# Molecular Shapes

VSEPR theory predicts molecular shapes based on repulsive forces between electron pairs. According to VSEPR theory, electron pairs around a central atom maximize the distance between them.

Valence shell electron-pair repulsion (VSEPR) theory is a covalent bond theory that uses the repulsive forces between single electrons and pairs of electrons about the central atom to predict their relative positions around the atomic nuclei. According to VSEPR theory, electron pairs around a central atom repel each other. Because of the repulsion between electrons, the shape with minimum energy is the one in which the distance between electron pairs is maximized. Therefore, the shape with the lowest energy is likely to be the actual form of the molecule. Electron pairs can either be inside a bond between atoms or exist as a lone pair.

There are generally between two and six electron pairs (either in the bonds of bonding atoms or as lone pairs) around any central atom, and the electron pairs can be arranged in either of five ways, each with a different geometry.

Electron-pair geometry is the shape description for all electron pairs (bonding and nonbonding) about a central atom. Electron-pair geometry maximizes the distance between every pair of electrons around a central atom. For example, if there are three pairs of electrons, increasing the angle between two electron pairs in a trigonal planar arrangement would push one of the pairs closer to the third pair, which would increase the potential energy of the third pair. When the distance between all of the electron pairs is maximized, the potential energy of the system is at a minimum. Systems always favor a position of minimum energy, so arrangements that maximize the distance between atoms and/or lone pairs are favored. Generally, large angles represent a greater degree of separation and are more favorable. Angles of 90° are least favorable.

### VSEPR Electron-Pair Configurations

Number of Electron Pairs Electron-Pair Geometry Geometry Geometry Definition Angle Between Electrons
2 Linear
Electron pairs form a straight line with the central atom. 180°
3 Trigonal planar
Electron pairs are arranged evenly around the central atom and are in the same plane as the central atom. 120°
4 Tetrahedral
Electron pairs are arranged around the central atom such that they are located at the four corners of a tetrahedron. 109.5°
5 Trigonal bipyramidal
Electron pairs are arranged such that they are located at the corners of a trigonal bipyramid around the central atom, like two triangular pyramids aligned base to base. 90° and 120°
6 Octahedral
Electron pairs are arranged such that they are located at the corners of an octahedron around the central atom, like two square pyramids aligned base to base. 90°

The number of electron pairs around the central atom determines the molecular geometry.

A special notation is used when describing geometries with VSEPR theory. The letter A is used to represent the central atom, X represents bonding pairs of electrons between the central atom and another atom, and E represents lone pairs. So the notation AX2E3 means there are two bonding pairs of electrons and three lone pairs of electrons arranged around a central atom, A. The angle formed between two terminal groups (X or E) and the central atom (A) is called the bond angle.
Note that the electron-pair geometry includes lone, unbound electron pairs. Lone pairs are not represented in molecular models. Because lone pairs are not shown, the molecular geometry may be different than the electron-pair geometry. The molecular geometry, also called molecular shape, is the shape formed by the central atom, considering only bonding electron pairs.

Consider a molecule with trigonal planar electron-pair geometry with one lone pair of electrons. The molecular geometry of this molecule is bent; it is not trigonal planar. Two atoms are bonded to a central atom at an angle. Furthermore, lone pairs are not constrained by a covalent bond. The electrons in a lone pair have higher energy, which means they can spread out more. Because of the higher energy, lone pairs have a greater repulsive force than bonding pairs. A lone pair–lone pair interaction is the most repulsive (has the highest energy), a bonding pair–bonding pair interaction is the least repulsive (has the lowest energy), and a lone pair–bonding pair interaction is in the middle.

When there is more than one possible location for a lone pair in the molecular geometry, the lone pair settles in the configuration that minimizes the repulsion between the lone pair and other electrons. Consider a central atom with four bonding pairs and one lone pair (AX4E). For this molecule the lone pair is located in the central plane so that there are two 90° lone pair–bonding electron interactions and two 120° lone pair–bonding electron interactions. If the lone pair were at the tip of the pyramid, there would be three 90° lone pair–bonding interactions and one 120° one. This arrangement would be a higher energy state and would be less stable.

### VSEPR Molecular Geometries

Electron Pairs Electron-Pair Geometry Lone Pairs VSEPR Notation Molecular Geometry Bond Angles
2 Linear
0 AX2 Linear
180°
3 Trigonal planar
0 AX3 Trigonal planar
120°
1 AX2E Bent
120°
4 Tetrahedral
0 AX4 Tetrahedral
109.5°
1 AX3E Trigonal pyramidal
109.5°
2 AX2E2 Bent
109.5°
5 Trigonal bipyramidal
0 AX5 Trigonal bipyramidal
90°, 120°
1 AX4E Sawhorse
90°, 120°
2 AX3E2 T-shape
90°
3 AX2E3 Linear
180°
6 Octahedral
0 AX6 Octahedral
90°
1 AX5E Square pyramidal
90°
2 AX4E2 Square planar
90°

