IChO 2003 · Theory problem 27

Hydrogen molecule

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Using the information provided on this graph, give numerical answers with appropriate units to the following questions:

Figure 1
  1. What are the equilibrium bond lengths of \(\ce{H2}\) and \(\ce{H2+}\)?
  2. What are the binding energies of \(\ce{H2}\) and \(\ce{H2+}\)?
  3. What is the ionisation energy of the \(\ce{H2}\) molecule?
  4. What is the ionisation energy of the \(\ce{H}\) atom?
  5. If we use electromagnetic radiation of frequency \(3.9\times10^{15}\ \mathrm{Hz}\) in order to ionise \(\ce{H2}\), what will be the velocity of the extracted electrons? (Ignore molecular vibrational energy.)

Official English text of the problem.

Answers

  1. Molecular hydrogen, \(\ce{H2}\) ([H][H]): \(r_e\approx0.75\ \text{Å}\). Dihydrogen cation, \(\ce{H2+}\) ([H][H+]): \(r_e\approx1.05\ \text{Å}\).
  2. \(D_e(\ce{H2})\approx460\ \mathrm{kJ\,mol^{-1}}\); \(D_e(\ce{H2+})\approx270\ \mathrm{kJ\,mol^{-1}}\).
  3. \(I(\ce{H2})\approx1500\ \mathrm{kJ\,mol^{-1}}\).
  4. \(I(\ce{H})\approx1310\ \mathrm{kJ\,mol^{-1}}\).
  5. Electron speed: \(v\approx4.5\times10^5\ \mathrm{m\,s^{-1}}\).

Solution

The approximate energy readings used below are:

Curve Minimum energy / \(\mathrm{kJ\,mol^{-1}}\) Energy at infinite separation / \(\mathrm{kJ\,mol^{-1}}\)
\(\ce{H2}\) \(-3080\) \(-2620\), corresponding to \(\ce{H + H}\)
\(\ce{H2+}\) \(-1580\) \(-1310\), corresponding to \(\ce{H + H+}\)

Here, \(\ce{H}\) is a hydrogen atom ([H]), and \(\ce{H+}\) is a proton ([H+]). The SMILES given for \(\ce{H2+}\) specifies connectivity and charge; it does not express its one-electron bond, whose bond order is \(1/2\).

1. Equilibrium bond lengths

The equilibrium separation is the position of the minimum of each potential-energy curve: shortening or lengthening the bond from this position increases its energy.

Reading the horizontal coordinates of the minima gives

\[ \boxed{r_e(\ce{H2})\approx0.75\ \text{Å}}, \qquad \boxed{r_e(\ce{H2+})\approx1.05\ \text{Å}}. \]

2. Binding energies

The binding energy is the positive energy required to separate the equilibrium molecule into the fragments shown at its dissociation limit.

For molecular hydrogen,

\[ \ce{H2 -> H + H}, \]

\[ D_e(\ce{H2}) =E(\ce{H + H})-E_{\min}(\ce{H2}) \approx-2620-(-3080) =\boxed{460\ \mathrm{kJ\,mol^{-1}}}. \]

For the dihydrogen cation,

\[ \ce{H2+ -> H + H+}, \]

\[ D_e(\ce{H2+}) =E(\ce{H + H+})-E_{\min}(\ce{H2+}) \approx-1310-(-1580) =\boxed{270\ \mathrm{kJ\,mol^{-1}}}. \]

3. Ionisation energy of the hydrogen molecule

The ionisation process is

\[ \ce{H2 -> H2+ + e-}. \]

Taking the molecule and molecular ion at their respective equilibrium geometries, and the removed electron at rest at infinity, the ionisation energy is the difference between the two curve minima:

\[ I(\ce{H2}) =E_{\min}(\ce{H2+})-E_{\min}(\ce{H2}) \approx-1580-(-3080) =\boxed{1500\ \mathrm{kJ\,mol^{-1}}}. \]

4. Ionisation energy of the hydrogen atom

Atomic ionisation is

\[ \ce{H -> H+ + e-}. \]

The two dissociation limits differ by precisely this process: one hydrogen atom remains unchanged while the other is ionised. Therefore,

\[ I(\ce{H}) =E(\ce{H + H+})-E(\ce{H + H}) \approx-1310-(-2620) =\boxed{1310\ \mathrm{kJ\,mol^{-1}}}. \]

5. Velocity of the extracted electrons

Photoionisation occurs according to

\[ \ce{H2 ->[h\nu] H2+ + e-}. \]

Ignoring molecular vibrational energy and the negligible kinetic energy of ion recoil, conservation of energy gives

\[ h\nu=\frac{I(\ce{H2})}{N_{\mathrm A}}+\frac12m_ev^2. \]

Using the standard constants

\[ h=6.626\times10^{-34}\ \mathrm{J\,s},\qquad N_{\mathrm A}=6.022\times10^{23}\ \mathrm{mol^{-1}},\qquad m_e=9.109\times10^{-31}\ \mathrm{kg}, \]

the photon energy is

\[ E_\gamma =(6.626\times10^{-34})(3.9\times10^{15}) =2.584\times10^{-18}\ \mathrm J. \]

The ionisation energy per molecule is

\[ E_{\mathrm{ion}} =\frac{1500\times10^3}{6.022\times10^{23}} =2.491\times10^{-18}\ \mathrm J. \]

Thus the electron receives kinetic energy

\[ K_e=E_\gamma-E_{\mathrm{ion}} \approx9.3\times10^{-20}\ \mathrm J, \]

and its speed is

\[ v=\sqrt{\frac{2K_e}{m_e}} \approx \sqrt{\frac{2(9.3\times10^{-20})}{9.109\times10^{-31}}} =\boxed{4.5\times10^5\ \mathrm{m\,s^{-1}}}. \]

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.