Question

The energy required to break one mole of $\text{ Cl}-\text{Cl }$bonds in the $\text{ C}{{\text{l}}_{\text{2}}}\text{ }$ is $\text{ 242 KJ mo}{{\text{l}}^{-1\text{ }}}\ $. The longest wavelength of light capable of breaking a single $\text{ Cl}-\text{Cl }$bond is:
A) $\text{ 594 nm }$
B) $\text{ 640 nm }$
C) $\text{ 700 nm }$
D) $\text{ 494 nm }$

Answer 1

Hint: The energy required to break a bond is known as the bond dissociation energy. We know that the energy of the radiation is related to the wavelength of the radiation for one mole of the molecule. The relation is stated as,
$\text{ E = }\dfrac{{{\text{N}}_{\text{A}}}\text{ h C}}{\lambda }\text{ }$
Where E is the bond dissociation energy, $\text{ }{{\text{N}}_{\text{A}}}\text{ }$ is the Avogadro's number, h is the Planck's constant $\text{ 6}\text{.626 }\times \text{ 10}-34\text{ J }{{\text{s}}^{-1}}\text{ }$ , c is the speed of light $\text{ 3}\times \text{1}{{\text{0}}^{\text{8}}}\text{ m }{{\text{s}}^{-1}}\text{ }$ and $\text{ }\lambda \text{ }$ is the wavelength of radiation required to break the bond.

Complete step by step solution:
Bond dissociation energy is the amount of energy required to break a chemical bond between A and B $\text{ A}-\text{B }$ . The bond dissociation energy depends on the extent of the bond formed between atoms A and B.
Here we have given the following data,
The energy required to break one mole of chlorine molecule is,$\text{ 242 KJ mo}{{\text{l}}^{-1\text{ }}}\ $
We are interested to find the longest wavelength of light capable of breaking a single $\text{ Cl}-\text{Cl }$bond.
We know that the energy of the radiation is related to the wavelengths of the radiation. The relation is stated as,
$\text{ E = h}\upsilon \text{ = }\dfrac{\text{ hC}}{\lambda }\text{ }$
However, we have considered the bond dissociation energy for the one-mole chlorine molecule. Thus the number of chlorine molecules in one mole is always equal to Avogadro's number $\text{ }{{\text{N}}_{\text{A}}}\text{ = 6}\text{.023 }\times \text{ 1}{{\text{0}}^{\text{23}}}\text{ mo}{{\text{l}}^{-1}}\text{ }$
Thus the energy relation is modified as,
$\text{ E = }\dfrac{{{\text{N}}_{\text{A}}}\text{ h C}}{\lambda }\text{ }$ (1)
Where E is the bond dissociation energy, $\text{ }{{\text{N}}_{\text{A}}}\text{ }$ is the Avogadro's number, h is the Planck's constant $\text{ 6}\text{.626 }\times \text{ 10}-34\text{ J }{{\text{s}}^{-1}}\text{ }$ , c is the speed of light $\text{ 3}\times \text{1}{{\text{0}}^{\text{8}}}\text{ m }{{\text{s}}^{-1}}\text{ }$ and $\text{ }\lambda \text{ }$ is the wavelength of radiation required to break the bond.
Let's substitute the values in the equation (1), we have
$\text{ 242 KJ mo}{{\text{l}}^{-1\text{ }}}\text{ = }\dfrac{\text{(6}\text{.023 }\times \text{1}{{\text{0}}^{\text{23}}}\text{ mo}{{\text{l}}^{-1}}\text{)}\left( 6.626\times {{10}^{-34}}\text{ J}{{\text{s}}^{-1}} \right)\text{ }\left( 3\times {{10}^{8}}\text{ m }{{\text{s}}^{-1}} \right)}{\lambda }\text{ }$
Rearrange equation with respect to wavelength. We have,
$\begin{align}
& \text{ }\lambda \text{ = }\dfrac{\text{(6}\text{.023 }\times \text{1}{{\text{0}}^{\text{23}}}\text{ mo}{{\text{l}}^{-1}}\text{)}\left( 6.626\times {{10}^{-34}}\text{ J}{{\text{s}}^{-1}} \right)\text{ }\left( 3\times {{10}^{8}}\text{ m }{{\text{s}}^{-1}} \right)}{\text{242 }\times \text{1}{{\text{0}}^{\text{3}}}\text{ mo}{{\text{l}}^{-1\text{ }}}}\text{ } \\
& \Rightarrow \text{ }\lambda \text{ = }\dfrac{\left( 1.19\times {{10}^{-1}} \right)}{\text{242 }\times \text{1}{{\text{0}}^{\text{3}}}\text{ }}\text{J} \\
& \Rightarrow \text{ }\lambda \text{ = 4}\text{.94 }\times \text{ 1}{{\text{0}}^{-7}}\text{m} \\
\end{align}$
The obtained value of the wavelength can be converted into nanometres. The relation between the one nanometre with the meter is,
$\text{ 1 nm = 1}{{\text{0}}^{-9\text{ }}}\text{nm }$
Therefore,
$\text{ }\lambda \text{ = 494 }\times \text{ 1}{{\text{0}}^{-9}}\text{m = 494 nm }$
Thus, the longest wavelength required to break a single $\text{ Cl}-\text{Cl }$is equal to $\text{ 494 nm }$ .

Hence, (D) is the correct option.

Note: Note that the wavelength is always calculated in the standard format, where the meter is converted into the suitable and required form. Some of the most used units of wavelengths are as follows,

	Unit	In meter
1.	Centimetre $\text{ }\left( \text{cm} \right)\text{ }$	$\text{ 1}{{\text{0}}^{-2}}\text{ }$
2.	millimetre$\text{ }\left( \text{mm} \right)\text{ }$	$\text{ 1}{{\text{0}}^{-3}}\text{ }$
3.	micrometre$\text{ }\left( \mu \text{m} \right)\text{ }$	$\text{ 1}{{\text{0}}^{-6}}\text{ }$
4	nanometres$\text{ }\left( \text{nm} \right)\text{ }$	$\text{ 1}{{\text{0}}^{-9}}\text{ }$
5.	Angstrom$\text{ }\left( \overset{\text{0}}{\mathop{\text{A}}}\, \right)\text{ }$	$\text{ 1}{{\text{0}}^{-10}}\text{ }$
6.	Picometer$\text{ }\left( \text{pm} \right)\text{ }$	$\text{ 1}{{\text{0}}^{-12}}\text{ }$.