HBrO4 Lewis structure

HBrO₄ (perbromic acid) has one hydrogen atom, one bromine atom, and four oxygen atoms.

In HBrO₄ Lewis structure, there are three double bonds and one single bond around the bromine atom, with four oxygen atoms attached to it. The oxygen atom with a single bond is attached with one hydrogen atom, and on each oxygen atom, there are two lone pairs.

Alternative method: Lewis structure of HBrO₄

Rough sketch

• First, determine the total number of valence electrons

In the periodic table, hydrogen lies in group 1, bromine lies in group 17, and oxygen lies in group 16.

Hence, hydrogen has one valence electron, bromine has seven valence electrons, and oxygen has six valence electrons.

Since HBrO₄ has one hydrogen atom, one bromine atom, and four oxygen atoms, so…

Valence electrons of one hydrogen atom = 1 × 1 = 1
Valence electrons of one bromine atom = 7 × 1 = 7
Valence electrons of four oxygen atoms = 6 × 4 = 24

And the total valence electrons = 1 + 7 + 24 = 32

Learn how to find: Hydrogen valence electrons, 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

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Now we have to choose the central atom from bromine and oxygen. 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 hydrogen and oxygen on either side.

• And finally, draw the rough sketch

Lone pair

Here, we have a total of 16 electron pairs. And five bonds are already marked. So we have to only mark the remaining eleven electron pairs as lone pairs on the sketch.

Also remember that hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell. 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 and hydrogen. But no need to mark on hydrogen, because hydrogen already has two electrons.

So for top oxygen, bottom oxygen, and left oxygen, there are three lone pairs. For right oxygen, there are two lone pairs, and for bromine, there is zero lone pair because all eleven 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 hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

For bromine atom, formal charge = 7 – 0 – ½ (8) = +3

For top oxygen, bottom oxygen, and left oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

For right oxygen atom, formal charge = 6 – 4 – ½ (4) = 0

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 left 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 bottom 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 top oxygen atom to make a new Br — O bond with the bromine atom as follows:

Final structure

The final structure of HBrO₄ comprises a central bromine atom linked to four oxygen atoms, one of which is additionally bonded to a hydrogen atom. In this configuration, the bromine atom utilizes an expanded valence shell to form three double covalent bonds with three oxygen atoms and a single covalent bond with the hydroxyl (OH) group, leaving no lone pairs on the bromine. Within this layout, each double-bonded oxygen atom satisfies the octet rule by maintaining two lone pairs, and the oxygen atom in the hydroxyl group reaches a stable octet by forming two single bonds and retaining two lone pairs. The hydrogen atom fulfills its stable duet through its single bond. This arrangement represents the most stable state for the molecule because it results in a formal charge of zero for every atom involved. Therefore, this specific electronic distribution serves as the definitive and most accurate Lewis representation of perbromic acid.

Next: PO₂^– Lewis structure

External links

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