Final answer:
Aluminum (Al) will have the greatest difference between its second and third ionization energies because the third electron is being removed from a closer and more tightly bound p orbital, which has a higher ionization energy than the s orbital.
Step-by-step explanation:
The element that would have the greatest difference between its second and third ionization energies among strontium (Sr), rubidium (Rb), bromine (Br), and aluminum (Al) is aluminum (Al). This is because after the removal of the first two electrons, the third electron is being removed from a closer and more tightly bound p orbital.
The electron configuration of Al is [Ne]3s23p1, after losing two electrons to become Al2+, the next electron to be removed will be from the 3p orbital which is closer to the nucleus and has a larger ionization energy jump compared to the other elements listed when removing the third electron.