For students taking Chemistry 101 at the University of Victoria in the Fall 2013 term.
Monday, September 23, 2013
Chapter 2
Any questions on Chapter 2 material? Ask them here.
53 comments:
Anonymous
said...
1. For the transition metals, many different ionized forms of them can exist. What makes these forms stable? Ex. Copper 2+ will have an electron config of [Ar]3d9
2. The Zeff of Sodium is actually +2.5 because of the 3s electron spending some time near the nucleus. Do we need to know about things like this, or can we always just use the simple formula you gave us (Zeff=Z-S)? Do we need to know about Slater's rules?
3. If we take Cobalt [Ar]4s2 3d7 and take away 2 electrons we get [Ar]3d7. If we then add 2 electrons to Cobalt 2+ we get [Ar]3d9. (this is according to the rules that say when we take away electrons we take from the shell with highest principal quantum number and when we add electrons we add to shell with least quantum number). So, if we take away 3 electrons and then add them back, we get something different? Is that right?
4. What type of trends for electron affinities do we need to know? They're sort of all over the place.
So...when we are considering the Zeff of electrons in the 3d orbital, do we say that the 4s is shielding it? Or do we assume only the 1s2s2p orbitals are shielding it.
Although the 4s is lower in energy, does it spend more than nearer to the nucleus than the 3d? or is the 4s bigger, and thus does a bad job screening the 3d?
^ for the first question #3 about cobalt... since Ionization energy= -electron affinity, would that still be the case if we took an electron out of the 4s and put it back into the 3d? When putting electrons back into an atom, 3d fills before 4s, right?
317: 1. They're made stable by the strengths of the bonds they form, i.e. you get the energy back that was spent on ionization energy. But there are limits; Os is sometimes found in the +8 state, but it is impossible to remove even that many from Ir, Pt, Au, etc., even though they have more valence electrons. As e- are removed from the d shell, the remaining d-electrons become more and more core-like. 2. You need to be aware that the formula Zeff = Z - S (S = no. of core electrons) is a VERY rough approximation. Slater's rules approximate Zeff better, but no, we don't expect you to know these rules. 3. The filling order is different from the removing order in this case, yes. That's because the energies of orbitals are not constant, and different orbitals are affected in different ways by removal of electrons. 4. I've not yet taught this in class. Ask me again after I have if it is still not clear. 921: No. 4s and 3d are both valence (non-core) electrons, and shield each other poorly. Assume only the 1s/2s/2p/3s/3p electrons are shielding. 1049: wait until I have taught electron affinity in class.
Assume valence electrons shield each other poorly (not at all, in the approximation Zeff = Z - core electrons). So no, 3d electrons do not shield 4s electrons, because they are both valence shells. So yes to "Ex. 4s, 3d, 4p all shield 5s but 5s,4d,5p do not shield each other?"
So then to just to clarify- Each new row in the period table is when we start counting the "S" in Zeff=Z-S. Therefore, in the 6th row of the periodic table, if we look at the 6p6 electron, it would have a Zeff of +32? Unshielded by all the electrons in the same period?
When we use the term "shell" are we refering to all subshells associated with quantum number n, or the entire row in the periodic table?
Ex 1. electron shell 4= 4s,3d,4p OR 2. electron shell 4= 4s,4p,4d,4f
If #2 is the correct usage of the word "shell," isn't that a bit ambiguous when we are determining Zeff, as S isn't screened by electrons in the same "shell", but 4s IS screened by 4f, as 4f is in a lower row in the periodic table
620, 621: The Zeff approximation is simply to teach you about trends across the periodic table, and it does a decent job of that. In elements with 6p electrons, the 5d and 4f orbitals are sufficiently contracted to be core-like (i.e. they are too low in energy to be chemically accessible), and so they are pretty effectively shielded. The approximation works OK for the first 20 elements or so. If you want a more accurate estimate, you can apply Slater's rules (not gone into in depth in your textbook and not examinable).
806, 807: n gives you shell, so yes, 2 is the correct definition (4s, 4p etc are subshells). 1038: Both of these are excited state electronic configurations. Ionization energies and electron affinities are measured from the ground state. Yes, it would be easier to remove an electron from a higher energy orbital.
is 4S or 3D higher in energy? When we ionize atoms, 4S is higher (that's why we take it first), but when we fill up electron shells left to right in the periodic table, we fill 4S first. What's going on? How can this be?
