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Ionisation of energy F^(ɵ) is 320 kJ mol...

Ionisation of energy `F^(ɵ)` is `320 kJ mol^(-1)`. The electronic gain enthalpy of fluorine would be

A

`-320 kJ mol^(-1)`

B

`-160 kJ mol^(-1)`

C

`+320 kJ mol^(-1)`

D

`+160 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`IE` and `EA` of an element are equal in magnitude but opposite in sign.
`:. IE = -EA`
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Ionisation energy of F is 320 kJ mol^(-1) . The electron affinity of fluorine would be:

Use the following data to calculate Delta_("lattice") H^(Θ) " for " NaBr. Delta_("Sub")H^(Θ) for sodium metal = 108.4 kJ mol^(-1) , ionisation enthalpy of sodium = 496 kJ mol^(-1) , electron gain enthalpy of bromine = -325 kJ mol^(-1) , bond dissociation enthalpy of bromine = 192 kJ mol^(-1), Delta_(f) H^(Θ) " for " NaBr(s) = - 360.1 kJ mol^(-1)

Knowledge Check

  • Ionisation energy of F^(-) is +320 mol^(-1) . The electron gain enthalpy of fluorine atom would be

    A
    `-320KJ mol^(-1)`
    B
    `-160KJ mol^(-1)`
    C
    `+320KJ mol^(-1)`
    D
    `+160KJ mol^(-1)`

  • Given Based on data provided, the value of electron gain enthalpy of fluorine would be :

    A
    ` - 300 kJ "mol"^(-1)`
    B
    ` - 350 kJ "mol"^(-1)`
    C
    ` - 328 kJ "mol"^(-1)`
    D
    ` - 228kJ "mol"^(-1)`

  • Given Based on data provided, the value of electron gain enthalpy of fluorine would be

    A
    `-300 kJ mol^(-1)`
    B
    `-328 kJ mol^(-1)`
    C
    `-350 kJ mol^(-1)`
    D
    `-228 kJ mol^(-1)`

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    Calculate the lattice enthalpy of KCl from the following data by Born- Haber's Cycle. Enthalpy of sublimation of K = 89 kJ mol^(–1) Enthalpy of dissociation of Cl = 244 kJ mol^(–1) Ionization enthalpy of potassium = 425 kJ mol^(–1) Electron gain enthalpy of chlorine = –355 kJ mol^(–1) Enthalpy of formation of KCl = –438 kJ mol^(-1)

    Use the following data to calculate Delta_("lattice") H^(@) for NaBr. Delta_("sub")H^(@) for sodium metal =108.4 kJ mol^(-1) , ionization enthalpy of sodium =496 kJ mol^(-1) ., electron gain enthalpy of bromine =-325 kJ mol^(-1) bond dissociation enthalpy of bromine =192 kJ mol^(-1) , Delta _f H^(@) for NaBr(s) - 360 kJ mol^(-1) .

    Use the following data to calculate Delta_("lattice")H^(@) for NaBr. Delta_("sub")H^(@) for sodium metal = 108.4 kJ "mol"^(-1) .Ionization enthalpy of sodium = 496 kJ "mol"^(-1) Electron gain enthalpy of bromine = -325 kJ "mol"^(-1) .Bond dissociation enthalpy of bromine = 192 kJ "mol"^(-1) . Delta_(f)^(H^(@)) for NaBr (s) = -360.1 kJ "mol"^(-1) .

    The ionization enthalpy of Na^+ formation from Nad is 495.8 kJ mol^(-1) , while the electron gain enthalpy of Bris -325.0 kJ mol^(-1) .Given the lattice enthalpy of NaBr is -728.4 kJ mol^(-1) The energy for the formation of NaBr ionic solid is (-)"______" xx 10^(-1)kJ "mol"^(-1)

    Given {:("Reaction","Energy Change (in KJ)",),(Li(s) rarr Li(g),161,),(Li(g) rarr Li^(+) (g),520,),((1)/(2) F_(2)(g) rarr F(g),77,),(F(g) + e^(-) rarr F(g),("Electron gain enthalpy"),),(Li^(+) (g) + F^(-)(g) rarr LiF(s),-1047,),(Li (s) + (1)/(2) f_(2) (g) rarr LiF(s),-617,):} Based on data provided, the value of electron gain enthalpy of fluorine would be :

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