Can any one tel me how nitrogen bonded to aromatic ring in Benzene diazonium chloride(BDC) got positive charge and to which atom is chlorine bonded in BDC??
I think any atom attains + charge if it looses electrons or when it shares lone pair of electrons to other atom in bonding.But to which atom does N attatched to aromatic ring in BDC shares its lone pair??

Thank you

It's the first time I see that compound. That's what I've got for its formation:

C_6H_5NH_2 + HNO_2 + HCl \rightarrow [C_6H_5N^+N][Cl^-]

This answers your second question. The Cl^- ion is bonded to the whole molecule. Apparently, the positive charge on the benzene diazonium is spread throughout the molecule.

For your first question, I think what happened is that a hydrogen from the aniline took the two electrons from the NH bond, leaving the nitrogen with one less electron, thus explaining the positive charge.

And for your third question, there isn't any lone pair if what I just said really happened. But in the case I made a wrong guess, the electron pair must have to be shared to the other nitrogen, since the positive nitrogen was initially bonded to the benzene ring.

Perito, some assistance here?

Thanks a lot for your answer.Now I've uderstood the -ve charge of Cl.
But Bonding in N is still unclear to me.
IF in case H has abstracted electron from N in aniline then the H attains -ve charge and N attains +ve charge.So they are only 2 electrons in P orbital (outer most orbital) of N .But in BDC the N with + charge forms a total of 4 baonds ,3 bonds with another N and 1 bond with benzene ring.How can it form 4 bonds if it is deficent of electrons in outermost orbit??
In most of the cases H get eliminated in form of proton but is it possible to get eliminated with -ve charge??

If in case N bonded to aromatic ring ,donates lone pair to another N then the other N should develop -ve charge.But actually there is no -ve charge on other N... So again the question arises how first N got +ve charge?

Ok, apparently, it's more complicated than that. This may be a hint, but I feel if we succeed in understanding the formation of the molecule from the reactants, we might find the answer.

The nitrogen in nitrous acid has a double bond with an oxygen atom and a single bond with an OH group and lastly has a lone pair.

I got that from wikipedia:


Nitrous acid is used to make diazides from amines; this occurs by nucleophilic attack of the amine onto the nitrite, reprotonation by the surrounding solvent, and double-elimination of water. The diazide can then be liberated as a carbene.

I didn't see what's reprotonation, but saw only deprotonation. Maybe that's a mistake in wikipedia.

I'll have more research...

Oh, I think I hit the right button! Yay! :p :D

Sandmeyer reaction - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Sandmeyer_reaction)

It explains it all! :)

It seems that the nitrogen from the benzene ring donates it's electrons to the nitrogen from the nitrous acid, which is already positively charged.

Also, the neutral nitrogen forms 3 bonds, that's normal.
The charged nitrogen forms 4 single bonds, leaving one electron out if it were to be neutral. It's absence makes it positively charged.

HI,now I've got it...
Thanks a lot!!
It is actually the electron in s orbital is forming the bond with other N initially I thought only P orbitals will form bonds.

You're welcome! :)

I learning too, you know :)

So, thank you too! :)