9.4 Group VII

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2 thoughts on “9.4 Group VII

  1. When going down group 7, reaction between H2 and X2 to form HX decreases. But, supposedly the X-X bond energy and H-X bond energy also decreases down the group. So, why isn’t the reaction faster but decreasing instead?

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    • Your simple question is very tough to answer…

      First of all, to get the right answer. we recall:
      “Going down gp7, reactivity decreases”
      hence
      “reaction rate of formation of HX decreases”

      This is observable. H2 reacts violently with F2, and H2 reacts a lot more slowly with I2.
      Therefore the bond strength of X-X alone cannot provide the answer to the rate of reactivity (or it would mean I2 should react with H2 faster!), so there must be something else happening here.

      The books seem to do a lousy job at explaining it, probably because they want to keep things simple. So let me suggest this scenario to you.

      Assume X-X breaks first. By far the easiest way THIS bond can break is by homolytic fission, producing X radicals. The general trend is going down gp 7 the X-X bond is weaker because the distance between the bonded pair and the nuclei is less, and importantly: the halogens further down the group have more inner e- shells shielding out the attraction between the nucleus and the bonded pair, so it’s easier to break. F2 is an exception because the atoms are so small that when bonded to each other, the e- in their lone pairs repel strongly with the lp on the other F atom, which ‘assists’ the splitting up of the molecule.

      OK. Now that we’ve broken up the molecule (X-X bond energy considered) we need to finish off the process and make our product. The radical must seek a source of electrons. With iodine radicals, the most available source of e- is an un-split I2 molecule, followed by the already split iodine radical. The very last place to act as a supply of e- is the H2 atom. So the iodine radical may bond to I2 to form an (I3.) radical in a propagation step.
      This New (I3.) radical can undergo a decomposition as a propagation step to reform the I2 and new iodine radical. Or, it could do a termination step when the iodine radical reacts with another iodine radical giving I2. This is the reactant we started with. It’s not going to be easy for the radical to ‘see’ the hydrogen molecules electron density in comparison to the I2 and (I.) species nearby. Also it will be difficult for the ‘poorly’ reactive iodine radical to react with and break the strong H2 bond. << This is perhaps the most important point. Its the radical reaction with the H2 which is the rate determining step and not the previous step of generating the halogen radical.

      With Cl2, the Cl radical produced can also do propagation and termination steps with Cl2 and (Cl.) respectively, but the Cl2 and (Cl.) radicals around have less electron density to “choose” from than was the case for iodine, hence are more likely to pick up on the e- density in the H2.

      No only that, but once the H-Cl bond is formed, it’s much harder for H-Cl to re-break again, so once the produce is formed, it stays formed and doesn’t go into an “equilibrium state” whereby it could easily go back to the reactants.

      So really, it’s the “interplay” between the reactivity of the halogen radical with H2 and the other surrounding X. and X2 species, plus the susceptibility of the H-X bond to re-break (due to any energies being supplied or given off in the process) which is the truer indication of the reaction rate is the way it is, and not simply the ability of the X2 to break up.

      To reiterate, the most important factor is the reactivity of the radical with the H2. Iodine radicals are the most stable, followed by Be radicals followed by Cl radicals followed by F radicals which are the most reactive. The H-H bond breaking step is the most difficult so it’s the species that interacts with H2 (i.e. the radical) which determines the rate of the reaction.

      Trouble is, all this is very complicated and way beyond the “simple” answers that an A-level course would demand. A module of chemical kinetics at degree level would be the suitable place to have to consider these things, but not at A-level. So, in order to “fake it” we talk about the energetics of the H-X bond breaking as a way to explain to reactivity between X2 and H2.

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