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Bonding in Diatomic Molecules

Bonding in Diatomic Molecules

A dihydrogen molecule (H2) forms from two hydrogen atoms. When the atomic orbitals of the two atoms combine, the electrons occupy the molecular orbital of lowest energy, the σ1s bonding orbital. A dihydrogen molecule, H2, readily forms because the energy of a H2 molecule is lower than that of two H atoms. The σ1s orbital that contains both electrons is lower in energy than either of the two 1s atomic orbitals.

A molecular orbital can hold two electrons, so both electrons in the H2 molecule are in the σ1s bonding orbital; the electron configuration is \({\left({\text{σ}}_{1s}\right)}^{2}.\) We represent this configuration by a molecular orbital energy diagram (see the figure below) in which a single upward arrow indicates one electron in an orbital, and two (upward and downward) arrows indicate two electrons of opposite spin.

A diagram is shown that has an upward-facing vertical arrow running along the left side labeled “E.” At the bottom center of the diagram is a horizontal line labeled, “sigma subscript 1 s,” that has two vertical half arrows drawn on it, one facing up and one facing down. This line is connected to the right and left by upward-facing, dotted lines to two more horizontal lines, each labeled, “1 s,” and each with one vertical half-arrow facing up drawn on it. These two lines are connected by upward-facing dotted lines to another line in the center of the diagram, but farther up from the first, and labeled, “sigma subscript 1 s superscript asterisk.” The left and right sides of the diagram have headers that read, ”Atomic orbitals,” while the center header reads, “Molecular orbitals.” The bottom left and right are labeled “H” while the center is labeled “H subscript 2.”

The molecular orbital energy diagram predicts that H2 will be a stable molecule with lower energy than the separated atoms.

A dihydrogen molecule contains two bonding electrons and no antibonding electrons so we have

\({\text{bond order in H}}_{2}=\phantom{\rule{0.2em}{0ex}}\frac{\left(2-0\right)}{2}\phantom{\rule{0.2em}{0ex}}=1\)

Because the bond order for the H–H bond is equal to 1, the bond is a single bond.

A helium atom has two electrons, both of which are in its 1s orbital. Two helium atoms do not combine to form a dihelium molecule, He2, with four electrons, because the stabilizing effect of the two electrons in the lower-energy bonding orbital would be offset by the destabilizing effect of the two electrons in the higher-energy antibonding molecular orbital.

We would write the hypothetical electron configuration of He2 as \({\left({\text{σ}}_{1s}\right)}^{2}{\left({\text{σ}}_{1s}^{*}\right)}^{2}\) as in the figure below. The net energy change would be zero, so there is no driving force for helium atoms to form the diatomic molecule. In fact, helium exists as discrete atoms rather than as diatomic molecules. The bond order in a hypothetical dihelium molecule would be zero.

\({\text{bond order in He}}_{2}=\phantom{\rule{0.2em}{0ex}}\frac{\left(2-2\right)}{2}\phantom{\rule{0.2em}{0ex}}=0\)

A bond order of zero indicates that no bond is formed between two atoms.

A diagram is shown that has an upward-facing vertical arrow running along the left side labeled, “E.” At the bottom center of the diagram is a horizontal line labeled, “sigma subscript 1 s,” that has two vertical half arrows drawn on it, one facing up and one facing down. This line is connected to the right and left by upward-facing, dotted lines to two more horizontal lines, each labeled, “1 s,” and each with one vertical half-arrow facing up and one facing down drawn on it. These two lines are connected by upward-facing dotted lines to another line in the center of the diagram, but farther up from the first, and labeled, “sigma subscript 1 s superscript asterisk.” This line has one upward-facing and one downward-facing vertical arrow drawn on it. The left and right sides of the diagram have headers that read, “Atomic orbitals,” while the center header reads, “Molecular orbitals.” The bottom left and right are labeled, “H e,” while the center is labeled, “H e subscript 2.”

The molecular orbital energy diagram predicts that He2 will not be a stable molecule, since it has equal numbers of bonding and antibonding electrons.

The Diatomic Molecules of the Second Period

Eight possible homonuclear diatomic molecules might be formed by the atoms of the second period of the periodic table: Li2, Be2, B2, C2, N2, O2, F2, and Ne2. However, we can predict that the Be2 molecule and the Ne2 molecule would not be stable. We can see this by a consideration of the molecular electron configurations (see the table below).