If the question, which is higher in energy, 4S or 3D, what would be the correct answer? Or does it really depend on the situation?
if we look at the element that has electron configuration [Ar]4s1 3p5 and take 1 electron it becomes [Ar]3p5. If we add an electron back to [Ar]3p5 we get [Ar]3p6.
Does electron affinity of [Ar]3p5 equal the negative of the ionization energy of [Ar]4p13s5 in this case?
Recall the clicker question you did in class with the chlorine gas, and how Eaffinity=-ionization energy
623: Yes, it depends. Orbitals are affected by increasing Zeff in different ways. So for the transition metals as elements, the ns and (n-1)d orbitals are similar in energy. But for transition metal *ions*, the ns orbitals are higher in energy, i.e. the (n-1)d orbitals are lowered more in energy by the increasing Zeff. 1007: I think you mean d when you're saying p. The electron affinity of an element is the negative ionization of that elements anion, i.e. it takes the same amount of energy to remove an electron from an anion as was released when you gave it in the first place.
re:1007: So it seems the energy to give and electron and take it back are equal and opposite. But, is the energy required to take an electron and then give it back also equal and opposite?
Keep in mind that when we take an electron from [Ar]4s13d5 and then give it back we end up with [Ar]3d6....different electron configs.
So does taking the first 4s electron and then putting in the 6th 3d electron use equal and opposite energies?
I don't follow your line of thinking. Why do you suggest you take an electron from the 4s orbital and then return it to the 3d? There is no question this issue is more complicated than is taught in first year texts; I will spend a little time in class explaining it in more detail.
^ we take electrons from the highest value n and put it back in the lowest value n. Those are the rules in the text. Thus, if we took an electron from [Ar]4s13d5 and put it back we'd end up with [ar]3d6.
^ page 58. "When electrons are removed from an atom to form a cation, they are always removed first from the occupied orbitals have the largest principal quantum number, n."
"Electrons added to an atom to form an anion are added to the empty or partially filled orbital have the lowest value of n."
Perhaps the rule doesn't work for ions? But then isn't that a bit ambiguous if we take 3 electrons from an atom? In reality we take 1 electron from the atom, and then take another electron from the ion. And so we wouldn't be able to use the rules for that ion...
I'm still not quite sure I understand what you're asking (it may be best for you to come to my office hours). The rule is that you fill according to the PT, then remove e- starting with the orbital of highest n. If you get asked (a rather peculiar!) question such as "what is the electron configuration of a 4+ ion that has had an electron added to it?", just work out the electron configuration of the 3+ ion. The second quote concerns itself with anions - these are typically easy to assign an e- configuration because the vast majority add enough e- to get a noble gas configuration.
I have been going through the homework assignment 2, chapter 2 and have been finding a lot of questions that ask "using Slater's rules...". Do we need to know this? and if so, where in the notebook was this taught?
You don't need to know Slater's rules. They're just a better approximation for Zeff, but the very rough method you were taught conveys all the important trends.
re: I'm still not quite sure I understand what you're asking (it may be best for you to come to my office hours).
When we add electrons (the number of protons is remaining constant), we always add to lowest value n. the text says this on page 58.
We therefore add to 3d before 4s. (again, protons remain constant. It is still the same element)
So... if we take an electron from [Ar]4s13p5 we get [Ar]3p5 and if we put an electron back into there we get [Ar]3p6.
Do you see what i'm getting at? Are the energies required to take that electron and put it back equal and opposite? Considering the electron is in a different orbital.
In the case you describe, you are removing an electron from a ground state atom and returning it to make a higher energy, excited state atom. So no, these energies would not be the same. You would get less energy back than you initially put in to remove it. The p58 quote relates to the formation of anions, and is awkwardly worded anyway (you don't add electrons to the empty 4f orbital when forming I- for example, you add it to the partially filled 5p).
^yeah, ok I get that now...so we always fill electrons according to the order of the periodic table? Not the lowest value of n as described in the book?
Why is the atomic radius of Galium bigger than Zinc? Doesn't Zeff decrease down a period? Is it because the 3d electrons become more corelike, and shield the 4s electrons?
Zeff increases across a row, yes, but you're right: by the time you have filled the d orbitals with electrons, they have become very core-like, and they shield the 4p electrons effectively.
I have read the text and I still don't get what Zeff is for and how do you find S? the core electrons. Like for Ne, aren't all the electrons core electrons because it is stable?
