Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The first ionization enthalpy of O is lower than that of N and F.
Reason R : The loss of an electron from O leads to stable half-filled $p$ orbital.
In light of the above statements, choose the most appropriate answer from the options given below:

The electronic configuration of $\text{N}$ is $1s^2 2s^2 2p^3$.

The electronic configuration of $\text{O}$ is $1s^2 2s^2 2p^4$.

The electronic configuration of $\text{F}$ is $1s^2 2s^2 2p^5$.

Generally, the first ionization enthalpy increases across a period due to an increase in effective nuclear charge. Therefore, the ionization enthalpy of $\text{F}$ is higher than that of $\text{O}$.

However, $\text{N}$ has a stable half-filled $2p^3$ configuration, which makes it difficult to remove an electron, resulting in a higher ionization enthalpy than $\text{O}$.

For $\text{O}$, the loss of one electron results in the stable half-filled $2p^3$ configuration ($\text{O}^+$ is $1s^2 2s^2 2p^3$). This extra stability of the resulting ion makes the removal of the first electron from $\text{O}$ easier compared to $\text{N}$.

Hence, the first ionization enthalpy of $\text{O}$ is lower than that of both $\text{N}$ and $\text{F}$.

Assertion A is correct. Reason R is also correct and provides the correct explanation for Assertion A.

Answer: Both $\text{A}$ and $\text{R}$ are correct and $\text{R}$ is the correct explanation of $\text{A}$.