The first ionisation potential of Na is 5.1 , | vedclass

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(B) The first ionization process is represented as: $Na \rightarrow Na^{+} + e^{-}$,where $IE = 5.1 \, eV$. The electron gain enthalpy of $Na^{+}$ cor

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$-5.1$

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(B) The first ionization process is represented as: $Na \rightarrow Na^{+} + e^{-}$,where $IE = 5.1 \, eV$. The electron gain enthalpy of $Na^{+}$ corresponds to the reverse process: $Na^{+} + e^{-} \rightarrow Na$. Since the second reaction is the exact inverse of the first reaction,the energy change is equal in magnitude but opposite in sign. Therefore,the electron gain enthalpy of $Na^{+}$ is $\Delta H_{eg} = -IE = -5.1 \, eV$.

The first ionisation potential of $Na$ is $5.1 \, eV$. The value of electron gain enthalpy of $Na^{+}$ will be : ............... $eV$

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