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Topic: Carbonium Ions  (Read 2844 times)

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Offline orgo814

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Carbonium Ions
« on: December 10, 2015, 07:30:30 PM »
The whole concept of a carbonium ion is confusing me. How does a carbon with 5 bonds have a positive charge? I would expect a negative charge? Insight appreciated.

Offline orgopete

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Re: Carbonium Ions
« Reply #1 on: December 11, 2015, 12:39:14 AM »
The whole concept of a carbonium ion is confusing me. How does a carbon with 5 bonds have a positive charge? I would expect a negative charge? Insight appreciated.

Five bonds? No. Four bonds is a completed octet, three bonds is an electron short and a net positive charge.
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Offline beardy

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Re: Carbonium Ions
« Reply #2 on: December 15, 2015, 11:33:34 AM »
Carb = carbon
Onium = cation

With a cation carbon in mind, how many bonds do you think carbon would need to render a positive charge?

Offline critzz

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Re: Carbonium Ions
« Reply #3 on: December 16, 2015, 10:21:11 PM »
I think that what he means is for example the methanium-ion (CH5+), which actually has 5 bonds and is considered a superacid.

So let's consider a hydroxonium-cation (H3O+).

To determine the formal charge on the oxygen you can use this formula: [formal charge = valence electrons - lone pair electrons - 1/2 × bonding electrons].
So if you fill in: formal charge = 6 valence electrons - 2 lone pair electrons - 1/2 × 6 bonding electrons = a formal charge of +1

For an ammonium-cation (NH4+) it works the same way: 5 - 0 - 4 = +1

However, in the case of methanium it doesn't really seem to work like that (as where your confusion probably comes from): 4 valence electrons - 0 lone pair electrons - 1/2 × 10 bonding electrons = -1 ???

What's going on here?

The methanium-cation can be seen as a CH3+-cation, where the empty p-orbital is coordinated to the bond of a H2-molecule, via a 3-center-2-electron bond.
Viewed in this way, the molecule still has a +1 charge.



Offline orgo814

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Re: Carbonium Ions
« Reply #4 on: December 17, 2015, 12:08:58 PM »
Thank you critzz!

Offline critzz

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Re: Carbonium Ions
« Reply #5 on: December 17, 2015, 01:49:34 PM »
You're welcome

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