IChO 2012 · Theory problem 8

Linear polycyclic aromatic hydrocarbons

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Polycyclic aromatic hydrocarbons (PAHs) are atmospheric pollutants, components of organic light emitting diodes and components of the interstellar medium. This problem deals with so-called linear PAHs, i.e., those being just one benzene ring wide whereas the length is varied. Specific examples are benzene, anthracene and pentacene whose structures are shown below. Their physical and chemical properties depend on the extent to which the \(\pi\) electron cloud is delocalized over the molecule.

Figure 1

8.1 The distance across the benzene ring is \(d = 240\ \mathrm{pm}\). Use this information to estimate the distances along the horizontal (\(x\)) axis for anthracene and pentacene, \(d_a\) and \(d_p\), respectively.

Assume for simplicity that the \(\pi\) electrons of benzene can be modelled as being confined to a square. Within this model, the conjugated \(\pi\) electrons of PAHs may be considered as free particles in a two dimensional rectangular box in the \(x\)-\(y\) plane.

For electrons in a two-dimensional box along the \(x\)- and \(y\)-axes, the quantized energy states of the electrons are given by

\[ E=\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right)\frac{h^2}{8m_e} \]

In this equation, \(n_x\) and \(n_y\) are the quantum numbers for the energy state and are integers between 1 and \(\infty\), \(h\) is Planck’s constant, \(m_e\) is the mass of the electron and \(L_x\) and \(L_y\) are the dimensions of the box.

For this problem, treat the \(\pi\) electrons of the PAHs as particles in a two dimensional box. In this case, the quantum numbers \(n_x\) and \(n_y\) are independent.

8.2 Assume that the benzene unit has \(x\) and \(y\) dimensions that are each of length \(d\). Derive a general formula for the quantized energies of linear PAHs as a function of quantum numbers \(n_x\) and \(n_y\), the length \(d\), the number of fused rings \(w\), and the fundamental constants \(h\) and \(m_e\).

The energy level diagram below for pentacene shows qualitatively the energies and quantum numbers \(n_x\), \(n_y\), for all levels occupied by \(\pi\) electrons and the lowest unoccupied energy level, with the electrons of opposite spins represented as the arrows pointing up or down. The levels are labelled with quantum numbers \((n_x; n_y)\).

Figure 2

8.3 The energy level diagram for anthracene is shown below. Note that some energy levels may have the same energy. Draw the correct number of up and down arrows to represent the \(\pi\) electrons in this diagram. Also, the blanks in parentheses within this diagram are the quantum numbers \(n_x\), \(n_y\), which you need to determine. Fill these blanks with the pertinent values of \(n_x\), \(n_y\) for each filled and the lowest unfilled energy level(s).

Figure 3

8.4 Use this model to create an energy level diagram for benzene and fill the pertinent energy levels with electrons. Include energy levels up to an including the lowest unoccupied energy level. Label each energy level in your diagrams with the corresponding \(n_x\), \(n_y\). Do not assume that the particle-in-a-square-box model used here gives the same energy levels as other models.

Often the reactivity of PAHs correlates inversely with the energy gap \(\Delta E\) between the highest energy level occupied by \(\pi\) electrons and the lowest unoccupied energy level.

8.5 Calculate the energy gap \(\Delta E\) (in joules) between the highest occupied and lowest unoccupied energy levels for benzene, anthracene and pentacene. Use your result from parts 8.3 and 8.4 for anthracene or benzene, respectively, or use \((2, 2)\) for the highest occupied energy level and \((3, 2)\) for the lowest unoccupied energy level for these two molecules (these may not be the true values).

8.6 Rank benzene (B), anthracene (A), and pentacene (P) in order of increasing reactivity by placing the corresponding letters from left to right in the box below.

8.7 The electronic absorption spectra (molar absorptivity vs. wavelength) for anthracene (A), benzene (B), and pentacene (P) are shown below. Based on a qualitative understanding of the particle in the box model, indicate which molecule corresponds to which spectrum by writing the appropriate letter in the box to its right.

Figure 4
Figure 5
Figure 6

Graphene is a sheet of carbon atoms arranged in a two-dimensional honeycomb pattern. It can be considered as an extreme case of a polyaromatic hydrocarbon with essentially infinite length in the two dimensions. The Nobel Prize for Physics was awarded in 2010 to Andrei Geim and Konstantin Novoselov for groundbreaking experiments on graphene.

