Consider the isoelectronic ions F- and Na+. (b) Using Equation 7....

Question

Consider the isoelectronic ions F- and Na+. (b) Using Equation 7.1 and assuming that core electrons
contribute 1.00 and valence electrons contribute 0.00
to the screening constant, S, calculate Zeff for the 2p electrons
in both ions.

Accepted Answer

hey everyone in this example we need to calculate effective nuclear charge of the three p electrons of potassium cat ion and the chloride an ion if the core electrons provide a value of one and the valence electrons provide zero to the screening constant. So we should recognize initially that our screening constant is representing our number of core electrons and it's represented by the symbol capital S. We want to recall our formula next for effective nuclear charge and that is going to be taking our atomic number, which we can say is c recall that it's represented by the simple Z. And it's taking the difference from our screening constant us for our given adam. So to start out we have the atom potassium as a cat ion and we should recall that our atomic number for neutral atoms only is going to equal our number of protons as well as equal our number of electrons. So what we would say for potassium is as a neutral atom it has an atomic value Or atomic number equal to 19. So it would still have this atomic number as an ion. However, it would have 19 protons. And instead of 19 electrons as well it's going to lose an electron. Because we recall that a positive caddy in charge means we lose that number of electrons. So we would have 18 electrons. So to find our effective nuclear charge of the potassium cat ion, we would begin by taking Our atomic number for potassium which we just agreed from the periodic table is going to be 19. So we'll use purple for this and we're going to subtract this from our screening constant value which we can get from finding our number of core electrons in our potassium ion here. So we want to analyze its atomic or sorry electron configuration. So we would see that the noble gas that comes before potassium on our periodic table is going to be argon. So we're going to begin its configuration with argon And actually we would end our configuration here because we lose that electron for our potassium ion meaning we would just have the configuration for argon. So we should write out organs configuration Which is going to be 1 S2 to us too two P 63 S two and then three P six to get to argon which is also our configuration for our potassium catalon. So we should recall that valence electrons are our outermost energy level electrons. So that would cover the third energy level here. So we would have eight valence electrons When we add our exponents six and 2. And then for our core electrons, that's going to be the energy levels that come before our outermost energy level. So that would be these electrons here where we have core electrons when we add the exponents. So this would be what we plug in for our screening constant. So we would have 19-10 core electrons. And this is going to give us a value for our effective nuclear charge equal to nine for our potassium castile. And this would be how much our protons in our nucleus of our potassium atom are pulling on the outer electrons Which creates discharge with a value of nine. So now we're going to compare to our effective nuclear charge for the an ion chloride as given in the prompt. So we would find chlorine on our periodic tables across period three and we would see that it has an atomic number equal to a value of 17 in group seven a Which we would recognize would correspond to protons and we won't have the same number of electrons because this is a charged atom. So we recognize we have a minus one charge. Recall that an ion charges or negative charges mean we gain that number of electrons. So we would have therefore 18 electrons for our chloride an ion which is the same so far as potassium as a cat ion. So to find its effective nuclear charge, we're going to take the atomic number of The chloride and I on which we agree is 17. So we'll use purple again and we're going to subtract from our number of core electrons. So we want to analyze for our configuration of the chloride ion, recognize that the chloride annan means we gained an electron. So because we have 18 electrons, we're going to have the same configuration as our potassium Catalan meaning that we still have a configuration of one S 22, S 22 P 63 S two Three piece six since we have 18 electrons for the chloride and ion as well. And so we would agree that we still have Also eight valence electrons in our outer shell the third energy level. And then we would also agree that we have core electrons in our inner shells that shield our nucleus of our chloride an ion. And so we would agree that this would be the number that we plug in for our screening constant. So we would have 17 -10 Which is going to give us a value equal to seven for our screening constant. Or sorry for our effective nuclear charge of the chloride an ion. So we would say that overall, therefore we have Effective nuclear charge for the potassium Catalan equal to a value of nine and an effective nuclear charge for the chloride an ion equal to a value of seven as our final answers to complete this example. And one thing to note is that because we have a higher value for our effective nuclear charge for our potassium catalon, our protons in our nucleus of our potassium ion have a stronger pull on the electrons making up our potassium atom in our outer energy levels. And so because the prompt is asking about the effective nuclear charge in regards to our three P or our valence electrons here, That means that our potassium Catalan also has a stronger pull on its electrons in the three P Sublevel. So they're technically going to be closer to the nucleus of our potassium atom, which is why we have this larger value for our effective nuclear charge since the electrons are closer to the nucleus. And this will actually decrease the size of our potassium atom in comparison to our chloride um are chloride atomic radius size, which is going to have a larger radius due to the fact that it's going to have a lower effective nuclear charge, meaning that the electrons are farther away from the nucleus of our chloride an ion because it has a less strong pull of those electrons. So that's just something to keep in mind for understanding effective nuclear charge. But overall what we have boxed in here is going to be our final answer to complete this example for the effective nuclear charges of our given Adams, the potassium catalon and the chloride an ion. I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

Consider the isoelectronic ions F- and Na+. (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant, S, calculate Zeff for the 2p electrons in both ions.

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