Alcohol-based fuels for automobiles lead to the production of formaldehyde \(\left(\mathrm{CH}_{2} \mathrm{O}\right)\) in exhaust gases. Formaldehyde undergoes photodissociation, which contributes to photochemical smog: $$ \mathrm{CH}_{2} \mathrm{O}+h \nu \longrightarrow \mathrm{CHO}+\mathrm{H} $$ The maximum wavelength of light that can cause this reaction is \(335 \mathrm{nm}\). (a) In what part of the electromagnetic spectrum is light with this wavelength found? (b) What is the maximum strength of a bond, in \(\mathrm{kJ} / \mathrm{mol}\), that can be broken by absorption of a photon of \(335-\mathrm{nm}\) light? (c) Compare your answer from part (b) to the appropriate value from Table 8.4 . What do you conclude about \(\mathrm{C}-\mathrm{H}\) bond energy in formaldehyde? (d) Write out the formaldehyde photodissociation reaction, showing Lewis-dot structures.

To determine the region of the electromagnetic spectrum in which the given wavelength (335 nm) is found, we can refer to the wavelength ranges for each region. Ultraviolet (UV) light has a wavelength range from about 10 nm to 400 nm. Therefore, the light with a wavelength of 335 nm falls within the ultraviolet region of the electromagnetic spectrum.

To calculate the maximum bond strength that can be broken by absorption of a photon of 335 nm light, first, we need to determine the energy of a single photon (E_photon) using the following equation: \(E_{\text {photon}}=\frac{hc}{\lambda}\), where h is Planck's constant (\(6.626 \times 10^{-34} \ \text{Js}\)), c is the speed of light (\(3.00 \times 10^{8} \ \text{m/s}\)), and λ is the wavelength in meters. Convert the wavelength to meters: \(\lambda = 335 \ \text{nm} \times \frac{1 \ \text{m}}{10^{9} \ \text{nm}} = 3.35 \times 10^{-7} \ \text{m}\). Now we can calculate the energy of one photon: \(E_{\text{photon}} = \frac{6.626 \times 10^{-34} \ \text{Js} \times 3.00 \times 10^{8} \ \text{m/s}}{3.35 \times 10^{-7} \ \text{m}} = 5.95 \times 10^{-19} \ \text{J}\). To find the energy per mole of photons, we'll multiply the energy of a single photon by Avogadro's number (\(6.022 \times 10^{23} \ \text{mol}^{-1}\)): \(E_{\text{photon/mol}} = 5.95 \times 10^{-19} \ \text{J} \times 6.022 \times 10^{23} \ \text{mol}^{-1} = 3.58 \times 10^{5} \ \text{J/mol}\). Finally, convert the energy to kJ/mol: \(E_{\text{photon/mol}} = 3.58 \times 10^{5} \ \text{J/mol} \times \frac{1 \ \text{kJ}}{1000 \ \text{J}} = 358 \ \text{kJ/mol}\). So, the maximum strength of a bond that can be broken by absorption of a photon of 335 nm light is 358 kJ/mol.

Unfortunately, without access to Table 8.4, I cannot complete part (c) of this exercise. However, you are instructed to compare the calculated bond strength (358 kJ/mol) to the appropriate value from Table 8.4, which should give insight into the C-H bond energy in formaldehyde. If the value is close to or equal to the C-H bond energy in the table, then the bond can be broken by the photons of that wavelength.

The formaldehyde photodissociation reaction is as follows: \( \begin{array}{ccc} \text{H} & \text{-} & \text{C}=\text{O}\\ & & |\phantom{l}\\ & & \text{H} \end{array} + h \nu \longrightarrow \begin{array}{cc} \text{H} & \text{-} \text{C}=\text{O}\\ \end{array} + \text{H} \) Where the left structure is formaldehyde and the right structures are the products after photodissociation, CHO radical, and H atom.