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An imaginary lattice is formed as given...

An imaginary lattice is formed as given
`Cl_(2) rarr 2Cl" "DeltaH = x_(1) " "Cl rarr Cl^(+) + e^(-) " "DeltaH = x_(2)`
`Cl + e^(-) rarr Cl^(-) " "Delta H = x_(3)" "Cl^(+) + Cl^(-) rarr Cl^(+)Cl^(-) " "Delta H = x_(4)`
Thus , enthalpy change when `Cl^(+)Cl^(-)` is formed is:

A

`(+(x_1)/2 + x_2 + x_3 + x_4)`

B

`(-(x_1)/2 + x_2 + x_3 + x_4)`

C

`(x_1 + x_2 - x_3 - x_4)`

D

`-(x_1 + x_2 - x_3 - x_4)`

Text Solution

Verified by Experts

The correct Answer is:
C
Adding all equation, we get
`x_1 + x_2 - x_3 - x_4 =`Enthalpy change when `Cl^(+)Cl^(-)` is formed.
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