Final answer:
The first and second ionizations of calcium involve removing a 4s electron from the outermost energy level, resulting in Ca2+. The 4s orbital is the highest energy level in ground state calcium and is thus the first to lose electrons during ionization, leading to a smaller ionic radius and higher charge density in Ca2+.
Step-by-step explanation:
The first and second ionizations of calcium involve removing a 4s electron from the outermost energy level or orbital. During the first ionization, a 4s electron is removed from calcium, making it Ca+. For the second ionization, another 4s electron is removed, resulting in Ca2+. The 4s electrons are removed because, in the ground state of calcium, the 4s orbital is the highest in energy and is therefore the first to lose electrons. After these electrons are removed, the Ca2+ ion ends up dropping to a lower energy level, the n = 3 level, which is much smaller in radius.
Comparing ionization energies, removing a 4s electron from calcium requires more energy than from potassium due to calcium's stronger nuclear attraction and the necessity to break the pairing of the 4s electrons. The second ionization potential for potassium (K) is higher because it involves removing an electron from a lower energy level, but for calcium, the second ionization only involves removing the remaining unpaired electron in the 4s orbital. In addition, the resulting Ca2+ ion has a higher charge density compared to the K+ ion, which influences lattice energies and other chemical properties.