Question

The correct order of ionization energy for comparing carbon, nitrogen, and oxygen atoms is

A) $\quad \mathrm{C} > \mathrm{N} > \mathrm{O}$

B) $\mathrm{C} > \mathrm{N} < \mathrm{O}$

C) $\mathrm{C} < \mathrm{N} > \mathrm{O}$

D) $\mathrm{C} < \mathrm{N} < \mathrm{O}$

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Answer

The correct option is C: $$ \mathrm{C} < \mathrm{N} > \mathrm{O} $$

Ionization energy typically increases as you move from left to right across a period in the periodic table. This trend occurs because the atomic radius generally decreases, which means the outer electrons are closer to the nucleus and more tightly bound to it. Therefore, more energy is required to remove an electron. As per the trend, one would expect: $$ \mathrm{C} < \mathrm{N} < \mathrm{O} $$

However, nitrogen ($\mathrm{N}$) possesses a half-filled $p$ orbital configuration ($2p^3$), which is notably more stable due to symmetry and exchange energy considerations. This half-filled structure makes electrons in nitrogen less willing to be removed compared to the usual trend, leading to an unusually high ionization energy for nitrogen.

Thus, due to this additional stability in nitrogen:

  • Carbon ($\mathrm{C}$), having a $2p^2$ configuration, is less stable compared to the half-filled configuration of nitrogen.

  • Oxygen ($\mathrm{O}$), having a $2p^4$ configuration, experiences inter-electron repulsion that slightly reduces its ionization energy compared to nitrogen.

As a result, the exception in the general trend due to the half-filled stability of nitrogen leads to the order: $$ \mathrm{C} < \mathrm{N} > \mathrm{O} $$ Thus, confirming that Option C is correct.


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