15
$\begingroup$

I have read that in VSEPR theory, multiple bonds are considered or treated as single bonds when predicting the geometry of a molecule? I've read in Yahoo! Answers that it is because only sigma bonds are used in determining their shape. What's the reasoning behind this?

$\endgroup$
5
$\begingroup$

VSEPR is used to predict highly idealized geometries corresponding to Platonic solids or combinations thereof (e.g., the trigonal bipyramidal geometry being essentially the result of fusing two faces of a tetrahedron). The basic rationale behind its use, and the reason for its success in many situations, is the recognition that electrons, being like-charged, repel one another electrostatically. Hence, as a first approximation, it's reasonable to assume that the most stable spatial configuration of any molecule would have the atoms spaced as uniformly and as distant from one another as possible. VSEPR achieves this by placing the central atom of a simple molecule (or some molecular sub-structure) at the center of a geometrically perfect polyhedron having as many vertices as non-central atoms and lone pairs, and then placing the non-central atoms and lone pairs at those vertices.

The reason that $\sigma$ bonds (and lone pairs) determine the geometry is that they form the basic skeleton of the molecule. Recall that $\sigma$ bonds are formed by head-on overlap of atomic orbitals, meaning that they are oriented along the imaginary axis connecting two atomic nuclei, and hence concentrate electron density in the region directly between the two nuclei. $\pi$ bonds, on the other hand, are essentially orthogonal to the $\sigma$ bond skeleton, and are substantially weaker. Moreover, $\pi$ bonds do not exist in isolation, meaning any $\pi$ bond between two given atoms is always formed secondarily to the $\sigma$ bond between said atoms. As such, $\pi$ bonds do not alter the basic idealized geometry of a molecule as dictated by $\sigma$ bonding, although in practice, because they do actually introduce additional electron density and require closer orbital overlap, bond multiplicity affects bond length and bond angles (as does, for that matter, the size of the atoms involved). The magnitude of those deviations from the idealized VSEPR geometry could, in principle, be quite large, although most often those deviations are slight enough that VSEPR remains useful as a first approximation.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Perhaps it's worth noting that pi bonds do contribute to the shape of a molecule and VSEPR takes this into consideration as a deviation from the first approximation. Examples would be $\ce{CF2O}$, $\ce{NO2}$ and $\ce{SO3^2-}$ where the change in bond order leads to repulsive effects and deviations from the 120 degree ideal angle. See Miessler & Tarr 5th edition pg 56. $\endgroup$ – bobthechemist Aug 7 '13 at 12:22
  • $\begingroup$ @bobthechemist, thanks for the reference. As I noted, the change in electron density with increasing bond multiplicity does distort bond angles and lengths, causing deviations from the ideal geometry. Would you agree that it's correct to say that the basic root geometry is solely determined by steric number? My understanding of VSEPR has always been that it doesn't distinguish between multiple and single bonds, and that all corrections to the ideal geometry predicted by VSEPR are the result of empirical observations of the real molecular structures (e.g., by X-ray diffraction). $\endgroup$ – Greg E. Aug 7 '13 at 22:00
  • $\begingroup$ @bobthechemist, I've read various guidelines that have suggested adjustments based on, e.g., the presence of lone pairs in place of bonded atoms (in those cases, some sources suggest that, as a correction to the first approximation, the bond angles between remaining atoms should be collapsed on the order of ~4 degrees, while others maintain angles equal to the angles of the "root" parent structures even for the "derived" shapes). Is there a standard practice for making those types of predictive corrections? $\endgroup$ – Greg E. Aug 7 '13 at 22:05
  • $\begingroup$ Yes, I agree that the heart of VSEPR is steric number. The presence of multiple bonds and lone pairs distorts angles from their ideal values. Personally, I've never seen any (semi)quantitative estimates of these changes and I only use these deviations to explain periodic trends (e.g. the decreasing angles upon going from methane, to ammonia to water). $\endgroup$ – bobthechemist Aug 8 '13 at 0:15
  • $\begingroup$ Actually, if there were a type of pi bonding that is the first bond between the binding atoms -without any sigma bonds- molecule shape wouldn't deviate from what VSEPR predicts more than it deviates now. $\endgroup$ – sencer Aug 8 '13 at 2:17
3
$\begingroup$

Question actually implies a misunderstanding. VSEPR has nothing to do with number of sigma bonds.

VSEPR is a primitive theory, relying only on simple electrostatic effects. In this view, nature of the bond is not relevant but only the number of groups surrounding the central atom is. This is the number of bound atoms, an number of unbound electron pairs.

All VSEPR says is that, groups surrounding the central atom should be separated as much as they can. This yields linear shape for two groups, triangular for three, tetrahedral for four etc...

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ You're obviously correct to note that the VSEPR geometry is determined by steric number. However, the OP's question is tantamount to the question of why multiple bonds and single bonds are treated identically in calculating said steric number. Clearly, the specific nature of sigma bonds is relevant to that. Moreover, any bonding between two given atoms will always include exactly one (and only one) sigma bond, so as a surrogate for counting bonded atoms, counting sigma bonds directly will always yield the same result. $\endgroup$ – Greg E. Aug 7 '13 at 4:28
  • 2
    $\begingroup$ Thank you for your kind note, Greg. You are right, with a few exceptional cases, when you are saying steric number is same with #sigma bonds. This is a neat -let me say- "coincidence". Though it is absurd to count sigma bonds when you are trying to figure out steric number, since only way to count sigma bonds is actually counting steric number. Also, relating VSEPR to MO theory is neither historically, nor logically reasonable. I understand what OP is asking, and I think the fair answer to this question is: What you read is not correct, bond symmetry is irrelevant. $\endgroup$ – sencer Aug 8 '13 at 2:12
  • $\begingroup$ I take your point, and I voted you up. As I think more about it, I think the fundamental point of the nature of bonding being irrelevant is correct. The gist, if I've understood you correctly: as it happens, sigma bonds are the primary form of covalent bonding, but even if some other form of covalent bonding were holding two atoms in close proximity, then the combination of internuclear and electron-electron repulsion would still demand maximum spatial separation, within the limits allowed by the bonding. That strikes me as an entirely accurate point. $\endgroup$ – Greg E. Aug 8 '13 at 3:24
1
$\begingroup$

Because any $\pi$ bond is always affiliated to a $\sigma$ bond, which participates the determination of the geometry. It is no way that a $\pi$ bond can be different orientation to its corresponding $\sigma$ bond.

| improve this answer | |
$\endgroup$

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