Consider a sheet of graphene with planar dimensions of \(L_x = 25\ \mathrm{nm}\) by \(L_y = 25\ \mathrm{nm}\). A section of this sheet is shown below. The area of one hexagonal 6-carbon unit is \(\sim52\,400\ \mathrm{pm^2}\).

8.8 Calculate the number of \(\pi\) electrons in a \((25\ \mathrm{nm}\times25\ \mathrm{nm})\) sheet of graphene. For this problem you can ignore edge electrons (i.e. those outside the full hexagons in the picture).

Figure 7

We can think about the \(\pi\) electrons in graphene as being free electrons in a 2-dimensional box. In systems containing large numbers of electrons, there is no single highest occupied energy level. Instead, there are many states of nearly the same energy above which the remaining are empty. These highest occupied states determine the so-called Fermi level. The Fermi level in graphene consists of multiple combinations of \(n_x\) and \(n_y\) quantum numbers.

8.9 Determine the energy of the Fermi level for the \(25\ \mathrm{nm}\times25\ \mathrm{nm}\) square of graphene relative to the lowest filled level. The lowest filled level has a non-zero energy; however, it is negligible, and can be assumed to be zero. To solve this problem it might be helpful to represent the \((n_x, n_y)\) quantum states as points on a 2-D grid (as shown below) and consider how the energy levels are filled with pairs of electrons. For the number of electrons use your result from part 8.8 or use a value of 1000 (this may not be the true value).

Figure 8

8.10 The conductivity of graphene-like materials correlates inversely with the energy gap between the lowest unoccupied and highest occupied energy levels. Use your analysis and understanding of \(\pi\) electrons in PAHs and graphene to predict whether the conductivity of a \(25\ \mathrm{nm}\times25\ \mathrm{nm}\) square of graphene, at a given temperature, is less than, equal to or greater than the conductivity of a \(1\ \mathrm{m}\times1\ \mathrm{m}\) square of graphene (which is the largest obtained to date). Circle the correct answer:

Figure 9

Official English text of the problem.

Answers

Molecular identifiers used throughout: benzene (B), c1ccccc1; anthracene (A), c1ccc2cc3ccccc3cc2c1; pentacene (P), c1ccc2cc3cc4cc5ccccc5cc4cc3cc2c1.

Define \(E_0=h^2/(8m_ed^2)\).

  • 8.1: \(\boxed{d_a=720\ \mathrm{pm},\quad d_p=1200\ \mathrm{pm}}\).
  • 8.2: \(\displaystyle \boxed{E_{n_x,n_y}=\frac{h^2}{8m_ed^2}\left(\frac{n_x^2}{w^2}+n_y^2\right)}\).
  • 8.3: In increasing energy, the doubly occupied states are \[ (1,1),\ (2,1),\ (3,1),\ (4,1),\ (5,1),\ (1,2),\ (2,2). \] The lowest unoccupied states are the degenerate pair \(\boxed{(6,1),\ (3,2)}\). The additional uppermost line in the printed diagram is \((4,2)\), also empty.
  • 8.4: \((1,1)\) at \(2E_0\) contains \(\uparrow\downarrow\); the degenerate states \((1,2)\) and \((2,1)\) at \(5E_0\) each contain \(\uparrow\downarrow\); \((2,2)\) at \(8E_0\) is empty.
  • 8.5: \(\boxed{\Delta E_B=3.14\times10^{-18}\ \mathrm{J}}\), \(\boxed{\Delta E_A=5.81\times10^{-19}\ \mathrm{J}}\), \(\boxed{\Delta E_P=1.26\times10^{-19}\ \mathrm{J}}\).
  • 8.6: Increasing reactivity: \(\boxed{\mathrm{B<A<P}}\).
  • 8.7: Spectra from top to bottom: \(\boxed{\mathrm{A,\ P,\ B}}\).
  • 8.8: \(\boxed{N_\pi\approx2.4\times10^4\text{ electrons}}\).
  • 8.9: \(\boxed{E_F-E_{1,1}\approx1.46\times10^{-18}\ \mathrm{J}}\).
  • 8.10: Circle \(\boxed{\text{less}}\).

