The first ionisation potential of `Na` is `5.1eV`. The value of electron gain enthalpy of `Na^(+)` will be

Options

A. `-2.55 eV`
B. `-5.1 eV`
C. `-10.2 eV`
D. `+2.55 eV`

Answer

B

Explanation

`{:(Na rarr Na^(+) +e^(-),,I^(st)I.E. = 5.1 eV),(Na^(+) rarr+e^(-) rarr Na,,"Electron gain enthalpy of" Na^(+)):}`
Because reaction is reverse, so `Delta_(eg)H =- 5.1 eV`.