Fifth ionization energy peaks when removing core electrons from stable configurations Steps: - Identify electron configurations: C ([He] 2s²2p²), N ([He] 2s²2p³), P ([Ne] 3s²3p³), Si ([Ne] 3s²3p²). - Determine ions before fifth IE: C⁴⁺ and Si⁴⁺ are noble gas cores ([He] and [Ne]); N⁴⁺ is [He] 2s¹; P⁴⁺ is [Ne] 3s¹. - Assess fifth IE: For C, it removes a 1s electron from [He] core (high energy due to poor shielding and high effective nuclear charge); others remove valence or less tightly bound core electrons. - Compare: Period 2 core (1s) removal requires more energy than period 3 (2p/3s) or valence shells. Why A is correct: - Carbon's fifth IE removes an electron from the 1s² core of C⁴⁺, where the helium-like configuration and Z=6 yield the highest energy per the trend of increasing IE for inner-shell electrons (Madelung rule). Why the others are wrong: - B (N): Fifth IE removes a 2s electron from N⁴⁺ ([He] 2s¹), screened by 1s², lower than C's 1s removal. - C (P): Fifth IE removes a 3s electron from P⁴⁺ ([Ne] 3s¹), …