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The first ionisation potential of Na is ...

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`

Text Solution

Verified by Experts

The correct Answer is:
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`
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