Consider the reaction between boron hydride and excess hydride ion. Draw the best Lewis structures for the products of the reaction include lone pairs and nonzero formal charges.

Answer 1

The reaction between boron hydride (BH3) and excess hydride ion (H-) would result in the formation of borane anion (BH4-).

To draw the Lewis structure for BH4-, we first start with boron (B) as the central atom. Boron only has 3 valence electrons, so it needs 3 additional electrons to complete its octet. Each hydrogen atom (H) contributes one valence electron, so in BH3, there are a total of 3 valence electrons from hydrogen.

Since hydride ion (H-) donates one additional electron, we now have 4 electrons around boron. We can place them as lone pairs in BH4- as follows:

H (lone pair) H (lone pair) H (lone pair)
\ | /
B H
/ |
H (lone pair) H (lone pair) H (lone pair)

Each hydrogen atom has a complete octet with two electrons as lone pairs, and the boron atom also now has a complete octet with 8 electrons around it.

All of the atoms have a formal charge of zero, as each hydrogen atom donates one electron, and boron accepts one electron from hydride ion.