Calculate the electronegativity of fluorine using the following data.
` E_(H-H) = 104.2 kcal "mol"^(-1) , E_(F-F) = 36.6 kcal "mol"^(-1) , E_(H-F) = 144.6 kcal "mol"^(-1)` and the electronegativity of H = 2.1.

Options

A. 2.53
B. 3.87
C. 4
D. 4.2

Answer

B

Explanation

`Delta = X_A - X_B= 0`
`.208 sqrt(E_(A-B) - sqrt(E_(A-A) xxE_(B-B)) )`
where,
`E_(A-B)` =bond enthalpy or bond energy of the A - B bond in kcal `"mol"^(-1)`
`E_(A-A)` =bond enthalpy or bond energy of the A - A bond in kcal `"mol"^(-1)` ,
`E_(B-B)` =bond enthalpy or bond energy of the B - B bond in kcal `"mol"^(-1)` .
` X_F - X_H =0 `
`.208 sqrt(134.6 - sqrt(104.2 xx 36.6))`
`X_F - 2.1 = 1.77`
`X_F = 2.1 + 1.77 = 3.87`