Oxygen has two stable allotropes, O2 (dioxygen) and O3 (ozone). — Bonding and Molecular Structure Chemistry Question
Problem Context
Oxygen has two stable allotropes, O2 (dioxygen) and O3 (ozone).
Explain why O2 has a higher normal boiling point than either N2 or F2.
Model Answer
In these nonpolar diatomic molecules, the only intermolecular forces are London dispersion forces. These interactions increase in strength with increasing numbers of electrons and with increasing polarizability of the electrons. O2 has two more electrons than N2, so it is expected that its boiling point will increase. The fact that F2 has a lower boiling point must indicate that its electrons are less polarizable than those in O2. This is characteristic of fluorine-containing molecules and is due to F's extremely high electronegativity resulting in a very low polarizability of its electrons.
Explain why both atomic oxygen (O) and O2 have two unpaired electrons in their ground states.
Model Answer
The ground-state electron configuration of atomic O is 1s22s22p4. Four electrons occupy the three 2p orbitals, so by Hund's rule, the lowest-energy configuration has two half-filled orbitals with electron spins aligned.
In O2, the occupied molecular orbitals are (s2s)2(s*2s)2(s2p)2(p2px)2(p2py)2(p*2px)1(p*2py)1. Again, with two electrons occupying the two degenerate π* orbitals, the lowest-energy configuration has the two electrons with the same spin in different orbitals.
Molecular oxygen has an excited state with no unpaired electrons, which emits light with a wavelength of 1270 nm to return to the ground state. What is the energy (in kJ mol-1) by which the excited state is higher than the ground state?
Model Answer
E = hc/l
E = (6.626 ´ 10-34 J s)(2.998 ´ 108 m s-1)/(1.270 ´ 10-6 m) = 1.564 ´ 10-19 J
This is the energy of one photon, the energy of a mole of photons is (1.564 ´ 10-22 kJ) ´ (6.022 ´ 1023 mol-1) = 94.19 kJ mol-1
Explain why O2 does not absorb light in the infrared region of the electromagnetic spectrum but O3 does.
Model Answer
To absorb light in the IR, a molecule must have a molecular vibration that causes a change in the molecule's dipole moment. O2 has only one possible vibration (stretching the O–O bond), but that does not cause the dipole moment to change from zero. O3 has three possible vibrations (symmetric stretch, asymmetric stretch, and bend). Because of the unsymmetrical distribution of charge in the molecule (see Lewis structures below), all of these motions cause a change in the dipole moment, so O3 absorbs at three fundamental frequencies in the IR region (701, 1042, and 1103 cm-1).
Dioxygen can be oxidized to form dioxygenyl cation (O2 +) or reduced to form superoxide ion (O2 –) or peroxide ion (O2 2–), while ozone can be reduced to form ozonide ion (O3 –). Among these six species, the O–O bond distances are 112 pm, 121 pm, 127 pm, 128 pm, 135 pm, and 154 pm. Fill in the table below by assigning the correct distances to the six species, and briefly justify your assignments. (You may consider 127 and 128 pm as essentially the same in this problem.)
Model Answer
According to the MO configuration in part b, the bond order in O2 is 2.0. Electrons are added to or removed from π* orbitals, so O2+ has a bond order of 2.5 while O2– has a bond order of 1.5 and O2 2- a bond order of 1.0. The Lewis structure of O3 (see part d) indicates that there are two O–O s bonds and one O–O π bond, so the overall bond order is 1.5 (same as that of superoxide). The extra electron in ozonide is added to a π* orbital, reducing the overall π bond order from 1.0 to 0.5. With the two s bonds, this means that there are a total of 2.5 bonds distributed over the two O-O bonds, giving an O–O bond order of 1.25. Aligning the stated bond lengths with these bond orders gives the assignments in the table.