The core electrons are the ones that shield the electrons you're considering from the nuclear charge. So the outermost electrons for Ne are the 2p electrons. They're not shielded from the nucleus by the 2p or 2s electrons, but the 1s electrons do shield them. So from the point of view of the 2p electrons, they see the nuclear charge as being +8, i.e. the 10 protons in the nucleus less the 2 shielding electrons. So the high EFFECTIVE nuclear charge means those outermost electrons are drawn in tightly to the nucleus.
Hello, Did we cover Slater's rules in class? I cannot find anything in my notes and the explanation in the text is confusing.
The first homework assignment question for chapter two asks: "What values do you estimate for Zeff using Slater's rules?" (for Na and K) How do we determine this?
You don't need to know Slater's rules. They're just a better approximation for Zeff, but the very rough method you were taught conveys all the important trends. Just skip that question.
Could you please explain this question from the chapter 2 homework? Thanks
Consider the isoelectronic ions F− and Na+. Using the equation Zeff=Z−S 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.
5:27 For F- the Zeff would be: Z-S where Z is the number of protons (9) and S is the number or core electrons (2 from the 1S orbital) so z-s would be 9-2=7 For Na+ the Zeff would be : Z-S where Z is the number of protons (11) and S is the number or core electrons (2 from the 1S orbital) so z-s would be 11-2=9
One of my adaptive follow up questions for the chapter 2 homework was about element 117. It had choice questions as follows. Could you please explain how to do each of these? I have no idea and resorted to blindly guessing... I'd rather not do that on a test! Thank you.
Consider element 117.
Estimate a value for its first ionization energy based on its position in the periodic table. (correct answer: 805 kJ/mol)
Estimate a value for its electron affinity based on its position in the periodic table. (correct answer: -235 kJ/mol)
Estimate a value for its atomic size based on its position in the periodic table. (correct answer: 1.65 Å)
Estimate a value for its common oxidation state based on its position in the periodic table. (correct answer: -1)
It's a periodic trends question, that requires you to establish that it is a Group 17 element and hence should have values following the trend F-Cl-Br-I-At. So you would need to look up these values for the other elements.
A question I got on mastering chemistry said "Which ion is smaller? Co^3+ or Co^4+ ?"
I thought as you moved from left to right radius decreased (as seen on the second slide in section 2.3), and Co^3+ is further right, so it would be smaller. Mastering chemistry gives Co^4+ as the correct answer.
Did I misunderstand the image or miss something else? Or is mastering chemistry incorrect? Thanks
943 As you move from left to right across the periodic table the radius does decrease, but when we are comparing Co+3 and Co+4 , they are both Cobalt so one is not further left or right than the other. The difference between the two is the number of electrons. Co+3 has one more electron than the 4+ form, so in the 4+ form the fewer electrons would be held more closely to the nucleus than in the 3+ form, thus it would be smaller.
Based on their positions in the periodic table, predict which atom of the following pair will have the smaller first ionization energy: Correct answer: Cu
Why is the answer Cu and not Fe? Cu is to the right of Fe so it would experience more Zeff and be harder to remove?
Cu is anomalous in this context, so that's not a good question. Your reasoning is sound; the 2nd & 3rd ionization energies are higher for Cu, for example. Removing the solitary 4s electron from Cu is easier than expected, probably due to the 10 3d electrons doing a pretty decent job of shielding the 4s from the nucleus.
do the dx^2-y^2 and dz^2 orbitals have 2 angular nodes? It sure doesn't look that way.... like the dx^2-y^2 orbital just looks like the combined px and py orbitals, so it should have no angular nodes?
Yes, they both do. The dx2-y2 is exactly the same as the dxy, just rotated 45 degrees. The dz2 has a nodal cone, which is equivalent to 2 nodal planes.
I was wondering if we need to know Slater's Rules for calculating Zeff, or just the generic Z-S, with all core electrons counting as '1' and valence electrons counting as '0'.
53 comments:
1. For the transition metals, many different ionized forms of them can exist. What makes these forms stable? Ex. Copper 2+ will have an electron config of [Ar]3d9
2. The Zeff of Sodium is actually +2.5 because of the 3s electron spending some time near the nucleus. Do we need to know about things like this, or can we always just use the simple formula you gave us (Zeff=Z-S)? Do we need to know about Slater's rules?
3. If we take Cobalt [Ar]4s2 3d7 and take away 2 electrons we get [Ar]3d7. If we then add 2 electrons to Cobalt 2+ we get [Ar]3d9. (this is according to the rules that say when we take away electrons we take from the shell with highest principal quantum number and when we add electrons we add to shell with least quantum number). So, if we take away 3 electrons and then add them back, we get something different? Is that right?
