Formal Charge Distribution vs Oxidation States?

Question

Please, look at the picture (linked image) to solve the question below..
http://rokin.mireene.com/no2cl.gif


For the Lewis diagram, above, determine:

The overall charge of the molecular species shown.
The formal oxidation number of the nitrogen atom.
The formal charge on the nitrogen atom.


Would some one please help me?
Thanks in advance!

Answer 1

Something seems to be goofy with your Lewis diagram. You show 2 covalent bonds between one O atom and N atom sharing 2 electrons from O, while you show no electron sharing from O atom in the bond between the 2nd O atom and N. I think that a single bond with each O sharing 1 electron with the N atom will be more reasonable and symmetrical. We can analyze two cases. In the 1st case below we assume that you have 1 single bond with each of the O atoms and 1 single bond with the Cl atom, the following are my thoughts. In the 2nd case below we assume that you have 2 double bonds between N and O and a single bond between N and Cl.

When this product exists(whatever temperature and pressure conditions), the overall charge on the molecule has to be zero because each atom is basically neutral. The covalent bonds between N and O involve 2s and 2p orbitals, while between N and Cl, the 3s and 3p orbitals are involved. Even though the electron clouds are deformed and are influenced by all these orbitals, formally only p orbitals are considered involved in the bonds. The electronegativities of Cl, N, and O are 3.0, 3.0, and 3.5. Therefore, the elctrons spend more time near oxygen atoms then at the N or Cl.
The bonding electrons of Cl being in the 3rd shell their N-Cl bond length entends further than the N-O bonds. The N-O bonds are the strongest because of the higher values of enthalpies of formation, Gibbs' free energies, as well as lower entropies, compared to N-Cl bond.

In the 1st case mentioned above, a total of 3 electrons are shared by N atom in the 3 single bonds. Two of those electrons spend lot more time near O atoms and therefore with respect to the N-O bonds, the oxidation number of N is +2. However, the 3 rd electron spends as much time near N atom as it does with the Cl atom and therefore N atom can be considered as losing one electron or gaining one. So it's oxidation number could be +1; and this is like in N2O or it could be -1 like in NH2OH. So one scenerio is N3+Cl-2O-and the other scenerio is N2+2O-Cl (least likely).

In the 2nd case mentioned above, with the doble bonds, following the same reasoning above, the oxidation number of N could be +5(as in HNO3) or +4(as in N2O3). So one scenerio is N5+2O2-Cl- and the other scenerio is N4+2O2-Cl(least likely). Since N is known to exist in oxidation states upt +5 in different other molecules, It’s most likely in this molecule it exist as +5 than +3.

Answer 2

Overall charge: difference between number of electrons and number of protons.. In this case, zero.

Formal oxidation number - share and share alike. So of the two electrons in each single bond, one goes to each of the atoms involved. For a double bond, each of the partners gets two electrons. Do this and you will find 7 electrons "belonging" to one oxygen, 6 to the other, 7 on the chlorine and only 4 on the N.

N had 5 valence electrons to start with, so it has formally lost one, giving you a formal charge of +1. Since the molecule is neutral, the total formal charges add up to 0. I leave it to you to work out where the -1 charge is.

Oxidation number: winner takes all. Pretend that all the valence electrons involved in bonding "belong" to the more electronegative partner, and see what that gives you. For example, give the two electrons in the N-Cl bond to Cl. That leaves Cl with 8 valence electrons. But it only had 7 to start with, so it has oxidation number -1.

If you do this properly, you will find that N "owns" no valence electrons at all. So what is its oxidation number?

Finally, you can check that when you add up all the oxidation numbers over all the atoms, you get the overall charge on the species (in this case, 0).