Molecular Orbital Theory

Valence Bond Model vs. Molecular Orbital Theory Forming Molecular Orbitals Why Some Molecules Do Not Exist
Molecular Orbitals of the Second Energy Level Bond Order
Valence Bond Model vs. Molecular Orbital Theory

Because arguments based on atomic orbitals focus on the bonds formed between valence electrons on an atom, they are often said to involve a valence-bond theory.
The valence-bond model can't adequately explain the fact that some molecules contains two equivalent bonds with a bond order between that of a single bond and a double bond. The best it can do is suggest that these molecules are mixtures, or hybrids, of the two Lewis structures that can be written for these molecules.
This problem, and many others, can be overcome by using a more sophisticated model of bonding based on molecular orbitals. Molecular orbital theory is more powerful than valence-bond theory because the orbitals reflect the geometry of the molecule to which they are applied. But this power carries a significant cost in terms of the ease with which the model can be visualized.
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Forming Molecular Orbitals

Molecular orbitals are obtained by combining the atomic orbitals on the atoms in the molecule. Consider the H2 molecule, for example. One of the molecular orbitals in this molecule is constructed by adding the mathematical functions for the two 1s atomic orbitals that come together to form this molecule. Another orbital is formed by subtracting one of these functions from the other, as shown in the figure below.
Diagram

One of these orbitals is called a bonding molecular orbital because electrons in this orbital spend most of their time in the region directly between the two nuclei. It is called a sigma () molecular orbital because it looks like an s orbital when viewed along the H-H bond. Electrons placed in the other orbital spend most of their time away from the region between the two nuclei. This orbital is therefore an antibonding, or sigma star (*), molecular orbital.
Diagram

The  bonding molecular orbital concentrates electrons in the region directly between the two nuclei. Placing an electron in this orbital therefore stabilizes the H2 molecule. Since the * antibonding molecular orbital forces the electron to spend most of its time away from the area between the nuclei, placing an electron in this orbital makes the molecule less stable.
Electrons are added to molecular orbitals, one at a time, starting with the lowest energy molecular orbital. The two electrons associated with a pair of hydrogen atoms are placed in the lowest energy, or  bonding, molecular orbital, as shown in the figure below. This diagram suggests that the energy of an H2 molecule is lower than that of a pair of isolated atoms. As a result, the H2 molecule is more stable than a pair of isolated atoms.
Diagram

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Using the Molecular Orbital Model to Explain Why Some Molecules Do Not Exist

This molecular orbital model can be used to explain why He2 molecules don't exist. Combining a pair of helium atoms with 1s2 electron configurations would produce a molecule with a pair of electrons in both the  bonding and the * antibonding molecular orbitals. The total energy of an He2 molecule would be essentially the same as the energy of a pair of isolated helium atoms, and there would be nothing to hold the helium atoms together to form a molecule.
The fact that an He2 molecule is neither more nor less stable than a pair of isolated helium atoms illustrates an important principle: The core orbitals on an atom make no contribution to the stability of the molecules that contain this atom. The only orbitals that are important in our discussion of molecular orbitals are those formed when valence-shell orbitals are combined. The molecular orbital diagram for an O2 molecule would therefore ignore the 1s electrons on both oxygen atoms and concentrate on the interactions between the 2s and 2p valence orbitals.
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Molecular Orbitals of the Second Energy Level

The 2s orbitals on one atom combine with the 2s orbitals on another to form a 2s bonding and a 2s* antibonding molecular orbital, just like the 1s and 1s* orbitals formed from the 1s atomic orbitals. If we arbitrarily define the Z axis of the coordinate system for the O2 molecule as the axis along which the bond forms, the 2pz orbitals on the adjacent atoms will meet head-on to form a 2p bonding and a 2p* antibonding molecular orbital, as shown in the figure below. These are called sigma orbitals because they look like s orbitals when viewed along the oxygen-oxygen bond.
Diagram

The 2px orbitals on one atom interact with the 2px orbitals on the other to form molecular orbitals that have a different shape, as shown in the figure below. These molecular orbitals are called pi () orbitals because they look like p orbitals when viewed along the bond. Whereas  and * orbitals concentrate the electrons along the axis on which the nuclei of the atoms lie,  and * orbitals concentrate the electrons either above or below this axis.
Diagram

