The successive ionization energies for a second-period element are given. What is the identity of the element? (LO 6.8) Ea1 = 1402 kJ/mol Ea2 = 2856 kJ/mol Ea3 = 4578 kJ/mol Ea4 = 7475 kJ/mol Ea5 = 9445 kJ/mol Ea6 = 53,266 kJ/mol Ea7 = 64,630 kJ/mol (a) Be (b) C (c) N (d) F

Video transcript

Hello everyone today. We have the following problem. Consider the following successive ionization energies for an element in the 3rd period of the periodic table determine the name of the element. So here we have a ton of ionization energies here listed below. And we were being asked to determine the name of the element that would possess these sorts of energies. So if you pay attention to them they follow a pretty steady rate. However, there is a big difference between the ionization energy of the third and the fourth process. And so we know that it takes Roughly four times as much energy. So it takes four times as much energy to remove an electron between the third ionization energy and the fourth ionization energy. And so this is going to indicate that we are dealing with Some element X. three plus since the first ionization energy deals with removing one electron, the second deals with another electron. And so the third one is going to deal with moving another electron. And so successfully these add up to removing three electrons and that's represented by X. Which is some element and then three plus. And so this just means that it required this high jump in energy to remove that third electron because it's already stable enough in its current state that it doesn't want to let go of its electron. And so an electron that follows the general path of our three plus is going to be a group three A element. And with our choices that's going to lead us with aluminum. So aluminum which has a symbol of a L. As our answer, which aligns with answer choice C. And with that we've answered the question overall, I hope it has helped until next time.