IChO 1998 · Theory problem 4

Molecular orbitals of ozone and the Walsh diagram

Download PDFIChO 1998

When two hydrogen atoms come together, the 1s atomic orbitals combine to form bonding and anti-bonding molecular orbitals:

Figure 1

In a similar way, we may combine the atomic orbitals of more complicated atoms to form molecular orbitals, taking into account the symmetry of the molecule.

Consider the ozone molecule, \(\ce{O3}\), which is shaped like an Australian boomerang. We can arrange the oxygens as follows (in the \(yz\) plane) and assume that there are \(1s\), \(2s\), \(2p_x\), \(2p_y\) and \(2p_z\) orbitals on each atom.

Figure 2

The atoms \(\ce{O}_{a}\) and \(\ce{O}_{c}\) are "related by symmetry" and the 1s orbitals on these atoms form symmetric and anti-symmetric combinations:

Figure 3

In this molecule the 1s atomic orbital on \(\ce{O}_{b}\) is classified as symmetric. It can combine with the symmetric combination of \(\ce{O}_{a}\) and \(\ce{O}_{c}\) given above (but not with the anti-symmetric combination) to form bonding and anti-bonding molecular orbitals. The anti-symmetric combination is non-bonding. The final three molecular orbitals are:

Figure 4

4.1 On the answer sheet, use a similar approach to construct the molecular orbitals arising from the separate interaction of the \(2s\), \(2p_x\), \(2p_y\), and \(2p_z\) atomic orbitals. (Remember to form the symmetric and anti-symmetric combinations of \(\ce{O}_{a}\) and \(\ce{O}_{c}\) first.)

We may now rearrange these molecular orbitals in order of increasing energy. This can be generalised to other triatomic molecules. The energy of these orbitals is different in a bent triatomic molecule (like ozone) compared to a linear molecule (like carbon dioxide). The variation in orbital energy may be represented in a "Walsh diagram" for \(\ce{XY2}\) molecules as shown on the answer sheet. It shows a plot of the energy of each orbital versus the \(\ce{Y-X-Y}\) bond angle. The orbitals have been given labels which we call “symmetry labels”.

The \(6a_1\) orbital referred to in the Walsh diagram is shown below.

Figure 5

4.2 Why does the energy of the \(6a_1\) orbital increase so rapidly as the bond angle changes from \(90^\circ\) to \(180^\circ\)?

Only occupied molecular orbitals affect the geometry, and a doubly occupied orbital has more influence than a singly occupied orbital. For example, \(\ce{O3}\) has 24 electrons and so at a bond angle of \(135^\circ\) the orbitals are doubly occupied up to \(6a_1\). Thus, the lowest-energy geometry of ozone (taking into account steric repulsion and the contrasting energy behaviour of the \(4b_2\), \(1a_2\) and \(6a_1\) orbitals) is probably towards the left of the Walsh diagram, which is consistent with the observed bond angle of \(116^\circ\).

4.3 At a bond angle of \(135^\circ\), what are the highest occupied orbitals for the molecules \(\ce{BO2}\), \(\ce{CO2}\), \(\ce{NO2}\) and \(\ce{FO2}\)?

4.4 The bond angles of \(\ce{BO2}\), \(\ce{CO2}\) and \(\ce{O3}\) are known experimentally to be \(180^\circ\), \(180^\circ\) and \(116^\circ\), respectively. Use the Walsh diagram on the answer sheet to predict whether \(\ce{NO2}\) and \(\ce{FO2}\) are more or less bent than \(\ce{O3}\).

Official English text of the problem.

Answers

  • 4.1: Each atomic-orbital set gives three molecular orbitals, making 12 orbitals. With coefficient signs listed in the order \((a,b,c)\), the bonding, non-bonding and anti-bonding patterns in the bent sketch are:
  • \(2s\) and \(2p_x\): \((+,+,+)\), \((+,0,-)\), \((+,-,+)\).
  • \(2p_y\) and \(2p_z\): \((+,-,+)\), \((+,0,-)\), \((+,+,+)\).

These describe ozone, \(\ce{O3}\), [O-][O+]=O. The \(2p_z\) overlap character changes upon straightening.

  • 4.2: The matching, facing lobes in \(6a_1\) lose their bonding overlap as the angle opens. At \(180^\circ\), adjacent \(2p_z\) orbitals overlap sideways with opposite phases, producing an anti-bonding interaction and a much higher energy.