The 2px atomic orbitals combine to form a x bonding molecular orbital and a x* antibonding molecular orbital. The same thing happens when the 2py orbitals interact, only in this case we get a y and a y* antibonding molecular orbital. Because there is no difference between the energies of the 2px and 2py atomic orbitals, there is no difference between the energies of the x and y or the x* and y* molecular orbitals.
The interaction of four valence atomic orbitals on one atom (2s, 2px, 2py and 2pz) with a set of four atomic orbitals on another atom leads to the formation of a total of eight molecular orbitals: 2s, 2s*, 2p, 2p*, x, y, x*, and y*.
There is a significant difference between the energies of the 2s and 2p orbitals on an atom. As a result, the 2s and *2s orbitals both lie at lower energies than the 2p, 2p*, x, y, x*, and y* orbitals. To sort out the relative energies of the six molecular orbitals formed when the 2p atomic orbitals on a pair of atoms are combined, we need to understand the relationship between the strength of the interaction between a pair of orbitals and the relative energies of the molecular orbitals they form.
Because they meet head-on, the interaction between the 2pz orbitals is stronger than the interaction between the 2px or 2py orbitals, which meet edge-on. As a result, the 2p orbital lies at a lower energy than the x and y orbitals, and the 2p* orbital lies at higher energy than the x* and y* orbitals, as shown in the figure below.
Diagram

Unfortunately an interaction is missing from this model. It is possible for the 2s orbital on one atom to interact with the 2pz orbital on the other. This interaction introduces an element of s-p mixing, or hybridization, into the molecular orbital theory. The result is a slight change in the relative energies of the molecular orbitals, to give the diagram shown in the figure below. Experiments have shown that O2 and F2 are best described by the model in the figure above, but B2, C2, and N2 are best described by a model that includes hybridization, as shown in the figure below.
Diagram

Practice Problem 9:
Construct a molecular orbital diagram for the O2 molecule.

Click here to check your answer to Practice Problem 9


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Bond Order

The number of bonds between a pair of atoms is called the bond order. Bond orders can be calculated from Lewis structures, which are the heart of the valence-bond model. Oxygen, for example, has a bond order of two.
Structure

When there is more than one Lewis structure for a molecule, the bond order is an average of these structures. The bond order in sulfur dioxide, for example, is 1.5  the average of an S-O single bond in one Lewis structure and an S=O double bond in the other.
Structures

In molecular orbital theory, we calculate bond orders by assuming that two electrons in a bonding molecular orbital contribute one net bond and that two electrons in an antibonding molecular orbital cancel the effect of one bond. We can calculate the bond order in the O2 molecule by noting that there are eight valence electrons in bonding molecular orbitals and four valence electrons in antibonding molecular orbitals in the electron configuration of this molecule. Thus, the bond order is two.
Equation

Although the Lewis structure and molecular orbital models of oxygen yield the same bond order, there is an important difference between these models. The electrons in the Lewis structure are all paired, but there are two unpaired electrons in the molecular orbital description of the molecule. As a result, we can test the predictions of these theories by studying the effect of a magnetic field on oxygen.
Atoms or molecules in which the electrons are paired are diamagnetic repelled by both poles of a magnetic. Those that have one or more unpaired electrons are paramagnetic  attracted to a magnetic field. Liquid oxygen is attracted to a magnetic field and can actually bridge the gap between the poles of a horseshoe magnet. The molecular orbital model of O2 is therefore superior to the valence-bond model, which cannot explain this property of oxygen.
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MCQs of VSEPR Theory

Question 1

Which one of the following statements concerning the length of carbon-carbon single, double, and triple covalent bonds is true?


a) The carbon-carbon single bond is shorter than either the carbon-carbon double or triple bond.

b) The carbon-carbon double bond is shorter than either the carbon-carbon single or triple bond.

c) The carbon-carbon triple bond is shorter than either the carbon-carbon single or double bond.

d) The carbon-carbon single, double, and triple bonds all have the same length.
Question 2

Which one of the following is the correct bond angle between atoms adopting a trigonal planar geometry?


a) 180°

b) 109.5°

c) 90°

d) 120°
Question 3

The atoms in a molecule of water adopt what kind of geometry?


a) Linear

b) Tetrahedral

c) Octahedral

d) Trigonal planar
Question 4

Ammonia, NH3, adopts a tetrahedral geometry. However, the non-bonding pair on the central nitrogen atom distorts the bond angle away from the expected 109.5°. Which of the following statements correctly describes how the bond angle is distorted?


a) The actual bond angle is reduced: it is less than 109.5°

b) The actual bond angle is increased: it is more than 109.5°
Question 5

About which of the bonds along the backbone of a polypeptide is rotation not possible?


