Why is it that the Lewis structure of $\ce{Cl_2SO}$ has a total of $24$ valence electrons? I thought that the number of valence electrons should be $2\times7 + 1\times6 + 1\times6 = 26$.
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Could you elaborate where you found this misleading statement? Is it in a textbook or other source? – Martin - マーチン May 27 '14 at 01:46
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26 is correct.. – DavePhD May 27 '14 at 01:50
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@DavePhD Thanks. It must be an error in my textbook. – okarin May 27 '14 at 01:53
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I'm guessing whoever wrote it did what I accidentally did the first time counting it up: forgot the lone pair on sulfur. – SendersReagent Mar 26 '16 at 23:17
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Sulfur is sometimes described as a hypervalent atom, or one that has more than eight valence electrons. This is possible to some extent; sulfur can indeed utilize its d-orbitals for bonding. However, recent quantum mechanical calculations suggest that the extent of this utilization is negligible. Therefore, your book is most likely representing the $\ce{Cl_2SO}$ molecule as a "charge-separated" molecule.
EDIT: Upon further thought this still makes no sense. Charge is conserved and thionyl chloride should always have 26 electrons (no matter how they are distributed). So perhaps, due to the high degree of ionic character in the $\ce{S-Cl}$ and $\ce{S-O}$ bonds, the missing two electrons are "dispersed" among these ligands. (This is speculation). Perhaps your book is genuinely in error. Perhaps they did not count a lone pair on the central sulfur atom or otherwise miscounted the number of electrons?
EDIT 2: Here's a picture of two possible Lewis structures of thionyl chloride; as we can see, the book agrees that the thionyl chloride molecule is better represented as a "charge-separated" molecule. Nonetheless, all the electrons are present.
EDIT 3: Here's a picture of thionyl chloride which makes it seem as if the molecule only has 24 valence electrons. NB: the lone pair is still there and its presence is implied through the lack of any indication of a non-zero formal charge on the sulfur atom.
Dissenter
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As the structure of thionyl chloride is pyramidal, there cannot be a $\pi$ orbital for the double bond. The charge separated version is therefore superior in representation. – Martin - マーチン May 27 '14 at 02:26
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Martin I think you've misinterpreted the second picture (the second picture doesn't show the lone pair on the sulfur). The LP's presence is implied through the lack of any indication of a formal charge on the sulfur. This still makes the molecule pyramidal even though it looks planar. – Dissenter May 27 '14 at 02:34
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2To be perfectly frank: There is no double bond, because there is no $\pi$ orbital, because the molecule is not planar. The charge separated structure is the best representation. The double bonded structure is often used, because it is more convenient for organic chemists to write no charges. – Martin - マーチン May 27 '14 at 03:05
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1Yes and no. A $\pi$ orbital needs to have a nodal plane and this is only possible with (at least local) planarity. In this case the sulphur is best described as $\ce{sp^3}$ hybridised and there is now way with this orbital configuration to have a $\pi$ bond. – Martin - マーチン May 27 '14 at 03:33
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- Lewis structure:
The properly way to determine the Lewis structure, based on this example, is:
- Total valence electrons: $7\cdot2 + 6\cdot2 = 26$
- Total electrons needed for octets/doublets: $8\cdot4 = 32$
- Total shared/bonding electrons: $32-26=6$ (In other words, there are only three bonds.)
- Total electrons in lone pairs: $\text{Step 1} - \text{Step 3} = 26 - 6 = 20$ (In other words, the are only 5 pairs of lone electrons (2 pairs for $\ce{O}$, 6 pairs to $\ce{Cl}$ and the big BUT of Lewis structure analysis: the remaining lone pairs correspond to the sulfur at bonding with oxygen.)
Furhter information on how to draw Lewis structures, please follow the link
Another.Chemist
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