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Hypothetically speaking, if a reagent in a bimolecular reaction has enough energy to break the bond of the other reagent but is orientated in the wrong direction, why does the bond not break? Does the reagent's orientation affect the activation energy of the reaction somehow? The point I'm trying to make is if you have enough energy to break a bond, why would molecular orientation matter when the same amount of energy is being used to distort bonds regardless of orientation?

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Suppose we have a simple reaction of a diatomic and a single atom $\ce{AB + C ->A +BC}$. So the reaction is breaking the bond between $\ce{A}$ and $\ce{B}$, which then bonds to $\ce{C}$. So we might expect the transition state to look like $\ce{A--B--C}$. This suggests the reaction is highly orientation dependent.

Looking at the process as a function of the angle formed by ABC we would expect that a 0 degree angle would vary likely lead to a reaction, since everything is lined up as we would expect, while a 180 degree angle would have almost no chance of forming the desired product as it would aligned more like this $\ce{C--A--B}$. While its not a perfectly linear relationship, we should expect that the reactivity will decrease monotonically as it goes from 0 to 180 degrees.

For more complex structures, the orientation becomes even more important. A large molecule may only have a very small reactive site, and if the other reactant collides with a completely different section of the molecule, there is virtually no chance the desired reaction will occur.

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  • $\begingroup$ I understand that if the nuclei of the reagents don't collide correctly the reaction won't occur. My gripe is when temperature remains constant why wouldn't the "AB" bond break since the same amount of energy is being transferred to the molecule. $\endgroup$ – henrey2306 Aug 10 '18 at 4:01
  • $\begingroup$ @henrey2306 well if it was ~180 degrees it might still break, it just wouldn't form the same products. If however, the angle was closer to 90 degrees, the bond would be unlikely to break at all because while the particles may be travelling with the same amount of energy in each case, the orientation of the collision determines how much of that energy goes into breaking the bond. If the particles only skim each other or collide at an odd angle, the energy of the collision doesn't go into breaking the bond. $\endgroup$ – Tyberius Aug 10 '18 at 4:08
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If you have two huge molecules, say, a sterol in you want to change an OH by some reaction, then clearly the way the reactants approach one another is important.

In the case of a reaction such as $\ce{Cl + OH \to HCl + O}$ then it does depend on the angle of approach. This is because the potential energy is not spherical around the diatomic but is different when approaching along the 'line of centres' compared to approaching at right angles to the OH bond.

Additionally the vibrational energy in a diatomic can have a vital role, not only in how many quanta are excited but at what point in the vibrational motion the collisions occurs. The effect is such that even if there is enough total energy to overcome the barrier to products, i.e. to get over the transition state, vibrational motion can easily frustrate this and the reaction heads back on itself after climbing up towards the transition state.

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You don't have enough energy to break the bond. The reaction's activation energy is (in most cases) lower than the energy level needed to first break one bond and only then form the other.

If your thermal energy is high enough, the two particles form a complex transition state IF the geometry is right. This transition state decays then forwards or backwards. If the geometry (orientation) is wrong, you don't get the transition state. Orbitals with wrong orientation don't overlap, energetically, and you absolutely need the energy from the new bond that is already forming to loosen the old one.

If your thermal energy was actually high enough to just break the bond, that could happen at every collision, also e.g. with inert gas/solvent molecules, or when the molecule bumps together with its own kind. That's called pyrolysis.

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