09 Molecular Geometry and Polarity

Count electron regions

VSEPR theory is a practical way to predict how atoms are arranged in three dimensions. Electron regions repel one another, so they tend to spread out around the central atom.

Count each single, double, or triple bond as one region. Count each lone pair on the central atom as one region as well. The number of regions gives the electron-domain geometry; the positions of the bonded atoms give the molecular geometry.

A convenient notation is $\mathrm{AX_nE_m}$: $\mathrm{A}$ is the central atom, $\mathrm{X}$ represents a bonded atom, and $\mathrm{E}$ represents a lone pair on the central atom. For example, $\mathrm{AX_3E}$ has three bonded atoms and one central lone pair.

Takeaway: Count electron regions first, then distinguish the geometry of all regions from the shape made by the atoms.

From electron regions to molecular shape

Once you have counted the electron regions, match their arrangement to a basic geometry. Lone pairs are included when naming the electron-domain geometry, but molecular geometry describes only the atom positions.

• Two regions: linear electron-domain geometry. With two bonded atoms and no lone pairs, the molecular geometry is linear, with an ideal bond angle of $180^\circ$.

• Three regions: trigonal planar electron-domain geometry. Three bonded atoms give a trigonal planar molecule with ideal angles of $120^\circ$; two bonded atoms and one lone pair give a bent molecule, usually with an angle below $120^\circ$.

• Four regions: tetrahedral electron-domain geometry. Four bonded atoms give a tetrahedral molecule with ideal angles of $109.5^\circ$. Three bonded atoms and one lone pair give a trigonal pyramidal shape; two bonded atoms and two lone pairs give a bent shape. Both shapes usually have smaller bond angles than the ideal tetrahedral angle.

• Five regions: trigonal bipyramidal electron-domain geometry. With one lone pair the shape is seesaw; with two it is T-shaped; with three it is linear. Lone pairs preferentially occupy equatorial positions.

• Six regions: octahedral electron-domain geometry. One lone pair gives a square pyramidal shape; two lone pairs opposite each other give a square planar shape.

Useful reference angles for the five-region arrangement are $90^\circ$, $120^\circ$, and $180^\circ$; for the six-region arrangement they are $90^\circ$ and $180^\circ$. These are ideal or common reference values. Lone pairs and differences among surrounding atoms can distort actual angles.

The effect of lone pairs is especially clear when comparing three molecules. In $\mathrm{CH_4}$, four bonding regions produce a tetrahedral shape and bond angles of about $109.5^\circ$. In $\mathrm{NH_3}$, one of four regions is a lone pair, producing a trigonal pyramidal shape and angles of about $107^\circ$. In $\mathrm{H_2O}$, two of four regions are lone pairs, producing a bent shape and an angle of about $104.5^\circ$.

Repulsions generally follow this order: lone pair–lone pair is strongest, followed by lone pair–bonding pair, then bonding pair–bonding pair. Stronger repulsion from lone pairs helps explain why nearby bond angles are often compressed.

Takeaway: Use the total region count to find the electron-domain geometry, then use the number of bonded atoms and lone pairs to name the molecular geometry.

Use geometry to determine polarity

A polar bond has an uneven distribution of bonding electrons. Its bond dipole points toward the more electronegative atom. To determine whether the whole molecule is polar, combine the bond dipoles as vectors:

A molecule is polar when this vector sum is nonzero. As a result, the presence of polar bonds alone does not establish molecular polarity; the molecule’s geometry determines whether the dipoles cancel.

• In linear $\mathrm{CO_2}$, the two polar $\mathrm{C=O}$ bond dipoles point in opposite directions and cancel. The molecule is nonpolar.

• In bent $\mathrm{H_2O}$, the $\mathrm{O-H}$ bond dipoles do not cancel, so the molecule is polar.

• In trigonal planar $\mathrm{BCl_3}$, three identical $\mathrm{B-Cl}$ bonds are arranged symmetrically. Their dipoles cancel, making the molecule nonpolar.

• In tetrahedral $\mathrm{CH_3Cl}$, the surrounding atoms are not all identical. Its bond dipoles do not cancel, so the molecule is polar.

A useful check is to identify polar bonds, determine the three-dimensional molecular geometry, and then consider whether the bond-dipole vectors cancel. Symmetry can lead to cancellation when surrounding atoms are identical, but shape alone is not enough: the identities and polarities of the bonds matter too.

Takeaway: Molecular polarity depends on both bond polarity and the vector arrangement of those bonds in the molecule.