BrO4– (perbromate) has one bromine atom and four oxygen atoms. In the lewis structure of BrO4–, there is one single bond and three double bonds around the bromine atom, with four oxygen atoms attached to it. The oxygen atom with a single bond has three lone pairs, and the oxygen atom with double bonds has two lone pairs.
Also, there is a negative (-1) charge on the oxygen atom with a single bond.
Here’s how you can draw the BrO4– lewis structure step by step.
Step #1: draw sketch
Step #2: mark lone pairs
Step #3: mark charges
Step #4: minimize charges
Step #5: minimize charges again (if there are)
Let’s break down each step in detail.
#1 Draw Sketch
- First, determine the total number of valence electrons
Hence, bromine has seven valence electrons and oxygen has six valence electrons.
Since BrO4– has one bromine atom and four oxygen atoms, so…
Valence electrons of one bromine atom = 7 × 1 = 7
Valence electrons of four oxygen atoms = 6 × 4 = 24
Now the BrO4– has a negative (-1) charge, so we have to add one more electron.
So the total valence electrons = 7 + 24 + 1 = 32
- Second, find the total electron pairs
We have a total of 32 valence electrons. And when we divide this value by two, we get the value of total electron pairs.
Total electron pairs = total valence electrons ÷ 2
So the total electron pairs = 32 ÷ 2 = 16
- Third, determine the central atom
We have to place the least electronegative atom at the center.
Since bromine is less electronegative than oxygen, assume that the central atom is bromine.
Therefore, place bromine in the center and oxygens on either side.
- And finally, draw the rough sketch
#2 Mark Lone Pairs
Here, we have a total of 16 electron pairs. And four Br — O bonds are already marked. So we have to only mark the remaining twelve electron pairs as lone pairs on the sketch.
Always start to mark the lone pairs from outside atoms. Here, the outside atoms are oxygens.
So for each oxygen, there are three lone pairs, and for bromine, there is zero lone pair because all twelve electron pairs are over.
Mark the lone pairs on the sketch as follows:
#3 Mark Charges
Use the following formula to calculate the formal charges on atoms:
Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons
For bromine atom, formal charge = 7 – 0 – ½ (8) = +3
For each oxygen atom, formal charge = 6 – 6 – ½ (2) = -1
Here, both bromine and oxygen atoms have charges, so mark them on the sketch as follows:
The above structure is not a stable lewis structure because both bromine and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.
#4 Minimize Charges
Convert a lone pair of the oxygen atom to make a new Br — O bond with the bromine atom as follows:
#5 Minimize Charges Again
Since there are charges on bromine and oxygen atoms, again convert a lone pair of the oxygen atom to make a new Br — O bond with the bromine atom as follows:
#6 Minimize Charges Again
There are still charges on bromine and oxygen atoms, so again convert a lone pair of the oxygen atom to make a new Br — O bond with the bromine atom as follows:
In the above structure, you can see that the central atom (bromine) forms an octet. Hence, the octet rule is satisfied.
Now there is still a negative (-1) charge on the oxygen atom.
This is okay, because the structure with a negative charge on the most electronegative atom is the best lewis structure. And in this case, the most electronegative element is oxygen.
Also, the above structure is more stable than the previous structures. Therefore, this structure is the most stable lewis structure of BrO4–.
And since the BrO4– has a negative (-1) charge, mention that charge on the lewis structure by drawing brackets as follows:
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