Three atoms have the following electron configurations: (a) 1s2 2...

Question

Three atoms have the following electron configurations: (a) 1s2 2s2 2p6 3s2 3p3 (b) 1s2 2s2 2p6 3s2 3p6 (c) 1s2 2s2 2p6 3s2 3p6 4s2 
Which of the three has the largest Ei2? Which has the smallest Ei7?

Accepted Answer

Hello everyone today. We have the following problem. Consider the following electron configurations for three atoms. And we have the following electron configurations identify the atom that will have the largest second ionization energy and the atom that will have the smallest seventh energy ionization. So we'll go through these answer choices one x 1. So for a The total number of electrons that we have is 36. And then we can do so by adding up each of these exponents here. And if we do so they will add up to 36. And so the number of electrons is going to Be correlated with the atomic number when a neutral species is present. And as this is this is going to be atomic number 36 which aligns with Krypton For b. The total number of electrons using the same process is going to be 38. Thus its atomic number, It's going to be 38. And this aligns with strontium and C using the same train of thought, It's gonna line with 33 electrons and it's Atomic number is going to be 33 which will be arsenic. And so we have to know our trend for ionization energy. And that trend states that ionization energy increases as we go from the left side of the periodic table to the right. And as we go down, our group's on the periodic table. And so we are left with these three elements here, we have to determine where there where they lie in this trend. So based on their position in the periodic table. The ionization energy of krypton is going to be higher than the ionization energy of arsenic, meaning it's easier to remove electrons from strontium. So if we put this in order we can say that the ionization energy of krypton is going to be the highest, followed by the ionization energy of strontium, followed by the ionization energy of arsenic. So it's the easiest to remove electrons from strontium. So the second highest in addition energy. So the second highest ionization energy will be krypton. So we go and highlight this and then we have to pay attention to the seventh electron because that is the seventh ionization energy to remove that seventh electron if we pay attention to the inner electrons for each of these. So the inner electrons Strontium has four p orbital And Arsenic has a three p orbital. This indicates that it's going to be harder to remove than our KR or krypton. And so Krypton We'll also have the 7th lowest or smallest seventh ionization energy. And with that we have answered our question overall, I hope this helped. And until next time

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