Formula –M 0 g/molCentre –
Drag to rotate · wheel or pinch to zoom · click an atom to bond a new atom to it

Covalent bonds

A covalent bond is a pair of electrons shared between two nuclei. Both nuclei are attracted to the shared density between them, which lowers the energy. Atoms in period 2 (C, N, O, F) can hold at most eight valence electrons (the octet rule). In neutral molecules this gives carbon 4 bonds, nitrogen 3, oxygen 2 and fluorine 1. The builder warns when an atom goes beyond that neutral valence. For carbon this would break the octet; for N or O it would need a formal charge (NH4+, H3O+), which the builder does not track.

Sharing two or three pairs gives double and triple bonds. Higher bond order means a shorter, stronger bond:

BondOrderTypical lengthMean bond enthalpy
C–C1154 pm347 kJ/mol
C=C2134 pm614 kJ/mol
C≡C3120 pm839 kJ/mol

Elements of period 3 and below (P, S, Cl, Br, I, Xe) can have more than eight electrons around them in Lewis structures (an expanded octet), as in PCl5, SF6, ClF3 and XeF4. Modern bonding theory explains these with delocalised (3-centre, 4-electron) bonding rather than d-orbital use, but the Lewis/VSEPR bookkeeping still predicts the shapes correctly. In the same way, the "P=O" in phosphate is better pictured as P+–O−, which keeps phosphorus at an octet.

Counting electrons: the Lewis structure

For a neutral molecule the total number of valence electrons is $\sum_i v_i$ (group number for main-group atoms). Each bond uses 2 electrons; the rest are placed as lone pairs. For an atom with $v$ valence electrons forming bonds of total order $b$, the number of lone pairs is

$$\text{LP} = \frac{v - b}{2}, \qquad \text{formal charge} = v - 2\,\text{LP} - b$$

VSEPR: Valence Shell Electron Pair Repulsion

Electron domains around a central atom (bonded atoms, counted once whatever the bond order, plus lone pairs) repel and spread as far apart as possible. The steric number is

$$\text{SN} = (\text{number of bonded atoms}) + (\text{number of lone pairs})$$

The domains take the electron geometry for that SN; the molecular shape is named from the atom positions only. Using AXmEn notation (A central atom, X bonded atoms, E lone pairs):

SNAXEShapeIdeal angle(s)HybridExample
2AX2Linear180°sp
3AX3Trigonal planar120°sp²
3AX2EBent<120° (SO₂ 119°)sp²
4AX4Tetrahedral109.5°sp³
4AX3ETrigonal pyramidal<109.5° (NH₃ 107°)sp³
4AX2E2Bent<109.5° (H₂O 104.5°)sp³
5AX5Trigonal bipyramidal90°, 120°, 180°sp³d
5AX4ESee-saw~90°, ~120°, ~180°sp³d
5AX3E2T-shaped~90°, ~180°sp³d
5AX2E3Linear180°sp³d
6AX6Octahedral90°, 180°sp³d²
6AX5ESquare pyramidal~90°sp³d²
6AX4E2Square planar90°, 180°sp³d²

Repulsion strength is ordered lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, because a lone pair is held by only one nucleus and spreads out closer to it. That is why H2O (104.5°) and NH3 (107°) are squeezed below 109.5°, and why lone pairs in a trigonal bipyramid always sit in the roomier equatorial positions (see-saw, T-shape, linear XeF2), while the two lone pairs of an octahedral centre sit opposite each other (square planar XeF4).

Hybridisation

Valence bond theory describes the domains with hybrid orbitals mixed from the atom's s, p (and, formally, d) orbitals. The number of hybrids equals the steric number:

$$\text{sp}:\ 2\ (180°)\quad \text{sp}^2:\ 3\ (120°)\quad \text{sp}^3:\ 4\ (109.5°)\quad \text{sp}^3\text{d}:\ 5\quad \text{sp}^3\text{d}^2:\ 6$$

Hybrids form σ bonds along the bond axis. The unhybridised p orbitals left on sp and sp² atoms overlap sideways to form π bonds: a double bond is σ + π, a triple bond σ + 2π. Because sideways overlap needs parallel p orbitals, the atoms around a C=C double bond lie in one plane (try ethene) and rotation about it is blocked. The angle between hybrids with fractional s character $s$ obeys $\cos\theta = -\dfrac{s}{1-s}$, so sp³ ($s = \tfrac14$) gives $\cos\theta = -\tfrac13$, $\theta = 109.47°$.

Polarity

When two atoms of different electronegativity share electrons, the pair is pulled towards the more electronegative atom, giving partial charges $\delta^+$ and $\delta^-$ and a bond dipole. Dipole moment is $\mu = q\,d$ (unit: debye, 1 D = 3.336×10−30 C m). Rough guide to bond type by $\Delta\chi$: below 0.4 essentially nonpolar covalent, 0.4–1.7 polar covalent, above about 1.7 largely ionic.

