = 0.200$ mol.

Step 2: moles of N-H bonds. Each molecule has three: $3 \times 0.200 = 0.600$ mol.

Step 3: energy for the sample. $$\text{energy} = 0.600 \times 390 = 234\ \text{kJ}$$

Step 4: energy for one bond. Convert to J, then divide by the Avogadro constant. $$\frac{390 \times 10^3\ \text{J mol}^{-1}}{6.022 \times 10^{23}\ \text{mol}^{-1}} = 6.48 \times 10^{-19}\ \text{J}$$

Answer: 234 kJ for the sample and $6.48 \times 10^{-19}$ J for one N-H bond (both 3 s.f.). As 390 kJ mol⁻¹ is an average, 234 kJ is an estimate.

Using bond energy and bond length to compare reactivity

Most reactions need at least one bond to break. In similar reactions, the weaker the bond that must break (usually the longer one), the less energy it takes to break, so the activation energy is usually lower and the reaction faster at a given temperature.

Comparison	Data / kJ mol⁻¹	What it explains
Nitrogen	N≡N 944	N₂ is very unreactive: its triple bond is one of the strongest in the Data Booklet, and the molecule is non-polar.
Hydrogen halides	H-F 562 to H-I 299	Thermal stability falls from HF to HI: the longer, weaker H-I bond breaks most easily on heating.
Halogenoalkanes	C-Cl 340, C-Br 280, C-I 240	Of these three, iodoalkanes hydrolyse fastest. C-Cl is the most polar, but C-I is about 29% weaker than C-Cl, and here bond strength matters more than polarity.
Alkenes and alkanes	π about 260, C-C 350, C-H 410	Alkenes react readily by breaking the weaker π bond (Worked example 3); alkanes, with only σ bonds, are much less reactive.

Bond strength is not the only factor: polarity (a δ+ atom attracts nucleophiles) and the stability of the products matter too. Use bond energies to compare similar molecules reacting in the same way.

Worked example 3: how strong is a π bond?

Question: Use these average bond energies to estimate the energy of the π bond in C=C and of each π bond in C≡C: C-C 350, C=C 610, C≡C 840 kJ mol⁻¹. Explain why ethene reacts with bromine in the dark at room temperature but ethane does not.

Step 1: assumption. The σ part of each multiple bond has the same energy as a C-C single bond, 350 kJ mol⁻¹.

Step 2: C=C is one σ plus one π. $$E(\pi) \approx 610 - 350 = 260\ \text{kJ mol}^{-1}$$

Step 3: C≡C is one σ plus two π. $$E(\pi) \approx \frac{840 - 350}{2} = 245\ \text{kJ mol}^{-1}$$

Step 4: compare. $\frac{260}{350} = 0.743$: a π bond is only about three-quarters as strong as a C-C σ bond.

Answer: about 260 kJ mol⁻¹ for the π bond in C=C and about 245 kJ mol⁻¹ for each π bond in C≡C. Ethene's π bond is weaker than any bond in ethane (C-C 350, C-H 410), and its electrons lie above and below the plane of the molecule, exposed to electrophiles such as Br₂. So ethene adds bromine by breaking only its π bond. Ethane has only strong σ bonds; it needs UV light, which splits Br₂ into atoms. These are estimates: the values are averages, and the σ part of a double bond is not exactly a C-C single bond.

Key Equations

Bonds in multiple bonds (each pair of bonded atoms has exactly one σ bond): $$\text{single} = 1\sigma \qquad \text{double} = 1\sigma + 1\pi \qquad \text{triple} = 1\sigma + 2\pi$$

Hybridisation of a C or N atom (count σ bonds and lone pairs on the atom; ignore π bonds): $$\sigma\ \text{bonds} + \text{lone pairs}:\quad 4 \Rightarrow \text{sp}^3 \qquad 3 \Rightarrow \text{sp}^2 \qquad 2 \Rightarrow \text{sp}$$

Estimated π bond energy (average bond energies in kJ mol⁻¹): $$E(\pi) \approx E(\text{C=C}) - E(\text{C-C})$$

Energy to break bonds in a sample ($n$ = moles of molecules; $b$ = number of that bond per molecule; $E$ = bond energy in kJ mol⁻¹; answer in kJ): $$\text{energy} = n \times b \times E$$

Energy per bond ($E$ in kJ mol⁻¹; $L = 6.022 \times 10^{23}$ mol⁻¹; answer in J): $$E_{\text{one bond}} = \frac{E \times 10^3}{L}$$

Common Mistakes to Avoid

• Mistake: defining a covalent bond as "sharing electrons". Fix: name the attraction and both parts: the electrostatic attraction between the nuclei of two atoms and a shared pair of electrons.
• Mistake: calling a dative bond a "transfer" of electrons, or drawing the arrow backwards. Fix: the donor shares its lone pair, and the arrow points from donor to acceptor. Al₂Cl₆ has two Cl→Al arrows and no Al-Al bond.
• Mistake: missing the C-H bonds, or counting a double bond as two σ bonds. Fix: each pair of bonded atoms has one σ bond. Ethene has 5 σ (four C-H, one C-C) and 1 π, not 1 σ (C-H missed) or 6 σ (C=C counted as two σ).
• Mistake: explaining hybridisation by the bonds formed, or calling a bond "sp²". Fix: it is mixing of orbitals on one atom: "each C in ethene is sp² hybridised: its 2s and two 2p orbitals mix to give three sp² orbitals".
• Mistake: describing a π bond as "overlap of p orbitals" with no direction, or drawing it between the nuclei. Fix: write "sideways overlap of parallel p orbitals, above and below the σ bond"; the overlap between the nuclei is the σ bond.
• Mistake: explaining the thermal stability of hydrogen halides with intermolecular forces, or with bond length alone. Fix: give the chain: H-I is longer than H-Cl, so weaker (299 against 431 kJ mol⁻¹), so it breaks more easily on heating.
• Mistake: multiplying by moles of molecules, not moles of bonds. Fix: in Worked example 2, 0.200 mol of NH₃ holds 0.600 mol of N-H bonds; 0.200 × 390 gives 78.0 kJ, not 234 kJ.

Exam Tips

• Define in full: a covalent bond needs "electrostatic attraction", "nuclei" and "shared pair"; bond energy needs "one mole", "particular covalent bond" and "gaseous state".
• Describe σ and π bonds by naming the orbitals and the type of overlap, and say that ethene is planar. In a diagram, mark the nuclei, draw the p orbitals parallel and label each bond.
• Explain reactivity or stability as a chain: longer bond, shared pair attracted less strongly, lower bond energy, bond breaks more easily. Quote values, and use those given in the question when it supplies them.
• Draw dative bonds as arrows from the donor's lone pair to the acceptor, show every lone pair, and put an ion in square brackets with its charge outside.