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Posted: 9/29/2014 8:47:20 PM EDT
Trying to find out how to arrive at the answer given.
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Posted: 9/29/2014 8:49:48 PM EDT
[#1]
If we do your homework for you, how will you learn it for the test?
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Posted: 9/29/2014 8:51:22 PM EDT
[#2]
Quote HistoryQuoted:
If we do your homework for you, how will you learn it for the test?
doing my homework for me? I provided the answer. 
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Posted: 9/29/2014 8:54:50 PM EDT
[#3]
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doing my homework for me? I provided the answer. Quote HistoryQuoted:
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If we do your homework for you, how will you learn it for the test?
doing my homework for me? I provided the answer. Are you sure thats the answer? I'd make sure. |
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Posted: 9/29/2014 8:56:27 PM EDT
[#4]
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Are you sure thats the answer? I'd make sure. Quote HistoryQuoted:
Quoted:
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If we do your homework for you, how will you learn it for the test?
doing my homework for me? I provided the answer. Are you sure thats the answer? I'd make sure. those are the answer's the instructor gave us to check our work. but its not matching. Im tryign to figure out what im doing wrong. otherwise, if i understand correctly . A. should be Pb=O(with a negative charge of 2). on the paper it looks like O to the power of negative two. |
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Posted: 9/29/2014 9:01:04 PM EDT
[#5]
I don't understand the question. I'm also having trouble following your notation with the subscripts or superscripts, can you take a picture of your handwritten work and post that? It would make more sense.
ETA: What is the verbatim question the professor is asking? Your quote never closes and I can't figure out where his question stops and your commentary starts. |
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Posted: 9/29/2014 9:10:03 PM EDT
[#6]
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I don't understand the question. I'm also having trouble following your notation with the subscripts or superscripts, can you take a picture of your handwritten work and post that? It would make more sense. ETA: What is the verbatim question the professor is asking? Your quote never closes and I can't figure out where his question stops and your commentary starts. posted. |
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Posted: 9/29/2014 9:19:43 PM EDT
[#7]
Okay, its an oxidation state question.
You need to find compounds that made from each metal combined with Oxygen in its oxidation state of (-2). Transition metals (and post-transition metals) don't always carry the same charge- they can be oxidized and reduced. In PbO for instance, the oxygen carries a (-2) charge, and the Pb carries a (+2) charge. On the other hand, in PbO2 oxygen still carries a (-2) charge, but Pb carries a (+4) charge. A lot of times the oxidation state of the metal is included in the name, like PbO2 is Lead(IV) Oxide; that way you know the oxidation state already. Some oxidation states are more stable than others depending on the molecule, but I think explaining that is above this level of chemistry by a bit. For now just understand that some atoms can exist in multiple oxidation states, and I guess memorize which ones have which. |
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Posted: 9/29/2014 9:25:55 PM EDT
[#8]
Pb can have different oxidative states
I think thsts one lesson he wants you to,learn. See above. 
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Posted: 9/29/2014 9:28:51 PM EDT
[#9]
Quote HistoryQuoted:
Okay, its an oxidation state question. You need to find compounds that made from each metal combined with Oxygen in its oxidation state of (-2). Transition metals (and post-transition metals) don't always carry the same charge- they can be oxidized and reduced. In PbO for instance, the oxygen carries a (-2) charge, and the Pb carries a (+2) charge. On the other hand, in PbO2 oxygen still carries a (-2) charge, but Pb carries a (+4) charge. A lot of times the oxidation state of the metal is included in the name, like PbO2 is Lead(IV) Oxide; that way you know the oxidation state already. Some oxidation states are more stable than others depending on the molecule, but I think explaining that is above this level of chemistry by a bit. For now just understand that some atoms can exist in multiple oxidation states, and I guess memorize which ones have which. Just to clarify the part bolded. because O has a (-2) charge. Pb in the compound has to be (+2). now, in the compound. we got PbO assuming the above. But why in the other case is it PbO2? how are we figuring that out from the provided question? we're sopposed to figure this out via the periodic table. |
