Question

1) If H-H , X-X and H-X bond energies are 104 Kcal/mole, 60 Kcal/mole, 101 Kcal/mole . Assuming the electronegative try of hydrogen is 2.1 the electronegative of unknown element X is ?

Answer 1

$$\begin{gathered} Actualbondenergy∆=E_{H-X}-\sqrt{E_{H-H}\times E_{X-X}} \\ =101-\sqrt{104\times 60} \\ =22.01Kcal/mole \\ Electronegativitydifferenceis \\ {X}_H-X_X=0.208\sqrt{∆} \\ 2.1-X_X=0.208\times \sqrt{22.01} \\ {X}_X=1.12 \end{gathered}$$
Electronegativity of unknown atom = 1.12