Final answer:
Iodine has the highest ionization energy among Mo, I, and K due to periodic trends. As we move across a period, ionization energy increases, and as we move down a group, it decreases. Successive ionization energies increase significantly after reaching a noble gas configuration.
Step-by-step explanation:
The student is asking which element among Mo (molybdenum), I (iodine), and K (potassium) has the highest ionization energy. Ionization energy is the energy required to remove an electron from an atom in its gaseous state. Iodine (I) has the highest ionization energy among these three elements, which is consistent with the periodic trend showing that ionization energy increases as one moves up a group and across a period from left to right.
For placing elements in order of increasing ionization energy, the order is Na < K < Ca < O < Ne, since ionization energy increases across a period and decreases down a group. Conversely, for decreasing ionization energy, the order is He > N > S > Si > Mg.
When we examine the successive ionization energies for an element, we find that these energies increase significantly after removing an electron that results in a noble gas electron configuration. This significant increase is attributed to the need to remove an electron from a closer, more tightly bound shell.
To determine the radius of ions, consider both the number of protons and electrons. A higher number of protons relative to electrons will lead to a smaller radius due to stronger electrostatic pull.
The exothermicity of a reaction involving pure elements depends on the energy change associated with their respective reactions. Generally, the reaction involving alkali metals tends to be more exothermic due to their higher reactivity, with Rb (rubidium) being a strong candidate for the most exothermic reaction.