Homework #4.13 (C)

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Jennifer Guzman 4C
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Joined: Wed Feb 14, 2018 3:01 am

Homework #4.13 (C)

Postby Jennifer Guzman 4C » Fri Nov 23, 2018 2:12 pm

According to the answer in the textbook, Iodine forms a single bond with oxygen. My question is why can't Iodine form double bonds with all three oxygen molecules? In the end, I still managed to have a VSEPR of Ax3E with a trigonal pyramid shape, less than 109.5
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Porus_Karwa_2E
Posts: 72
Joined: Fri Sep 28, 2018 12:24 am

Re: Homework #4.13 (C)

Postby Porus_Karwa_2E » Fri Nov 23, 2018 2:22 pm

It is more stable in terms of formal charge like this. Plus this doesn't break the octet rule.

Michael Novelo 4G
Posts: 64
Joined: Fri Sep 28, 2018 12:28 am

Re: Homework #4.13 (C)

Postby Michael Novelo 4G » Fri Nov 23, 2018 3:58 pm

It has to do with the formal charge stability. In the photo the charge of I is 2 and the charge of O is -1 for a total charge of -1. In the case of double bonds I is a charge of -1 and O is 0 which is a total charge of -1. The reason why Iodine should not have a charge of -1 and is less stable is because Oxygen is more electronegative than Iodine so Oxygen is the preferred atom in this case to hold a negative charge, in this case -1. It's more stable if Iodine does not have a charge of -1.


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