In \(\mathrm{CH}_{3} \mathrm{I}\) the \(\mathrm{C}\) - I bond-dissociation energy is \(241 \mathrm{~kJ} / \mathrm{mol}\). In \(\mathrm{C}_{6} \mathrm{H}_{5}\) I the \(\mathrm{C}-\) I bond-dissociation energy is \(280 \mathrm{~kJ} / \mathrm{mol}\). What is the range of wavelengths of photons that can cause \(\mathrm{C}-\mathrm{I}\) bond rupture in one molecule but not in the other?

First, we need to convert the given bond dissociation energies from kJ/mol to J/molecule. We will use Avogadro's number (6.022 x 10^23) for this purpose. For CH3I: \( 241 \ \frac{kJ}{mol} * \frac{10^3 J}{1 kJ} * \frac{1 mol}{6.022\times10^{23} \ molecules} \) For C6H5I: \( 280 \ \frac{kJ}{mol} * \frac{10^3 J}{1 kJ} * \frac{1 mol}{6.022\times10^{23} \ molecules} \)

Now that we have the energy of the bonds, we can find the photon energy responsible for breaking the bond. This can be done using Planck's equation: \( E = h\nu \) Where h is Planck's constant, \(6.626\times10^{-34} \ Js\), and ν is the frequency of the photon. Since the frequency (ν) and the wavelength (λ) are related by the speed of light (c) as follows: \( \nu = \frac{c}{\lambda} \) We can rewrite the Planck's equation as: \( E = hc \times \frac{1}{\lambda} \) Now, we can find the wavelength (λ) for each molecule by rearranging the equation: \( \lambda = \frac{hc}{E} \) Substitute the energy and Planck's constant values for both molecules: For CH3I: \( \lambda_{CH_{3}I} = \frac{6.626 \times 10^{-34} Js \cdot 3\times10^8 \frac{m}{s}}{241 \times 10^3 \frac{J}{mole} \times \frac{1 mole}{6.022\times10^{23} \ molecules}} \) For C6H5I: \( \lambda_{C_{6}H_{5}I} = \frac{6.626 \times 10^{-34} Js \cdot 3\times10^8 \frac{m}{s}}{280 \times 10^3 \frac{J}{mole} \times \frac{1 mole}{6.022\times10^{23} \ molecules}} \)

Now that we have found the wavelengths that cause the bonds to rupture, we can find the range of wavelengths that can cause bond rupture in one molecule but not in the other. The range of wavelengths is between the wavelengths calculated for CH3I and C6H5I: \( \lambda_{range} = (\lambda_{CH_{3}I}, \lambda_{C_{6}H_{5}I}) \) This range represents the photons that can cause bond rupture in one molecule but not in the other. Make sure to compare the values obtained for each molecule and order them accordingly to find the correct wavelength range.