The molecular dipole is the vector sum of bond dipoles (plus a contribution from lone pairs). Symmetric shapes cancel: CO2, BF3, CH4, PCl5, SF6 and XeF4 have polar bonds but are nonpolar molecules. H2O, NH3 and ClF3 do not cancel and are polar. The builder estimates this as $\vec\mu \approx \sum \Delta\chi\,\hat{\mathbf u}$, each unit vector pointing from the less to the more electronegative atom, and draws the result with the chemist's crossed arrow (pointing to the δ− end).

Model limits. The 3D shapes come from a simplified force field (bond springs, VSEPR domain templates with lone-pair compression for SN 3–4, planar double bonds, soft non-bonded repulsion); it is not a quantum calculation. Stereochemistry (R/S, cis/trans, chair conformers) is not controlled. Resonance is not modelled: nitrogen next to C=O or in aromatic rings (amides, caffeine), and the single-bonded oxygen of esters and carboxylic acids, are really close to planar sp², though simple VSEPR counts them as sp³. The net-dipole estimate ignores lone-pair dipoles, so it under-estimates NH3 and misses the small dipole of molecules like CO.

1. Molar mass of caffeine, C8H10N4O2

$$M = 8(12.011) + 10(1.008) + 4(14.007) + 2(15.999)$$
$$= 96.088 + 10.080 + 56.028 + 31.998 = 194.19\ \text{g mol}^{-1}$$

Load the caffeine preset: the molecule panel shows the same value. One 95 mg dose (a strong cup of coffee) is $0.095/194.19 = 4.9\times10^{-4}$ mol.

2. Predicting the shape of ClF3

  1. Valence electrons: Cl 7 + 3 × F 7 = 28.
  2. Three Cl–F single bonds use 6. Each F takes 3 lone pairs (18 electrons). The remaining 28 − 6 − 18 = 4 go on Cl as 2 lone pairs.
  3. Cl: 3 bonded atoms + 2 lone pairs, SN = 5, AX3E2. Electron geometry trigonal bipyramidal, hybridisation sp³d.
  4. Lone pairs take equatorial positions (each then has only two 90° neighbours instead of three). The atoms form a T-shape with F–Cl–F angles near 90° (measured 87.5°) and 180° (175°).

Load ClF3, select the Cl and compare the VSEPR panel.

3. Why water is polar and CO2 is not

CO2: $\Delta\chi(\text{C–O}) = 3.44 - 2.55 = 0.89$. The molecule is linear, the two bond vectors are equal and opposite: $\vec\mu = 0.89\hat{\mathbf u} - 0.89\hat{\mathbf u} = 0$. Nonpolar.

H2O: $\Delta\chi(\text{O–H}) = 3.44 - 2.20 = 1.24$, bent at $\theta = 104.5°$. The components along the bisector add:

$$|\vec\mu| = 2(1.24)\cos\left(\tfrac{104.5°}{2}\right) = 2(1.24)(0.612) = 1.52$$

Turn on Bond dipoles with the water preset; the panel shows a net value of about 1.52 Δχ units. (The measured dipole moment of water is 1.85 D.)

4. Hybridisation in aspirin

Aspirin, C9H8O4 (M = 180.16 g/mol), has 9 carbons: the six ring carbons and the two C=O carbons have SN = 3 (sp², trigonal planar, 120°); the methyl carbon has SN = 4 (sp³, tetrahedral). The ester oxygen and the O–H oxygen have 2 bonds and 2 lone pairs: SN = 4, sp³, bent in simple VSEPR counting (resonance with the neighbouring C=O makes the ester oxygen closer to sp² in reality). Click the centres list in the VSEPR panel to check each one.

5. Counting σ and π bonds in ethyne, C2H2

H–C≡C–H: two C–H σ bonds, plus the triple bond (1 σ + 2 π). Total 3 σ and 2 π. Each carbon has SN = 2 (sp), so the molecule is linear with 180° angles.

Build in the lab above. Challenges are checked automatically every time the molecule changes; there must be no valence warnings.

Investigations

  1. Start from methane, then click one H and replace it with Cl, then with F, I and so on. Watch the net dipole change. Why is CCl4 nonpolar while CH3Cl is polar?
  2. Build NH3, then use the Bond tool to change N–H bonds, or add a fourth atom to N. Read the warning. Does NH4+ break the octet, or does N carry a formal charge? Work it out with the formal-charge formula in the Theory tab.
  3. Load SF6, then delete F atoms one at a time with Automatic hydrogens off. Follow the shapes: octahedral → square pyramidal (SF5, a radical) → see-saw (SF4) → bent (SF2).
  4. Load ethene and ethane (build C–C). Drag to rotate and compare: why are all six atoms of ethene in one plane?
  5. Load glucose. How many sp³ centres does it have? Which atoms carry lone pairs?