### Using VSEPR Theory

VSEPR theory and Lewis structures can be used to predict shapes of molecules. The bonding pairs of electrons and lone pairs of electrons of a molecule determine the overall shape.
Consider how VSEPR theory predicts the shape of nitrogen triiodide (NI3). The first step is to draw the Lewis structure for this molecule. The nitrogen atom has five valence electrons, and each iodine atom has seven valence electrons.
The Lewis structure reveals that the central atom is nitrogen and that it is surrounded by four electron pairs, which corresponds to a tetrahedral electron-pair geometry. But the molecule has three bonding pairs and one lone pair (a VSEPR notation of AX3E), so its molecular geometry is not the same as the electron-pair geometry. Instead, the top of the pyramid is occupied by the lone pair, and VSEPR predicts a molecular shape that is trigonal pyramidal with 109.5° bond angles. Note that in reality the bond angles of nitrogen triiodide are slightly less than 109.5° because the forces of repulsion between the two iodine atoms and an iodine atom and a lone pair are different from each other. A more complicated structure is the tetrachloroiodate ion (ICl4).
In the case of the tetrachloroiodate ion, the central atom, iodine, has four bonding pairs and two lone pairs, so the structure is AX4E2. The iodine atom in this molecule contains an expanded octet. The electron-pair geometry of this structure is octahedral, but the two lone pairs mean that the molecular geometry is different. Remember that lone pair–lone pair interactions are the most repulsive, so the two lone pairs are arranged on opposite sides of the central atom, 180° apart. The molecular structure is therefore square planar, and the bond angles are all 90°.
VSEPR theory can also be used to predict molecular geometries of molecules with more than one central atom. The technique is similar, but it is applied to each central atom individually, and the two resulting shapes are simply joined together. Consider the structure of methanol (CH3OH).
There are two central atoms: carbon (C) and oxygen (O). The carbon atom has four bonding pairs of electrons and no lone pairs (AX4), so the molecular geometry is the same as the electron-pair geometry and is tetrahedral. The oxygen atom has two bonding atoms and two lone pairs (AX2E2), so its electron-pair geometry is tetrahedral but its molecular geometry is bent. The structure for the entire molecule can be pictured as a tetrahedron with an angled arm branching off one corner.

### VSEPR Theory and Polar Covalent Bonds

VSEPR theory can also predict shapes due to polar covalent bonds. A dipole moment is a vector that describes the charge magnitude and direction on a polar covalent bond. When dipole moments of a molecule cancel each other, the molecule is nonpolar. If there is a net dipole moment, the molecule is polar.

A polar covalent bond is a covalent bond in which the electron density is more localized on one end of the bond. One end is slightly positive, and one end is slightly negative. A dipole moment ($\mu$) is a vector quantity that defines the extent of the charge on either side of a polar covalent bond, with the direction that points from the positive side of the bond toward the negative side. The negative side is the more electronegative atom, and the positive side is the less electronegative atom. The dipole moment is the product of the charge ($\delta{+}$ and $\delta{-}$) and distance (d). The distance is simply the bond length. The dipole moment is measured in debyes (D), where $1\;\rm D=3.335641\times10^{-30}\;\rm C\cdot \rm m$.

The polarity of a molecule is related to its shape. Consider the carbon dioxide (CO2) molecule. Each ${\rm{C{-}O}}$ bond has a dipole moment that has a greater electron density on the oxygen side of the bond, because oxygen is more electronegative than carbon. Schematically, the individual ${\rm{C{-}O}}$bond dipole moments are represented by a crossed arrow that points toward the side of greater negativity. The plus end of the arrow indicates the positively charged side of the molecule, and the tip of the arrow represents the negatively charged side.
Because the dipole moments in carbon dioxide are equal in magnitude but opposite in direction, they cancel each other, and $\mu$ is equal to zero when the molecule is considered as a whole. A molecule with a dipole moment equal to zero is called a nonpolar molecule. Carbon dioxide is nonpolar, and VSEPR theory predicts that carbon dioxide is a linear molecule. If the molecule were bent, there would be a net partial charge on the oxygen atoms and the carbon atom, and the dipole moment would not equal zero.
Water, on the other hand, has a central oxygen atom that is more electronegative than the hydrogen atoms bound to it.
In this case, the dipole moments are equal in magnitude but have components that point toward each other, so the entire molecule has a dipole moment greater than zero. A polar molecule, such as water, has a specific dipole moment not equal to zero. Because a water molecule is polar, VSEPR theory predicts that it has a bent structure. The two lone pairs and the hydrogen atoms repel each other, resulting in a bend in the shape of the molecule. In a water molecule, there are net partial charges on the two hydrogen atoms and the oxygen atom.
Step-By-Step Example
Calculation of Molecular Dipole Moment
Given that the dipole moment of the ${\rm{O{-}H}}$ bond in a water molecule is ${\mu_{\rm{O{-}H}}}=1.5\;\rm{D}$, calculate the overall dipole moment $\mu_{\rm{H}_2\rm{O}}$ of a water molecule using the measured ${\rm{H}{-}\rm{O}{-}\rm{H}}$ bond angle of the molecule, 104.5°.
Step 1

Consider the known information about the molecule.

The center of the overall dipole moment is located at a point equidistant between the two hydrogen atoms, at the base of an isosceles triangle with an obtuse angle of 104.5°.

Step 2
Use the known information to determine how to calculate the dipole moment.
Consider the water molecule oriented with the oxygen atom at the top and the two hydrogen atoms below, symmetrically on either side.
The x-components of the $\rm{O}{-}\rm{H}$ dipole moments point in opposite directions and cancel. The y-components of the $\rm{O}{-}\rm{H}$ dipole moments both point in the positive y-direction and do not cancel. The overall dipole moment is therefore the sum of the y-components of the $\rm{O}{-}\rm{H}$ dipole moments.
Step 3
Use trigonometry to calculate the magnitude of the dipole moment. First calculate the contribution from each hydrogen atom, which is the y-component of the $\rm{O{-}H}$ dipole moment.
$\mu_{{\rm{O{-}H}},y}=\mu_{\rm{O{-}H}}\cos\left(\frac{104.5\degree}2\right)=\rm{(1.5\;D)\cos(52.25\degree)=0.918\;D}$
Solution
Calculate the total dipole moment for H2O by adding the contributions from both hydrogen atoms.
$\mu_{\rm{H_2O}}=2\mu_{{\rm{O{-}H}},y}=\rm{2(0.918\;D)=1.8\;D}$