Practicing Success
The correct order of second ionization potential of \(C, N, O\) and \(F\) is |
\(C > N > O > F\) \(O > N > F > C\) \(O > F > N > C\) \(F > O > N > C\) |
\(O > F > N > C\) |
The correct answer is option 3. \(O > F > N > C\). Let us re-evaluate the second ionization potential (IP2) order for \(C, N, O, F\) to see why \(O > F > N > C\) might be the correct order. Second Ionization Potential: The energy required to remove an electron from a singly charged cation (\(X^+\)). Effective Nuclear Charge \((Z_{eff})\): The net positive charge experienced by an electron in a multi-electron atom. It generally increases across a period. Electron Configuration: Atoms with stable or half-filled configurations tend to have higher ionization potentials. Detailed Analysis Carbon (C): Configuration: \(1s^2 2s^2 2p^2\) First Ionization: \(C \rightarrow C^+ + e^-\) gives \(C^+\) with \(1s^2 2s^2 2p^1\) Second Ionization: \(C^+ \rightarrow C^{2+} + e^-\) removes an electron from the \(2p\) orbital, resulting in \(1s^2 2s^2\). Second Ionization Potential: Relatively lower because removing a second electron still leaves a stable \(2s^2\) core. Nitrogen (N): Configuration: \(1s^2 2s^2 2p^3\) First Ionization: \(N \rightarrow N^+ + e^-\) gives \(N^+\) with \(1s^2 2s^2 2p^2\) Second Ionization: \(N^+ \rightarrow N^{2+} + e^-\) removes an electron from a half-filled stable \(2p\) orbital. Second Ionization Potential: Higher than carbon because the half-filled \(2p\) subshell is relatively stable. Oxygen (O): Configuration: \(1s^2 2s^2 2p^4\) First Ionization: \(O \rightarrow O^+ + e^-\) gives \(O^+\) with \(1s^2 2s^2 2p^3\) Second Ionization: \(O^+ \rightarrow O^{2+} + e^-\) removes an electron from a half-filled stable \(2p\) orbital. Second Ionization Potential: Higher than nitrogen because it ends up with a half-filled \(2p\) subshell, making it very stable. Fluorine (F): Configuration: \(1s^2 2s^2 2p^5\) First Ionization: \(F \rightarrow F^+ + e^-\) gives \(F^+\) with \(1s^2 2s^2 2p^4\) Second Ionization: \(F^+ \rightarrow F^{2+} + e^-\) removes an electron from \(2p^4\), resulting in \(2p^3\). Second Ionization Potential: Higher than nitrogen but lower than oxygen because it moves from a less stable to a stable half-filled \(2p\) configuration. Given the analysis: Oxygen: Removing the second electron results in a stable half-filled \(2p^3\) configuration, which is energetically favorable. Fluorine: Although it has a high effective nuclear charge, the second ionization leads to a half-filled \(2p^3\) configuration, which is less stable compared to oxygen losing to \(2p^3\) state but still higher than nitrogen. Nitrogen: Removing the second electron disrupts the stable half-filled \(2p^3\) configuration. Carbon: The second ionization leads to \(1s^2 2s^2\) configuration, which is the most stable but starts from a lower \(Z_{eff}\). Thus, the second ionization potential order is \(O > F > N > C \). The correct answer is \(O > F > N > C\). |