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Posted: 9/29/2014 9:30:11 PM EDT
[#10]
Quote HistoryQuoted:
Just to clarify the part bolded. because O has a (-2) charge. Pb in the compound has to be (+2). now, in the compound. we got PbO assuming the above. But why in the other case is it PbO2? how are we figuring that out from the provided question? Quote HistoryQuoted:
Quoted:
Okay, its an oxidation state question. You need to find compounds that made from each metal combined with Oxygen in its oxidation state of (-2). Transition metals (and post-transition metals) don't always carry the same charge- they can be oxidized and reduced. In PbO for instance, the oxygen carries a (-2) charge, and the Pb carries a (+2) charge. On the other hand, in PbO2 oxygen still carries a (-2) charge, but Pb carries a (+4) charge. A lot of times the oxidation state of the metal is included in the name, like PbO2 is Lead(IV) Oxide; that way you know the oxidation state already. You have to know that lead has more than one oxidation state Some oxidation states are more stable than others depending on the molecule, but I think explaining that is above this level of chemistry by a bit. For now just understand that some atoms can exist in multiple oxidation states, and I guess memorize which ones have which. Just to clarify the part bolded. because O has a (-2) charge. Pb in the compound has to be (+2). now, in the compound. we got PbO assuming the above. But why in the other case is it PbO2? how are we figuring that out from the provided question? You have to know that Pb has more than one Ox state. Look at periodic table |
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Posted: 9/29/2014 9:32:31 PM EDT
[#11]
Quote HistoryQuoted:
Just to clarify the part bolded. because O has a (-2) charge. Pb in the compound has to be (+2). now, in the compound. we got PbO assuming the above. But why in the other case is it PbO2? how are we figuring that out from the provided question? we're sopposed to figure this out via the periodic table. Quote HistoryQuoted:
Quoted:
Okay, its an oxidation state question. You need to find compounds that made from each metal combined with Oxygen in its oxidation state of (-2). Transition metals (and post-transition metals) don't always carry the same charge- they can be oxidized and reduced. In PbO for instance, the oxygen carries a (-2) charge, and the Pb carries a (+2) charge. On the other hand, in PbO2 oxygen still carries a (-2) charge, but Pb carries a (+4) charge. A lot of times the oxidation state of the metal is included in the name, like PbO2 is Lead(IV) Oxide; that way you know the oxidation state already. Some oxidation states are more stable than others depending on the molecule, but I think explaining that is above this level of chemistry by a bit. For now just understand that some atoms can exist in multiple oxidation states, and I guess memorize which ones have which. Just to clarify the part bolded. because O has a (-2) charge. Pb in the compound has to be (+2). now, in the compound. we got PbO assuming the above. But why in the other case is it PbO2? how are we figuring that out from the provided question? we're sopposed to figure this out via the periodic table. It looks crossed out. Are you sure your prof is correct? They have been wrong before. Maybe you should ask someone to make sure. |
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Posted: 9/29/2014 9:33:46 PM EDT
[#12]
Quote HistoryQuoted:
It looks crossed out. Are you sure your prof is correct? They have been wrong before. Maybe you should ask someone to make sure. Quote HistoryQuoted:
Quoted:
Quoted:
Okay, its an oxidation state question. You need to find compounds that made from each metal combined with Oxygen in its oxidation state of (-2). Transition metals (and post-transition metals) don't always carry the same charge- they can be oxidized and reduced. In PbO for instance, the oxygen carries a (-2) charge, and the Pb carries a (+2) charge. On the other hand, in PbO2 oxygen still carries a (-2) charge, but Pb carries a (+4) charge. A lot of times the oxidation state of the metal is included in the name, like PbO2 is Lead(IV) Oxide; that way you know the oxidation state already. Some oxidation states are more stable than others depending on the molecule, but I think explaining that is above this level of chemistry by a bit. For now just understand that some atoms can exist in multiple oxidation states, and I guess memorize which ones have which. Just to clarify the part bolded. because O has a (-2) charge. Pb in the compound has to be (+2). now, in the compound. we got PbO assuming the above. But why in the other case is it PbO2? how are we figuring that out from the provided question? we're sopposed to figure this out via the periodic table. It looks crossed out. Are you sure your prof is correct? They have been wrong before. Maybe you should ask someone to make sure. No. that's a piece of eraser shaving. |
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Posted: 9/29/2014 9:34:41 PM EDT
[#13]
Quote HistoryQuoted:
