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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 eV`C. `-10.2 eV`D. `+2.55 eV` |
Answer» Correct Answer - B `{:(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`. |
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