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Calculate the electronegativity of fluor...

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`

Text Solution

AI Generated Solution

To calculate the electronegativity of fluorine (X_F) using the given bond energies and the electronegativity of hydrogen (X_H), we will use the formula for electronegativity difference: \[ X_A - X_B = 0.208 \times \left( E_{AB} - \frac{1}{2}(E_{AA} + E_{BB}) \right)^{1/2} \] Where: - \( E_{AB} \) is the bond energy of the AB bond (H-F bond in this case). - \( E_{AA} \) is the bond energy of the AA bond (H-H bond). - \( E_{BB} \) is the bond energy of the BB bond (F-F bond). ...
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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 .

Calculate the enthalpy of formation of ammonia from the following bond energy data: (N-H) bond = 389 kJ mol^(-1), (H-H) bond = 435 kJ mol^(-1) , and (N-=N)bond = 945.36kJ mol^(-1) .

Knowledge Check

  • The enthalpy of formation of ammonia when calculated from the following bond energy data is (B.E. of N-H, H-H, N-=N is "389 kJ mol"^(-1), "435 kJ mol"^(-1),"945.36 kJ mol"^(-1) respectively)

    A
    `-"41.82 kJ mol"^(-1)`
    B
    `+"83.64 kJ mol"^(-1)`
    C
    `-"945.36 kJ mol"^(-1)`
    D
    `-"833 kJ mol"^(-1)`

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