You have to know that Pb has more than one Ox state. Look at periodic table Quote HistoryQuoted:
Quoted:
Quoted:
Okay, its an oxidation state question. You need to find compounds that made from each metal combined with Oxygen in its oxidation state of (-2). Transition metals (and post-transition metals) don't always carry the same charge- they can be oxidized and reduced. In PbO for instance, the oxygen carries a (-2) charge, and the Pb carries a (+2) charge. On the other hand, in PbO2 oxygen still carries a (-2) charge, but Pb carries a (+4) charge. A lot of times the oxidation state of the metal is included in the name, like PbO2 is Lead(IV) Oxide; that way you know the oxidation state already. You have to know that lead has more than one oxidation state Some oxidation states are more stable than others depending on the molecule, but I think explaining that is above this level of chemistry by a bit. For now just understand that some atoms can exist in multiple oxidation states, and I guess memorize which ones have which. Just to clarify the part bolded. because O has a (-2) charge. Pb in the compound has to be (+2). now, in the compound. we got PbO assuming the above. But why in the other case is it PbO2? how are we figuring that out from the provided question? You have to know that Pb has more than one Ox state. Look at periodic table then i guess the correct question would be. How do i determine the different oxidation states of a transition metal from the periodic table? Edit: you mean the one with Roman numerals? because mine doesn't not have roman numerals on it. |
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Posted: 9/29/2014 9:39:42 PM EDT
[#14]
Quote HistoryQuoted:
then i guess the correct question would be. How do i determine the different oxidation states of a transition metal from the periodic table? Quote HistoryQuoted:
Quoted:
Quoted:
Quoted:
Okay, its an oxidation state question. You need to find compounds that made from each metal combined with Oxygen in its oxidation state of (-2). Transition metals (and post-transition metals) don't always carry the same charge- they can be oxidized and reduced. In PbO for instance, the oxygen carries a (-2) charge, and the Pb carries a (+2) charge. On the other hand, in PbO2 oxygen still carries a (-2) charge, but Pb carries a (+4) charge. A lot of times the oxidation state of the metal is included in the name, like PbO2 is Lead(IV) Oxide; that way you know the oxidation state already. You have to know that lead has more than one oxidation state Some oxidation states are more stable than others depending on the molecule, but I think explaining that is above this level of chemistry by a bit. For now just understand that some atoms can exist in multiple oxidation states, and I guess memorize which ones have which. Just to clarify the part bolded. because O has a (-2) charge. Pb in the compound has to be (+2). now, in the compound. we got PbO assuming the above. But why in the other case is it PbO2? how are we figuring that out from the provided question? You have to know that Pb has more than one Ox state. Look at periodic table then i guess the correct question would be. How do i determine the different oxidation states of a transition metal from the periodic table? The tsble has changed since 1989 I suspect. but not Pb or that series
But look at the Roman numeral at top. IV it offers clues But. The answer I is this. When you see that he is asking you about PB and O Open your textbook to the discussion of the pertinent discussion of thst series of chemicals Likely, it will discuss their oxidatiion ststes along with other qualities |
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Posted: 9/29/2014 9:40:26 PM EDT
[#15]
You know that Transition metals have multiple oxidation states. You need to know which oxidation states are common for each metal. The chemistry behind which ones are stable in nature is fairly complex so lets ignore that and assume you can memorize common oxidation states.
Pb exists in nature with oxidation states of (+2) and (+4). You also know that oxygen always has an oxidation state of (-2), so the only two combinations of lead and oxygen are PbO and PbO2. Side note: stability of oxidation states is based on paired electrons in the valence shell and shell hybridization- this is more physical or inorganic chemistry material. ETA: To find oxidation state on the periodic table, start with finding the number of electrons needed to fill the valence shell (count towards the right until you hit the end of a section, whether thats the end of the D block for transition metals, or the noble gases for post-transition metals and non-metals). For lead, that number is 4- meaning it needs 4 electrons in order to be stable, its oxidation state is (+4); all the group 14 elements have an oxidation state of (+4). But some of them have other oxidation states too. Lead is also stable at (+2) due to electrons shifting from one orbital to another. |
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Posted: 9/29/2014 9:43:14 PM EDT
[#16]
Lol, high school chemistry.