We predict valence molecular orbital electron configurations just as we predict electron configurations of atoms. Valence electrons are assigned to valence molecular orbitals with the lowest possible energies. Consistent with Hund’s rule, whenever there are two or more degenerate molecular orbitals, electrons fill each orbital of that type singly before any pairing of electrons takes place.

As we saw in valence bond theory, σ bonds are generally more stable than π bonds formed from degenerate atomic orbitals. Similarly, in molecular orbital theory, σ orbitals are usually more stable than π orbitals. However, this is not always the case. The MOs for the valence orbitals of the second period are shown in the figure below. Looking at Ne2 molecular orbitals, we see that the order is consistent with the generic diagram shown in the previous section.

However, for atoms with three or fewer electrons in the p orbitals (Li through N) we observe a different pattern, in which the σp orbital is higher in energy than the πp set. Obtain the molecular orbital diagram for a homonuclear diatomic ion by adding or subtracting electrons from the diagram for the neutral molecule.

A graph is shown in which the y-axis is labeled, “E,” and appears as a vertical, upward-facing arrow. Across the top, the graph reads, “L i subscript 2,” “B e subscript 2,” “B subscript 2,” “C subscript 2,” “N subscript 2,” “O subscript 2,” “F subscript 2,” and “Ne subscript 2.” Directly below each of these element terms is a single pink line, and all lines are connected to one another by a dashed line, to create an overall line that decreases in height as it moves from left to right across the graph. This line is labeled, “sigma subscript 2 p x superscript asterisk”. Directly below each of these lines is a set of two pink lines, and all lines are connected to one another by a dashed line, to create an overall line that decreases in height as it moves from left to right across the graph. It is consistently lower than the first line. This line is labeled, “pi subscript 2 p y superscript asterisk,” and, “pi subscript 2 p z superscript asterisk.” Directly below each of these double lines is a single pink line, and all lines are connected to one another by a dashed line, to create an overall line that decreases in height as it moves from left to right across the graph. It has a distinctive drop at the label, “O subscript 2.” This line is labeled, “sigma subscript 2 p x.” Directly below each of these lines is a set of two pink lines, and all lines are connected to one another by a dashed line to create an overall line that decreases very slightly in height as it moves from left to right across the graph. It is consistently lower than the third line until it reaches the point labeled, “O subscript 2.” This line is labeled, “pi subscript 2 p y,” and, “pi subscript 2 p z.” Directly below each of these lines is a single blue line, and all lines are connected to one another by a dashed line to create an overall line that decreases in height as it moves from left to right across the graph. This line is labeled, “sigma subscript 2 s superscript asterisk.” Finally, directly below each of these lines is a single blue line, and all lines are connected to one another by a dashed line to create an overall line that decreases in height as it moves from left to right across the graph. This line is labeled. “sigma subscript 2 s.”

This shows the MO diagrams for each homonuclear diatomic molecule in the second period. The orbital energies decrease across the period as the effective nuclear charge increases and atomic radius decreases. Between N2 and O2, the order of the orbitals changes.

Resource:

You can practice labeling and filling molecular orbitals with this interactive tutorial from the University of Sydney.

This switch in orbital ordering occurs because of a phenomenon called s-p mixing. s-p mixing does not create new orbitals; it merely influences the energies of the existing molecular orbitals. The σs wavefunction mathematically combines with the σp wavefunction, with the result that the σs orbital becomes more stable, and the σp orbital becomes less stable (see the figure below). Similarly, the antibonding orbitals also undergo s-p mixing, with the σs* becoming more stable and the σp* becoming less stable.

A diagram is shown. At the bottom left of the diagram is a horizontal line that is connected to the right and left by upward-facing, dotted lines to two more horizontal lines. Those two lines are connected by upward-facing dotted lines to another line in the center of the diagram but farther up from the first. Each of the bottom two central lines has a vertical downward-facing arrow. Above this structure is a horizontal line that is connected to the right and left by upward-facing, dotted lines to two sets of three horizontal lines and those two lines are connected by upward-facing dotted lines to another line in the center of the diagram, but further up from the first. In between the horizontal lines of this structure are two pairs of horizontal lines that are above the first line but below the second and connected by dotted lines to the side horizontal lines. The bottom and top central lines each have an upward-facing vertical arrow. These two structures are redrawn on the right side of the diagram, but this time, the central lines of the bottom structure are moved downward in relation to the side lines. The upper portion of the structure has its central lines shifted upward in relation to the side lines. This structure also shows the bottom line appearing above the set of two lines.

