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The first ionisation potential of `Na` is `5.1eV`. The value of eectrons gain enthalpy of `Na^(+)` will beA. `-2.55 eV`B. `-5.1` eVC. `-10.2 eV`D. 2.55 eV |
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Answer» Correct Answer - 2 `Na to Na^(+)+e^(-), I.P. =+5.1 eV` `Na^(+) + e to Na to DeltaH_(eg) =-5.1 eV` |
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