Draw the best lewis structure for bro4- and determine the formal charge on bromine
The Correct Answer and Explanation is:
The best Lewis structure for the bromate ion (\text{BrO}_4^-) involves arranging the atoms and electrons in such a way that satisfies the octet rule for most atoms while also considering the formal charges to be as close to zero as possible. Let’s go step by step to draw the Lewis structure and determine the formal charge on bromine.
Step 1: Count the total valence electrons
- Bromine ((\text{Br})) is in Group 17 (halogens), so it has 7 valence electrons.
- Oxygen ((\text{O})) is in Group 16, so each oxygen has 6 valence electrons.
- The ion has a -1 charge, meaning we add 1 electron to the total.
Total valence electrons = (7 (\text{from Br}) + 4 \times 6 (\text{from O}) + 1 (\text{extra electron for the negative charge}) = 7 + 24 + 1 = 32) electrons.
Step 2: Arrange the atoms
- Bromine is less electronegative than oxygen, so it is placed in the center of the structure, with the four oxygen atoms surrounding it.
- The oxygen atoms are connected to the bromine atom by single bonds initially.
Step 3: Distribute the electrons
- Place single bonds between the bromine and the four oxygen atoms. Each single bond consists of 2 electrons, so 4 bonds will use up 8 electrons.
- Distribute the remaining 24 electrons (32 total – 8 used for bonds) as lone pairs on the oxygen atoms. Since each oxygen atom needs 8 electrons to satisfy the octet rule, each oxygen will get 6 more electrons, except for one oxygen that will form a double bond with the bromine.
Step 4: Form double bonds
- One of the oxygen atoms will form a double bond with bromine, using 4 electrons (2 electrons from the bond and 2 lone pairs on oxygen), so that oxygen now satisfies the octet rule. This leaves the other three oxygen atoms each with 3 lone pairs of electrons and a single bond to bromine.
Step 5: Determine the formal charge on bromine
The formal charge ((FC)) on an atom is calculated using the formula:
[
FC = \text{Valence electrons} – \left(\text{Lone electrons} + \frac{\text{Bonded electrons}}{2}\right)
]
For bromine in (\text{BrO}_4^-), bromine is connected to 4 oxygens and has no lone electrons. The number of electrons it shares (bonded electrons) is (2 \times 4 = 8).
Thus:
[
FC = 7 (\text{valence electrons}) – \left(0 (\text{lone electrons}) + \frac{8 (\text{bonded electrons})}{2}\right) = 7 – 4 = +3
]
So the formal charge on bromine is +3.
Final Lewis structure:
The final structure will show bromine at the center, with three single bonds to oxygen atoms (each with lone pairs) and one double bond to another oxygen (also with lone pairs). The negative charge is distributed over the oxygens, where each oxygen has a formal charge of -1 (except the one double-bonded to bromine, which has a formal charge of 0).
Thus, the best Lewis structure for (\text{BrO}_4^-) involves a central bromine atom with a formal charge of +3 and the oxygens with formal charges that balance out to -1 overall.