Home
Class 11
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

Verified by Experts

The correct Answer is:
3.82
Suppose the electronegativity of H and F are
reprsented as `chi_(H) and chi_(F)` respectively . Appling
Pauling's formula
`|chi_(H)- chi_(F) |=0.208sqrt(Delta) `
`Delta= BE (H-F) - sqrt(BE(H-H)xxBE(F-F))`
`= 134.6 - sqrt(104.2xx36.6)`
`= 134.6 - 61.8` kcal mol `^(-1)`
72. 8 kcal mol `^(-1)`
`therefore |chi_(H)- chi_(F) |=0.208sqrt(72.8) = 1.77`
As `chi_(F)gt chi_(H),chi_(F) = 1.77 + chi_(H) = 1.77+2.1 = 3.82`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CHEMICAL BONDING AND MOLECULAR STRUCTURE

    PRADEEP|Exercise COMPETITION FOCUS JEE (Main and Advanced)/ MEDICAL ENTRANCE SPECIAL (VIII. ASSERTION-REASON TYPE QUESTIONS TYPE I)|30 Videos

  • CHEMICAL BONDING AND MOLECULAR STRUCTURE

    PRADEEP|Exercise ADVANCED PROBLEMS|11 Videos

  • CHEMICAL BONDING AND MOLECULAR STRUCTURE

    PRADEEP|Exercise COMPETITION FOCUS JEE (Main and Advanced)/ MEDICAL ENTRANCE SPECIAL (VI. INTEGER TYPE QUESTIONS)|15 Videos

  • APPENDIX

    PRADEEP|Exercise MODEL TEST PAPER <br> (Section C )|9 Videos

  • CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

    PRADEEP|Exercise Competition Focus (Jee Main and Advanced / Medical Entrance ) ( Assertion - Reason Type Question ) (Type II)|12 Videos

Similar Questions

Explore conceptually related problems

Calculate the electronegativity of fluorine from following data : F_(H-H) =104. 2 k cal mol ^(-1) E _(F-F) =36.6 kcal mol ^(-1) E _(H-F) = 134. 6 kcal mol ^(-1) Electronegativity of H is 2.05

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

Knowledge Check

  • 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.

    A
    2.53
    B
    3.87
    C
    4
    D
    4.2

  • 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

    A
    `3.0`
    B
    `2.1`
    C
    `2.5`
    D
    `3.1`

  • 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)

    A
    `+14.1 kcal mol^(-1)`
    B
    `-14.1 kcal mol^(-1)`
    C
    `-141 kcal mol^(-1)`
    D
    `+141 kcal mol^(-1)`

    • Similar Questions

    Explore conceptually related problems

    Calculate the electronegativity of fluorine from the following data: E_(H-H)=10.4.2 kcal mol^(-1) E_(F-F)= 36.6 kcal mol^(-1) E_(H-P)= 134.6 kcal mol^(-1) Electronegativity of H= 2.05

    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 .

    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) .

    From the following data, calculate the standard enthalpy of formation of propane Delta_(f)H^(Theta) CH_(4) = - 17 kcal mol^(-1) Delta_(f)H^(Theta)C_(2)H_(6) =- 24 kcal mol^(-1), BE (C-H) = 99 kcal mol^(-1) (C- C) = 84 kcal mol^(-1) .

    Home
    Profile