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Jeevnich of the following elements would you expect to have the lowest first ionization energy?...

Question

Jeevnich of the following elements would you expect to have the lowest first ionization energy?

Jeevnich of the following elements would you expect to have the lowest first ionization energy?



Answers

Which main group atom would be expected to have the lowest second ionization energy?

When determining the trend in ionization energy, we can refer to the periodic table. Here's a representation of the periodic table with the pack group a numbering identification in black and the american group identification in blue. You and we will see that ionization energy increases as we go up a group or across a period. So a general trend ionization energy would be an increase along this diagonal toward the upper right corner. Therefore the general trend of ionization energy for a group increases, proceeding up the group of the periodic table. And the general trend of ionization energy along a period would be an increase also proceeding left to right on the periodic table. So which group of elements has the lowest ionization energy? Well, because ionization energy decreases as we go right to left, then it must be group one or group one A. That has the elements collectively with the lowest ionization energies. And then as we go farther to the right increasing ionization energy, it must be group ate A. Or group 18 that has the elements collectively, with the highest ionization energy.

The working arts organization. Angie on. We spoke about before. I'm going across a period on. We increase the organization energy right on. The reason for that is because we were decreasing the atomic radius. We are increasing the attraction between the nucleus and electoral elections are closer to the nucleus and dose That requires more energy to remove the electron because there's a great attraction between positively charged nucleus on the to those. If we put this in order in terms of the period, then we can say, uh, this would be decreasing right in terms of the group is well, when you go down a group, your you will juice the analyzation energy is your adding that comes to a higher sublevel skirt away from the nucleus It's further away from. The increase has a lower attraction between the positive charge increase in the electron, which is further away in a highest level, and therefore it's easiest removed. So we'll see. Um, okay. Being the lowest

So often has want us to To us too. To before four electrons. Um huh. And all the subsequent, um, versions are bonded versions off this oxygen atom will be stable than the oxygen atom itself because bonding stabilizes by structure. Remember, brands like cause in this case is 12. Here you lose any electronics, we have 11 and hear you gin in that plan, so you have 30. So let's write off the laconic configurations for this. It's a star singular and PC two i m. P X equals of crime. And he won 212 and then High Star and he eggs. He was a high star. And why one? And on the long order. In this case, if you subtract the number of bonding from the anti wanding and then divide by two movie, too, that's the lobby electron configuration. In this case, start to Sigma MPC, too. Hi and P s. He was supply and P y. It's two and two and then high star and P X equals to Christ R and P one is you. The bond ponder will be eight minus 32 by two five divided by to dupe one fine And in this case, I want you to the electronic configuration you have on slide change. When it comes to the last orbital that they're going out, you have an extra anti bonding the lexicon. Um, so the bar, now that changes and you have eight minus five. I went by to see if I do this one for fun. So based on the bond orders, we can say that this stabilised, um, molecule here is you is or two plus, um, less than that is, or two then on to Marnie on the least stable. It's the oxygen in, based on stability and stability is based on barn ardor. So the more stable it is, the harder it just remove an electron, so the ionization energy would be highs. In the same marker, you'll be heist for two plus less for or to even less afford to minus, at least for oxygen atom. Because we're moving in that drawing, you're actually getting it in half filled confrontation, which is very desirable

So hey, we're looking at ionization energies, and so I've got a small diagram drawn up on the screen. After all, the molecules we're looking at is well as the order of ionization energy already. And so if we have a low ionization energy, that tends to mean that we are populating on type awning orbital's, and that's wherever it we're removing our electrons from now. Father to this, if our electrons are paired in an orbital, are ionization. Energy will also be lower because of that pairing repulsion energy. And so it is more favorable to remove the electron. So as our ionization energies increase, it tends to be because we are then removing electrons from our bonding. Orbital announced removed electrons from your bonding. Orbital is unfavorable because you are destabilizing the molecule. So again, like I said, we have our ionization energy order here for our reference. And then it is rationalized using the trends I just discussed


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