Chemistry

Question Analysis
This question involves comparing first ionization energies of isoelectronic cations from adjacent Period 4 elements.
The main focus is on linking nuclear charge and electron configuration to ionization energy for ions with identical electron counts.

Key Concept Explanation
All three ions are isoelectronic with the electron configuration [Ar] (1s²2s²2p⁶3s²3p⁶).
Isoelectronic species follow the trend: lower nuclear charge = weaker attraction to valence electrons = lower ionization energy.
Nuclear charges: K⁺ (19⁺) < Ca²⁺ (20⁺) < Sc³⁺ (21⁺).

Step-by-Step Solution
1. 2. List the nuclear charges of the parent atoms (determines the ion’s nuclear charge). 4. Conclude K⁺ has the lowest first ionization energy.
Identify the electron configurations of the ions (all are isoelectronic with Ar, 1s²2s²2p⁶3s²3p⁶):
K⁺ → 19⁺, 18 e⁻ → [Ne]3s²3p⁶
Ca²⁺ → 20⁺, 18 e⁻ → [Ne]3s²3p⁶