To apply Molecular Orbital Theory to the diatomic homonuclear molecule from the elements in the second period.

You are watching: Determine the bond order in a molecule or ion with 4 valence electrons.

If we combine the splitting shistoricsweetsballroom.comes for the 2s and 2p orbitals, we can predict bond order in all of the diatomic molecules and ions composed of elements in the first complete row of the periodic table. Remember that only the valence orbitals of the atoms need be considered; as we saw in the cases of lithium hydride and dilithium, the inner orbitals remain tightly bound and retain their localized atomic character. We now describe examples of systems involving period 2 homonuclear diatomic molecules, such as N2, O2, and F2.

four key points to remember when drawing molecular orbital diagrams:

The number of molecular orbitals produced is the same as the number of atomic orbitals used to create them (the "law of conservation of orbitals"). As the overlap between two atomic orbitals increases, the difference in energy between the resulting bonding and antibonding molecular orbitals increases. When two atomic orbitals combine to form a pair of molecular orbitals, the bonding molecular orbital is stabilized about as much as the antibonding molecular orbital is destabilized. The interaction between atomic orbitals is greatest when they have the same energy.

We illustrate how to use these points by constructing a molecular orbital energy-level diagram for F2. We use the diagram in part (a) in Figure $$\PageIndex{1}$$; the n = 1 orbitals (σ1s and σ1s*) are located well below those of the n = 2 level and are not shown. As illustrated in the diagram, the σ2s and σ2s* molecular orbitals are much lower in energy than the molecular orbitals derived from the 2p atomic orbitals because of the large difference in energy between the 2s and 2p atomic orbitals of fluorine. The lowest-energy molecular orbital derived from the three 2p orbitals on each F is $$\sigma _{2p_{z}}$$ and the next most stable are the two degenerate orbitals, $$\pi _{2p_{x}}$$ and $$\pi _{2p_{y}}$$. For each bonding orbital in the diagram, there is an antibonding orbital, and the antibonding orbital is destabilized by about as much as the corresponding bonding orbital is stabilized. As a result, the $$\sigma ^{\star }_{2p_{z}}$$ orbital is higher in energy than either of the degenerate $$\pi _{2p_{x}}^{\star }$$ and $$\pi _{2p_{y}}^{\star }$$ orbitals. We can now fill the orbitals, beginning with the one that is lowest in energy.

Each fluorine has 7 valence electrons, so there are a total of 14 valence electrons in the F2 molecule. Starting at the lowest energy level, the electrons are placed in the orbitals according to the Pauli principle and Hund’s rule. Two electrons each fill the σ2s and σ2s* orbitals, 2 fill the $$\sigma _{2p_{z}}$$ orbital, 4 fill the two degenerate π orbitals, and 4 fill the two degenerate π* orbitals, for a total of 14 electrons. To determine what type of bonding the molecular orbital approach predicts F2 to have, we must calculate the bond order. According to our diagram, there are 8 bonding electrons and 6 antibonding electrons, giving a bond order of (8 − 6) ÷ 2 = 1. Thus F2 is predicted to have a stable F–F single bond, in agreement with experimental data.

Example $$\PageIndex{3}$$: Diatomic Sulfur

Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in S2, a bright blue gas at high temperatures.

Given: historicsweetsballroom.comical species

Asked for: molecular orbital energy-level diagram, bond order, and number of unpaired electrons

Strategy:

Write the valence electron configuration of sulfur and determine the type of molecular orbitals formed in S2. Predict the relative energies of the molecular orbitals based on how close in energy the valence atomic orbitals are to one another. Draw the molecular orbital energy-level diagram for this system and determine the total number of valence electrons in S2. Fill the molecular orbitals in order of increasing energy, being sure to obey the Pauli principle and Hund’s rule. Calculate the bond order and describe the bonding.

Solution:

A Sulfur has a 3s23p4 valence electron configuration. To create a molecular orbital energy-level diagram similar to those in Figure $$\PageIndex{1}$$ and Figure $$\PageIndex{3}$$, we need to know how close in energy the 3s and 3p atomic orbitals are because their energy separation will determine whether the $$\pi _{3p_{x,y}}$$ or the $$\sigma _{3p_{z}}$$> molecular orbital is higher in energy. Because the nsnp energy gap increases as the nuclear charge increases (Figure $$\PageIndex{3}$$), the $$\sigma _{3p_{z}}$$ molecular orbital will be lower in energy than the $$\pi _{3p_{x,y}}$$ pair.

B The molecular orbital energy-level diagram is as follows:

## Molecular Orbitals for Heteronuclear Diatomic Molecules

Diatomic molecules with two different atoms are called heteronuclear diatomic molecules. When two nonidentical atoms interact to form a historicsweetsballroom.comical bond, the interacting atomic orbitals do not have the same energy. If, for example, element B is more electronegative than element A (χB > χA), the net result is a “skewed” molecular orbital energy-level diagram, such as the one shown for a hypothetical A–B molecule in Figure $$\PageIndex{4}$$. The atomic orbitals of element B are uniformly lower in energy than the corresponding atomic orbitals of element A because of the enhanced stability of the electrons in element B. The molecular orbitals are no longer symmetrical, and the energies of the bonding molecular orbitals are more similar to those of the atomic orbitals of B. Hence the electron density of bonding electrons is likely to be closer to the more electronegative atom. In this way, molecular orbital theory can describe a polar covalent bond.

