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Chasney913
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Joined: 28 Aug 2007
Posts: 117
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 Posted: 25 May 2008 01:33:01 pm Post subject: |
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So I have some chemistry stuff, and I'd like to understand it a bit better, because it's somewhat confusing to me, so if you can help that'd be great.
We're doing Lewis structures and formal charges, and I've been given the wonderful task of drawing a couple. Most are fairly simple, you just satisfy the octet rule. However, some are not so simple.
I know that 2nd period elements must satisfy the octet rule, because they have no d shell. I also know that having small formal charges are preferable to larger ones, and that the charge should try to reside on the more electronegative atom.
I have two questions, which I will try to lay out. One, is it preferable to have a complete octet, or low or non-existent formal charges? I assume that a complete octet is necessary for the 2nd period elements, but I'm not sure about the others. Also, I'd like to know what to do when I don't have nearly enough electrons to complete an octet, and all the atoms are second period, so they must have a complete octet. Example: an odd compound, ON2O5. I tried writing it, but it always seems that I need 4 more electrons. Similar situation with (OH)5IO.
If anyone can help, I'd really appreciate it. |
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thornahawk
μολών λαβέ
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Joined: 27 Mar 2005
Posts: 569
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| Posted: 25 May 2008 08:06:36 pm Post subject: |
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Quote: I have two questions, which I will try to lay out. One, is it preferable to have a complete octet, or low or non-existent formal charges? I assume that a complete octet is necessary for the 2nd period elements, but I'm not sure about the others. Also, I'd like to know what to do when I don't have nearly enough electrons to complete an octet, and all the atoms are second period, so they must have a complete octet. Example: an odd compound, ON2O5. I tried writing it, but it always seems that I need 4 more electrons. Similar situation with (OH)5IO.
Second-period elements are always said to strictly adhere to the octet rule; one glaring exception would be the electron-deficient atom of boron, and thus its binary compounds would have boron in a trivalent state.
Below the second period, as you have mentioned, things get crazier. There are no hard-and-fast rules, but here's a quick set that can get you in a pinch:
Group 1 - +1
Group 2 - +2
Group 13 - +3 (Al, Ga); +1 & +3 (In, Tl)
Group 14 - +4 (Si, Ge); +2 & +4 (Sn, Pb)
Group 15 - +/-3 & +5
Group 16 - -2, +4, +6. (S, Se, Te)
Group 17 - -1, +3, +5, +7 (Cl, Br, I)
(I won't deal with the transition metals, for their (complicated!) valence behavior is usually not covered in basic chemistry)
More often than not, if you must go with at least two atoms possessing a formal charge, you should usually try to keep the charges low. :)
ON2O5, and (OH)5IO look... weird. Are you sure that's what's written in the book?
thornahawk |
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Chasney913
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Joined: 28 Aug 2007
Posts: 117
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| Posted: 25 May 2008 08:45:47 pm Post subject: |
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I'm fairly sure. I've attached the scanned assignment. Question 4 E and 4 H. Unless I'm misreading them. The only structures that satisfy the octet rule for all second period elements, and use all the electrons are (that I've found)
Code:
:O: :Ö: :O:
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N---O---N
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:O: :O:
With a positive charge on the nitrogens, and the oxygen with one lone pair, and a negative charge on the other oxygens, excluding the double bonded ones.
For (HO)5IO, I have six oxygens coming of a central iodine, with a hydrogen bonded to five of them. All the oxygens are double bonded to the iodine, giving it 24 valence electrons, which seems a bit much. The oxygens also have one lone pair if with a hydrogen, two on the one that isn't. This gives neutral charges on everything except the iodine.
Take a look at that assignment. I thought some of the structures, especially in #4 were odd. It's due tomorrow, so it's a moot point if I get something wrong. |
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