VSEPR Theory and Electron Dot Structures
Why is a water molecule bent, while carbon dioxide is perfectly straight? Both have three atoms, yet their shapes are different, and that difference decides whether water is a liquid at room temperature and why CO₂ is a gas in your soda. By the end of this page you will be able to predict the shape of a molecule from its formula, and you will come back to this question about water and CO₂.
Electron dot structures
To predict a shape, you first need to know where the electrons are. An electron dot (Lewis) structure shows the valence electrons of every atom in a molecule. Bonding pairs are drawn as lines (or pairs of dots) between atoms. Lone pairs are pairs of electrons that stay on one atom and are not shared.
Atoms usually follow the octet rule: they end up with 8 electrons around them (hydrogen is happy with 2). Here is a reliable recipe.
Draw the electron dot (Lewis) structure of methane, $\text{CH}4$. State how many bonding pairs and how many lone pairs surround the central carbon atom, and name the…
Draw the electron dot (Lewis) structure of methane, $\text{CH}_4$. State how many bonding pairs and how many lone pairs surround the central carbon atom, and name the molecular geometry.
For $\text{CH}2\text{O}$ (formaldehyde, with carbon as the central atom bonded to two H atoms and one O atom), count the total valence electrons, draw the electron dot…
For $\text{CH}_2\text{O}$ (formaldehyde, with carbon as the central atom bonded to two H atoms and one O atom), count the total valence electrons, draw the electron dot structure including any double bond, and predict the molecular geometry. Explain how a double bond is treated as one region of electron density in VSEPR.
VSEPR: electron pairs push each other apart
Imagine tying balloons together at their knots. The balloons spread out as far from each other as they can. Electron pairs do the same thing, because they all carry a negative charge and repel each other.
A region of electron density (also called an electron domain) is any one of these: a single bond, a double bond, a triple bond, or a lone pair. A double or triple bond counts as one region, because all its electrons sit between the same two atoms and push in the same direction.
The regions can be arranged in three basic ways:
| Regions of electron density | Electron-pair geometry | Angle between regions |
|---|---|---|
| 2 | linear | $180^\circ$ |
| 3 | trigonal planar | $120^\circ$ |
| 4 | tetrahedral | $109.5^\circ$ |
In your own words, state what VSEPR stands for and what the theory says about how electron pairs around a central atom arrange themselves. Then name the geometry of a…
In your own words, state what VSEPR stands for and what the theory says about how electron pairs around a central atom arrange themselves. Then name the geometry of a molecule with two bonding pairs and no lone pairs on the central atom, such as $\text{CO}_2$.
From electron pairs to molecular shape
Here is the key distinction. The electron-pair geometry describes where all regions of electron density point, including lone pairs. The molecular geometry describes only where the atoms are. You cannot "see" a lone pair in the shape, but it still pushes on the other pairs.
| Regions | Bonding pairs | Lone pairs | Molecular geometry | Bond angle | Example |
|---|---|---|---|---|---|
| 2 | 2 | 0 | linear | $180^\circ$ | $\text{CO}_2$ |
| 3 | 3 | 0 | trigonal planar | $120^\circ$ | $\text{BF}_3$ |
| 3 | 2 | 1 | bent | slightly less than $120^\circ$ | $\text{SO}_2$ |
| 4 | 4 | 0 | tetrahedral | $109.5^\circ$ | $\text{CH}_4$ |
| 4 | 3 | 1 | trigonal pyramidal | $107^\circ$ | $\text{NH}_3$ |
| 4 | 2 | 2 | bent | $104.5^\circ$ | $\text{H}_2\text{O}$ |
Notice that methane, ammonia, and water all have 4 regions, so all three have a tetrahedral electron-pair geometry. Their molecular shapes differ because they have 0, 1, and 2 lone pairs.
Lone pairs take up more room than bonding pairs, since they are held by only one nucleus. They squeeze the bonded atoms closer together, which is why the angles shrink: $109.5^\circ \to 107^\circ \to 104.5^\circ$.
Note that boron in $\text{BF}_3$ is an exception to the octet rule: it has only 6 electrons around it (3 bonding pairs), which is fine because it has no lone pair.
Now go back to the opening question. $\text{CO}_2$ has two double bonds and no lone pairs on C, so 2 regions: linear. Water has 2 lone pairs on O, so it is bent. That is the whole difference.
Classify each molecule as linear, bent, trigonal planar, trigonal pyramidal, or tetrahedral: (a) $\text{H}2\text{O}$, (b) $\text{NH}3$, (c) $\text{BF}3$.
Classify each molecule as linear, bent, trigonal planar, trigonal pyramidal, or tetrahedral: (a) $\text{H}_2\text{O}$, (b) $\text{NH}_3$, (c) $\text{BF}_3$.
Ammonia, $\text{NH}3$, and methane, $\text{CH}4$, both have four electron pairs around the central atom, yet their molecular shapes differ. Draw the electron dot…
Ammonia, $\text{NH}_3$, and methane, $\text{CH}_4$, both have four electron pairs around the central atom, yet their molecular shapes differ. Draw the electron dot structure of each, give the electron-pair geometry and the molecular geometry of each, and explain why the shapes differ.
Emma is studying for a chemistry quiz. She draws the structure of $\text{H}2\text{O}$ as a straight line, H-O-H, with an angle of $180^\circ$. Explain what is wrong with…
Emma is studying for a chemistry quiz. She draws the structure of $\text{H}_2\text{O}$ as a straight line, H-O-H, with an angle of $180^\circ$. Explain what is wrong with her prediction, using the number of lone pairs on oxygen. Then state the approximate bond angle in real water and compare it to the angle in methane.
Shape decides polarity
A bond is polar when the two atoms share electrons unequally (different electronegativity). We can draw each polar bond as an arrow, called a bond dipole, pointing toward the more electronegative atom.
A whole molecule is polar only if the bond dipoles do not cancel. Shape decides this.
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In $\text{CO}_2$ (linear), the two C=O dipoles point in opposite directions with equal strength. They cancel, so the molecule is nonpolar.
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In $\text{H}_2\text{O}$ (bent), the two O-H dipoles point toward O at an angle. They add up to a net dipole, so the molecule is polar.
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In $\text{CCl}_4$ (tetrahedral), the four C-Cl dipoles are balanced in all directions and cancel. The molecule is nonpolar, even though every bond is polar.
Compare $\text{BF}3$, $\text{NF}3$, and $\text{CF}4$. For each one, count the valence electrons, draw the electron dot structure, and identify the number of bonding and…
Compare $\text{BF}_3$, $\text{NF}_3$, and $\text{CF}_4$. For each one, count the valence electrons, draw the electron dot structure, and identify the number of bonding and lone pairs on the central atom. Then give each molecular geometry and approximate bond angle, and argue why $\text{BF}_3$ is planar while $\text{NF}_3$ is not, even though both have three fluorine atoms.
A student claims, "Any molecule with polar bonds must be a polar molecule." Test this claim using $\text{CO}2$, $\text{H}2\text{O}$, and $\text{CCl}4$. For each…
A student claims, "Any molecule with polar bonds must be a polar molecule." Test this claim using $\text{CO}_2$, $\text{H}_2\text{O}$, and $\text{CCl}_4$. For each molecule, draw the electron dot structure, give the geometry, describe the direction of the bond dipoles, and decide whether the molecule is polar. State whether the claim is correct and justify your answer.