I actually need the review... |
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Posted: 9/29/2014 9:44:10 PM EDT
[#17]
I kept the periodic table and somthing similar to this tacked to the wall in front of my desk
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Posted: 9/29/2014 9:49:43 PM EDT
[#18]
I would listen to Sonoran TJ
I took chemistry before many on here were born And my Alzheimer's isnt helping |
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Posted: 9/29/2014 9:50:46 PM EDT
[#19]
Quote HistoryQuoted:
I would listen to Sonoran TJ I took chemistry before many on here were born And my Alzheimer's isnt helping Possibly before I was born 
I'm a senior Chemistry student, so this is my bread and butter. |
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Posted: 9/29/2014 9:58:39 PM EDT
[#20]
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Possibly before I was born
I'm a senior Chemistry student, so this is my bread and butter. Quote HistoryQuoted:
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I would listen to Sonoran TJ I took chemistry before many on here were born And my Alzheimer's isnt helping Possibly before I was born
I'm a senior Chemistry student, so this is my bread and butter. First Chem class was 1986 Last undergraduate 1990 ( one year gen, one year o chem and one elective ) The elective? The chemistry and phRmacology of illicit drugs 5 students. Invitation only State crime lab We analyzed coke, marijuana etc. We made crack from cocaine powder It was a great class. Dr Raymond Poore at jacksonville state u. I took some Chem and physics in grad school but it was all biological medical in nature |
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Posted: 9/29/2014 9:59:44 PM EDT
[#21]
Quote HistoryQuoted:
Okay, its an oxidation state question. You need to find compounds that made from each metal combined with Oxygen in its oxidation state of (-2). Transition metals (and post-transition metals) don't always carry the same charge- they can be oxidized and reduced. In PbO for instance, the oxygen carries a (-2) charge, and the Pb carries a (+2) charge. On the other hand, in PbO2 oxygen still carries a (-2) charge, but Pb carries a (+4) charge. A lot of times the oxidation state of the metal is included in the name, like PbO2 is Lead(IV) Oxide; that way you know the oxidation state already. Some oxidation states are more stable than others depending on the molecule, but I think explaining that is above this level of chemistry by a bit. For now just understand that some atoms can exist in multiple oxidation states, and I guess memorize which ones have which. This ^ ^ ^ For instance, Manganese has +2 (MnO), +4 (MnO2), +7 (Mn2O7). It also has +3 and +6. |
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Posted: 9/29/2014 10:00:16 PM EDT
[#22]
Quote HistoryQuoted:
You know that Transition metals have multiple oxidation states. You need to know which oxidation states are common for each metal. The chemistry behind which ones are stable in nature is fairly complex so lets ignore that and assume you can memorize common oxidation states. Pb exists in nature with oxidation states of (+2) and (+4). You also know that oxygen always has an oxidation state of (-2), so the only two combinations of lead and oxygen are PbO and PbO2. Side note: stability of oxidation states is based on paired electrons in the valence shell and shell hybridization- this is more physical or inorganic chemistry material. ETA: To find oxidation state on the periodic table, start with finding the number of electrons needed to fill the valence shell (count towards the right until you hit the end of a section, whether thats the end of the D block for transition metals, or the noble gases for post-transition metals and non-metals). For lead, that number is 4- meaning it needs 4 electrons in order to be stable, its oxidation state is (+4); all the group 14 elements have an oxidation state of (+4). But some of them have other oxidation states too. Lead is also stable at (+2) due to electrons shifting from one orbital to another. oh. not quite gone through the shells and electrons material.. |
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Posted: 9/29/2014 10:01:18 PM EDT
[#23]
Quote HistoryQuoted: Possibly before I was born I'm a senior Chemistry student, so this is my bread and butter. Quote HistoryQuoted: Quoted: I would listen to Sonoran TJ I took chemistry before many on here were born And my Alzheimer's isnt helping Possibly before I was born I'm a senior Chemistry student, so this is my bread and butter. |
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Posted: 9/29/2014 10:08:47 PM EDT
[#24]
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Senior? Quote HistoryQuoted:
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I would listen to Sonoran TJ I took chemistry before many on here were born And my Alzheimer's isnt helping Possibly before I was born
