Question

The electronegativity of fluorine obtained from the following data is x.
$\begin{aligned} & \text{ }{{\text{E}}_{\text{H}-\text{H }}}=104.2\text{ kcal mo}{{\text{l}}^{-1}}\text{ } \\\ & \text{ }{{\text{E}}_{\text{F}-\text{F }}}=36.6\text{ kcal mo}{{\text{l}}^{-1}}\text{ } \\\ & \text{ }{{\text{E}}_{\text{H}-\text{F }}}=134.6\text{ kcal mo}{{\text{l}}^{-1}}\text{ } \\\ \end{aligned}$
Electronegativity of the hydrogen atom is $\text{ = 2}\text{.1 }$
Find the value of x to the nearest integer.

Answer 1

The electronegativity is a tendency of an atom to attract the bonding pair electrons. According to Pauling, the electronegativity between the covalent bond between two different atoms $\text{ }\left( \text{A}-\text{B} \right)\text{ }$ is found to be stronger than the average of $\text{ }\left( \text{A}-\text{A} \right)\text{ }$ and $\text{ }\left( \text{B}-\text{B} \right)\text{ }$bonds. The difference in electronegativity between atom A and B is given as:
$\text{ }{{\text{ }\\!\\!\chi\\!\\!\text{ }}_{\text{A}}}-{{\text{ }\\!\\!\chi\\!\\!\text{ }}_{\text{B}}}\text{ = }{{\left( e\text{V} \right)}^{-{1}/{2}\;}}\sqrt{{{\text{E}}_{\text{A}-\text{B}}}-\dfrac{\left( {{\text{E}}_{\text{A}-\text{A}}}\text{+}{{\text{E}}_{\text{B}-\text{B}}} \right)}{2}}\text{ }$

Complete Solution :
Electronegativity is defined as the tendency of an element of an atom to accept the electrons to itself when combined in a compound. In other words, it is a property of an atom to attract the bonding pair electrons.
The electronegativity is expressed in terms of $\text{ }\\!\\!\chi\\!\\!\text{ }$ .
The electronegativity of fluorine atoms can be determined through the electronegativity of $\text{ H}-\text{H }$ $\text{F}-\text{F }$ and $\text{H}-\text{F }$ bond.
We are provided with the following data:
$\begin{aligned} & \text{ }{{\text{E}}_{\text{H}-\text{H }}}=104.2\text{ kcal mo}{{\text{l}}^{-1}}\text{ } \\\ & \text{ }{{\text{E}}_{\text{F}-\text{F }}}=36.6\text{ kcal mo}{{\text{l}}^{-1}}\text{ } \\\ & \text{ }{{\text{E}}_{\text{H}-\text{F }}}=134.6\text{ kcal mo}{{\text{l}}^{-1}}\text{ } \\\ \end{aligned}$
Electronegativity of the hydrogen atom is $\text{ }{{\text{ }\\!\\!\chi\\!\\!\text{ }}_{\text{H}}}\text{= 2}\text{.1 }$ .
We are interested to determine the electronegativity of the fluorine atom.
The electronegativity of a $\text{H}-\text{F }$ bond is calculated as follows:
$\text{ }{{\text{ }\\!\\!\chi\\!\\!\text{ }}_{\text{H}}}-{{\text{ }\\!\\!\chi\\!\\!\text{ }}_{\text{F}}}\text{ = 0}\text{.208}\sqrt{{{\text{E}}_{\text{H}-\text{F}}}-\dfrac{\left( {{\text{E}}_{\text{H}-\text{H}}}\text{+}{{\text{E}}_{\text{F}-\text{F}}} \right)}{2}}\text{ }$ (1)
Here, $\text{ }{{\left( e\text{V} \right)}^{-{1}/{2}\;}}\text{ = 0}\text{.208 }$ .
Let’s substitute the values of electronegativity in equation (1) We have,
$\begin{aligned} & \text{ 2}\text{.1}-{{\text{ }\\!\\!\chi\\!\\!\text{ }}_{\text{F}}}\text{ = 0}\text{.208}\sqrt{134.6-\dfrac{\left( \text{104}\text{.2+36}\text{.6} \right)}{2}}\text{ } \\\ & \Rightarrow \text{ 2}\text{.1}-{{\text{ }\\!\\!\chi\\!\\!\text{ }}_{\text{F}}}\text{ = 0}\text{.208}\sqrt{134.6-\text{70}\text{.4}}\text{ } \\\ & \Rightarrow \text{2}\text{.1}-{{\text{ }\\!\\!\chi\\!\\!\text{ }}_{\text{F}}}\text{ = 0}\text{.208}\sqrt{64.2}\text{ } \\\ & \Rightarrow \text{2}\text{.1}-{{\text{ }\\!\\!\chi\\!\\!\text{ }}_{\text{F}}}\text{ = 0}\text{.208}\times \left( 8.012 \right) \\\ & \Rightarrow {{\text{ }\\!\\!\chi\\!\\!\text{ }}_{\text{F}}}\text{ = 2}\text{.1+1}\text{.66 = 3}\text{.76 = x } \\\ \end{aligned}$
Therefore, the electronegativity of the fluorine atom is $\text{ 3}\text{.76 }$ .

Note: Note that, Pauling’s formula is an approximate method to determine the electronegativity of the bond but it gives the right intuition about the polarity of the bond. If atoms have similar electronegativities they have some tendency to attract electrons and it is a covalent bond. However, the large difference in electronegativity leads to a higher degree of polar character making it an ionic bond. Here the $\text{H}-\text{F }$ bond is more polar.