Find hybridisation of Carbon compounds

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Discussion Overview

The discussion centers on the methods for calculating the hybridization of carbon compounds, exploring various approaches and their effectiveness. Participants share their experiences and challenges with different techniques, particularly in relation to carbon's bonding behavior.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions the effectiveness of the "usual" method of calculating hybridization by adding valence electrons and dividing by 8, suggesting it often leads to incorrect answers.
  • Another participant describes their method, which involves calculating the total valence electrons and deriving hybridization based on the quotient and remainder, but this method is challenged by others.
  • Some participants express skepticism about the validity of the number-based method, arguing it lacks logical foundation and does not accurately reflect the distribution of lone pairs in molecules like NF3.
  • There is a discussion about the importance of considering geometry and electron pair arrangements when determining hybridization, particularly for carbon compounds.
  • One participant mentions that hybridization patterns involving 5 to 7 are not applicable in main group compounds, suggesting that such methods may be outdated or incorrect.
  • Another participant states that carbon can form up to 4 sigma bonds, implying a limit to its bonding capabilities.
  • There is a mention of free radicals and how they affect the counting of sigma bonds and hybridization, with examples provided for clarification.

Areas of Agreement / Disagreement

Participants express differing opinions on the validity of various methods for determining hybridization. There is no consensus on a single approach, as some methods are challenged while others are defended. The discussion remains unresolved regarding the most effective technique.

Contextual Notes

Some participants note that the methods discussed may depend on specific definitions and assumptions about hybridization and bonding, which are not universally agreed upon.

Saitama
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Hello!
How to calculate the hybridisation in carbon compounds? Most of the time whenever i try to find out the hybridisation i get a wrong answer.
I think the usual way of finding hybridisation doesn't work with carbon compounds. Is there any other way?
 
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May be you could describe what you consider to be the "usual" way?
 
DrDu said:
May be you could describe what you consider to be the "usual" way?

By "Usual way" i mean adding the valence electrons of the atoms of the molecule and dividing by 8. If the value comes:-
2- sp
3- sp2
4- sp3
5- sp3d
6- sp3d2
7- sp3d3

For example, NF3
Total valence electrons=26
Dividing by 8 gives 3 as quotient and 2 as remainder.
"3" is the number of sigma bonds.
We will divide the remainder by 2 for finding out the lone pairs.
2/2=1 <-------Number of lone pairs.

I hope you understand. :)
 
This is certainly not the usual way to look at hybridization. I have never seen this before and in fact it seems to make not the least sense at all. Where did you find this number mysticism?
 
DrDu said:
This is certainly not the usual way to look at hybridization. I have never seen this before and in fact it seems to make not the least sense at all. Where did you find this number mysticism?

Why doesn't it make sense?
I think that's the most easiest way to find the hybridisation. This is just a trick to find the hybridisation. Which way do you use to find hybridisation?

I forgot to mention what's the hybridisiation in the example.
NF3 has sp3 hybridisation. 3 + 1= 4.
4 indicate sp3 hybridisation.
 
Pranav-Arora said:
Why doesn't it make sense?
I think that's the most easiest way to find the hybridisation. This is just a trick to find the hybridisation. Which way do you use to find hybridisation?

I forgot to mention what's the hybridisiation in the example.
NF3 has sp3 hybridisation. 3 + 1= 4.
4 indicate sp3 hybridisation.
And why should this one lone pair which you calculate belong to N and not to one of the F?
 
DrDu said:
And why should this one lone pair which you calculate belong to N and not to one of the F?

Because the least electronegative element is placed in the center. :)

I know that this number method is only applicable in finding out the hybridisation but it does not tell of which atom this hybridisation is. Therefore, we need to adjust and place the atoms properly.

In my textbook its stated:-
"A discussion of the valence bond theory is based on the knowledge of atomic orbitals, electronic configuration of elements, the overlap of elements, the overlap criteria of atomic orbitals, the hybridisation of atomic orbitals and the principles of variation and superposition. A rigorous treatment of VB theory in terms of these aspects is beyond the scope of this book. Therefore, for the sake of convenience, valence bond theory has been discussed in terms of qualitative and non-mathematical treatment only."