I'm a senior Chemistry student, so this is my bread and butter. Technically in my 5th year, but yes. |
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Posted: 9/29/2014 10:12:55 PM EDT
[#25]
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Technically in my 5th year, but yes. Quote HistoryQuoted:
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I would listen to Sonoran TJ I took chemistry before many on here were born And my Alzheimer's isnt helping Possibly before I was born
I'm a senior Chemistry student, so this is my bread and butter. Technically in my 5th year, but yes. Just tell,people you are staying an extra year to do research
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Posted: 9/29/2014 10:21:12 PM EDT
[#26]
Quote HistoryQuoted:
oh. not quite gone through the shells and electrons material.. Quote HistoryQuoted:
Quoted:
You know that Transition metals have multiple oxidation states. You need to know which oxidation states are common for each metal. The chemistry behind which ones are stable in nature is fairly complex so lets ignore that and assume you can memorize common oxidation states. Pb exists in nature with oxidation states of (+2) and (+4). You also know that oxygen always has an oxidation state of (-2), so the only two combinations of lead and oxygen are PbO and PbO2. Side note: stability of oxidation states is based on paired electrons in the valence shell and shell hybridization- this is more physical or inorganic chemistry material. ETA: To find oxidation state on the periodic table, start with finding the number of electrons needed to fill the valence shell (count towards the right until you hit the end of a section, whether thats the end of the D block for transition metals, or the noble gases for post-transition metals and non-metals). For lead, that number is 4- meaning it needs 4 electrons in order to be stable, its oxidation state is (+4); all the group 14 elements have an oxidation state of (+4). But some of them have other oxidation states too. Lead is also stable at (+2) due to electrons shifting from one orbital to another. oh. not quite gone through the shells and electrons material.. Get through it, it's what the entire table is based upon. It makes much more sense, and is a much more useful tool, once you understand valence. |
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Posted: 9/29/2014 10:22:23 PM EDT
[#27]
Quote HistoryQuoted: Just tell,people you are staying an extra year to do research |
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Posted: 9/29/2014 10:28:02 PM EDT
[#28]
Quote HistoryQuoted:
In what, anthropogenic climate change? Quote HistoryQuoted:
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Just tell,people you are staying an extra year to do research Human sexuality, the effect of alcohol on the human body, and football.
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Posted: 9/29/2014 10:39:19 PM EDT
[#29]
Quote HistoryQuoted:
In what, anthropogenic climate change? Quote HistoryQuoted:
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Just tell,people you are staying an extra year to do research They would give him a fucking million dollar,Grant
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Posted: 9/29/2014 10:50:58 PM EDT
[#30]
d. 3O2 + 4Fe --> 2Fe2O3
2Fe + O2 --> 2FeO |
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Posted: 9/29/2014 11:19:48 PM EDT
[#31]
alright, so i went back in my notes to try and understand valence electrons . I'm still having trouble with transition metal elements though.
the numbers across the top of the Periodic table indicate charge, from 1-8 and the number of electrons on its outer shell. but at 4. it can go either way, positive or negative. The numbers across the left side, or row numbers also tell you the number of shells they have. is that correct so far? The Atomic number is the number of protons, and is also the number of electrons. so, Sodium [Na] has an atomic number os 11. so it has 11 electrons, 3 sheels, with 1 electron in its outer shell, and the first shell only holds 2. so the remaining 8 electrons are going to be in its second shell. so that's pretty much the gist of it from 1-2, and 3-8 columns. what im having trouble is how to apply the same to the transition metals. Scandium is in row 4, column three. so it has 3 electrons in its outer shell. 2 in the middle. 8 in its second shell, and the rest of it in the 3rd shell. but Titanium to its right has three different states. why? how do i figure that out? Wait. You're saying. Unless you MEMORIZE. that that particular element has different states. simply looking at a periodic table isnt going to tell you that? there's no formula, or trick on the table to get that detail? |
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Posted: 9/30/2014 12:46:39 AM EDT
[#32]
The shells are a bit more complicated than that, especially for D-block elements.
