(a) Draw the Lewis structure for hydrogen peroxide, $\mathrm{H}_{2} \mathrm{O}_{2}$. (b) What is the weakest bond in hydrogen peroxide? (c) Hydrogen peroxide is sold commercially as an aqueous solution in brown bottles to protect it from light. Calculate the longest wavelength of light that has sufficient energy to break the weakest bond in hydrogen peroxide.

To draw the Lewis structure for H2O2, first count the total number of valence electrons. Oxygen has 6 valence electrons and hydrogen has 1 valence electron. Since there are 2 oxygen atoms and 2 hydrogen atoms, the total number of valence electrons is 2(6) + 2(1) = 14. Now, arrange the electrons around the oxygen and hydrogen atoms to satisfy the octet rule. The oxygen atom forms a single bond with one hydrogen atom and another oxygen atom. The remaining oxygen atom forms a single bond with the other hydrogen atom. The Lewis structure for H2O2 is: ``` H O O H \ || || / ```

To determine the weakest bond, we need to consider the bond dissociation enthalpy (BDE) for each bond. Bonds with a higher BDE are stronger and require more energy to break. The bond dissociation enthalpies for the O-H and O-O bonds in hydrogen peroxide are: - O-H: 467 kJ/mol - O-O: 207 kJ/mol From these values, we can conclude that the O-O bond is the weakest bond in hydrogen peroxide.

Having identified the weakest bond with a BDE of 207 kJ/mol, we can calculate the energy required to break the bond. To convert this value to J/mol, use the conversion factor: \(207~\text{kJ/mol} * \left(\frac{1000~\text{J}}{1~\text{kJ}}\right) = 207000~\text{J/mol}\) Next, use the Planck's equation to determine the energy of a photon: \(E = h\nu\), where \(E\) is the energy of a photon, \(h\) is Planck's constant (\(6.626 * 10^{-34}~\text{J} \cdot \text{s}\)), and \(\nu\) is the frequency of the light. We can also relate the frequency to the wavelength using the speed of light equation, \(c = \lambda\nu\), where \(c\) is the speed of light (\(3 * 10^8~\text{m/s}\)) and \(\lambda\) is the wavelength. Substituting the wavelength equation into the energy equation, we get: \(E = \cfrac{hc}{\lambda}\) We want to calculate the longest wavelength (\(\lambda\)) of light capable of breaking the weakest bond with an energy of 207000 J/mol. Rearrange the equation and solve for the wavelength: \(\lambda = \cfrac{hc}{E}\) Plugging in the values, we have: \(\lambda = \cfrac{(6.626 * 10^{-34}~\text{J} \cdot \text{s}) (3 * 10^8~\text{m/s})}{207000~\text{J/mol}}\) Now, we should convert the energy from J/mol to J/photon to have the same units. To do this, we utilize Avogadro's number which is \(6.022 * 10^{23}~\text{mol}^{-1}\): \(\cfrac{207000~\text{J/mol}}{6.022 * 10^{23}~\text{mol}^{-1}} = 3.45 * 10^{-19}~\text{J/photon}\) Finally, we can solve for the wavelength: \(\lambda = \cfrac{(6.626 * 10^{-34}~\text{J} \cdot \text{s}) (3 * 10^8~\text{m/s})}{3.45 * 10^{-19}~\text{J/photon}} = 5.73 * 10^{-6}~\text{m}\) The longest wavelength of light that has sufficient energy to break the weakest bond in hydrogen peroxide is approximately \(5.73 * 10^{-6}~\text{m}\) or 5730 nm.