Question

In which of the following, elements are correctly arranged in the decreasing order of their second ionization enthalpies.

• A. O>F>N>C
• B. N>O>F>C
• C. F>O>N>C
  % (Note: Image check shows option (a) is O>F>N>C but marked correct is (a), which is O>F>N>C) % The image for options shows 1. O>F>N>C; 2. N>O>F>C; 3. F>O>N>C; 4. C>N>O>F % The marked correct answer is 1.
• D. C>N>O>F % This should be based on image

Hint:

Second ionization enthalpy (IE$_2$) refers to removing an electron from M$^+$ ion.
Consider the electronic configuration of the M$^+$ ion.
Half-filled ($p^3, d^5$) and fully-filled ($p^6, s^2$) subshells are particularly stable and require more energy to remove an electron from.
For O$^+ (2p^3)$, removing an electron disrupts a stable half-filled $p$ subshell, leading to a very high IE$_2$ for Oxygen. This often makes IE$_2$(O)>IE$_2$(F).
General trend is increasing IE across a period. Combine this with stability considerations.

Answer 1

The Correct Option is A

Second ionization enthalpy (IE$_2$) is the energy required to remove an electron from a unipositive ion (M$^+$) to form a dipositive ion (M$^{2+}$). M$^+$(g) $\rightarrow$ M$^{2+}$(g) + e$^-$ The elements are C, N, O, F. They are all in the second period. Electronic configurations of the neutral atoms: C: $1s^2 2s^2 2p^2$ N: $1s^2 2s^2 2p^3$ O: $1s^2 2s^2 2p^4$ F: $1s^2 2s^2 2p^5$ Electronic configurations of their unipositive ions (M$^+$), from which the second electron will be removed: C$^+$: $1s^2 2s^2 2p^1$ (removing an electron from $2p^1$) N$^+$: $1s^2 2s^2 2p^2$ (removing an electron from $2p^2$) O$^+$: $1s^2 2s^2 2p^3$ (removing an electron from $2p^3$ - a half-filled stable configuration) F$^+$: $1s^2 2s^2 2p^4$ (removing an electron from $2p^4$) 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$_2$ for C$^+$, N$^+$, O$^+$, F$^+$: All these ions have electrons in the $2p$ subshell. Effective nuclear charge increases from C$^+$ to F$^+$. \begin{itemize} \item C$^+$ ($2p^1$): Removing this $2p$ electron. \item N$^+$ ($2p^2$): Removing one of two $2p$ electrons. Higher nuclear charge than C$^+$. \item O$^+$ ($2p^3$): Removing an electron from a half-filled $2p^3$ configuration. This configuration is relatively stable, so removing an electron from it requires significantly more energy. \item F$^+$ ($2p^4$): Removing an electron from $2p^4$. Higher nuclear charge than O$^+$. After removal, it becomes $2p^3$ (stable). \end{itemize} Expected order based on general trend (increasing nuclear charge): C$^+$<N$^+$<O$^+$<F$^+$. However, stability of electronic configurations plays a crucial role. O$^+$ has a $2p^3$ (half-filled) configuration. Removing an electron from this stable configuration requires a large amount of energy. Therefore, IE$_2$ of Oxygen (i.e., for O$^+$) is exceptionally high. F$^+$ has $2p^4$. Removing an electron leads to $2p^3$. This might be easier than from O$^+$. N$^+$ has $2p^2$. C$^+$ has $2p^1$. Let's consider the elements: \begin{itemize} \item Oxygen (O): O$^+$ is $1s^2 2s^2 2p^3$. Removing an e$^-$ from half-filled $2p^3$ requires high energy. So, IE$_2$(O) will be very high. \item Fluorine (F): F$^+$ is $1s^2 2s^2 2p^4$. Removing an e$^-$ to get $2p^3$. Nuclear charge is higher than O. General trend suggests IE$_2$(F)>IE$_2$(O), but O$^+$ is $2p^3$. Actual values (in kJ/mol): IE$_2$(O) = 3388, IE$_2$(F) = 3374. So IE$_2$(O)>IE$_2$(F). This is because breaking the $2p^3$ stability of O$^+$ dominates over the higher nuclear charge of F$^+$. \item Nitrogen (N): N$^+$ is $1s^2 2s^2 2p^2$. IE$_2$(N) = 2856 kJ/mol. \item Carbon (C): C$^+$ is $1s^2 2s^2 2p^1$. IE$_2$(C) = 2353 kJ/mol. \end{itemize} So the actual decreasing order of IE$_2$ is: O>F>N>C. $3388 (\text{O})>3374 (\text{F})>2856 (\text{N})>2353 (\text{C})$. This matches option (a). \[ \boxed{\text{O>F>N>C}} \]