Classification Of Elements And Periodicity In Properties

We need to identify the element with the highest second ionization enthalpy (IE ) among Carbon, Phosphorus, Nitrogen, and Oxygen. Second ionization enthalpy is the energy required to remove an electron from the unipositive ion (M ). M (g) M (g) + e Let's look at the electronic configurations of the M ions: \begin{itemize} \item Carbon (C): Neutral C: . C : . Removing an electron from . \item Phosphorus (P): Neutral P: (or [Ne] ). P : [Ne] . Removing an electron from . \item Nitrogen (N): Neutral N: . N : . Removing an electron from . \item Oxygen (O): Neutral O: . O : . Removing an electron from a half-filled configuration. \end{itemize} Key factors affecting ionization enthalpy: 1. Nuclear Charge: Higher nuclear charge generally leads to higher IE. 2. Atomic/Ionic Size: Smaller size generally leads to higher IE. 3. Electronic Configuration: Electrons from stable (half-filled or fully-filled) subshells are harder to remove. 4. Screening Effect: Inner electrons screen outer electrons from the nucleus. Comparing the M ions: \begin{itemize} \item C ): Second period. \item N ): Second period. Higher nuclear charge than C . \item O ): Second period. Higher nuclear charge than N . Crucially, O has a half-filled subshell. This configuration is relatively stable, so removing an electron from it requires a significant amount of energy. \item P ([Ne] ): Third period. Electrons are in the shell, which are further from the nucleus and better screened than electrons. Thus, IE for P is generally expected to be lower than for second-period elements like N or O with similar valence configurations for their ions. \end{itemize} Between C, N, O (all in the second period): O needs to lose an electron from a stable configuration. This makes IE for Oxygen exceptionally high for its period. N ( ) and C ( ) do not have this stability for the ion. Generally, IE increases across a period. So, for M ions , the IE would generally increase. C ( ), N ( ), O ( ), F ( ). The exceptional stability of the configuration of O means that IE (O) is very high. As seen in the previous question (Q.123), IE (O)>IE (F)>IE (N)>IE (C). Actual values (kJ/mol): IE (C) = 2353 IE (N) = 2856 IE (O) = 3388 IE (P): P is [Ne] . IE (P) = 1907 kJ/mol. This is lower than for the second-period elements due to larger size and increased screening. So, comparing 2353 (C), 1907 (P), 2856 (N), 3388 (O). Oxygen has the highest second ionization enthalpy among the given options.

Second ionization enthalpy (IE ) is the energy required to remove an electron from a unipositive ion (M ) to form a dipositive ion (M ). M (g) M (g) + e The elements are C, N, O, F. They are all in the second period. Electronic configurations of the neutral atoms: C: N: O: F: Electronic configurations of their unipositive ions (M ), from which the second electron will be removed: C : (removing an electron from ) N : (removing an electron from ) O : (removing an electron from - a half-filled stable configuration) F : (removing an electron from ) General trends for ionization enthalpy: Increases across a period (due to increasing nuclear charge and decreasing size). Decreases down a group. Exceptions occur due to electronic configurations (e.g., half-filled or fully-filled subshells are more stable). Comparing IE for C , N , O , F : All these ions have electrons in the subshell. Effective nuclear charge increases from C to F . \begin{itemize} \item C ( ): Removing this electron. \item N ( ): Removing one of two electrons. Higher nuclear charge than C . \item O ( ): Removing an electron from a half-filled configuration. This configuration is relatively stable, so removing an electron from it requires significantly more energy. \item F ( ): Removing an electron from . Higher nuclear charge than O . After removal, it becomes (stable). \end{itemize} Expected order based on general trend (increasing nuclear charge): C . However, stability of electronic configurations plays a crucial role. O has a (half-filled) configuration. Removing an electron from this stable configuration requires a large amount of energy. Therefore, IE of Oxygen (i.e., for O ) is exceptionally high. F has . Removing an electron leads to . This might be easier than from O . N has . C has . Let's consider the elements: \begin{itemize} \item Oxygen (O): O is . Removing an e from half-filled requires high energy. So, IE (O) will be very high. \item Fluorine (F): F is . Removing an e to get . Nuclear charge is higher than O. General trend suggests IE (F)>IE (O), but O is . Actual values (in kJ/mol): IE (O) = 3388, IE (F) = 3374. So IE (O)>IE (F). This is because breaking the stability of O dominates over the higher nuclear charge of F . \item Nitrogen (N): N is . IE (N) = 2856 kJ/mol. \item Carbon (C): C is . IE (C) = 2353 kJ/mol. \end{itemize} So the actual decreasing order of IE is: O>F>N>C. . This matches option (a).