Question

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

• A. $- 2.55\, eV$
• B. $- 5.1\, eV$
• C. $- 10.2\, eV$
• D. $+ 2.55\, eV$

Answer 1

The Correct Option is B

To find the electron gain enthalpy of $Na^+$, we first need to understand the relationship between ionization potential and electron gain enthalpy. The ionization potential of an element is the energy required to remove an electron from a neutral atom. The electron gain enthalpy is the energy change when an electron is added to an ion.

The given first ionization potential of sodium (Na) is 5.1\, \text{eV}. This means:

• Na \rightarrow Na^+ + e^- \quad (\Delta H = 5.1\, \text{eV})

Since the process of forming Na^+ from Na requires energy equal to the ionization potential, the reverse process (adding an electron back to Na^+) would release energy of the same magnitude but opposite in sign:

• Na^+ + e^- \rightarrow Na \quad (\Delta H = -5.1\, \text{eV})

This is the electron gain enthalpy for Na^+.

Let's analyze why the correct answer is - 5.1\, \text{eV}:

• - 2.55\, \text{eV}: This value is incorrect as it is half of the ionization energy and suggests a misunderstanding of the relationship between ionization and gain enthalpy.
• - 5.1\, \text{eV}: This value directly correlates to the fact that adding an electron to a cation results in the release of the same amount of energy that ionization had originally required.
• - 10.2\, \text{eV}: This value is double the ionization energy and unrelated to the principles of electron addition.
• + 2.55\, \text{eV}: This suggests energy absorption rather than release, which contradicts the exothermic nature of electron gain.

Therefore, the correct value of the electron gain enthalpy of Na^+ is - 5.1\, \text{eV}, which matches the ionization potential but takes the opposite sign.