  • 4.3: At \(135^\circ\), the highest occupied orbitals are:

  • Boron dioxide, \(\ce{BO2}\), O=B[O]: \(4b_2\), singly occupied.
  • Carbon dioxide, \(\ce{CO2}\), O=C=O: \(4b_2\), doubly occupied.
  • Nitrogen dioxide, \(\ce{NO2}\), [O-][N+]=O: \(6a_1\), singly occupied.

  • Fluorine dioxide in the question’s \(\ce{O-F-O}\) model, \(\ce{FO2}\), [O-][F+][O]: \(2b_1\), singly occupied.

  • 4.4: Nitrogen dioxide is predicted to be less bent than ozone, with an angle greater than \(116^\circ\). Fluorine dioxide in the stated model is predicted to be more bent, with an angle less than \(116^\circ\).

Solution

The SMILES above specify connectivity and representative localized electronic structures. They do not show molecular-orbital phases, resonance delocalization or bond angles. In particular, the fluorine dioxide representation follows the central-fluorine \(\ce{O-F-O}\) arrangement implicit in the question’s \(\ce{XY2}\) model.

4.1 — Constructing the molecular orbitals

Let \(\phi_{a,q}\), \(\phi_{b,q}\) and \(\phi_{c,q}\) denote the atomic orbitals of type \(q\) on the three atoms. Use the same global axes on every atom: the positive lobe of a \(p_q\) orbital points along \(+q\).

First form the two terminal-atom combinations, without specifying normalization:

\[ T_q^+=\phi_{a,q}+\phi_{c,q}, \qquad T_q^-=\phi_{a,q}-\phi_{c,q}. \]

Consider reflection in the \(xz\) plane, which exchanges the terminal atoms.

  • For \(2s\), \(2p_x\) and \(2p_z\), this reflection does not reverse the individual orbital’s phase. Thus \(T_q^+\) is symmetric and \(T_q^-\) is anti-symmetric.
  • For \(2p_y\), reflection reverses the orbital’s phase because \(y\) changes sign. Consequently, \(T_y^+\) is anti-symmetric and \(T_y^-\) is symmetric.

The central \(2p_y\) orbital is also anti-symmetric under this reflection. The other three central orbitals are symmetric. Therefore, in every set, \(T_q^+\) has the symmetry needed to interact with the corresponding central orbital.

For completeness, their full symmetry labels in the stated coordinate system are:

Atomic-orbital set Terminal sum \(T_q^+\) Terminal difference \(T_q^-\) Central orbital
\(2s\) \(a_1\) \(b_2\) \(a_1\)
\(2p_x\) \(b_1\) \(a_2\) \(b_1\)
\(2p_y\) \(b_2\) \(a_1\) \(b_2\)
\(2p_z\) \(a_1\) \(b_2\) \(a_1\)

Combining \(T_q^+\) with the central orbital in its two possible relative phases produces two molecular orbitals. The terminal difference \(T_q^-\) has zero central contribution and is non-bonding in the separate-interaction construction requested.

The required drawings can be specified by the following phase patterns. The signs refer to whole atomic-orbital coefficients, in the order \((a,b,c)\); they do not prescribe their magnitudes.

Atomic-orbital set Bonding pattern in the bent sketch Non-bonding pattern Anti-bonding pattern in the bent sketch
\(2s\) \((+,+,+)\) \((+,0,-)\) \((+,-,+)\)
\(2p_x\) \((+,+,+)\) \((+,0,-)\) \((+,-,+)\)
\(2p_y\) \((+,-,+)\) \((+,0,-)\) \((+,+,+)\)
\(2p_z\) \((+,-,+)\) \((+,0,-)\) \((+,+,+)\)

To draw these precisely:

  • \(2s\): Use three spheres of the same phase for the bonding orbital. Reverse the central sphere for the anti-bonding orbital.
  • \(2p_x\): Draw all three orbitals perpendicular to the molecular plane. Corresponding lobes on the same side of the plane have the same phase in the bonding orbital. Reverse the central orbital for the anti-bonding orbital.
  • \(2p_y\): Draw horizontal \(p\) orbitals. Reverse the central orbital relative to the terminal orbitals so that the lobes facing one another along both bonds have matching phases.
  • \(2p_z\): Draw vertical \(p\) orbitals. In the bent bonding arrangement, the central downward lobe and both terminal upward lobes have the same phase. This is the phase arrangement illustrated for \(6a_1\).

