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(C) $1$. Electronic configuration of $N$ $(Z=7)$ is $1s^2 2s^2 2p^3$. It has a stable half-filled $p$-orbital, which makes its first ionization enthal

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$A$ is correct but $R$ is not correct.

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(C) $1$. Electronic configuration of $N$ $(Z=7)$ is $1s^2 2s^2 2p^3$. It has a stable half-filled $p$-orbital, which makes its first ionization enthalpy higher than that of $O$ ($Z=8$, $1s^2 2s^2 2p^4$).$2$. $F$ ($Z=9$, $1s^2 2s^2 2p^5$) has a higher effective nuclear charge than $O$, leading to a higher first ionization enthalpy.$3$. Thus, Assertion $A$ is correct.$4$. Reason $R$ states that the loss of an electron from $O$ leads to a stable half-filled $p$-orbital. The configuration of $O^+$ is $1s^2 2s^2 2p^3$, which is indeed a stable half-filled $p$-orbital. However, this explains the second ionization enthalpy of $O$, not the first. Therefore, Reason $R$ is incorrect as an explanation for the first ionization enthalpy trend.$5$. Hence, $A$ is correct but $R$ is not correct.

Similar Questions

Amongst the elements with the following electronic configurations,which one of them may have the highest ionisation energy?

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