Q39 / 180 · KCET 2012 · Chemistry
The correct order of ionisation energy of $\mathrm{C}, \mathrm{N}, \mathrm{O}$ and $\mathrm{F}$ is
- A$\mathrm{C} < \mathrm{N} < \mathrm{O} < \mathrm{F}$
- Correct answerB$\mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}$
- C$\mathrm{F} < \mathrm{O} < \mathrm{N} < \mathrm{C}$
- D$\mathrm{F} < \mathrm{N} < \mathrm{C} < \mathrm{O}$
Solution
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 $\mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}$.
Hence, the correct answer is option: 'B'.