4. What type of trends for electron affinities do we need to know? They're sort of all over the place.
Zeff=Z-S
So...when we are considering the Zeff of electrons in the 3d orbital, do we say that the 4s is shielding it? Or do we assume only the 1s2s2p orbitals are shielding it.
Although the 4s is lower in energy, does it spend more than nearer to the nucleus than the 3d? or is the 4s bigger, and thus does a bad job screening the 3d?
^ for the first question #3 about cobalt...
since Ionization energy= -electron affinity, would that still be the case if we took an electron out of the 4s and put it back into the 3d?
When putting electrons back into an atom, 3d fills before 4s, right?
317: 1. They're made stable by the strengths of the bonds they form, i.e. you get the energy back that was spent on ionization energy. But there are limits; Os is sometimes found in the +8 state, but it is impossible to remove even that many from Ir, Pt, Au, etc., even though they have more valence electrons. As e- are removed from the d shell, the remaining d-electrons become more and more core-like.
2. You need to be aware that the formula Zeff = Z - S (S = no. of core electrons) is a VERY rough approximation. Slater's rules approximate Zeff better, but no, we don't expect you to know these rules.
3. The filling order is different from the removing order in this case, yes. That's because the energies of orbitals are not constant, and different orbitals are affected in different ways by removal of electrons.
4. I've not yet taught this in class. Ask me again after I have if it is still not clear.
921: No. 4s and 3d are both valence (non-core) electrons, and shield each other poorly. Assume only the 1s/2s/2p/3s/3p electrons are shielding.
1049: wait until I have taught electron affinity in class.
When we want the Zeff of the 4s, do we assume 3d is shielding it, even though 3d is higher in energy than 4s?
If no, then are we saying that in the periodic table, the shielding effect starts up after every row in the table?
Ex. 4s, 3d, 4p all shield 5s but 5s,4d,5p do not shield each other?
Or should it be 4s4p4d4f all shield 5s but 5s5p5d5f and theoretically 5g don't shield eachother?
Assume valence electrons shield each other poorly (not at all, in the approximation Zeff = Z - core electrons). So no, 3d electrons do not shield 4s electrons, because they are both valence shells. So yes to "Ex. 4s, 3d, 4p all shield 5s but 5s,4d,5p do not shield each other?"
So then to just to clarify-
Each new row in the period table is when we start counting the "S" in Zeff=Z-S.
Therefore, in the 6th row of the periodic table, if we look at the 6p6 electron, it would have a Zeff of +32? Unshielded by all the electrons in the same period?
^ the 6p6 electron is unsheided by the 6s, 4f, 5d, and 6p orbitals?
When we use the term "shell" are we refering to all subshells associated with quantum number n, or the entire row in the periodic table?
Ex 1. electron shell 4= 4s,3d,4p
OR 2. electron shell 4= 4s,4p,4d,4f
If #2 is the correct usage of the word "shell," isn't that a bit ambiguous when we are determining Zeff, as S isn't screened by electrons in the same "shell", but 4s IS screened by 4f, as 4f is in a lower row in the periodic table
^ 4f is screened by 4s, my bad
does [Ar]4s1 3p5 have the same ionization energy and affinity as [Ar]3p6?
[Ar]4p13d5 would be easier to take an electron from because of the 4s being bigger right? Also, is [Ar]4p13d5 bigger than [Ar]3p6?
620, 621: The Zeff approximation is simply to teach you about trends across the periodic table, and it does a decent job of that. In elements with 6p electrons, the 5d and 4f orbitals are sufficiently contracted to be core-like (i.e. they are too low in energy to be chemically accessible), and so they are pretty effectively shielded. The approximation works OK for the first 20 elements or so. If you want a more accurate estimate, you can apply Slater's rules (not gone into in depth in your textbook and not examinable).
806, 807: n gives you shell, so yes, 2 is the correct definition (4s, 4p etc are subshells).
1038: Both of these are excited state electronic configurations. Ionization energies and electron affinities are measured from the ground state. Yes, it would be easier to remove an electron from a higher energy orbital.
Are we given the same data sheet as the one on the back of the chem 101 notebook on midterms and exams?
Yes
is 4S or 3D higher in energy? When we ionize atoms, 4S is higher (that's why we take it first), but when we fill up electron shells left to right in the periodic table, we fill 4S first. What's going on? How can this be?