Without mixing, the MO pattern occurs as expected, with the σp orbital lower in energy than the σp orbitals. When s-p mixing occurs, the orbitals shift as shown, with the σp orbital higher in energy than the πp orbitals.

s-p mixing occurs when the s and p orbitals have similar energies. The energy difference between 2s and 2p orbitals in O, F, and Neis greater than that in Li, Be, B, C, and N. Because of this, O2, F2, and Ne2 exhibit negligible s-p mixing (not sufficient to change the energy ordering), and their MO diagrams follow the normal pattern, as shown in the figure above. All of the other period 2 diatomic molecules do have s-p mixing, which leads to the pattern where the σp orbital is raised above the πp set.

Using the MO diagrams shown in the figure above, we can add in the electrons and determine the molecular electron configuration and bond order for each of the diatomic molecules. As shown in The table below, Be2 and Ne2 molecules would have a bond order of 0, and these molecules do not exist.

Electron Configuration and Bond Order for Molecular Orbitals in Homonuclear Diatomic Molecules of Period Two Elements
Molecule Electron Configuration Bond Order
Li2 \({\left({\text{σ}}_{2s}\right)}^{2}\) 1
Be2 (unstable) \({\left({\text{σ}}_{2s}\right)}^{2}{\left({\text{σ}}_{2s}^{*}\right)}^{2}\) 0
B2 \({\left({\text{σ}}_{2s}\right)}^{2}{\left({\text{σ}}_{2s}^{*}\right)}^{2}{\left({\text{π}}_{2py},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}\right)}^{2}\) 1
C2 \({\left({\text{σ}}_{2s}\right)}^{2}{\left({\text{σ}}_{2s}^{*}\right)}^{2}{\left({\text{π}}_{2py},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}\right)}^{4}\) 2
N2 \({\left({\text{σ}}_{2s}\right)}^{2}{\left({\text{σ}}_{2s}^{*}\right)}^{2}{\left({\text{π}}_{2py},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}\right)}^{4}{\left({\text{σ}}_{2px}\right)}^{2}\) 3
O2 \({\left({\text{σ}}_{2s}\right)}^{2}{\left({\text{σ}}_{2s}^{*}\right)}^{2}{\left({\text{σ}}_{2px}\right)}^{2}{\left({\text{π}}_{2py},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}\right)}^{4}{\left({\text{π}}_{2py}^{*},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}^{*}\right)}^{2}\) 2
F2 \({\left({\text{σ}}_{2s}\right)}^{2}{\left({\text{σ}}_{2s}^{*}\right)}^{2}{\left({\text{σ}}_{2px}\right)}^{2}{\left({\text{π}}_{2py},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}\right)}^{4}{\left({\text{π}}_{2py}^{*},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}^{*}\right)}^{4}\) 1
Ne2 (unstable) \({\left({\text{σ}}_{2s}\right)}^{2}{\left({\text{σ}}_{2s}^{*}\right)}^{2}{\left({\text{σ}}_{2px}\right)}^{2}{\left({\text{π}}_{2py},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}\right)}^{4}{\left({\text{π}}_{2py}^{*},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}^{*}\right)}^{4}{\left({\text{σ}}_{2px}^{*}\right)}^{2}\) 0

The combination of two lithium atoms to form a lithium molecule, Li2, is analogous to the formation of H2, but the atomic orbitals involved are the valence 2s orbitals. Each of the two lithium atoms has one valence electron. Hence, we have two valence electrons available for the σ2s bonding molecular orbital. Because both valence electrons would be in a bonding orbital, we would predict the Li2 molecule to be stable. The molecule is, in fact, present in appreciable concentration in lithium vapor at temperatures near the boiling point of the element. All of the other molecules in the table above with a bond order greater than zero are also known.

