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I've just learned how to predict the shapes of molecules in class today using VSEPR theory. I would like to ask is there anyway to find the number of bond pairs and lone pairs without drawing a dot-and-cross diagram?

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  • $\begingroup$ I assume you're using VSEPR theory to predict the shapes? $\endgroup$ – bon Aug 11 '15 at 17:43
  • $\begingroup$ Yeap. I'm using VSEPR @bon $\endgroup$ – Mathxx Aug 11 '15 at 17:43
  • $\begingroup$ @Mathxx number of bond pairs and lone pairs for atom, molecule or what? Also drawing a dot-and-cross diagram is only for very basic training imo, it's not a real tool or sth imo $\endgroup$ – Mithoron Aug 11 '15 at 21:21
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There is a simple, four-step calculation that you can perform:

  1. count the valence electrons the atoms you are bonding have.

  2. count the number of valence electrons the atoms would like to have for a noble gas’ valence shell. (i.e. eight for everything main group, two for hydrogen.)

  3. substract $2.-1.$, i.e. the first (existing) from the second (desired). This is your number of bonding electrons. Divide by two for bonding electron pairs.

  4. substract $1.-3.$, i.e. the third (bonding) from the first (existing). This is your number of free electrons. Divide by two for lone pairs.

Using a simple example such as sulphur dioxide:

  1. $6\ (\ce{S}) + 2 \cdot 6\ (\ce{2 O}) = 18$

  2. $8\ (\ce{S}) + 2 \cdot 8\ (\ce{2 O}) = 24$

  3. $24 - 18 = 6$, i.e. three bonding electron pairs.

  4. $18 - 6 = 12$ i.e. six lone pairs.

Unfortunately, that by itself does not allow you to write the structure; you need to know which element is in the centre and how they are connected. Here, we have a central sulphur bonded to two oxygens. The final result is:

$$\ce{O=S+-O-}$$

With two lone pairs on the left-hand oxygen, one on the sulphur and three on the right hand oxygen. The formal charges are important.

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Terminal atom must obey octet (or duplet) rule. Using this, you can determine the number of bond pair around central atom, the rest are lone pair

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    $\begingroup$ I don't quite see how you can jump from the first statement to the second statement. Perhaps you could elaborate. $\endgroup$ – bon Aug 12 '15 at 10:28
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    $\begingroup$ This is not a very precise or an advanced question. And based on what he said about"he just learnt VSEPR theory", I think it is more beneficial to leave some space for him to think at this stage instead of helping him to do everything. I don't want those best answer stuff, I intended to guide him to think and try out some examples himself. $\endgroup$ – y chung Aug 12 '15 at 10:35
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There is a very simple way to do it, but it is not recommended at all. Also there are a lot of assumptions in this method.

  1. Detect the central atom along with the peripheral atoms. For Example:- In Methane, the Carbon atom is the central atom.

  2. Count the valance electrons of the central atom and the peripheral atoms. Exception: For Hydrogen , using "1" won't yield results. So specifically for Hydrogen use "7". This is one of the assumptions we have to make.

  3. Divide the above value obtained by 8. The quotient gives the value of the number of sigma bond pairs and the remainder divided by 2 gives the number of lone pairs.

Explanation:

The number of valance electrons counted divided by 8 will give the number of sigma bonds formed. This is just like counting the number of atoms which are getting complete octets, i.e. forming sigma bonds to get octet complete and becoming stable. The remainder is the no. of non bonded electrons, which when divided by 2 gives the number of lone pairs.

Example:

In Sulfur tetrafluoride:
Number of Valence electrons$$1\times(6)+4\times(7)=34$$
No. Of Bond Pairs$$\frac{34}{8}=4$$
No. Of Lone Pairs$$\frac{34\%8}{2}=1$$

To be very clear, this method is not fool proof. It is just a workaround to save time if you do not want to make the structures as it works on majority of compounds.

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