6
$\begingroup$

Is chlorine in vinyl chloride sp2 or sp3 hybridized?

Calculating using steric number it is found to be sp3 hybridized. But according to I. L. Finar* it is sp2.

*I.L. Finar: Organic chemistry Vol.1 Fundamental principles. Sixth Edition. Orient Longmans: 1973. Page 328.

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ Let's put it this way: what observable difference would that make? $\endgroup$
    – Ivan Neretin
    Commented Jun 27, 2017 at 16:07
  • 3
    $\begingroup$ Terminal, electronegative atoms are in 99% of the cases sp hybridised. (There are no bunny ears.) $\endgroup$
    – Martin - マーチン ♦
    Commented Jun 28, 2017 at 7:10

1 Answer 1

Reset to default
10
$\begingroup$

Disclaimer: An important point to remember is, that hybridisation always follows the molecular structure, it is never the cause for a certain structure. As such, hybridisation is an interpretation tool, nothing more.

As a rule of thumb, terminal atoms (excluding hydrogen) are almost always (maximally) approximately sp hybridised. While other hybridisation schemes can be applied, they are usually not a good representation.*

One popular example is water. In many texts the central oxygen is described as having (approximately) four sp3 orbitals, which essentially makes the lone pairs equivalent. That view, however, is not in agreement with the photoelectron spectrum, which clearly shows that the lone pairs are not equivalent. At this point I'd like to refer you to Michael Laing's article: "No rabbit ears on water. The structure of the water molecule: What should we tell the students?" (J. Chem. Educ. 1987, 64 (2), 124.).

The same principle described there applies to terminal atoms. Because of the local $C_\mathrm{\infty}$ symmetry the notion of three equivalent lone pairs is likely to be false. The most likely (main resonance contributor) structure for these atoms are (if hybridisation is at all feasible, i.e. mostly in the second period) one sp orbital forming the bond, one sp lone pair, two (perpendicular) p lone pairs.

In this particular example, I have calculated the molecule on the DF-BP86/def2-SVP level of theory and analysed it with the natural bond orbital theory (NBO6). Below you find the localised orbitals which best fit with the common Lewis structure of the molecule.
I have ordered the orbitals starting with the carbon-carbon σ- and carbon-chlorine σ-bonds on the bottom. Following that is the carbon-carbon π-bond. Continuing with the three carbon-hydrogen σ-bonds, and finally ending with the three chlorine lone pairs. The occupied orbitals are in blue and orange, while the corresponding virtual orbitals are next to it in red and yellow. The detailed analysis in numbers is at the end of the post.

nbo of vinyl chloride

In ast's answer there is the claim that chlorine must be sp2 hybridised because of a resonance contributor. While this explanation is extremely tempting because it easy to understand, you cannot judge from the possibility of resonance to the electronic structure. As another general rule of thumb you can remember: the lesser the orbitals are hybridised, the more likely the structure is.
The calculation also provides an analysis in terms of natural resonance theory. On the DF-BP86/def2-SVP level the contribution of the second configuration to the total electronic structure is only 6%. The main configuration contributes with about 90%, other configurations (ionic) make up the remaining 4%.

In summary, I'm afraid, but your book is incorrect, or at least incomplete. The description with two equivalent sp2 lone pairs is unnecessary complicated and does not accurately reproduce the electronic structure.

Truncated results of the NBO analysis (skipping core orbitals)

