[cliverlong]

Hello,

   Does anyone know of a site that explains why/how covalently bonded anions such as

Sulphate   SO₄^2-
Nitrate      NO₃^-
Nitrite      NO₂^-
Hydroxide      OH^-
Carbonate      CO₃^-


etc. etc.

become charged ions?

I don't know where to look or what search term to put into Google.


Thanks for any pointers


Clive

[LeFowl]

the sulfate ion is a polyatomic ion made up of one sulfur atom, and four oxygen atoms, that exists only if it can gain 2 electrons (from some metal cation or cations).
polyatomic ions exist as a unit, acting with a single charge.

that was wikipedia on so4 2-....i cant find anything else..see if that helps

[cliverlong]

Thanks for the reply.

I feel you have restated the "what" that was in my question but I am still looking for an answer to the "how/why".


Thanks

Clive

[sjb]

I guess it's very much the same concept as individual atoms getting ionised, formation of a more stable state, be it a complete octet (as in hydroxide) or something more advanced.

[Astrokel]

a good question to give a thought on 

Perhaps this help bit: http://www.chemicalforums.com/index.php?topic=27899.msg105716#msg105716 (last post)

[cliverlong]

Hello

    I think hybridization of orbitals is the key. The following works – doesn't mean it is right

I will get to sulphate via sulphur trioxide. This would be easier with a diagram but I will try words

Consider sulphur's electronic structure. It is period 3, group 6 (old style). Its outer orbitals are

3s₂, 3p₄

The 3s orbital is full (electron pair) and 3p contains one full (pair) and two single-electron orbitals

In addition, sulphur has an empty 3d orbital.


Consider oxygen's electronic structure. It is period 2, group 6 (old style). Its outer orbitals are

2s₂, 2p₄

The 2s orbital is full (electron pair) and 2p contains one full (pair) and two single electron orbitals.


Let's promote one of sulphur's 3s₂ electrons and one of the 3p₂ electrons into empty 3d orbitals. This creates 6 (1 x s, 3 x p, 2 x d) 3-orbitals occupied by single electrons. Let us hybrize the 6 orbitals into an sp³d² where each orbital contains one electron. (There are three unused 3d orbitals that will come into play later).

Now each oxygen has 2 2s orbitals containing an unpaired electron. So these can form covalent bonds with 2 of the sp³d² hybrids in sulphur. Since there are 6 sp³d² hybrids in sulphur we get the three oxygens forming 6 bonds with sulphur – we get SO₃.

Now let's look at those unused 3d orbitals – three of them. If we want to introduce another oxygen atom we need to hybridize all of sulphur's 3-orbitals (3s₂, 3p₆, 3d₁₀). Moving from SO₃ we now have six available “electron spaces” in sulphur's fully hybridized sp³d⁵ orbitals. The fourth oxygen can provide 4 electrons from its 2p orbital. This gives us SO₄ All bonds are covalent. There are still 2 “electron spaces” left in this “fully hybridized” sulphur sp³d⁵ orbital. So if we introduce another two electrons the sulphur hybrid orbital is now completely full of electrons. These introduced electrons aren't “bonding electrons" as such, they are there to fill all the spaces in the hybrid orbital. These extra two electrons have provided the 2- charge that the sulphate anion SO₄^2- has.

Job done.

I'll try and apply the idea to the other anions and only post if my method falls apart.

I'll go and lie down now.

[cliverlong]

Some of my original charges on the anions were wrong.

Corrected now

Sulphate   SO₄^2-
Nitrate      NO₃^2-
Nitrite      NO₂^-
Hydroxide      OH^-
Carbonate      CO₃^2-

[cliverlong]

The "method" works for

hydroxide
carbonate
carbon dioxide (to check carbonate)

the "method" doesn't work for nitrate


Hmmmm .

[sjb]

Some of my original charges on the anions were wrong.

Corrected now

Nitrate      NO₃^2-

No, nitrate is NO₃^-

Any good now?

[cliverlong]

No, nitrate is NO₃^-

Any good now?

Thanks. Doh!

Well I can get the numbers to work but I think that's all I'm doing on this example, playing with numbers. Here's how it could work for nitrate - can someone let me know if I am wrong what the correct configuration is?

Promote one of the nitrogen 2s² electrons to the empty 3s orbital (* See below)

This results in the following nitrogen orbitals all having one electron: 2s, 2p, 3s

Now hybridize the 2s, 2p, 3s orbitals to make a s²p³s² () orbital - which can contain 5 electron pairs.

I hybridize to make all the N-O bonds identical.

Now the oxygens have full 2s orbital, one full 2p orbital and 2 2s orbitals with one electron in each.

Two oxygens can place their 2 unpaired electrons in 4 of the nitrogen s²p³s² orbitals (I know, a diagram would really help here).

This gives us NO₂

These leaves one nitrogen s²p³s²  orbital containing one unpaired electron.

The third oxygen provides one electron from its 2p and has one "spare" electron. This creates the -1 charge on the nitrate NO₃ ^- ion . (** see second point below).



Clive

* problem 1. The energy gap between 2s and 3s may be (is?) far too high to promote an electron and then hybridize the 2s 2p and 3s orbitals

** Although the electron "accounting" works I find it a bit discomforting that I have combined two neutral species NO2 and O and created an NO₃ ^- ion which has a charge.

[cliverlong]

I found this site

http://www.chem1.com/acad/webtext/chembond/cb07.html

which suggests that the hybridization (if such a thing exists !) happens on the oxygen orbitals and pi-bonding of the unbonded oxygen p-orbitals.


Clive