Question

The first ionisation potential of `Na` is `5.1eV`. The value of eectrons gain enthalpy of `Na^(+)` will be
A. `-2.55eV`
B. `-2.1 eV`
C. `-10.2 eV`
D. `+2.55 eV`

Answer 1

Correct Answer - B::C::D
`Na to Na^(+)+e^(-)` First IE
`Na^(+)+e^(-) to Na`
Electron gain enthalopy of `Na^(+)` is reverse of (IE)
Because reaction is reverse so
`/_H(eq)=-5.1 eV`