Question

The first ionization potential of $\mathrm{Na}$ is $5.1\mathrm{eV}$, the value of electron gain enthalpy of $\mathrm{Na}$ will be?

• A. $-51\mathrm{eV}$
• B. $-5.1\mathrm{eV}$
• C. $-255\mathrm{eV}$
• D. $-2.55\mathrm{eV}$

Answer 1

The correct option is B

$-5.1\mathrm{eV}$

Explanation for correct option:

B. $-5.1\mathrm{eV}$

Ionization potential:

• “The term ionization energy is a reference to the amount of energy necessary to expel an electron from the gaseous form of an atom or molecule.”

The ionization potential of Sodium will take place like this:

$\mathrm{Na}\left(\mathrm{s}\right)\to {\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}$

The value for ionization potential is given which is $5.1\mathrm{eV}$.

Electron gain enthalpy:

• “Electron gain enthalpy is defined as the amount of energy released when an electron is added to an isolated gaseous atom.”
• The electron gain enthalpy in an inverse of ionization potential.

Thus, the electron gain enthalpy of sodium is $-5.1\mathrm{eV}$ which is inverse of $5.1\mathrm{eV}$.

Explanation for incorrect options:

The electron gain enthalpy of sodium is $-5.1\mathrm{eV}$ which is inverse of $5.1\mathrm{eV}$. Thus, options A,C and D stand incorrect.

Hence, the correct option is B i.e. $-5.1\mathrm{eV}$