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1. Let’s start off with a question: Based on information in the Periodic Table, which do you think is bigger: Na or Mg? 11 12 Na Mg 23.0 24.3 Predict on paper (1) before continuing, and explain why you picked the one you did.

2. 11 12 Na Mg 23.0 24.3 If you predicted that Mg is bigger, and you gave as your reason that it has more protons and neutrons, keep in mind that this might make the nucleus of Mg bigger, but the nucleus is only a very, very tiny portion of the whole atom.

3. 11 12 Na Mg 23.0 24.3 If you predicted that Mg is bigger because it has more electrons, now you might be on to something, but remember that those outermost electrons in Na and Mg are being placed into the same exact sublevel (the 3s).

4. 11 12 Na Mg 23.0 24.3 So perhaps this led you to predict that because they both have the same number of occupied electron energy levels, the two atoms should be the same size.

5. 11 12 Na Mg 23.0 24.3 That may be closer to the truth, but remember that each atom has its own unique spacing of the energy levels... So... let’s learn more about the forces at play before we draw any conclusions...

6. Let’s consider a lithium atom… + + + Since its atomic number is 3, we know it has three protons in the nucleus. This means that its nucleus has a charge of 3+. And Li’s electron configuration is 1s2 2s1 This means it has 2 electrons in the first level (1s2 2s1) … and 1 electron in the second level (1s22s1) So this is what Li’s Bohr diagram would look like:

7. + + + Now let’s consider how much each electron is attracted by the nucleus… First, the two electrons on the lowest level… First, the two electrons on the lowest level… since they have no electrons beneath them to shield them from the nucleus, they “feel” the full 3+ charge pulling them in toward the nucleus. For these two electrons, the effective nuclear charge (z*) is 3+. Now, what about the one electron on the 2nd level… Now, what about the one electron on the 2nd level… beneath it, there are two electrons that act to shield it from the charge on the nucleus. Thus, this electron only “feels” a charge of 1+ pulling it in toward the nucleus. For this outermost electron, z* = 1+ z* = 1+ z* = 3+ In other words, these two electrons come between the 3+ nucleus and the outermost electron. In other words, these two electrons come between the 3+ nucleus and the outermost electron. These two electrons effectively cancel out two of the protons. In other words, these two electrons come between the 3+ nucleus and the outermost electron. These two electrons effectively cancel out two of the protons. And this leaves just one proton In other words, these two electrons come between the 3+ nucleus and the outermost electron. These two electrons effectively cancel out two of the protons. And this leaves just one proton to attract the outermost electron. So the effective nuclear charge “felt” by the outermost electron is only 1+. This blocking out of the nuclear charge by lower level electrons is known as “shielding.”

8. + + + + + + + + + + + + Now let’s consider Bohr diagrams for the next two elements on the periodic table: Be and B How many protons would they each have in their nuclei? (Write down your prediction before continuing.(2) How many protons would they each have in their nuclei? (Write down your prediction before continuing Is this what you predicted? And what about their electron configurations? (Write them down before continuing Is this what you wrote down? And what about their electron configurations? (Write them down before continuing. (3) And so what would their Bohr diagrams look like? (Write them down before continuing Is this what you were thinking? And so what would their Bohr diagrams look like? (Write them down before continuing. (4) Finally, what effective nuclear charges will be experienced by each of these electrons? (Write them down before continuing (5) .) Finally, what effective nuclear charges will be experienced by each of these electrons? (Write them down before continuing (5) .) Is this what you had? z* = 3+ z* = 2+ z* = 1+ z* = 5+ z* = 4+ z* = 3+ Li Be B 1s2 2s2 2p1 1s2 2s2