     (Occupancy)   Bond orbital / Hybrids
 ------------------ Lewis ------------------------------------------------------
   8. (1.99378) LP ( 1)Cl  6            s( 81.55%)p 0.23( 18.44%)d 0.00(  0.01%)
   9. (1.97199) LP ( 2)Cl  6            s(  0.18%)p99.99( 99.79%)d 0.18(  0.03%)
  10. (1.90237) LP ( 3)Cl  6            s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
  11. (1.99642) BD ( 1) C  1- C  2
               ( 48.61%)   0.6972* C  1 s(  0.00%)p 1.00( 99.94%)d 0.00(  0.06%)
               ( 51.39%)   0.7169* C  2 s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
  12. (1.99535) BD ( 2) C  1- C  2
               ( 49.24%)   0.7017* C  1 s( 40.16%)p 1.49( 59.76%)d 0.00(  0.08%)
               ( 50.76%)   0.7124* C  2 s( 44.47%)p 1.25( 55.47%)d 0.00(  0.06%)
  13. (1.96941) BD ( 1) C  1- H  3
               ( 61.15%)   0.7820* C  1 s( 29.81%)p 2.35( 70.15%)d 0.00(  0.04%)
               ( 38.85%)   0.6233* H  3 s( 99.91%)p 0.00(  0.09%)
  14. (1.98105) BD ( 1) C  1- H  4
               ( 61.33%)   0.7832* C  1 s( 30.09%)p 2.32( 69.87%)d 0.00(  0.04%)
               ( 38.67%)   0.6218* H  4 s( 99.91%)p 0.00(  0.09%)
  15. (1.98492) BD ( 1) C  2- H  5
               ( 62.36%)   0.7897* C  2 s( 32.79%)p 2.05( 67.18%)d 0.00(  0.03%)
               ( 37.64%)   0.6135* H  5 s( 99.89%)p 0.00(  0.11%)
  16. (1.99056) BD ( 1) C  2-Cl  6
               ( 43.14%)   0.6568* C  2 s( 22.75%)p 3.39( 77.00%)d 0.01(  0.25%)
               ( 56.86%)   0.7541*Cl  6 s( 18.31%)p 4.43( 81.12%)d 0.03(  0.57%)
 ---------------- non-Lewis ----------------------------------------------------
  17. (0.09516) BD*( 1) C  1- C  2
               ( 51.39%)   0.7169* C  1 s(  0.00%)p 1.00( 99.94%)d 0.00(  0.06%)
               ( 48.61%)  -0.6972* C  2 s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
  18. (0.01232) BD*( 2) C  1- C  2
               ( 50.76%)   0.7124* C  1 s( 40.16%)p 1.49( 59.76%)d 0.00(  0.08%)
               ( 49.24%)  -0.7017* C  2 s( 44.47%)p 1.25( 55.47%)d 0.00(  0.06%)
  19. (0.00909) BD*( 1) C  1- H  3
               ( 38.85%)   0.6233* C  1 s( 29.81%)p 2.35( 70.15%)d 0.00(  0.04%)
               ( 61.15%)  -0.7820* H  3 s( 99.91%)p 0.00(  0.09%)
  20. (0.01032) BD*( 1) C  1- H  4
               ( 38.67%)   0.6218* C  1 s( 30.09%)p 2.32( 69.87%)d 0.00(  0.04%)
               ( 61.33%)  -0.7832* H  4 s( 99.91%)p 0.00(  0.09%)
  21. (0.02582) BD*( 1) C  2- H  5
               ( 37.64%)   0.6135* C  2 s( 32.79%)p 2.05( 67.18%)d 0.00(  0.03%)
               ( 62.36%)  -0.7897* H  5 s( 99.89%)p 0.00(  0.11%)
  22. (0.03245) BD*( 1) C  2-Cl  6
               ( 56.86%)   0.7541* C  2 s( 22.75%)p 3.39( 77.00%)d 0.01(  0.25%)
               ( 43.14%)  -0.6568*Cl  6 s( 18.31%)p 4.43( 81.12%)d 0.03(  0.57%)

Summary of Natural Population Analysis

                                     Natural Population
             Natural    ---------------------------------------------
  Atom No    Charge        Core      Valence    Rydberg      Total
 --------------------------------------------------------------------
    C  1   -0.43992      1.99995     4.42868    0.01129     6.43992
    C  2   -0.22801      1.99996     4.20978    0.01827     6.22801
    H  3    0.22758      0.00000     0.76996    0.00246     0.77242
    H  4    0.22498      0.00000     0.77165    0.00337     0.77502
    H  5    0.23463      0.00000     0.76237    0.00300     0.76537
   Cl  6   -0.01926      9.99998     7.00353    0.01575    17.01926
 ====================================================================
 * Total *  0.00000     13.99989    17.94598    0.05414    32.00000

Optimised Geometry DF-BP86/def2-SVP

6
symmetry cs ( E = -538.027155015 A.U. )
C        1.309538928      1.048660302      0.000000000
C        0.000000000      0.759346920      0.000000000
H        1.630332304      2.102388803      0.000000000
H        2.083175437      0.265507033      0.000000000
H       -0.793639761      1.523771533      0.000000000
Cl      -0.633947150     -0.867041806      0.000000000

* It is also important to understand, that atoms are never hybridised, only orbitals are. The presence of an sp hybrid orbital does not exclude the presence of a sp3 hybrid orbital.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Would u elaborate on what's on the second column of the picture you added? $\endgroup$
    – Mockingbird
    Commented Jun 28, 2017 at 12:09
  • 1
    $\begingroup$ @Mockingbird These are the corresponding (anti-bonding) virtual orbitals that belong to the bonding orbitals. Note that they all have a nodal plane perpendicular to the bonding axis. $\endgroup$
    – Martin - マーチン ♦
    Commented Jun 28, 2017 at 12:10
  • $\begingroup$ Why do lone pairs don't have corresponding antibonding orbital? Or, you just didn't mention? $\endgroup$
    – Mockingbird
    Commented Jun 28, 2017 at 12:19
  • $\begingroup$ @Mockingbird They don't form a bond. There is only one combination, that's why lone pairs are usually considered as non-bonding. $\endgroup$
    – Martin - マーチン ♦
    Commented Jun 28, 2017 at 12:21
  • $\begingroup$ But is it possible to rationalise this using some simply chemistry explanation, instead of just basing this theoretically on a computational calculation? For example, I was thinking, it is likely terminal atoms are sp-hybridised because they often only make one sigma bond to the central atom and so to maximise the strength of this bond, it would be reasonable that the terminal atom uses an sp-hybrid for bonding since the sp-hybrid orbital is directional and has high s character. Would that be a reasonable simple chemistry rationalisation of your computational result? $\endgroup$
    – Tan Yong Boon
    Commented Jul 5, 2018 at 10:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.