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.
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.
To draw the Lewis structures for the products of the reaction between boron hydride (BH3) and excess hydride ion (H-), we need to follow these steps:
Step 1: Determine the number of valence electrons for each molecule/ion.
- Boron hydride (BH3) has 3 valence electrons for boron (group 3A) and 1 valence electron for each hydrogen, totaling 3 + 3(1) = 6 valence electrons.
- Hydride ion (H-) has 1 valence electron for hydrogen.
Step 2: Assign a central atom.
- In this case, boron (B) will be the central atom.
Step 3: Connect the atoms using a single bond (electron pair) between the central atom and the surrounding atoms.
- Since boron only has three available valence electrons, it can only form three single bonds with hydrogen atoms.
Step 4: Distribute the remaining electrons to satisfy the octet rule for all atoms, except hydrogen.
- Place the remaining electrons on each atom as lone pairs.
Step 5: Check if all atoms have an octet (except hydrogen) and whether formal charges are minimized.
- If necessary, move lone pairs from adjacent atoms to form double or triple bonds to satisfy the octet rule.
Let's now draw the Lewis structures for the products of the reaction:
Product 1: BH4-
- Start by connecting boron (B) with four hydrides (H).
H H
\ /
B
/ \
H H
- Place a negative charge (extra electron) on boron (B-) to satisfy the octet rule.
H H
| /
B- H
| \
H H
Product 2: H2B-HBH2
- Connect one boron (B) atom with two hydrides (H), then connect the other boron (B) atom with one hydride (H).
H B H
/ \ /
H B
\
H
- Place a double bond between the two boron atoms (B=B) to satisfy the octet rule.
H B H
/ \ /
H B
\ /
H
Please note that these Lewis structures indicate the best arrangement of atoms and electrons based on formal charge minimization and octet rule satisfaction.
To draw the Lewis structures for the products of the reaction between boron hydride (BH3) and excess hydride ion (H-), we first need to determine the products.
The reaction between BH3 and H- can be represented as follows:
BH3 + H- -> B + H2
The products of this reaction are a boron atom (B) and hydrogen gas (H2).
Now, let's draw the Lewis structures for each of the products:
1. Boron atom (B):
The Lewis structure for boron (B) is:
H
|
H - B - H
|
H
Boron has three valence electrons, and it forms three single bonds with hydrogen atoms to complete its octet.
2. Hydrogen gas (H2):
The Lewis structure for hydrogen gas (H2) is a simple diatomic molecule:
H - H
Each hydrogen atom shares a single electron with the other, forming a covalent bond.
These are the Lewis structures for the products of the reaction between BH3 and H-.