There are 3 common shells for electrons- S, P, and D shells. There are also F shells, but let's not worry about them for now. S shells fill first, and they can only hold 2 electrons. Atoms with valent S shells are in groups 1 and 2. P shells can hold 6 electrons. Atoms with valent P shells are in groups 13 through 18. D shells can hold 10 electrons. Atoms with valent D shells are in groups 3 through 12. The first row, or period, as only one shell; the 1s shell. The second row has 3 shells total, and two valence shells; the 1s is full, and depending on the element, the 2s and 2p are filled next. Each new row has a full set of shells from the row above it, and then whatever new electrons it gains. I use this periodic table: Dynamic Periodic Table Click on the tab at the top for orbitals, and play around with that. It also shows oxidation states for each element in the box on the orbitals tab. |
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Posted: 9/30/2014 4:02:08 PM EDT
[#33]
Quote HistoryQuoted:
alright, so i went back in my notes to try and understand valence electrons . I'm still having trouble with transition metal elements though. the numbers across the top of the Periodic table indicate charge, from 1-8 and the number of electrons on its outer shell. but at 4. it can go either way, positive or negative. The numbers across the left side, or row numbers also tell you the number of shells they have. is that correct so far? The Atomic number is the number of protons, and is also the number of electrons. so, Sodium [Na] has an atomic number os 11. so it has 11 electrons, 3 sheels, with 1 electron in its outer shell, and the first shell only holds 2. so the remaining 8 electrons are going to be in its second shell. so that's pretty much the gist of it from 1-2, and 3-8 columns. what im having trouble is how to apply the same to the transition metals. Scandium is in row 4, column three. so it has 3 electrons in its outer shell. 2 in the middle. 8 in its second shell, and the rest of it in the 3rd shell. but Titanium to its right has three different states. why? how do i figure that out? Wait. You're saying. Unless you MEMORIZE. that that particular element has different states. simply looking at a periodic table isnt going to tell you that? there's no formula, or trick on the table to get that detail? Memorizing the periodic table was week one We started immediately Flash cards for oxidstion states, atomic mass and orbitals. I had shit tons of flash cards. Some time in, we were given a blank periodic table and filled it in along with orbitals of each element Needless to say, attrition was very high in Chemistry 105/106
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Posted: 10/1/2014 10:19:36 PM EDT
[#34]
okay, so i sat with my professor during office hours.
Specifically. we aren't using "orbitals". From the notes. here's what he posted. C. Distribution of electrons per element from the periodic table.
1. Period number corresponds to the shell numner. 2.1 shell contains 2 electrons in all elemetns except H. 3.eletrons are added to everyu shell above shell #1 when its an outer shell .. : . O . : <-outer shell. .. 4. shells larger than shell number 2 contain more than 8 electrons only when they are inner shells of a given element. 11 5.electrons of transition elements are added to their next inner shell. Ex. If the outer shell is 4, the tansitions are adding electrons to shell #3. 6. Electrons of inner transition elements are added 2 shells in. ex. if the outer is is 6, the inner transitions are adding to shell #4. 7.we will consider “orbitals” when we understand quantum mechanics. shell# Electrons# AS-33 1 2 2 8 3 18 4 5 Hg-80 Shell# Electrons# 1 2 2 8 3 18 4 32 5 18 6 2 IV.outer shell electrons. A. involved in chemicals bonding B.2 types of simple chemical bonding 1.ionic bonding – always contains a metal 2.covalent bonding-always contains a metal 2.covalent bonding-never contains metal. I. Ionic Bonding. A. Requires the formation of ions. B.Ion- an atom with a charge. 1.ions form by the loss or gain of electrons 2. if electrons are gained the ion is negative. 3.iF “ “LOST “ “ “POSITIVE. c. the elements in the “ A series” has predictable ions 1. 1A-IIA-positive ions of +1 to +3 – cations 2.VA-VIIA-negative ions of -3 to -1 – Anions. He said i was over thinking it. What he wants us to know is how to find the number of electrons in each shell via this method. max number of electrons able to be in each shell is 2(N) sqaured. N being shell number. 
Carbons atomic number is 6. therefore it has 6 electrons. It sits in row 2. Therefore it has 2 shells. Shell 1's maximum number of electrons is 2. Shell 2's maximum number of electrons is 8. There fore.. Shell 1. has 2 electrons. Shell 2 has 4 electrons. When using this method. it seems to work for me..until. the transition metals. b. Platinum (Pt) number 78 Shell number...................Electrons 1................................. 2 2..................................8 3 ................................18 4.................................32 5.................................16 6..................................2 i get the idea that on all of the transition metals. the outer shell or valence shell only has 2 electrons, (for the purpose of this class) and the remainder fill into shell just beneath it. but on things with more than 6 shells. what would go between shells 5 and and up to the valence shell? |
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Posted: 10/1/2014 11:02:26 PM EDT
[#35]
bump?
