Question

From the following data: $E_{H-H}=104.2kcalmo{l}^{-1}$; $E_{C-C}=83.1kcalmo{l}^{-1}$; $E_{C-H}=98.8kcalmo{l}^{-1}$.
The electronegativity of carbon is $128.5x\times {10}^{-2}$. Then, the value of $x$ is: (up to nearest integer)

Answer 1

According to Pauling's scale of electronegativity, the difference between heteronuclear bond ($A-B$) and arithmetic mean of homonuclear bond ($A-A$ and $B-B$), is related to the electronegativity difference between $A$ (${\chi }_A$) and $B$ (${\chi }_B$) as:

$E_{A-B}=\frac{1}{2}(E_{A-A}+E_{B-B})+{\Delta }_{AB}or{E}_{C-H}=\frac{1}{2}(E_{C-C}+E_{H-H})+{\Delta }_{C-H}$

$and{\chi }_C-{\chi }_H=0.208\sqrt{{\Delta }_{C-H}}for{\chi }_H=2.1\left(arbitraryvaluegivenbyPauling\right){\therefore \chi }_C=2.1+0.208\sqrt{{\Delta }_{C-H}}$

Upon substitution we get:

$E_{C-H}=98.9kcal/mol{E}_{C-C}=83.1kcal/mol,E_{H-H}=104.2kcal/mol$

${\Delta }_{C-H}=5.15{\therefore \chi }_C=257\times {10}^{-2}$

Therefore $x=2$ is the answer.