9. z* = 3+ z* = 2+ z* = 5+ z* = 4+ + + + + + + + + + + + + So what do these Bohr Diagrams and effective nuclear charges tell us about these atoms? First, compared to Li, Be’s electrons are being pulled in with a stronger charge, both at the first level and the second. First, compared to Li, Be’s electrons are being pulled in with a stronger charge, both at the first level and the second. This stronger attractive force would act to pull these levels in closer to the nucleus. This means that Be is actually smaller than Li. One way to describe how big atoms are is by the “atomic radius” -- the distance from the nucleus to the outermost electrons. Li’s atomic radius is 145 pm while Be’s is only 105 pm. (A pm is a trillionth of a meter.) Now, how would B’s size compare with the other two? Make a prediction (6) and explain why. This means that Be is actually smaller than Li. One way to describe how big atoms are is by the “atomic radius” -- the distance from the nucleus to the outermost electrons. This means that Be is actually smaller than Li. One way to describe how big atoms are is by the “atomic radius” -- the distance from the nucleus to the outermost electrons. Li’s atomic radius is 145 pm while Be’s is only 105 pm. (A pm is a trillionth of a meter.) z* = 1+ z* = 3+ 145 pm 105 pm Li Be B

10. z* = 3+ z* = 2+ z* = 5+ z* = 4+ + + + + + + + + + + + + Is this what you predicted? Is this what you predicted? And did you explain that B should be smaller than either Li or Be because it has the greatest effective nuclear charge (z*) pulling its electrons in? If Li’s atomic radius is 145 pm and Be’s is 105 pm, make a guess as to what B’s atomic radius might be (7). Is this what you predicted? And did you explain that B should be smaller than either Li or Be because it has the greatest effective nuclear charge (z*) pulling its electrons in? Were you close? z* = 1+ z* = 3+ 145 pm 105 pm 85 pm Li Be B

11. C N O F Ne + + + + + + + + + + + + Li Be B And this pattern of getting smaller and smaller as you move to the right continues across the entire periodic table.

12. Ne z* = 8+ z* = 10+ + + + + + + + + + + Ne And this pattern of getting smaller and smaller as you move to the right continues across the entire periodic table. But what happens after Ne??? And this pattern of getting smaller and smaller as you move to the right continues across the entire periodic table. Ne’s Bohr Diagram looks like this: Ne’s Bohr Diagram looks like this: With a 10+ charge on the nucleus and effective nuclear charges of 10+ and 8+ pulling the electrons in very close… Li Be B C N O F

13. z* = 1+ z* = 9+ z* = 11+ + + + + + + + + + + + + + + + + + + + + + Na Ne After Ne comes Na. Predict what Na’s Bohr diagram would look like (8). It has an electron configuration: 1s2 2s2 2p6 3s1 (or [Ne] 3s1), which means it has all the same electrons as Ne, plus one more electron in the 3rd level. Is this what you predicted? Is this what you predicted? And now predict what the z* values would be for each level (9). Is this what you predicted? And now predict what the z* values would be for each level (9). Is this what you were thinking? Li Be B C N O F z* = 8+ z* = 10+

14. z* = 1+ z* = 9+ z* = 11+ + + + + + + + + + + + + + + + + + + + + + Ne And what would you predict for the size of an Na atom? Its outermost electron is only being held in place with a very weak 1+ effective nuclear charge. So how far out do you think that valence (outermost) level would be (10) ? Is this what you were picturing? Its outermost electron is only being held in place with a very weak 1+ effective nuclear charge. So how far out do you think that valence (outermost) level would be (10) ? Li Be B C N O F z* = 8+ z* = 10+ 180 pm Na

15. Ne Na Mg Al Si P S Cl Ar + + + + + + + + + + + + + + + + + + + + + Na Ne Placing these back into the series… shows an obvious break in the trend for this period. Thus it would seem to make more sense to have Na start a new period. Placing these back into the series… shows an obvious break in the trend for this period. Thus it would seem to make more sense to have Na start a new period. After all, Na does have just one valence electron (just like Li), and it has a large radius (just like Li). In fact Na is bigger! Write down your prediction as to why you think Na is bigger than Li (11) . Placing these back into the series… Li Be B C N O F Did you say that Na is bigger because it has a whole extra level compared to Li? If so, good job. But what happens next? After Na is Mg, which has two electrons in its valence level (just like Be), each of which experiences a z* of 2+ (just like Be). This would make Mg smaller than Na, just as Be was smaller than Li. After Na is Mg, which has two electrons in its valence level (just like Be), each of which experiences a z* of 2+ (just like Be). After that, the pattern continues for period 3 just as it did for period 2.

