2E.11 Part B

(Polar molecules, Non-polar molecules, etc.)

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Letty Liu 2C
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Joined: Fri Aug 09, 2019 12:16 am

2E.11 Part B

Postby Letty Liu 2C » Sun Nov 17, 2019 11:19 pm

The question asks for a prediction of the shape of iodine tetrachloride. After drawing the Lewis structure, there are three bonds to chlorine and two lone pairs on iodine.I understand that all the bonds form a trigonal bypyramidal, but why is the ending shape T-shaped and not trigonal planar?

Posts: 100
Joined: Fri Aug 30, 2019 12:18 am

Re: 2E.11 Part B

Postby ShravanPatel2B » Sun Nov 17, 2019 11:27 pm

This is because the lone pairs of electrons prefer to lie in the equatorial positions due to the fact that this causes less electron repulsions. This leaves the atoms in the T Shape.

Liliana Aguas 3G
Posts: 59
Joined: Wed Sep 18, 2019 12:20 am

Re: 2E.11 Part B

Postby Liliana Aguas 3G » Sun Nov 17, 2019 11:29 pm

Yes, you have to determine which to take off from the electron arrangement shape to get the least amount of repulsion for the molecule shape.

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