4.19 6th Edition

[Kim Tran 1J]

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?

[Rehan Chinoy 1K]

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]

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 120^o)
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 <120^o)