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Posted: 10/1/2014 11:24:57 PM EDT
[#36]
can you turn lead into gold yet??
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Posted: 10/1/2014 11:35:40 PM EDT
[#37]
can someone correct me on this.
for nickel, 28 electrons. D block. 1s2 2s2 2s6 3s2 3s6 4s2 4s8 (in D block, so you ad those to the 3rd shell instead. 3s8 so you get. electrons in- shell1....2 shell2....8 shell3...16 shell4.....2 |
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Posted: 10/2/2014 1:42:42 AM EDT
[#38]
bump?
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Posted: 10/2/2014 1:50:51 AM EDT
[#39]
I don't like how you think of it in shells, think in terms of orbitals.
Row one consists of the 1s orbital. Row two has the 2s and 2p orbitals. When you get to Row four, there is the 4s, 3d, then 4p orbitals. s orbitals can have 2 electrons, p orbitals can have 6, and d orbitals can have 10. You can count electrons to get to the orbital of your atom. Example: Fe 1s(2) 2s(2) 2p(6) 3s(2) 3p(6) 4s(2) 3d(6) for Fe with zero charge. That can be abbreviated [Ar] 4s(2) 3d(6). The [Ar] is shorthand for every electron orbital that an Argon atom has. The last orbital to fill, the "bonding" orbital for Fe is the 3d orbital. It can hold up to 10 electrons, when it has ten it is happy. Iron doesn't usually exist at a zero charge though, it commonly exists at (2) or (3). Which means the orbital has 2 or 3 fewer electrons in it. |
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Posted: 10/2/2014 1:53:34 AM EDT
[#40]
Quote HistoryQuoted:
can someone correct me on this. for nickel, 28 electrons. D block. 1s2 2s2 2s6 3s2 3s6 4s2 4s8 (in D block, so you ad those to the 3rd shell instead. 3s8 so you get. electrons in- shell1....2 shell2....8 shell3...16 shell4.....2 Ni is 1s(2) 2s(2) 2p(6) 3s(2) 3p(6) 4s(2) 3d(8) 2 e- in shell 1 8 e- in shell 2 8 e- in shell 3 10 e- in shell 4 |
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Posted: 10/2/2014 2:21:59 AM EDT
[#41]
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I don't like how you think of it in shells, think in terms of orbitals. Row one consists of the 1s orbital. Row two has the 2s and 2p orbitals. When you get to Row four, there is the 4s, 3d, then 4p orbitals. s orbitals can have 2 electrons, p orbitals can have 6, and d orbitals can have 10. You can count electrons to get to the orbital of your atom. Example: Fe 1s(2) 2s(2) 2p(6) 3s(2) 3p(6) 4s(2) 3d(6) for Fe with zero charge. That can be abbreviated [Ar] 4s(2) 3d(6). The [Ar] is shorthand for every electron orbital that an Argon atom has. The last orbital to fill, the "bonding" orbital for Fe is the 3d orbital. It can hold up to 10 electrons, when it has ten it is happy. Iron doesn't usually exist at a zero charge though, it commonly exists at (2) or (3). Which means the orbital has 2 or 3 fewer electrons in it. this is what im trying to answer..
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Posted: 10/2/2014 2:44:06 AM EDT
[#42]
bump
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Posted: 10/2/2014 2:50:35 AM EDT
[#43]
1s2
2s2 2p6 3s2 3p6 4s2 3d10 4p6 I'm having nightmares now... Thanks a lot... Posted Via AR15.Com Mobile |
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Posted: 10/2/2014 2:53:42 AM EDT
[#44]
alright. thanks. but how do i do something like uranium?how are we getting 32 electrons in the 4th shell? and the others?