a) 1

b) 2

c) 3
Question 6

sp3 hybridisation involves the hybridisation of how many atomic orbitals?


a) 1

b) 2

c) 3

d) 4
Question 7

Four sp3 hybrid orbitals adopt what kind of geometry?


a) Linear

b) Trigonal planar

c) Octahedral

d) Tetrahedral
Question 8

When applying VSEPR theory to predict molecular shape, which of the following do we not need to take into account?


a) Valence electrons occupying sigma bonding orbitals

b) Valence electrons occupying pi bonding orbitals

c) Valence electrons occupying non-bonding orbitals
Question 9

Which of the following statements regarding the measurement of the atomic radius are correct? Please select all that apply.


a) The atomic radius is measured between atoms of different elements

b) The atomic radius is measured between atoms of the same element

c) The atomic radius is half the distance between the nuclei of two joined atoms

d) The atomic radius is the distance between the nuclei of two joined atoms

e) The atomic radius is only measured between two covalently-bonded atoms

f) The atomic radius can be measured between both covalently- and ionically-bonded atoms
Question 10

From the following possible responses, select those responses that give the combination of bonds that makes up a triple covalent bond.


a) Two sigma bonds

b) One sigma bond

c) Two pi bonds

d) One pi bond

e) Three sigma bonds

VSEPR THEORY

Valence-Shell Electron-Pair Repulsion Theory (VSEPR)

Predicting the Shapes of Molecules Incorporating Double and Triple Bonds
The Role of Nonbonding Electrons Table Summarizing VSEPR Theory
Predicting the Shapes of Molecules

There is no direct relationship between the formula of a compound and the shape of its molecules. The shapes of these molecules can be predicted from their Lewis structures, however, with a model developed about 30 years ago, known as the valence-shell electron-pair repulsion (VSEPR) theory.
The VSEPR theory assumes that each atom in a molecule will achieve a geometry that minimizes the repulsion between electrons in the valence shell of that atom. The five compounds shown in the figure below can be used to demonstrate how the VSEPR theory can be applied to simple molecules.
Table of Geometries

There are only two places in the valence shell of the central atom in BeF2 where electrons can be found. Repulsion between these pairs of electrons can be minimized by arranging them so that they point in opposite directions. Thus, the VSEPR theory predicts that BeF2 should be a linear molecule, with a 180o angle between the two Be-F bonds.
Structure

There are three places on the central atom in boron trifluoride (BF3) where valence electrons can be found. Repulsion between these electrons can be minimized by arranging them toward the corners of an equilateral triangle. The VSEPR theory therefore predicts a trigonal planar geometry for the BF3 molecule, with a F-B-F bond angle of 120o.
Structure

BeF2 and BF3 are both two-dimensional molecules, in which the atoms lie in the same plane. If we place the same restriction on methane (CH4), we would get a square-planar geometry in which the H-C-H bond angle is 90o. If we let this system expand into three dimensions, however, we end up with a tetrahedral molecule in which the H-C-H bond angle is 109o28'.
Structure

Repulsion between the five pairs of valence electrons on the phosphorus atom in PF5 can be minimized by distributing these electrons toward the corners of a trigonal bipyramid. Three of the positions in a trigonal bipyramid are labeled equatorial because they lie along the equator of the molecule. The other two are axial because they lie along an axis perpendicular to the equatorial plane. The angle between the three equatorial positions is 120o, while the angle between an axial and an equatorial position is 90o.
Structure

There are six places on the central atom in SF6 where valence electrons can be found. The repulsion between these electrons can be minimized by distributing them toward the corners of an octahedron. The term octahedron literally means "eight sides," but it is the six corners, or vertices, that interest us. To imagine the geometry of an SF6 molecule, locate fluorine atoms on opposite sides of the sulfur atom along the X, Y, and Z axes of an XYZ coordinate system.
Structure

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Interactive tutorial on VSEPR theory by John Nash

Interactive tutorial on chemical bonds, molecular shapes, and molecular models by Dr. Anna Cavinato and Dr. David Camp, Eastern Oregon University,

Incorporating Double and Triple Bonds Into the VSEPR Theory

Compounds that contain double and triple bonds raise an important point: The geometry around an atom is determined by the number of places in the valence shell of an atom where electrons can be found, not the number of pairs of valence electrons. Consider the Lewis structures of carbon dioxide (CO2) and the carbonate (CO32-) ion, for example.
Structures