The O2 molecule has enough electrons to half fill the \(\left({\text{π}}_{2py}^{*},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}^{*}\right)\) level. We expect the two electrons that occupy these two degenerate orbitals to be unpaired, and this molecular electronic configuration for O2 is in accord with the fact that the oxygen molecule has two unpaired electrons (the figure in the example below). The presence of two unpaired electrons has proved to be difficult to explain using Lewis structures, but the molecular orbital theory explains it quite well. In fact, the unpaired electrons of the oxygen molecule provide a strong piece of support for the molecular orbital theory.

Example: Molecular Orbital Diagrams, Bond Order, and Number of Unpaired Electrons

Draw the molecular orbital diagram for the oxygen molecule, O2. From this diagram, calculate the bond order for O2. How does this diagram account for the paramagnetism of O2?

Solution

We draw a molecular orbital energy diagram similar to that shown in this figure above. Each oxygen atom contributes six electrons, so the diagram appears as shown in the figure below.

A diagram is shown that has an upward-facing vertical arrow running along the left side labeled, “E.” At the bottom center of the diagram is a horizontal line labeled, “sigma subscript 2 s,” that has two vertical half arrows drawn on it, one facing up and one facing down. This line is connected to the right and left by upward-facing, dotted lines to two more horizontal lines, each labeled, “2 s,” and with two vertical half arrows drawn on them, one facing up and one facing down. These two lines are connected by upward-facing dotted lines to another line in the center of the diagram, but farther up from the first and labeled, “sigma subscript 2 s superscript asterisk.” This horizontal line has two vertical half-arrow drawn on it, one facing up and one facing down. Moving further up the center of the diagram is a horizontal line labeled, “sigma subscript 2 p subscript x,” which lies below two horizontal lines, lying side-by-side, and labeled “pi subscript 2 p subscript y,” and “pi subscript 2 p subscript z.” Both the bottom and top lines are connected to the right and left by upward-facing, dotted lines to three more horizontal lines, each labeled, “2 p,” on either side. These sets of lines each hold three upward-facing and one downward-facing half-arrow. They are connected by upward-facing dotted lines to another single line and then pair of double lines in the center of the diagram, but farther up from the lower lines. They are labeled, “sigma subscript 2 p subscript x superscript asterisk,” “pi subscript 2 p subscript y superscript asterisk,” and “pi subscript 2 p subscript z superscript asterisk,” respectively. The lower of these two central, horizontal lines each contain one upward-facing half-arrow. The left and right sides of the diagram have headers that read, ”Atomic orbitals,” while the center header reads, “Molecular orbitals.”

The molecular orbital energy diagram for O2 predicts two unpaired electrons.

We calculate the bond order as

\({\text{O}}_{2}=\phantom{\rule{0.2em}{0ex}}\frac{\left(8-4\right)}{2}\phantom{\rule{0.2em}{0ex}}=2\)

Oxygen's paramagnetism is explained by the presence of two unpaired electrons in the (π2py, π2pz)* molecular orbitals.

Example: Ion Predictions with MO Diagrams

Give the molecular orbital configuration for the valence electrons in \({\text{C}}_{2}{}^{\text{2−}}.\) Will this ion be stable?

Solution

Looking at the appropriate MO diagram, we see that the π orbitals are lower in energy than the σp orbital. The valence electron configuration for C2 is \({\left({\text{σ}}_{2s}\right)}^{2}{\left({\text{σ}}_{\text{2}s}^{*}\right)}^{2}{\left({\text{π}}_{2py},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}\right)}^{4}.\) Adding two more electrons to generate the \({\text{C}}_{2}{}^{\text{2−}}\) anion will give a valence electron configuration of \({\left({\text{σ}}_{2s}\right)}^{2}{\left({\text{σ}}_{\text{2}s}^{*}\right)}^{2}{\left({\text{π}}_{2py},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}\right)}^{4}{\left({\text{σ}}_{2px}\right)}^{2}.\) Since this has six more bonding electrons than antibonding, the bond order will be 3, and the ion should be stable.

Resource:

Creating molecular orbital diagrams for molecules with more than two atoms relies on the same basic ideas as the diatomic examples presented here. However, with more atoms, computers are required to calculate how the atomic orbitals combine. See three-dimensional drawings of the molecular orbitals for C6H6.

This lesson is part of:

Advanced Theories of Covalent Bonding

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