16. Li Be B C N O F Ne Mg Al Si P S Cl Ar Na Notice how as you go left to right across the periodic table, the atomic radius steadily decreases. Notice how as you go left to right across the periodic table, the atomic radius steadily decreases. And as you move down the periodic table, the atomic radius gradually increases.

17. Using these trends, and the periodic table below… Ans: Si Predict (12) which would be larger: C or Si? Ans: As (13) Which would be larger: As or Kr? Ans: S (14) Which would be larger: S or Ne?

18. z* = 3+ z* = 2+ z* = 5+ z* = 4+ + + + + + + + + + + + + As it turns out, size isn’t the only characteristic of the atom influenced by the effective nuclear charge. “Ionization energy” is also affected. Ionization energy is the energy required to remove an electron from an atom. The stronger the force holding the electron in place, the more energy it will take to remove it. Try to remove Li’s outermost electron below (see arrow) by clicking on it repeatedly. Try to remove Li’s outermost electron below (see arrow) by clicking on it repeatedly. That didn’t take very much energy, did it? Try to remove Li’s outermost electron below (see arrow) by clicking on it repeatedly. That didn’t take very much energy, did it? Now predict (15) how hard it will be to try to remove one of Be’s outermost electrons (see arrow). Click on it and see if you were right. That was a little harder, wasn’t it? How about B? How will its ionization compare with the other two (16)? Try to remove Li’s outermost electron below (see arrow) by clicking on it repeatedly. That didn’t take very much energy, did it? Now predict (15) how hard it will be to try to remove one of Be’s outermost electrons (see arrow). Click on it and see if you were right. z* = 1+ z* = 3+ Li Be B

19. z* = 3+ z* = 2+ z* = 5+ z* = 4+ + + + + + + + + + + + + Go ahead and click on the electron (see arrow) and find out. Was your prediction correct? With an effective nuclear charge of 3+ acting on the valence electrons, they are especially difficult to remove. Was your prediction correct? So what do you think the general trend is for ionization energy as you go from left to right across the periodic table: Increase or decrease? And why (17) ? So what do you think the general trend is for ionization energy as you go from left to right across the periodic table: Increase or decrease? And why (17) ? If you said increase, you were right! And if you also said that it increases because there is an increase in the effective nuclear charge holding these valence electrons in place, well then you are a true genius! z* = 1+ z* = 3+ Li Be B

20. z* = 8+ z* = 10+ + + + + + + + + + So, with an effective nuclear charge of 8+ holding it in place, how hard would it be to remove a valence electron from an atom of Ne (18)? Go ahead and click on the electron (see arrow) to see just how hard… So… are you exhausted!!?? Ne

21. z* = 1+ z* = 1+ + + + + + + + + + + + + + + And how do you think Na’s ionization energy would compare with Li’s? Na has a lot more protons than Li and therefore a much greater positive charge on the nucleus (11+ compared to only 3+), but all that really matters is the charge “felt” by the valence electron (the z*). What are the z*’s on the valence electrons in these two atoms (19) ? Li If you said z* = 1+ for both good job! So how will their ionization energies compare? Write down your prediction before continuing (20). Now click on Li’s valence electron (see arrow) to see how much energy it takes to remove it Pretty easy, huh? Now try Na’s (see arrow) Wow, that was even easier! Why (21)? Even though they both have the same z*, since Na is bigger, the electron is a little farther away from the nucleus, and that means the force holding it in place is a little weaker. Since the force is a little weaker, the ionization energy is a little lower. Na

22. Fill in the following blanks (22 & 23): As you go from left to right across the periodic table, ionization energy generally tends to ______ because the ______________ tends to _______. increase effective nuclear charge (z*) increase As you go down the per. table, ionization energy generally tends to _______ . The z* _____________, but the ___________ tends to ________ and this causes a ________ force of attraction. decrease atomic radius remains the same increase weaker

23. Using these trends, and the periodic table below… Ans: C (24) Which would have a greater I.E.: C or Si? Ans: Kr (25) Which would have a greater I.E.: As or Kr? Ans: Ne (26) Which would have a greater IE? S or Ne

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