There are four pairs of bonding electrons on the carbon atom in CO2, but only two places where these electrons can be found. (There are electrons in the C=O double bond on the left and electrons in the double bond on the right.) The force of repulsion between these electrons is minimized when the two C=O double bonds are placed on opposite sides of the carbon atom. The VSEPR theory therefore predicts that CO2 will be a linear molecule, just like BeF2, with a bond angle of 180o.
The Lewis structure of the carbonate ion also suggests a total of four pairs of valence electrons on the central atom. But these electrons are concentrated in three places: The two C-O single bonds and the C=O double bond. Repulsions between these electrons are minimized when the three oxygen atoms are arranged toward the corners of an equilateral triangle. The CO32- ion should therefore have a trigonal-planar geometry, just like BF3, with a 120o bond angle.
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The Role of Nonbonding Electrons in the VSEPR Theory

The valence electrons on the central atom in both NH3 and H2O should be distributed toward the corners of a tetrahedron, as shown in the figure below. Our goal, however, isn't predicting the distribution of valence electrons. It is to use this distribution of electrons to predict the shape of the molecule. Until now, the two have been the same. Once we include nonbonding electrons, that is no longer true.
Diagram

The VSEPR theory predicts that the valence electrons on the central atoms in ammonia and water will point toward the corners of a tetrahedron. Because we can't locate the nonbonding electrons with any precision, this prediction can't be tested directly. But the results of the VSEPR theory can be used to predict the positions of the nuclei in these molecules, which can be tested experimentally. If we focus on the positions of the nuclei in ammonia, we predict that the NH3 molecule should have a shape best described as trigonal pyramidal, with the nitrogen at the top of the pyramid. Water, on the other hand, should have a shape that can be described as bent, or angular. Both of these predictions have been shown to be correct, which reinforces our faith in the VSEPR theory.
Practive Problem 6:
Use the Lewis structure of the ICl2+ ion shown in the figure below to predict the shape of this ion.

Structure

Click here to check your answer to Practice Problem 6


Practice Problem 7:
Use the Lewis structure of the NO2 molecule shown in the figure below to predict the shape of this molecule.

Structure

Click here to check your answer to Practice Problem 7

When we extend the VSEPR theory to molecules in which the electrons are distributed toward the corners of a trigonal bipyramid, we run into the question of whether nonbonding electrons should be placed in equatorial or axial positions. Experimentally we find that nonbonding electrons usually occupy equatorial positions in a trigonal bipyramid.
To understand why, we have to recognize that nonbonding electrons take up more space than bonding electrons. Nonbonding electrons need to be close to only one nucleus, and there is a considerable amount of space in which nonbonding electrons can reside and still be near the nucleus of the atom. Bonding electrons, however, must be simultaneously close to two nuclei, and only a small region of space between the nuclei satisfies this restriction.
Because they occupy more space, the force of repulsion between pairs of nonbonding electrons is relatively large. The force of repulsion between a pair of nonbonding electrons and a pair of bonding electrons is somewhat smaller, and the repulsion between pairs of bonding electrons is even smaller.
The figure below can help us understand why nonbonding electrons are placed in equatorial positions in a trigonal bipyramid.
Diagram

If the nonbonding electrons in SF4 are placed in an axial position, they will be relatively close (90o) to three pairs of bonding electrons. But if the nonbonding electrons are placed in an equatorial position, they will be 90o away from only two pairs of bonding electrons. As a result, the repulsion between nonbonding and bonding electrons is minimized if the nonbonding electrons are placed in an equatorial position in SF4.
The results of applying the VSEPR theory to SF4, ClF3, and the I3- ion are shown in the figure below.
Diagram

When the nonbonding pair of electrons on the sulfur atom in SF4 is placed in an equatorial position, the molecule can be best described as having a see-saw or teeter-totter shape. Repulsion between valence electrons on the chlorine atom in ClF3 can be minimized by placing both pairs of nonbonding electrons in equatorial positions in a trigonal bipyramid. When this is done, we get a geometry that can be described as T-shaped. The Lewis structure of the triiodide (I3-) ion suggests a trigonal bipyramidal distribution of valence electrons on the central atom. When the three pairs of nonbonding electrons on this atom are placed in equatorial positions, we get a linear molecule.
Molecular geometries based on an octahedral distribution of valence electrons are easier to predict because the corners of an octahedron are all identical.
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To view a table summarizing VSEPR theory, click here.