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The first ionisation potential of `Na` is `5.1eV`. The value of eectrons gain enthalpy of `Na^(+)` will beA. `+ 10.2 eV`B. ` –5.1 eV `C. `–10.2 eV`D. `+ 2.55 eV` |
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Answer» Correct Answer - B `Moverset(M^(+))toM^++e^_ " "IE_1=5.1eV` `M^+e^(-)toM " "DeltaH_(eg)=-5.1eV` |
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