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In which of the dimerization process, the achievement of the octet is not the driving force

  1. $\ce{2AlCl3 \to Al2Cl6}$

  2. $\ce{BeCl2 \to BeCl2}$ (solid)

  3. $\ce{2ICl3 \to I2Cl6} $

  4. $\ce{2NO2 \to N2O4}$

Answer: (3) $\ce{2ICl3 \to I2Cl6} $

The answer was given without explanation from the problem book I was solving, so I ask then, what exactly is the reason that $\ce{ICl3}$ dimerize?

From this website , I find it is because it is odd electron and wants to complete the octet, but the question asks why it dimerizes disregarding that rule.

Related

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  • $\begingroup$ Your own reasoning – based on searching, reading and thinking – is supposed to be present to avoid the question closure for lack of own explicit effort. How do I ask a good question. // Why not to analyze, where reaching the octet is and is not achieved ? $\endgroup$
    – Poutnik
    Aug 11 at 7:21
  • $\begingroup$ I understood what you said now, the thing is they specifically asked which one is not driven by octet rule/ odd electron atom rule $\endgroup$
    – Buraian
    Aug 11 at 7:34
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    $\begingroup$ P.S.: Note that I was refering just to the partial question, if reaching octet is the case or not. I was not referring to the second part what is the cause if it is not. $\endgroup$
    – Poutnik
    Aug 11 at 7:35
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    $\begingroup$ It's a dreadful question. Achievement of an octet is never the driving force. The driving force is that dimerisation leads to a lower free energy. Achievement of an octet is merely a hand-waving rationalisation of that observation, it is not the cause. $\endgroup$
    – Ian Bush
    Aug 11 at 11:02
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    $\begingroup$ The real difference is that the others dimerize/polymerize to form more bonds, whereas ICl3 dimerizes (which, by the way, is true) to strengthen its existing bonds (2c-2e -> 3c-4e). $\endgroup$ Aug 11 at 15:32

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