4.19 6th Edition

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

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Kim Tran 1J
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Joined: Fri Sep 28, 2018 12:24 am

4.19 6th Edition

Postby Kim Tran 1J » Mon Nov 26, 2018 5:24 pm

The question asks to predict the shapes and estimate the bond angles. For BH2- the e- arrangement is tetrahedral and the molecular shape is bent/angular (AX2E2) so I thought the bond angle for H-B-H would be less than 109.5 degrees. But the solution manual says the H-B-H angle is slightly less than 120 degrees, so I am confused. What would be the correct answer?
Last edited by Kim Tran 1J on Mon Nov 26, 2018 5:29 pm, edited 1 time in total.

Rehan Chinoy 1K
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Joined: Fri Sep 28, 2018 12:25 am

Re: 4.19 6th Edition

Postby Rehan Chinoy 1K » Mon Nov 26, 2018 5:28 pm

It would be less than 120 degrees because the electron geometry is trigonal planar not tetrahedral. Thus, when one of the regions of electron density is a lone pair, the bond angles become less than 120 degrees. You would be correct if the electron geometry was tetrahedral i.e. there was one more region of electron density.

Andonios Karas 4H
Posts: 30
Joined: Fri Sep 28, 2018 12:27 am

Re: 4.19 6th Edition

Postby Andonios Karas 4H » Mon Nov 26, 2018 5:34 pm

Boron is an exception to the octet rule and is content with having only 6 electrons in its valence shell.
Therefore, when drawing the Lewis Structure of BH2-, you should have 3 regions of electron density. Two in the form of B-H bonds and one as a lone pair on Boron.
Because it has 3 regions of electron density, the electron geometry must be trigonal planar (bond angle of 120o)
And because it has 2 bonding pairs and 1 lone pair (AX2E), the molecular geometry must be bent or angular with the lone pair pushing down more on the bonding pairs (leading to bond angle <120o)

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