2A.10-2A.12 - Shapes And Bond Angles

Why Molecules Have A Shape

Carbon dioxide and water are both small triatomic molecules, but they are not the same shape: $\ce{CO2}$ is linear, while $\ce{H2O}$ is bent. The difference is not the number of atoms. It is the electron pairs around the central atom.

Electron-pair repulsion theory, often called VSEPR, begins with one idea:

The central atom matters because the shape is described around it. In $\ce{BeCl2(g)}$ the central atom is Be; in $\ce{BCl3}$ it is B; in $\ce{CH4}$, $\ce{CO2}$ and simple organic molecules it is usually C. Hydrogen is almost always an outer atom because it can form only one bond.

There are two kinds of electron pair to count around the central atom.

The name of the molecular shape describes the positions of the atoms, not the lone pairs. That is why $\ce{NH3}$ is trigonal pyramidal, not tetrahedral, even though there are four electron pairs around nitrogen. One of the four pairs is a lone pair, so only three atoms are attached to the central atom.

For multiple bonds, keep the distinction between a pair count and a direction count. A C=O double bond contains two shared pairs, but both pairs lie between the same two atoms, so for the gross shape around carbon it is treated as one region of electron density in that direction. This is why $\ce{CO2}$ has two electron-density directions around carbon and is linear.

Shapes With Only Bonding Pairs

Start with the cleanest cases: every outer electron pair around the central atom is a bonding pair. The repulsions are then as equal as possible, so the ideal shapes and angles are the ones students usually learn first.

[DIAGRAM: asset_name: Electron Domain Geometries; asset_slug: edexcel_a_level_chemistry_l010_electron_domain_geometries; recommended_method: matplotlib; description: Pale-colour linear, trigonal planar and tetrahedral molecules with labelled angles and wedge/dash perspective; five- and six-domain shapes appear in the following diagram.]

Notice two details in that table.

First, $\ce{NH4+}$ is tetrahedral. In formation, a lone pair on $\ce{NH3}$ can be donated to $\ce{H+}$, but once the N-H bond has formed it behaves as an ordinary covalent bond for shape. Around N in $\ce{NH4+}$ there are four bonding pairs and no lone pairs.

Second, $\ce{PCl5}$ is specified as $\ce{PCl5(g)}$. The gaseous molecule has trigonal bipyramidal molecules. Solid phosphorus(V) chloride has ionic character and is outside the shape claim being made here.

Worked example: explaining the shape of $\ce{NH4+}$

Question: Predict the shape and bond angle of $\ce{NH4+}$.

Route:

1. The central atom is nitrogen.
2. There are four N-H bonds around nitrogen.
3. There are no lone pairs on nitrogen in $\ce{NH4+}$.
4. Four bonding pairs repel equally and arrange tetrahedrally.

Answer: $\ce{NH4+}$ is tetrahedral with H-N-H bond angles of about 109.5 degrees. The positive charge does not make the ion flat or pyramidal; the shape follows the four bonding pairs around nitrogen.

[DIAGRAM: asset_name: Five And Six Electron Domains; asset_slug: edexcel_a_level_chemistry_l010_five_six_domains; recommended_method: matplotlib; description: Trigonal bipyramidal PCl5 and octahedral SF6 shown with axial and equatorial positions and angles.]

How Lone Pairs Change Bond Angles

Lone pairs repel more strongly than bonding pairs.

This happens because a bonding pair is attracted by two nuclei, while a lone pair is held mainly near the central atom. The lone pair occupies more space around the central atom, so it pushes bonding pairs closer together. The shape name changes because lone pairs are counted for repulsion but not named as atoms in the molecular shape.

The classic four-pair sequence is the one you will use constantly.

The sequence is not "one lone pair removes one atom from the name" in a mechanical way. The reason is still repulsion. $\ce{NH3}$ has one lone pair, so the three N-H bonding pairs are pushed slightly closer together than in $\ce{CH4}$. $\ce{H2O}$ has two lone pairs, so the two O-H bonding pairs are pushed even closer together.

The same logic works from other electron-pair arrangements.

For six electron domains with two lone pairs, the pair positions are opposite each other. This avoids a 90-degree lone pair–lone pair interaction, minimising the strongest repulsion. The remaining four bonding positions lie in one plane, giving square planar shape with 90- and 180-degree bond angles.

For five electron domains, lone pairs prefer equatorial positions in a trigonal bipyramidal arrangement. An equatorial lone pair has fewer 90-degree interactions than an axial lone pair, so it gives lower repulsion. You usually do not need exact experimental angles for these extension shapes; you do need the reasoning direction: lone pairs occupy space and compress nearby bond angles.

Common-error contrast

Weak answer: "$\ce{NH3}$ is 107 degrees because it has three hydrogens."

Stronger answer: "$\ce{NH3}$ has four electron pairs around nitrogen: three bonding pairs and one lone pair. The electron pairs arrange approximately tetrahedrally, but the lone pair repels bonding pairs more strongly, so the H-N-H angle is compressed from 109.5 degrees to about 107 degrees."

[DIAGRAM: asset_name: Lone Pairs And Molecular Shapes; asset_slug: edexcel_a_level_chemistry_l010_lone_pair_shapes; recommended_method: matplotlib; description: NH3, H2O and XeF4 show explicit lone pairs and molecular shape; opposite lone pairs in octahedral XeF4 leave a square plane.]

A Reliable Prediction Method

For unfamiliar molecules and ions, use a route rather than memory. Questions often ask students to predict the shape of a related species, so the working must show how the shape follows from the electron pairs.

Use this five-step method.

1. Choose the central atom. It is usually the atom that can make the most bonds; hydrogen is not central.
2. Draw enough of a dot-and-cross or structural diagram to know what is bonded to the central atom.
3. Count bonding regions and lone pairs around the central atom. Remember that a double or triple bond counts as one direction of electron density for the shape.
4. Use the total number of electron domains to choose the electron-pair arrangement.
5. Name the molecular shape from the atoms only, then give the bond angle and any lone-pair compression.

Worked example 1: $\ce{PF3}$

Predict the shape and bond angle in $\ce{PF3}$.

Phosphorus is the central atom. Phosphorus has five outer-shell electrons and forms three P-F single bonds. That leaves one lone pair on phosphorus, so there are four electron pairs around the central atom: three bonding pairs and one lone pair.

Four electron pairs arrange approximately tetrahedrally. The molecular shape is trigonal pyramidal because only the three P-F bonds are visible in the shape. The F-P-F angle is less than 109.5 degrees because the lone pair repels bonding pairs more strongly than bonding pairs repel each other.

Worked example 2: $\ce{SO2}$

Predict the shape around sulfur in $\ce{SO2}$ using electron-domain reasoning.

Sulfur is the central atom. The simplest shape model has two S-O bonding regions and one lone pair around sulfur. Three electron domains arrange trigonal planar, but only two bonded atoms are attached to sulfur. The molecular shape is bent, with an O-S-O angle slightly less than 120 degrees because the lone pair compresses the angle between the bonding regions.

This example is a good place to avoid overcounting a double bond. Each S=O direction is one bonding region for the gross shape, even though the bonding description is more detailed than a single shared pair.

Organic transfer

Simple organic molecules use the same rules.

The organic molecule may be larger, but the shape prediction is local: choose one central atom and count electron domains around that atom.

Quick Recap

Count bonding regions and lone pairs, arrange them for minimum repulsion, then name the positions of atoms. Multiple bonds count as one direction for gross shape. Lone pairs alter the repulsions and can compress angles; quote ideal angles or a justified less-than prediction when an exact experimental value is not supplied.