Solution

8.1

Adjacent rings share a vertical edge, so each additional ring adds another horizontal distance \(d\). Therefore,

\[ d_a=3d=3(240\ \mathrm{pm})=\boxed{720\ \mathrm{pm}}, \]

\[ d_p=5d=5(240\ \mathrm{pm})=\boxed{1200\ \mathrm{pm}}. \]

8.2

For a linear PAH containing \(w\) rings, the stipulated box dimensions are

\[ L_x=wd,\qquad L_y=d. \]

Substituting into the given energy expression gives

\[ E_{n_x,n_y} =\frac{h^2}{8m_e} \left(\frac{n_x^2}{w^2d^2}+\frac{n_y^2}{d^2}\right) =\boxed{\frac{h^2}{8m_ed^2} \left(\frac{n_x^2}{w^2}+n_y^2\right)}. \]

Thus, with \(E_0=h^2/(8m_ed^2)\),

\[ \frac{E_{n_x,n_y}}{E_0}=\frac{n_x^2}{w^2}+n_y^2. \]

8.3

Each carbon atom contributes one \(\pi\) electron. The first ring contains six carbon atoms, and each additional fused ring adds four, so

\[ N_\pi=6+4(w-1)=4w+2. \]

Anthracene, \(\ce{C14H10}\), has \(w=3\) and therefore 14 \(\pi\) electrons. Each spatial state holds two electrons of opposite spin, so the seven lowest spatial states are doubly occupied.

For anthracene,

\[ \frac{E}{E_0}=\frac{n_x^2}{9}+n_y^2. \]

Fill the printed diagram as follows, reading this table from top to bottom:

\(E/E_0\) State \((n_x,n_y)\) Electron arrows
\(52/9\) \((4,2)\) Empty
\(5\) \((6,1)\) and \((3,2)\), side by side Both empty
\(40/9\) \((2,2)\) \(\uparrow\downarrow\)
\(37/9\) \((1,2)\) \(\uparrow\downarrow\)
\(34/9\) \((5,1)\) \(\uparrow\downarrow\)
\(25/9\) \((4,1)\) \(\uparrow\downarrow\)
\(2\) \((3,1)\) \(\uparrow\downarrow\)
\(13/9\) \((2,1)\) \(\uparrow\downarrow\)
\(10/9\) \((1,1)\) \(\uparrow\downarrow\)

The two lowest empty states have equal energy because

\[ \frac{E_{6,1}}{E_0}=\frac{36}{9}+1=5, \qquad \frac{E_{3,2}}{E_0}=\frac{9}{9}+4=5. \]

Hence the highest occupied state is \((2,2)\), and the lowest unoccupied states are \((6,1)\) and \((3,2)\). Their left–right order in the diagram is interchangeable.

8.4

Benzene, \(\ce{C6H6}\), has six \(\pi\) electrons. Since \(w=1\),

\[ \frac{E}{E_0}=n_x^2+n_y^2. \]

The required diagram has three energy heights:

\(E/E_0\) State(s) Electron arrows
\(8\) \((2,2)\) Empty
\(5\) \((1,2)\) and \((2,1)\), side by side \(\uparrow\downarrow\) in each
\(2\) \((1,1)\) \(\uparrow\downarrow\)

Draw one occupied line at \(2E_0\), two occupied lines at the same height \(5E_0\), and one empty line at \(8E_0\). The six electrons completely fill the first three spatial states.

8.5

Using the standard constants

\[ h=6.626\times10^{-34}\ \mathrm{J\,s}, \qquad m_e=9.109\times10^{-31}\ \mathrm{kg}, \]

and \(d=240\times10^{-12}\ \mathrm{m}\),

\[ E_0=\frac{h^2}{8m_ed^2} \approx1.046\times10^{-18}\ \mathrm{J}. \]

For benzene, the highest occupied energy is \(5E_0\), and the lowest unoccupied energy is \(8E_0\). Thus,

\[ \Delta E_B=(8-5)E_0=3E_0 =\boxed{3.14\times10^{-18}\ \mathrm{J}}. \]

For anthracene, the highest occupied energy is \((40/9)E_0\), and the lowest unoccupied energy is \(5E_0\). Therefore,

\[ \Delta E_A=\left(5-\frac{40}{9}\right)E_0 =\frac{5}{9}E_0 =\boxed{5.81\times10^{-19}\ \mathrm{J}}. \]

Pentacene, \(\ce{C22H14}\), has \(w=5\). Its supplied diagram identifies the highest occupied state as \((9,1)\) and the lowest unoccupied state as \((3,2)\):

\[ E_{\mathrm{H},P} =\left(\frac{81}{25}+1\right)E_0 =\frac{106}{25}E_0, \]

\[ E_{\mathrm{L},P} =\left(\frac{9}{25}+4\right)E_0 =\frac{109}{25}E_0. \]

Hence,

\[ \Delta E_P =\left(\frac{109-106}{25}\right)E_0 =\frac{3}{25}E_0 =\boxed{1.26\times10^{-19}\ \mathrm{J}}. \]

8.6

The calculated gaps satisfy

\[ \Delta E_B>\Delta E_A>\Delta E_P. \]

Using the stated inverse correlation between reactivity and the energy gap, the order of increasing reactivity is

\[ \boxed{\text{benzene}<\text{anthracene}<\text{pentacene}}. \]

Write B A P from left to right.

