KCET · Chemistry · Classification of Elements and Periodicity in Properties
The correct order of ionisation energy of \(\mathrm{C}, \mathrm{N}, \mathrm{O}\) and \(\mathrm{F}\) is
- A \(\mathrm{C} < \mathrm{N} < \mathrm{O} < \mathrm{F}\)
- B \(\mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}\)
- C \(\mathrm{F} < \mathrm{O} < \mathrm{N} < \mathrm{C}\)
- D \(\mathrm{F} < \mathrm{N} < \mathrm{C} < \mathrm{O}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}\)
Step-by-step Solution
Detailed explanation
On moving from left to right in a period, ionisation energy increases. Thus, correct order of ionisation energy of \(C, N, O, F\) should be
\[
\mathrm{C} < \mathrm{N} < \mathrm{O} < \mathrm{F}
\]
But the ionisation energy of \(\mathrm{N}\) is found to be greater than that of O. This is due to the presence of half-filled p-orbitals in \(\mathrm{N}\) which provide it extra stability. Thus, the correct order is \(\mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}\).
\[
\mathrm{C} < \mathrm{N} < \mathrm{O} < \mathrm{F}
\]
But the ionisation energy of \(\mathrm{N}\) is found to be greater than that of O. This is due to the presence of half-filled p-orbitals in \(\mathrm{N}\) which provide it extra stability. Thus, the correct order is \(\mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}\).
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