For every non-bonding drawing, omit the central orbital and give the two terminal orbitals opposite coefficient signs. Reversing every sign in any drawing leaves the physical orbital unchanged.

Thus,

\[ 4\text{ atomic-orbital sets}\times3\text{ molecular orbitals per set} =\boxed{12\text{ molecular orbitals}}. \]

The bonding descriptions for \(2p_z\) refer to the bent geometry; they do not remain valid unchanged as the molecule becomes linear.

4.2 — Why \(6a_1\) rises rapidly in energy

In the illustrated \(6a_1\) orbital, the terminal \(2p_z\) orbitals have the same coefficient sign, while the central orbital has the opposite sign:

\[ \psi_{6a_1}\sim \phi_{a,z}-\lambda\phi_{b,z}+\phi_{c,z}, \qquad \lambda>0. \]

In the bent geometry, the central downward lobe faces the terminal upward lobes, and those facing lobes have matching phases. This gives a bonding contribution.

If the bond angle is \(\theta\), the magnitude of the projection of a bond direction onto the \(z\) axis is

\[ \cos\!\left(\frac{\theta}{2}\right). \]

As \(\theta\) increases towards \(180^\circ\), this projection falls to zero: the favourable end-on component of the overlap disappears.

At \(180^\circ\), the atoms lie along the \(y\) direction. Their \(2p_z\) orbitals are then parallel and overlap sideways. Because the central orbital has the opposite coefficient sign from the terminal orbitals, adjacent lobes overlap out of phase. The interaction is now \(\pi\)-anti-bonding.

The combination of losing bonding overlap and developing anti-bonding overlap explains the rapid increase in energy.

4.3 — Highest occupied orbitals at \(135^\circ\)

Each neutral molecule contains

\[ N_e=Z_X+2Z_{\mathrm O}, \]

where \(Z_{\mathrm O}=8\).

In the Walsh diagram of the answer sheet, the frontier orbitals at 135° are ordered

\[ 1a_2<4b_2<6a_1<2b_1. \]

The problem states that ozone’s 24 electrons doubly occupy all orbitals through \(6a_1\). Therefore, the successive frontier levels accommodate:

Orbital Electrons entering this level
\(1a_2\) 19th and 20th
\(4b_2\) 21st and 22nd
\(6a_1\) 23rd and 24th
\(2b_1\) 25th and 26th

Filling these levels gives:

Molecule Total electrons Highest occupied orbital Occupancy of that orbital
\(\ce{BO2}\) \(5+2(8)=21\) \(4b_2\) 1
\(\ce{CO2}\) \(6+2(8)=22\) \(4b_2\) 2
\(\ce{NO2}\) \(7+2(8)=23\) \(6a_1\) 1
\(\ce{FO2}\) \(9+2(8)=25\) \(2b_1\) 1

For the odd-electron species, the highest occupied orbital is also the singly occupied molecular orbital.

4.4 — Comparing the bending with ozone

The Walsh argument compares how the energies of the occupied orbitals change with angle, weighted by their occupancies.

Nitrogen dioxide: Relative to ozone, it has one fewer electron in \(6a_1\). Since \(6a_1\) is strongly stabilized by bending, single occupation provides a weaker preference for bending than double occupation. The competing preference of other occupied orbitals for a larger angle therefore has greater relative influence.

Hence,

\[ \boxed{\theta(\ce{NO2})>116^\circ: \quad \text{nitrogen dioxide is less bent than ozone}.} \]

Fluorine dioxide: It retains the doubly occupied \(6a_1\) orbital and adds one electron to \(2b_1\).

The \(2b_1\) orbital is the anti-bonding combination of the out-of-plane \(2p_x\) orbitals:

\[ \psi_{2b_1}\sim \phi_{a,x}-\mu\phi_{b,x}+\phi_{c,x}, \qquad \mu>0. \]

Although its central–terminal interactions are anti-bonding, its two terminal orbitals have the same phase. For fixed central–terminal distance \(r\), the terminal separation is

\[ d_{ac}=2r\sin\!\left(\frac{\theta}{2}\right). \]

Bending decreases \(d_{ac}\), strengthening the favourable interaction between the same-phase terminal lobes. Thus \(2b_1\) gains stabilization on bending. Occupying it adds a further preference for a smaller angle.

The qualitative prediction is therefore

\[ \boxed{\theta(\ce{FO2})<116^\circ: \quad \text{fluorine dioxide is more bent than ozone}.} \]

The solution was checked against the official solution.