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Based on their positions in the periodic table, predict which has the largest first ionization energy: Mg_ Ba, B, 0, Te:Nd Pm sml Eupositions in the periodic Pu{Ani...

Question

Based on their positions in the periodic table, predict which has the largest first ionization energy: Mg_ Ba, B, 0, Te:Nd Pm sml Eupositions in the periodic Pu{Anicm Based on their predict - which has the smallest first table, ionization energy: Li, Cs, N, F,! larger atom in each pair ipossible): basis of periodic trends, choose the On the Cr or For Se Ge or Po Sn or radius: $, Ca, F, Rb, Si. elements in order of decreasing Arrange the or ion from each pair: Choose the larger ator Ca?e or Ch b

Based on their positions in the periodic table, predict which has the largest first ionization energy: Mg_ Ba, B, 0, Te: Nd Pm sml Eu positions in the periodic Pu{Anicm Based on their predict - which has the smallest first table, ionization energy: Li, Cs, N, F,! larger atom in each pair ipossible): basis of periodic trends, choose the On the Cr or For Se Ge or Po Sn or radius: $, Ca, F, Rb, Si. elements in order of decreasing Arrange the or ion from each pair: Choose the larger ator Ca?e or Ch b. F or F- Kor K



Answers

Based on their positions in the periodic table, predict which atom of the following pairs will have the larger first ionization energy: (a) $\mathrm{Cl}, \mathrm{Ar} ;$ (b) Be, Ca; (c) $\mathrm{K}, \mathrm{Co}$; (d) S, Ge; (e) Sn, Te.

This question says. Based on their positions in the periodic table, predict which Adam the following pairs will have a smaller first ionization energy. We should remember that smaller first ionization energy means easier to remove the first Valence Electron. Now it's like Look at the bears. The 1st 1 is chlorine and argon. Chlorine is 17 are gone is 18. So both of these elements will have their valence electrons in the three p orbital, but are gone with a higher atomic number will have one more proton in the nucleus. That means a greater nuclear charge of greater pull on the electrons and therefore making it harder to remove the electrons. So chlorine, one less proton, less pull on the electrons. Easier to remove chlorine. We'll have the lower first ionization energy number comparing beryllium, which is four, and calcium which is 20. Hear beryllium valence, electrons, Aaron, the two s orbital while calcium czar in the four s orbital for us is higher energy farther away from the nucleus experiences less pulled from the nucleus. Easier to remove those vans electrons. So calcium would have the lower first ionization energy. Then we're going going to compare potassium and cobalt. That's potassium is 19 coupons 27 both of these Electra. Both of these elements will have their valence electrons in the forest orbital, even though kobo will also have three D electrons. But Cobalt has many more protons in its nucleus has come to be seen by its higher atomic number. And just like in a more protons and nucleus means greater nuclear charge. Greater pull on the electrons harder to remove a veil, its electrons. So potassium would be easier and would have a lower first ionization energy. They were gonna look at Sulphur, 16 and Jermaine Iam, 32 suffers Vance, Electrons, Aaron, the three p Orbital's Romanians, Aaron, the four p four p s higher and energy farther away from nucleus, less pull from the nucleus. Easier to remove. So Germaine Ian would have a lower first ionization energy. And finally comparing 10 which is 50 and Tillery um, which is 52. Just like in a delirium, has a greater atomic number. So has more protons in the nucleus. Therefore, it attracts the electrons more closely, makes them harder to remove the 10. Having fewer protons would be easier to remove. I would have a lower first ionization energy

Problem, 69 from Chapter six is asking us to predict which of the Adams have the largest first ionization energy, and the ions that are in question are magnesium, barium, oxygen, boron and delirium. So it's asking us to predict based on their positions in the periodic table on this period table I extracted from the textbook. So it's important to know that ionization energy mostly decreases down a group. Um, so I'm going to draw an upward facing arrow, noting that the, um, elements that are higher up the periodic table have a higher ionization energy than those lower in the periodic table. Saw ionization energy mostly decreases down a group, but it increases across a period. So John Farrow, across a period so those that are further on the right hand side have a greater ionization energy than those in the left hand side of the periodic table. So this, on the question which of these atoms have a larger first ionization energy? So that's things that are higher up on the periodic table and further to the right, so out of the options, oxygen would be the best as it is, um, higher up in the peacock table and furthest to the right, even though it is on the same in the same period as boron. Because it is further to the right, it gives us a higher ionization energy.

