Why does iodine trichloride dimerize?

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

• 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 ? Aug 11 at 7:21
• I understood what you said now, the thing is they specifically asked which one is not driven by octet rule/ odd electron atom rule Aug 11 at 7:34
• 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. Aug 11 at 7:35
• 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. Aug 11 at 11:02
• 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). Aug 11 at 15:32