Question

Correct order of ionisation enthalpy is:

• A. \( \mathrm{F>Cl>Cl^->F^-} \)
• B. \( \mathrm{F^->Cl^->F>Cl} \)
• C. \( \mathrm{Cl>F>Cl^->F^-} \)
• D. \( \mathrm{F>Cl>F^->Cl^-} \)

Hint:

Always remember:
Smaller size \(\Rightarrow\) higher ionisation enthalpy
Anions lose electrons more easily than neutral atoms

Answer 1

The Correct Option is D

Concept: Ionisation enthalpy is the energy required to remove an electron from a gaseous species. Key points:
Smaller atomic size \(\Rightarrow\) higher ionisation enthalpy
Anions have lower ionisation enthalpy than their neutral atoms due to extra electron–electron repulsion
Step 1: Compare neutral atoms \( \mathrm{F} \) and \( \mathrm{Cl} \). Fluorine is smaller than chlorine, so its valence electron is held more strongly. \[ \mathrm{F>Cl} \]
Step 2: Compare anions \( \mathrm{F^-} \) and \( \mathrm{Cl^-} \). Although both are anions:
\( \mathrm{F^-} \) is smaller
Extra electron experiences less shielding Hence, removing an electron from \( \mathrm{F^-} \) requires more energy than from \( \mathrm{Cl^-} \). \[ \mathrm{F^->Cl^-} \]
Step 3: Compare atoms with their respective anions. Anions always have lower ionisation enthalpy than their neutral atoms due to increased electron–electron repulsion. \[ \mathrm{F>F^-} \quad \text{and} \quad \mathrm{Cl>Cl^-} \]
Step 4: Final order. Combining all comparisons: \[ \mathrm{F>Cl>F^->Cl^-} \] \[ \boxed{\mathrm{F>Cl>F^->Cl^-}} \]