Molecular shapes (VSEPR), polarity and dipole moment, hybridization

The Lewis structure shows the connections of the molecule, but on flat paper. Real molecules are three-dimensional, and their shapes determine everything: polarity, boiling point, solubility, biological effect. This week you will learn the method for going from a Lewis structure to a three-dimensional shape (VSEPR), going from shape to polarity, and the model that explains the shape with orbitals (hybridization).

Golden rules:

1. First the Lewis structure, then count the groups. The number of groups determines everything.
2. Double and triple bonds are a single group.
3. Only the atoms give the shape its name. Lone pairs bend the shape but do not appear in the name.
4. Number of hybrid orbitals = number of groups.

VSEPR theory

Valence-shell electron-pair repulsion (VSEPR): the electron groups around the central atom repel each other and arrange themselves as far apart as possible. An electron group is one of the following:

• a single bond,
• a double or triple bond (counts as one group),
• a lone pair,
• (rarely) a single unpaired electron.

To find the number of groups quickly: first count the total valence electrons, subtract the octets of the terminal atoms (2 for H), and divide the rest by two. This is the number of lone pairs on the center ($E$). Together with the number of bonded atoms $X$, $X + E$ gives the number of groups. This shortcut works as long as all the terminal atoms satisfy the octet.

Electron geometry and molecular shape

Separate the two concepts:

• Electron-group geometry: the arrangement of all the groups (bonds and pairs).
• Molecular geometry: the arrangement of only the atoms.

If there are no lone pairs, the two are the same.

Why do the angles narrow?

A lone pair is attached to only one nucleus, so it takes up more room around the central atom. The order of repulsion:

$\text{lone pair–lone pair} > \text{lone pair–bonding pair} > \text{bonding pair–bonding pair}$

That is why 109.5° → 107° in $\mathrm{NH_3}$ → 104.5° in $\mathrm{H_2O}$. Multiple bonds also take up more room than single bonds; in $\mathrm{H_2CO}$ the H–C–H angle is a little less than 120°.

Where does a lone pair go with five groups?

In a trigonal bipyramid there are two kinds of positions: axial (3 neighbors at 90°) and equatorial (2 neighbors at 90°). To reduce 90° repulsions, lone pairs choose the equatorial position. The result: AX₄E seesaw, AX₃E₂ T-shaped, AX₂E₃ linear.

With six groups, the second lone pair goes directly opposite the first: AX₄E₂ becomes square planar.

Example 1: $\mathrm{ClF_3}$.

total e−: 7 + 3(7) = 28
octets of terminal atoms: 3 × 8 = 24  → Cl has 4 e− = 2 lone pairs
X = 3, E = 2 → 5 groups, AX3E2
electron geometry: trigonal bipyramidal;  both pairs equatorial
molecular shape: T-shaped, F–Cl–F angles ~87.5° (a little less than 90°)

Example 2: $\mathrm{ClO_3^-}$.

total e−: 7 + 3(6) + 1 = 26
octets of terminal atoms: 24 → 1 pair on Cl
AX3E → trigonal pyramidal (like NH3)

PITFALL · counting a multiple bond as two groups. In $\mathrm{CO_2}$, C has four bonding pairs but only two groups. If you think 4 groups you get tetrahedral; the right answer is linear.

Polarity and dipole moment

If the electrons in a bond are not shared equally, a bond dipole arises. Its size is measured by the dipole moment:

$\mu = Q \cdot r$

$Q$ is the separated charge, $r$ the distance between the charges. The unit is the debye: $1\ \text{D} = 3.336\times10^{-30}$ C·m.

Bond dipoles are vectors. The net dipole of a molecule is the vector sum of the bond dipoles. So:

• Polar bonds + symmetric shape + identical terminal atoms → nonpolar molecule. $\mathrm{CO_2}$: the two C=O dipoles point in opposite directions, the total is 0. $\mathrm{BF_3}$, $\mathrm{CCl_4}$, $\mathrm{SF_6}$, $\mathrm{XeF_4}$ are also nonpolar.
• Polar bonds + asymmetric shape → polar molecule. $\mathrm{H_2O}$ (bent), $\mathrm{NH_3}$ (pyramidal), $\mathrm{SO_2}$ (bent).
• Symmetric shape but different terminal atoms → generally polar: $\mathrm{CH_2Cl_2}$, $\mathrm{CHCl_3}$.

A quick check: if there is a lone pair on the center (except for perfectly symmetric exceptions like AX₄E₂ and AX₂E₃), the molecule is almost always polar.

Percent ionic character

If the bond were fully ionic, the charge would be $e$ and $\mu_{\text{ionic}} = e \cdot r$. The ratio of the measured dipole moment to this value gives the ionic character of the bond.

Example 3. For HF, $r = 92$ pm, measured $\mu = 1.82$ D.

$\mu_{\text{ionic}} = \frac{1.602\times10^{-19} \times 92\times10^{-12}}{3.336\times10^{-30}} = 4.42\ \text{D} \qquad \frac{1.82}{4.42}\times100 = 41\%$

The same calculation for HCl gives 17.7% (in the questions section). The ionic character increases as the electronegativity difference grows.

Hybridization

VSEPR predicts the shape but does not explain how it forms. The valence orbitals of the C atom are $2s$ and three $2p$; the $p$ orbitals are at 90° to each other. So how do four equivalent bonds at 109.5° form in $\mathrm{CH_4}$?

The answer of valence bond theory: when bonding, the atom mixes its valence orbitals to form new, equivalent hybrid orbitals. The rules:

• If $n$ atomic orbitals mix, $n$ hybrid orbitals result.
• Hybrid orbitals point toward the electron geometry of VSEPR.
• Number of hybrid orbitals = number of electron groups.

Lone pairs also sit in hybrid orbitals. In $\mathrm{NH_3}$, N is sp³: three hybrids bond with H, and the lone pair sits in the fourth. In $\mathrm{H_2O}$, O is also sp³.

Example 4. The two C atoms in acetaldehyde $\mathrm{CH_3CHO}$:

CH3 carbon: 4 single bonds   → 4 groups → sp3, tetrahedral
CHO carbon: C–C, C–H, C=O    → 3 groups → sp2, trigonal planar (1 pure p left → for the π bond)
O atom:     C=O + 2 lone pairs → 3 groups → sp2

What the pure (unhybridized) p orbitals are for is the topic of next week: π bonds.

TIP · the hybrid order. After finding the number of groups, to write the name of the hybrid add up the superscripts: sp (1+1 = 2), sp² (1+2 = 3), sp³ (1+3 = 4), sp³d (1+3+1 = 5), sp³d² (1+3+2 = 6).

Common mistakes