Predict the correct order of bond angle. Give valid reasons.
\(\sf Cl_2O, ~ ClO_2^-, ~ ClO_2\)
Stacey Warren - Expert brainly.com
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I guess, I will start with Cl2O since that's probably the most straightforward.
there are two chlorine atoms, and one oxygen atom. (6 valence electrons) in oxygen, while there are 7 valence electrons in the two chlorines to give (2x7) = 14 + 6 = 20 valence electrons in total.
I guess that the structure of this molecule would be something like water, because we can see oxygen has two lone pairs, so I guess because of VESPR the lone pairs would cause the two chlorine atoms to push a bit closer together causing a bond angle to be smaller. maybe the angle has to be maybe 107 don't remember off the top of my head.
I don't know for sure how to reason this out, since I know some molecules aren't "pretty" like people assume. Specifically thin this case, I would be almost willing to believe that all 3 of these compounds have a \(Cl\) in the center, since \(Cl_2\) is common, I think adding an oxygen to the end might be how it really looks: |dw:1439681403988:dw|
Ok why would I suspect this? Well I'm not sure I don't know, but what I do know is Nitrous oxide \(N_2O\) has this structure: |dw:1439681444695:dw|
Plus if \(Cl\) is in the center, it makes all the molecules more comparable in my opinion which I think is the point of the question: |dw:1439681485600:dw|
So those are my main 2 arguments for why I believe \(Cl\) could be at the center... But like I said I really don't know how to reason that out.
Thank you so much guys for the help. I did some research too and think that the correct order should be
\(\sf ClO2 > ClO_2^- > Cl_2O\)
Reason why I think so is because bond angle decreases by increase in number of lone pair electrons on central atom due to increased repulsion. Also if the central atom is more electronegative or is shorter then the electrons are more closer and hence the repulsion is greater so bond angle will be less.