Figure $$\PageIndex{4}$$: Molecular Orbital Energy-Level Diagram for a Heteronuclear Diatomic Molecule AB, Where χB > χA. The bonding molecular orbitals are closer in energy to the atomic orbitals of the more electronegative B atom. Consequently, the electrons in the bonding orbitals are not shared equally between the two atoms. On average, they are closer to the B atom, resulting in a polar covalent bond.

A molecular orbital energy-level diagram is always skewed toward the more electronegative atom.

### An Odd Number of Valence Electrons: NO

Nitric oxide (NO) is an example of a heteronuclear diatomic molecule. The reaction of O2 with N2 at high temperatures in internal combustion engines forms nitric oxide, which undergoes a complex reaction with O2 to produce NO2, which in turn is responsible for the brown color we associate with air pollution. Recently, however, nitric oxide has also been recognized to be a vital biological messenger involved in regulating blood pressure and long-term memory in mammals.

Because NO has an odd number of valence electrons (5 from nitrogen and 6 from oxygen, for a total of 11), its bonding and properties cannot be successfully explained by either the Lewis electron-pair approach or valence bond theory. The molecular orbital energy-level diagram for NO (Figure $$\PageIndex{5}$$) shows that the general pattern is similar to that for the O2 molecule (Figure $$\PageIndex{3}$$). Because 10 electrons are sufficient to fill all the bonding molecular orbitals derived from 2p atomic orbitals, the 11th electron must occupy one of the degenerate π* orbitals. The predicted bond order for NO is therefore (8-3) ÷ 2 = 2 1/2 . Experimental data, showing an N–O bond length of 115 pm and N–O bond energy of 631 kJ/mol, are consistent with this description. These values lie between those of the N2 and O2 molecules, which have triple and double bonds, respectively. As we stated earlier, molecular orbital theory can therefore explain the bonding in molecules with an odd number of electrons, such as NO, whereas Lewis electron structures cannot.

Figure $$\PageIndex{5}$$: Molecular Orbital Energy-Level Diagram for NO. Because NO has 11 valence electrons, it is paramagnetic, with a single electron occupying the $$\left ( \pi ^{\star }_{2p_{x}},\; \pi ^{\star }_{2p_{y}} \right )$$ pair of orbitals.

Molecular orbital theory can also tell us something about the historicsweetsballroom.comistry of $$NO$$. As indicated in the energy-level diagram in Figure $$\PageIndex{5}$$, NO has a single electron in a relatively high-energy molecular orbital. We might therefore expect it to have similar reactivity as alkali metals such as Li and Na with their single valence electrons. In fact, $$NO$$ is easily oxidized to the $$NO^+$$ cation, which is isoelectronic with $$N_2$$ and has a bond order of 3, corresponding to an N≡O triple bond.

## Nonbonding Molecular Orbitals

Molecular orbital theory is also able to explain the presence of lone pairs of electrons. Consider, for example, the HCl molecule, whose Lewis electron structure has three lone pairs of electrons on the chlorine atom. Using the molecular orbital approach to describe the bonding in HCl, we can see from Figure $$\PageIndex{6}$$ that the 1s orbital of atomic hydrogen is closest in energy to the 3p orbitals of chlorine. Consequently, the filled Cl 3s atomic orbital is not involved in bonding to any appreciable extent, and the only important interactions are those between the H 1s and Cl 3p orbitals. Of the three p orbitals, only one, designated as 3pz, can interact with the H 1s orbital. The 3px and 3py atomic orbitals have no net overlap with the 1s orbital on hydrogen, so they are not involved in bonding. Because the energies of the Cl 3s, 3px, and 3py orbitals do not change when HCl forms, they are called nonbonding molecular orbitals. A nonbonding molecular orbital occupied by a pair of electrons is the molecular orbital equivalent of a lone pair of electrons. By definition, electrons in nonbonding orbitals have no effect on bond order, so they are not counted in the calculation of bond order. Thus the predicted bond order of HCl is (2 − 0) ÷ 2 = 1. Because the σ bonding molecular orbital is closer in energy to the Cl 3pz than to the H 1s atomic orbital, the electrons in the σ orbital are concentrated closer to the chlorine atom than to hydrogen. A molecular orbital approach to bonding can therefore be used to describe the polarization of the H–Cl bond to give $$H^{\delta +} -- Cl^{\delta -}$$.

Figure $$\PageIndex{6}$$: Molecular Orbital Energy-Level Diagram for HCl. The hydrogen 1s atomic orbital interacts most strongly with the 3pz orbital on chlorine, producing a bonding/antibonding pair of molecular orbitals. The other electrons on Cl are best viewed as nonbonding. As a result, only the bonding σ orbital is occupied by electrons, giving a bond order of 1.

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Electrons in nonbonding molecular orbitals have no effect on bond order.