what does the lewis structure of (IO3) look like??
It looks marvelous!
o thanks, didnt really help, suppose i should be more specific..
i count up the valence electrons and get 6 for each O and 7 for the I, this gives me an odd number,
how do i deal with this
First you need to determine the charge on the polyanion. Then, assume that the oxygens are at a valence of -2. From those two bits of information, determine the oxidation state of the iodine.
Now, start at the iodine and put seven electrons around it. That is the ground state for iodine. Take away as many electrons as you need to get to the oxidation state of the iodine that you determined above. (it isn't 7....) Do the same for the oxygens but add two electrons to it's ground state.
Put it together... You should end up with one excess electron charge which gives the complex it's -1 valency.
so it NEEDS to be -1, cannot be neutral??
another question, what is a "stronger" rule? the octet or spreading the formal charge, more specifically, inPO4(-3) i can either have a symetric structure, p in the middle 4 O's around it, 4 pairs each, seems perfect, but the formal charge of the O's is -1 each and of the P is 1, so each has a formal charge,
P in the middle with 4 O' s aroung 3 of them with single bond and 3 pairs and 1 with double bond an 2 pairs, this giving P 10 electrons instead of 8 around it, formal charge comes to -1 for each of the single bond O's and 0 for the P and double bonded O,
another option is
p in the middle 4 O's around each with double bonds and 2 lone pairs
i learned that i must try and get the FC of each closest to 0
You can do the same kind of analysis with phosphate as well.
sorry, but i dont really understan what to do, i am studying chemistry in another language so the terminology you are using isnt very understood to me. could you show me an example of how it is meant to look or maybe try in simpler english. sorry bout this
Let's do sulfite, SO3-2. First, I assume the oxidation # (valence) of oxygen is -2 for each oxygen. The overall charge of the ion is -2 and there are three oxygens.
The oxidation # (valence) for sulfur is equal but opposite in sign to the difference between the oxidation #'s on oxygen and the overall ion's charge .... Sulfur's valence = [-1]*[3(-2)-(-2)] = 4
Zero valent sulfur has 6 electrons in it's outer shell. Sulfur in an oxidation state of +4 would have 4 fewer electrons and therefore has only 2 electrons.
Zero valent oxygen also has 6 electrons in it's outer shell. Oxygen in an oxidation state of -2 would have two additional electrons and therefore has 8.
The Lewis structure therefore has a central sulfur, with it's two electrons, and it shares three pairs of electrons with the oxygens... one pair each with the individual oxygen atoms. The overall charge on the molecule is -2. There is a complete octet around each atom within the complex.
Now, you do it for phosphate and IO3-...
according to what do you assume the valence for oxygen is -2, what does this mean, does it not have 6 valence electrons? do you mean the 2 non bonding electrons?
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