The first ionisation potential of `Na` is `5.1eV`. The value of electrons 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 to Na^(+) + e^(-) 1st I.E.`
`Na^(+) + e^(-) to Na` Electron gain enthalpy of `Na^(+)`
Because reaction is reverse so then.
`DeltaH_(eg) = - 5.1 ev.`