If the question, which is higher in energy, 4S or 3D, what would be the correct answer? Or does it really depend on the situation?
if we look at the element that has electron configuration [Ar]4s1 3p5 and take 1 electron it becomes [Ar]3p5. If we add an electron back to [Ar]3p5 we get [Ar]3p6.
Does electron affinity of [Ar]3p5 equal the negative of the ionization energy of [Ar]4p13s5 in this case?
Recall the clicker question you did in class with the chlorine gas, and how Eaffinity=-ionization energy
623: Yes, it depends. Orbitals are affected by increasing Zeff in different ways. So for the transition metals as elements, the ns and (n-1)d orbitals are similar in energy. But for transition metal *ions*, the ns orbitals are higher in energy, i.e. the (n-1)d orbitals are lowered more in energy by the increasing Zeff.
1007: I think you mean d when you're saying p. The electron affinity of an element is the negative ionization of that elements anion, i.e. it takes the same amount of energy to remove an electron from an anion as was released when you gave it in the first place.
re:1007: So it seems the energy to give and electron and take it back are equal and opposite. But, is the energy required to take an electron and then give it back also equal and opposite?
Keep in mind that when we take an electron from [Ar]4s13d5 and then give it back we end up with [Ar]3d6....different electron configs.
So does taking the first 4s electron and then putting in the 6th 3d electron use equal and opposite energies?
I don't follow your line of thinking. Why do you suggest you take an electron from the 4s orbital and then return it to the 3d?
There is no question this issue is more complicated than is taught in first year texts; I will spend a little time in class explaining it in more detail.
Actually, I won't do it in class, because it will take too long and probably confuse more people than it helps. Look out for a future blog posting.
^ we take electrons from the highest value n and put it back in the lowest value n. Those are the rules in the text. Thus, if we took an electron from [Ar]4s13d5 and put it back we'd end up with [ar]3d6.
Do these rules not apply for ions or something?
Can you point me to where it says that in the text?
^ page 58. "When electrons are removed from an atom to form a cation, they are always removed first from the occupied orbitals have the largest principal quantum number, n."
"Electrons added to an atom to form an anion are added to the empty or partially filled orbital have the lowest value of n."
Perhaps the rule doesn't work for ions? But then isn't that a bit ambiguous if we take 3 electrons from an atom? In reality we take 1 electron from the atom, and then take another electron from the ion. And so we wouldn't be able to use the rules for that ion...
I'm still not quite sure I understand what you're asking (it may be best for you to come to my office hours). The rule is that you fill according to the PT, then remove e- starting with the orbital of highest n. If you get asked (a rather peculiar!) question such as "what is the electron configuration of a 4+ ion that has had an electron added to it?", just work out the electron configuration of the 3+ ion.
The second quote concerns itself with anions - these are typically easy to assign an e- configuration because the vast majority add enough e- to get a noble gas configuration.
I have been going through the homework assignment 2, chapter 2 and have been finding a lot of questions that ask "using Slater's rules...". Do we need to know this? and if so, where in the notebook was this taught?
You don't need to know Slater's rules. They're just a better approximation for Zeff, but the very rough method you were taught conveys all the important trends.
re: I'm still not quite sure I understand what you're asking (it may be best for you to come to my office hours).
When we add electrons (the number of protons is remaining constant), we always add to lowest value n.
the text says this on page 58.
We therefore add to 3d before 4s. (again, protons remain constant. It is still the same element)
So... if we take an electron from [Ar]4s13p5 we get [Ar]3p5 and if we put an electron back into there we get [Ar]3p6.
Do you see what i'm getting at?
Are the energies required to take that electron and put it back equal and opposite? Considering the electron is in a different orbital.
^ replace all the p's with d's. sorry
In the case you describe, you are removing an electron from a ground state atom and returning it to make a higher energy, excited state atom. So no, these energies would not be the same. You would get less energy back than you initially put in to remove it.
The p58 quote relates to the formation of anions, and is awkwardly worded anyway (you don't add electrons to the empty 4f orbital when forming I- for example, you add it to the partially filled 5p).
^yeah, ok I get that now...so we always fill electrons according to the order of the periodic table? Not the lowest value of n as described in the book?
Yes.
Why is the atomic radius of Galium bigger than Zinc?
Doesn't Zeff decrease down a period?
Is it because the 3d electrons become more corelike, and shield the 4s electrons?
Zeff increases across a row, yes, but you're right: by the time you have filled the d orbitals with electrons, they have become very core-like, and they shield the 4p electrons effectively.
