[McCoy]

a) Draw a structure of nitromethane and explain why it must have formal charges on N and O.(My question: how do you tell that the double bonds is going to between N and O ,instead of having lone pair on N?)
b)Draw a structure of dimethyl sulphoxide and determine the formal charge on S and O.( i can darw the structure because i don't have to worry about the multiple bonds since it's obvious that there's none here and i can tell there's a formal negative charge on O(because it has 7 electrons insteads of 8 inorder to reach an octet ),but I've problem recognising that S have a formal positive charge(it has 6 Plus lone pair and that makes 8;the book says it has lost an electron,but I don't ''see'' it))
so is there a common sense way of helping me?
Thanks guys

[Rabn]

consider the electronegativity. Sulphur cannot have a lone pair in the presence of oxygen, not to mention that oxygen will never have only 7 electrons.  It may share an electron in resonance, but Oxygen is a chemical bully, it takes what it wants.

[Dan]

Sulphur cannot have a lone pair in the presence of oxygen

What do you mean? If you draw the Lewis structure you can see that DMSO, along with all sulfoxides, has a lone pair on sulfur.

how do you tell that the double bonds is going to between N and O ,instead of having lone pair on N?

So the possibilites if you keep a lone pair on N are:
         ..
1. Me-N-O^-
         |
         O⁺
         ..
2. Me-N-O^.
         |
         O^.

As Rabn said, consider the electronegativity of O. in the case of 1, we have an O atom with only 6 electrons, and in the diradical 2 we have two O atoms with 7 electrons each. These would be very unstable structures, O is very electronegative so it will pick up a full octet if it can.

If we look back at structure 1, you might see that if we use the lone pair on N to donate into the positively charged O then we satisfy the octet for O.

         ..                          +
1. Me-N-O^-  <-----> 3. Me-N-O^-
         |                           ll
         O⁺                         O

So 3 is a much more stable resonance structure, and closer to the truth.

[McCoy]

Thanks for your help.
Yea, I've no problem with asigning a +ve charge on S when taking into acount electronegativity(becausei just assume that the less electronegative element would have a negative chargeon it(which is N in this case), but my problem is using the valence electrons to predict a formal charge.
i think the multiple bond issue is solve...I think you're saying to have a double bond or a lone pair is determined  by the stability of the resultant structure(or precisely whether each element involved has its outermost shell full...i.e satisfies octet rule).

so how can i asign partial charges apart from using electronegativity? can i always assume that the one with more lone pairs carries the -ve charge?
thank you very much for your time.

[Dan]

so how can i asign partial charges apart from using electronegativity? can i always assume that the one with more lone pairs carries the -ve charge?

This depends on the structure. Assigning partial charges based on electronegativity is fine for some cases - those in which inductive effects (which arise due to electronegativity differences) are the dominant (and usually only) contributor to polarisation. You need to watch out for resonance.

For molecules for which you can draw relatively low energy resonance canonicals the "true" electronic distribution is taken to be a weighted average of these forms. Compounds in which resonance is an important effect have a series of pi orbitals which overlap or are arranged to allow overlapping. resonance effects generally dominate over indctive effects.

An example of each case is given below

[bluenote]

First, for dimethyl sulfoxide:  It is usually drawn with a double bond between O and S and no formal charge.  S STILL HAS a lone pair!  This makes S hypervalent with 10 electrons.  It can do this by using a relatively low energy d orbital (same thing with P compounds.)  The structure with a single bond and a negative charge on O and a positive charge on S is a much better description of its reactivity.  Just remember that atoms with low energy d orbitals can be hypervalent (more than 8 electrons) but atoms without a d orbital will never be hypervalent.  (That is not actually true, but for our discussion here it will suffice.)

As to your more general question, review the priorities for ranking resonance structures.  You will see that electronegativity is important but it is not the most important priority. 

I hope this helps.

[Rabn]

I would like to formally charge myself with late night posting derived ignorance to which I plead guilty. As my punishment I will be forced to renig my statement regarding Suplhur with a lone pair and then do ten push ups. done and done.

[azmanam]

The structure with a single bond and a negative charge on O and a positive charge on S is a much better description of its reactivity

That reminds me of the Grossman rule for predicting reactivity.  It's useful here and throughout organic chemistry:

The second best resonance structure is often a good indication of a compounds reactivity.

It explains why carbonyls are electrophilic at carbon, why terminal alkenes are nucleophilic at the terminal carbon (in predicting the mechanism with Markovnikov's rule) and why DMSO is electrophilic at sulfur (and nucleophilic at oxygen if you're doing a swern:) )

[McCoy]

I want to thank all of you guys for your wonderful contribution to this thread. Understanding the formal charges together with the ''resonance thing'' shouldn't  have been easy without your help.

cheers.

[McCoy]

so how can i asign partial charges apart from using electronegativity? can i always assume that the one with more lone pairs carries the -ve charge?

This depends on the structure. Assigning partial charges based on electronegativity is fine for some cases - those in which inductive effects (which arise due to electronegativity differences) are the dominant (and usually only) contributor to polarisation. You need to watch out for resonance.

For molecules for which you can draw relatively low energy resonance canonicals the "true" electronic distribution is taken to be a weighted average of these forms. Compounds in which resonance is an important effect have a series of pi orbitals which overlap or are arranged to allow overlapping. resonance effects generally dominate over indctive effects.

An example of each case is given below

wow this is brilliant. Thank you very much.Your help is highly appreciated.