BrO4- Lewis structure

BrO₄^– (perbromate) has one bromine atom and four oxygen atoms.

In the BrO₄^– Lewis structure, 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.

Alternative method: Lewis structure of BrO₄^–

Rough sketch

• First, determine the total number of valence electrons

In the periodic table, bromine lies in group 17, and oxygen lies in group 16.

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

Learn how to find: Bromine valence electrons and Oxygen valence electrons

• 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

Lone pair

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.

Also remember that bromine is a period 4 element, so it can keep more than 8 electrons in its last shell. And oxygen is a period 2 element, so it can not keep more than 8 electrons in its last shell.

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:

Formal charge

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.

Convert a lone pair of the oxygen atom to make a new Br — O bond with the bromine atom as follows:

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:

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:

Final structure

The final structure of BrO₄^– involves a central bromine atom connected to three oxygen atoms through double covalent bonds and to a fourth oxygen atom via a single covalent bond. In this arrangement, the bromine atom utilizes an expanded octet to accommodate fourteen valence electrons, while the three double-bonded oxygens satisfy their octets with two lone pairs each, and the single-bonded oxygen completes its octet with three lone pairs. This configuration is the most stable because it minimizes formal charges across the ion, resulting in a formal charge of zero on the bromine and the three double-bonded oxygens, while the -1 formal charge is localized on the more electronegative single-bonded oxygen atom. Consequently, this specific electronic distribution represents the definitive and most accurate Lewis representation of the perbromate ion.

To complete the representation, draw square brackets around the entire Lewis structure and place a “-” or “-1” sign as a superscript outside the upper right bracket. This notation signifies that the negative charge is a property of the whole BrO₄^– ion.

Next: SeCl₄ Lewis structure

External video

• How to Draw the Lewis Dot Structure for BrO4- (See note in description!) – YouTube • Wayne Breslyn

External links

• https://lambdageeks.com/bro4-lewis-structure/