I have read the text and I still don't get what Zeff is for and how do you find S? the core electrons. Like for Ne, aren't all the electrons core electrons because it is stable?
The core electrons are the ones that shield the electrons you're considering from the nuclear charge. So the outermost electrons for Ne are the 2p electrons. They're not shielded from the nucleus by the 2p or 2s electrons, but the 1s electrons do shield them. So from the point of view of the 2p electrons, they see the nuclear charge as being +8, i.e. the 10 protons in the nucleus less the 2 shielding electrons. So the high EFFECTIVE nuclear charge means those outermost electrons are drawn in tightly to the nucleus.
So from what orbital do you remove an electron if both s and p have the same value for n?
From the one with the highest value of l, i.e. the p orbital.
Hello,
Did we cover Slater's rules in class? I cannot find anything in my notes and the explanation in the text is confusing.
The first homework assignment question for chapter two asks:
"What values do you estimate for Zeff using Slater's rules?" (for Na and K)
How do we determine this?
Thank you
You don't need to know Slater's rules. They're just a better approximation for Zeff, but the very rough method you were taught conveys all the important trends. Just skip that question.
Could you please explain this question from the chapter 2 homework? Thanks
Consider the isoelectronic ions F− and Na+.
Using the equation Zeff=Z−S 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.
5:27
For F- the Zeff would be: Z-S where Z is the number of protons (9) and S is the number or core electrons (2 from the 1S orbital) so z-s would be 9-2=7
For Na+ the Zeff would be : Z-S where Z is the number of protons (11) and S is the number or core electrons (2 from the 1S orbital) so z-s would be 11-2=9
One of my adaptive follow up questions for the chapter 2 homework was about element 117. It had choice questions as follows. Could you please explain how to do each of these? I have no idea and resorted to blindly guessing... I'd rather not do that on a test!
Thank you.
Consider element 117.
Estimate a value for its first ionization energy based on its position in the periodic table. (correct answer: 805 kJ/mol)
Estimate a value for its electron affinity based on its position in the periodic table. (correct answer: -235 kJ/mol)
Estimate a value for its atomic size based on its position in the periodic table. (correct answer: 1.65 Å)
Estimate a value for its common oxidation state based on its position in the periodic table. (correct answer: -1)
It's a periodic trends question, that requires you to establish that it is a Group 17 element and hence should have values following the trend F-Cl-Br-I-At. So you would need to look up these values for the other elements.
A question I got on mastering chemistry said "Which ion is smaller? Co^3+ or Co^4+ ?"
I thought as you moved from left to right radius decreased (as seen on the second slide in section 2.3), and Co^3+ is further right, so it would be smaller. Mastering chemistry gives Co^4+ as the correct answer.
Did I misunderstand the image or miss something else? Or is mastering chemistry incorrect?
Thanks
943
As you move from left to right across the periodic table the radius does decrease, but when we are comparing Co+3 and Co+4 , they are both Cobalt so one is not further left or right than the other. The difference between the two is the number of electrons. Co+3 has one more electron than the 4+ form, so in the 4+ form the fewer electrons would be held more closely to the nucleus than in the 3+ form, thus it would be smaller.
Based on their positions in the periodic table, predict which atom of the following pair will have the smaller first ionization energy:
Correct answer: Cu
Why is the answer Cu and not Fe?
Cu is to the right of Fe so it would experience more Zeff and be harder to remove?
Cu is anomalous in this context, so that's not a good question. Your reasoning is sound; the 2nd & 3rd ionization energies are higher for Cu, for example. Removing the solitary 4s electron from Cu is easier than expected, probably due to the 10 3d electrons doing a pretty decent job of shielding the 4s from the nucleus.
do the dx^2-y^2 and dz^2 orbitals have 2 angular nodes? It sure doesn't look that way.... like the dx^2-y^2 orbital just looks like the combined px and py orbitals, so it should have no angular nodes?
Yes, they both do. The dx2-y2 is exactly the same as the dxy, just rotated 45 degrees. The dz2 has a nodal cone, which is equivalent to 2 nodal planes.
I was wondering if we need to know Slater's Rules for calculating Zeff, or just the generic Z-S, with all core electrons counting as '1' and valence electrons counting as '0'.
When n intial is inifity do we use the n=5 for the equation 1/wavelength =RH(1/nf^2)-(1/ni^2)?
150: No. The simpler rule gives you the trends you need.
103: no, use n = infinity. 1/infinity^2 = 0.
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