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Posted: 10/2/2014 2:55:17 AM EDT
[#45]
Quote HistoryQuoted:
alright. thanks. but how do i do something like uranium?how are we getting 32 electrons in the 4th shell? and the others? When you hit uranium, you're starting to fill f orbitals... To be more precise, when you're in the lanthanide/actinium series, you're filling f orbitals. Posted Via AR15.Com Mobile |
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Posted: 10/2/2014 2:56:28 AM EDT
[#46]
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When you hit uranium, you're starting to fill f orbitals... Posted Via AR15.Com Mobile Quote HistoryQuoted:
Quoted:
alright. thanks. but how do i do something like uranium?how are we getting 32 electrons in the 4th shell? and the others? When you hit uranium, you're starting to fill f orbitals... Posted Via AR15.Com Mobile which ones are F orbitals on the periodic table? i cant find one with those labled. 
this? so do we count the D orbitals and teh F orbitals in row 6 and 7? so in row 6. it would be 6s(2) 5d(10) 6p(6) AND 6f(15) ? |
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Posted: 10/2/2014 3:03:02 AM EDT
[#47]
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which ones are F orbitals on the periodic table? i cant find one with those labled. http://www.mikeblaber.org/oldwine/chm1045/notes/Struct/EPeriod/IMG00011.GIF this? so do we count the D orbitals and teh F orbitals in row 6 and 7? so in row 6. it would be 6s(2) 5d(10) 6p(6) AND 6f(15) ? Quote HistoryQuoted:
Quoted:
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alright. thanks. but how do i do something like uranium?how are we getting 32 electrons in the 4th shell? and the others? When you hit uranium, you're starting to fill f orbitals... Posted Via AR15.Com Mobile which ones are F orbitals on the periodic table? i cant find one with those labled. http://www.mikeblaber.org/oldwine/chm1045/notes/Struct/EPeriod/IMG00011.GIF this? so do we count the D orbitals and teh F orbitals in row 6 and 7? so in row 6. it would be 6s(2) 5d(10) 6p(6) AND 6f(15) ? Nope. 6s2, 4fX, then 5d, then 6p, as you move across the row |
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Posted: 10/2/2014 3:06:43 AM EDT
[#48]
Quote HistoryQuoted:
<a href="http://s2.photobucket.com/user/mall-ninja/media/IMG00011_zps8dc0be3f.gif.html" target="_blank">http://i2.photobucket.com/albums/y30/mall-ninja/IMG00011_zps8dc0be3f.gif</a> Quote HistoryQuoted:
Quoted:
Quoted:
Quoted:
alright. thanks. but how do i do something like uranium?how are we getting 32 electrons in the 4th shell? and the others? When you hit uranium, you're starting to fill f orbitals... Posted Via AR15.Com Mobile which ones are F orbitals on the periodic table? i cant find one with those labled. <a href="http://s2.photobucket.com/user/mall-ninja/media/IMG00011_zps8dc0be3f.gif.html" target="_blank">http://i2.photobucket.com/albums/y30/mall-ninja/IMG00011_zps8dc0be3f.gif</a> when dealing with elements 57-71, and 89-103, are you counting F orbital on top of D in 6s and 7s? or separately. I'm sorry for being question intensive, I've been shown to do this the wrong way. |
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Posted: 10/2/2014 3:12:35 AM EDT
[#49]
Quote HistoryQuoted:
when dealing with elements 57-71, and 89-103, are you counting F orbital on top of D in 6s and 7s? or separately. I'm sorry for being question intensive, I've been shown to do this the wrong way. moving across 57-71, you are filling the 4f orbitals. At this point, the 5d's are not yet filled. Once you hit 72, then you start filling 5d's. At 81, you've started filling the 6p's |
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Posted: 10/2/2014 3:28:15 AM EDT
[#50]
Quote HistoryQuoted:
moving across 57-71, you are filling the 4f orbitals. At this point, the 5d's are not yet filled. Once you hit 72, then you start filling 5d's. At 81, you've started filling the 6p's Quote HistoryQuoted:
Quoted:
when dealing with elements 57-71, and 89-103, are you counting F orbital on top of D in 6s and 7s? or separately. I'm sorry for being question intensive, I've been shown to do this the wrong way. moving across 57-71, you are filling the 4f orbitals. At this point, the 5d's are not yet filled. Once you hit 72, then you start filling 5d's. At 81, you've started filling the 6p's edit. Uranium 92 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 5d10 6p6 AND 4f10 7s2 5f4 shells and electrons 1.....2 2.....8 3.....18 4.....32 5.....22 6.....8 7.....2 is this correct? |
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