An important reaction in the formation of photochemical smog is the photodissociation of \(\mathrm{NO}_{2}\) : $$ \mathrm{NO}_{2}+h \nu \longrightarrow \mathrm{NO}(g)+\mathrm{O}(g) $$ The maximum wavelength of light that can cause this reaction is $420 \mathrm{nm} .$ (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 420 -nm light? \(?\) (c) Write out the photodissociation reaction showing Lewis-dot structures.

First, we need to know the ranges of the electromagnetic spectrum to see where 420 nm falls. The electromagnetic spectrum can be classified into seven categories according to their wavelengths: 1. Radio waves: wavelength (> 1 m) 2. Microwaves: wavelength (1 mm - 1 m) 3. Infrared: wavelength (700 nm - 1 mm) 4. Visible light: wavelength (400 nm - 700 nm) 5. Ultraviolet: wavelength (10 nm - 400 nm) 6. X-rays: wavelength (0.01 nm - 10 nm) 7. Gamma rays: wavelength (< 0.01 nm) Given that our wavelength is 420 nm, it falls within the visible light range of the electromagnetic spectrum.

To calculate the maximum bond strength, we will use the Planck-Einstein relation and the concept of energy required to break a bond: \(E = h \cdot \nu\) Where: E = energy of the photon (Joules) h = Planck's constant (\(6.626 \times 10^{-34}\) Js) \(\nu\) = frequency (Hz) We can relate frequency (\(\nu\)) and wavelength (\(\lambda\)) with the speed of light (c) using the following equation: \(\nu = \frac{c}{\lambda}\) Now, we can substitute and solve for E: \(E = h \cdot \frac{c}{\lambda}\) To obtain energy per mole of photons, we can multiply E by the Avogadro's number (NA) and convert the energy to kJ/mol. \(E_{mol} = E \cdot N_A\)

1. Convert the wavelength to meters: \(\lambda = 420 \cdot 10^{-9} m\) 2. Calculate the frequency: \(\nu = \frac{3 \times 10^8 \, \text{m/s}}{420 \cdot 10^{-9}\, \text{m}} = 7.14 \times 10^{14} \, \text{Hz}\) 3. Calculate the energy of one photon: \(E = (6.626 \times 10^{-34} \, \text{Js}) \cdot (7.14 \times 10^{14} \, \text{Hz}) = 4.73 \times 10^{-19} \, \text{J}\) 4. Calculate the energy per mole: \(E_{mol} = (4.73 \times 10^{-19}\, \text{J}) \cdot (6.022 \times 10^{23}\, \text{mol}^{-1}) = 2.85 \times 10^5 \, \text{J/mol}\) 5. Convert the energy to kJ/mol: \(E_{mol} = 2.85 \times 10^5 \, \text{J/mol} \cdot \frac{1\, \text{kJ}}{1000\, \text{J}} = 285\, \text{kJ/mol}\) Thus, the maximum strength of a bond that can be broken by the absorption of a photon of 420-nm light is 285 kJ/mol.

To write the photodissociation reaction using Lewis-dot structures, we need to demonstrate the electron distribution in each molecule of the reaction: NO₂: :N═O: , Oxygen is double bonded to Nitrogen and one unpaired electron on Nitrogen. NO: :N≡O: , Oxygen is triple bonded to Nitrogen. O: ·O , Oxygen has one unpaired electron. Now, we can write the photodissociation reaction using the Lewis-dot structures: :NO₂ + hν -> :N≡O + ·O