This question is going to take a look at determining an important periodic trend of ionization energy. So the ionization energy is the energy required to remove an electron from a substance. And so we're going to do some comparisons of ionization energy through some different sets of of atoms or elements. The first set we're going to look at is comparing bro mean and Krypton general rule that we want to follow for ionization energy is as we move left to right across the periodic table, we see an increase in ionization energy and as you move top to bottom down any family on the periodic table, we see a decrease and ionization energy. So we kind of keep that on the back burner as we do some of these comparisons, Including this very 1st 1. When we compare bro mean to Krypton, we see that krypton is to the right of roaming on the periodic table and that means it should have a higher ionization energy, then bro mean, suffering to put these in order of ionization energy. Krypton than would have the higher ionization energy then bro mean according to our trend the same thing with another set of examples we're going to compare here silicon to let that's not what I meant to write. Let's try that again here. Silicon and lead. And what is their relationship on the periodic table there in the same group or family? And we know that as we move down a family, we see a decrease in ionization energy. So that means lead being below silicon, Silicon should have a higher ionization energy. Then what does another set that will take a look at compares again two atoms or elements that are from the same group of the periodic table. four on an aluminum. Once again, we see that boron is above aluminum in terms of its placement on the periodic table. And like the previous example, that means then that boron should have a higher ionization energy than aluminum in the same family. The next set that will compare Is for two Group 1 or Alkali metals, lithium and rubidium. And again, the trend of being in the same group applies here. So that as we're looking at lithium, it's above rubidium in the periodic table, meaning it should have the higher ionization energy in the group of lithium and rubidium. The final example will look at here compares carbon to calcium and so if we look again at the placement of the periodic table, carbon is to the right of calcium, but it's also above calcium in the periodic table. For both of those reasons, carbon than should have the higher ionization entry energy when compared to calcium.

So this problem We're given a list of atoms and we wantto identify the one with the smallest ionization energy to do that which first consider the general trends of ionization energy. Then the periodic table and those air that ionization energy is going to increase as you move from bottom to top within a group. That has a lot to do with the fact that the the Adams at the bottom of a group have a larger atomic radius, which means that those electrons and Vaillant shell or held pretty far away from the nucleus and their attraction to the positive charge of the nucleus is pretty weak. It was much easier to pull off Ah seven s electron from France IEM than it is to pull off this six s s museum electron. That same principle is actually applicable to the other trend of the periodic table where ionization energy increases from left to right. So by the same rules, the size of the Adam plays someone of a role. But another thing to consider with ease is the fact that they have a larger number of protons condensed within a the same energy level, so within so we look? Att The second period here, for example, All of these valence electrons are held within the n equal to energy level and the number of protons air increasing from left to right. Because Florian has nine protons condensed in its nucleus, all holding onto an unequal to energy level of electrons, it's much more difficult to pull an electron from the outermost shell of flooring than it is from the outermost show of nitrogen, for example. And again, that has to do with that that increasing proton counts. There are two groups that break this trend. It's gonna be Group 13 and group 16 through 13. If we remove an electron, it yields a fully filled s orbital. And for Group 16 if we remove an electron that yields off, ah, half filled P orbital, you should remember from instruction is that fully filled. Orbital's and half filled orbital's are actually very energetically favourable and what that means because removing electrons from each of these groups results and either a fully filled orbital or half filled orbital. It means that these groups are going to have lower ionization energies again because there's processes were more favorable. So that's not necessarily directly important for this problem, but I wanted to talk about it just to make sure that you're keeping that in mind as you're doing other problems. That's the atoms in question for this problem are going to be lithium cesium nitrogen Florida. Narita So what we're going to do to look at these, really just figure out because we're gonna identify the one with the smallest ionization energy. We're just going to look whichever one falls to the bottom, left or closest to the bottom left of the periodic table because our trends imply that's where the atom with the lowest ionization energy should be. When we look at France, iam right above Francie, um is actually cesium. So what that means is that cesium is going to have the lowest ionization energy in the group listed here.


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