8.7

Increasing the extent of the conjugated system generally reduces the relevant excitation energies. Since the energy of an absorbed photon is

\[ E_{\mathrm{photon}}=\frac{hc}{\lambda}, \]

smaller excitation energies correspond to longer wavelengths.

The spectra therefore have the following assignments:

Spectrum Assignment Reason
Top A — anthracene Absorption extends to intermediate wavelengths, approximately \(300\)–\(400\ \mathrm{nm}\).
Middle P — pentacene Absorption extends farthest into the visible region, approximately \(600\)–\(700\ \mathrm{nm}\).
Bottom B — benzene Absorption occurs at the shortest wavelengths, in the ultraviolet.

Thus the letters from top to bottom are \(\boxed{\mathrm{A,\ P,\ B}}\).

8.8

Graphene is an extended planar honeycomb network of carbon atoms. The problem does not specify a finite molecular boundary, so there is no unique finite molecular SMILES for the sheet.

Each interior carbon atom belongs to three hexagons. Consequently, each hexagon accounts for

\[ 6\left(\frac13\right)=2 \]

carbon atoms, and hence two \(\pi\) electrons.

The sheet area is

\[ A=(25\,000\ \mathrm{pm})^2 =6.25\times10^8\ \mathrm{pm^2}. \]

The approximate number of hexagons is

\[ N_{\mathrm{hex}} =\frac{6.25\times10^8\ \mathrm{pm^2}} {52\,400\ \mathrm{pm^2}} =1.19275\times10^4. \]

Therefore,

\[ N_\pi=2N_{\mathrm{hex}} =2.38550\times10^4 \approx\boxed{2.4\times10^4\text{ electrons}}. \]

8.9

For a square of side \(L\),

\[ E_{n_x,n_y} =\frac{h^2}{8m_eL^2}(n_x^2+n_y^2). \]

Thus equal-energy states lie on circles in the \((n_x,n_y)\) plane. Because both quantum numbers are positive, the occupied states lie in the positive quadrant.

Define the radius of the occupied region by

\[ R_F^2=n_x^2+n_y^2 \]

at the Fermi energy. For many occupied states, their number is approximately the area of a quarter-circle:

\[ N_{\mathrm{spatial}}\approx\frac{\pi R_F^2}{4}. \]

Each spatial state accommodates two electrons, so

\[ N_\pi\approx2\frac{\pi R_F^2}{4} =\frac{\pi R_F^2}{2}, \]

and therefore

\[ R_F^2\approx\frac{2N_\pi}{\pi}. \]

Substitution into the energy expression gives

\[ E_F\approx \frac{h^2}{8m_eL^2}\frac{2N_\pi}{\pi} =\boxed{\frac{h^2N_\pi}{4\pi m_eL^2}}. \]

Using the electron count before rounding,

\[ E_F\approx \frac{(6.626\times10^{-34}\ \mathrm{J\,s})^2 (2.38550\times10^4)} {4\pi(9.109\times10^{-31}\ \mathrm{kg}) (25\times10^{-9}\ \mathrm{m})^2} \approx1.464\times10^{-18}\ \mathrm{J}. \]

The lowest state is \((1,1)\), with energy

\[ E_{1,1}=\frac{h^2}{4m_eL^2} \approx1.93\times10^{-22}\ \mathrm{J}. \]

This is negligible compared with \(E_F\), as stipulated. Thus,

\[ \boxed{E_F-E_{1,1}\approx1.46\times10^{-18}\ \mathrm{J}}. \]

8.10

The state-counting result in 8.9 can be written as

\[ N(E)\approx\frac{4\pi m_eA}{h^2}E. \]

The density of available states is therefore proportional to the sheet area. A larger sheet has much more closely spaced levels near the Fermi energy and a smaller characteristic gap between occupied and unoccupied levels.

Although the Fermi energy remains approximately constant at fixed electron density, the level spacing decreases as the sheet grows. Under the inverse relationship between gap and conductivity specified in the question, the smaller sheet has lower conductivity:

\[ \boxed{ \sigma_{25\,\mathrm{nm}\times25\,\mathrm{nm}} < \sigma_{1\,\mathrm{m}\times1\,\mathrm{m}} }. \]

Circle less.

The solution was checked against the official solution.