So maybe that's the reason my teacher didn't teach us the way you are talking about. :)
 
Bump :(
 
Did your teacher show you this strange calculation or is it from the book?
I had hoped that you understood that it gives the wrong number of lone electron pairs.
In NF3 there is one lone pair at N and 3 at each F.
 
  • #10
It is taught by my teacher and in some of the refreshers and supplement books available in market here shows this method.

I already said that this method is not correct, we have to arrange the bonds and lone pairs accordingly. Can you please tell me your way? :)
 
  • #11
On an elementary level it is not difficult to work out the number of bonds and lone pairs in a given molecule. Just make sure that every main group element has 8 electrons around at least in "normal" compounds eventually introducing some formal charges on the atoms.
The hybridisation which is preferably to be used to describe a molecule depends not only on the number of electron pairs but also on the geometry, that's why you are having probably problems with carbon compounds. Finally, your hhybridization paterns 5 to 7 don't appear at least in main group compounds, although this is still repeated in old or bad books.
A still very readable introduction to the principles (which however still contains d orbital contribution in main group elemental compounds) is the classic book by Linus Pauling, General chemistry.
 
  • #12
DrDu said:
On an elementary level it is not difficult to work out the number of bonds and lone pairs in a given molecule. Just make sure that every main group element has 8 electrons around at least in "normal" compounds eventually introducing some formal charges on the atoms.
The hybridisation which is preferably to be used to describe a molecule depends not only on the number of electron pairs but also on the geometry, that's why you are having probably problems with carbon compounds. Finally, your hhybridization paterns 5 to 7 don't appear at least in main group compounds, although this is still repeated in old or bad books.
A still very readable introduction to the principles (which however still contains d orbital contribution in main group elemental compounds) is the classic book by Linus Pauling, General chemistry.

Thanks for the book suggestion. I will definitely look into that. :)
 
  • #13
Coming back to the original question, how many possibilities does the C atom to form covalent bonds ?
 
  • #14
dextercioby said:
Coming back to the original question, how many possibilities does the C atom to form covalent bonds ?

I don't understand your question. :confused:
At the most it can form 4 sigma bonds.
 
  • #15
And these sigma bonds are they identical ?
 
  • #16
dextercioby said:
And these sigma bonds are they identical ?

No idea. :)

Wouldn't that depend on hybridisation?
 
  • #17
Carbon forms compounds simply to complete its octet and so finding out the hybridisation in organic compounds is not much of a hassle. Count the number of sigma bonds and lone pairs of the atom whose hybridisation you need to find. 2,3 and 4 correspond to sp, sp2 and sp3 respectively. The lone electron in a free radical exists in and unhybridised orbital so for example a CH3 free radical would be sp2.
 
  • #18
I don't understand your last statement. :confused:

mishrashubham said:
The lone electron in a free radical exists in and unhybridised orbital so for example a CH3 free radical would be sp2.
 
  • #19
A Free Radical contains an unpaired electron. So after counting the number of sigma bonds you wouldn't count that electron if it is a free radical.
 
  • #20
mishrashubham said:
A Free Radical contains an unpaired electron. So after counting the number of sigma bonds you wouldn't count that electron if it is a free radical.

Can you elucidate it further using an example? :)
I am really poor at chemistry. :P

Btw, are you accessing PF from a mobile because you seem to have given a mobile wiki link. :)
 
  • #21
Pranav-Arora said:
Can you elucidate it further using an example? :)
I am really poor at chemistry. :P

Assume PCl4 exists (neutral). The it would have sp3 hybridisation.

Pranav-Arora said:
Btw, are you accessing PF from a mobile because you seem to have given a mobile wiki link. :)

Yes sorry forgot to remove the 'm' from the url, I was accessing from a mobile device.
 
Last edited:

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