Home
Class 12
CHEMISTRY
Calculate the electronegativity of fluor...

Calculate the electronegativity of fluorine from following data :
`E_(H - H) = 104.2` kcal `mol^(-1)`
`E_(F - F) = 36.6` kcal `mol^(-1)`
`E_(H -F) = 134.6` kcal `mol^(-1)`
Electronegativity of H is 2.05.

Text Solution

AI Generated Solution

To calculate the electronegativity of fluorine (denoted as \( X_F \)) using the provided bond enthalpy data and the electronegativity of hydrogen, we can apply the Pauling equation. Here are the steps to derive the electronegativity of fluorine: ### Step-by-Step Solution: 1. **Identify the Given Data:** - Bond enthalpy of \( H-H \) ( \( E_{H-H} \) ) = 104.2 kcal/mol - Bond enthalpy of \( F-F \) ( \( E_{F-F} \) ) = 36.6 kcal/mol - Bond enthalpy of \( H-F \) ( \( E_{H-F} \) ) = 134.6 kcal/mol ...
Promotional Banner

Similar Questions

Explore conceptually related problems

Calculate the electronegativity of fluorine from the following data: E_(H-H) = 104.4 kcal mol^(-1), E_(F-F) = 36.6 kcal mol^(-1) E_(H-F) = 134.3 kcal mol^(-1), chi_(H) = 2.1

Calculate the electronegativity of fluorine form the following data : E_(H-H) = 104.2 kcal "mol"^(-1), E_(C-C) =83.1 kcal "mol"^(-1) . E_(C-H) =9.8.s kcal "mo"l^(-1) Electronegativity of H=2.1 .

The electronegativity of carbon from the following data is nearly E_(H - H) = 104.2 "kcal" mol^(-1) , E_(C-C) = 83.1 kcal mol^(-1) , E_(C-H) = 98.8 kcal mol^(-1) , X_(H) = 2.1

Calculate electronegativity of carbon at Pauling scale Given that : E_(H-H) =104 .2 kcal "mol"^(-1) E_(C-C) =83.1 kcal "mol"^(-1) , E_(C-H) =98.8 kcal "mol"^(-1) . Electronegativity of hydrogen =2.1 .

Find the electron affinity of chlorine from the following data. Enthalpy of formation of LiCI is -97.5 kcal mol^(-1) , lattice energy of LiCI =- 197.7 kcal mol^(-1) . Dissociation energy of chlirine is 57.6 kcal mol^(-1) , sublimation enthalpy of lithium =+ 38.3 kcal mol^(-1) , ionisation energy of lithium =123.8 kcal mol^(-1) .

Calculate the lattice energy from the following data (given 1 eV = 23.0 kcal mol^(-1) ) i. Delta_(f) H^(ɵ) (KI) = -78.0 kcal mol^(-1) ii. IE_(1) of K = 4.0 eV iii. Delta_("diss")H^(ɵ)(I_(2)) = 28.0 kcal mol^(-1) iv. Delta_("sub")H^(ɵ)(K) = 20.0 kcal mol^(-1) ltbvrgt v. EA of I = -70.0 kcal mol^(-1) vi. Delta_("sub")H^(ɵ) of I_(2) = 14.0 kcal mol^(-1)