# TeO3 Lewis structure

TeO3 (tellurium trioxide) has one tellurium atom and three oxygen atoms.

In TeO3 Lewis structure, there are three double bonds around the tellurium atom, with three oxygen atoms attached to it, on each oxygen atom, there are two lone pairs.

Contents

## Steps

Use these steps to correctly draw the TeO3 Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required
#4 Convert lone pairs of the atoms, and minimize formal charges
#5 Repeat step 4 if needed, until all charges are minimized, to get a stable Lewis structure

Let’s discuss each step in more detail.

### #1 First draw a rough sketch

• First, determine the total number of valence electrons

In the periodic table, both tellurium and oxygen lie in group 16.

Hence, both tellurium and oxygen have six valence electrons.

Since TeO3 has one tellurium atom and three oxygen atoms, so…

Valence electrons of one tellurium atom = 6 × 1 = 6
Valence electrons of three oxygen atoms = 6 × 3 = 18

And the total valence electrons = 6 + 18 = 24

• Second, find the total electron pairs

We have a total of 24 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 = 24 ÷ 2 = 12

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since tellurium is less electronegative than oxygen, assume that the central atom is tellurium.

Therefore, place tellurium in the center and oxygens on either side.

• And finally, draw the rough sketch

### #2 Mark lone pairs on the atoms

Here, we have a total of 12 electron pairs. And three Te — O bonds are already marked. So we have to only mark the remaining nine electron pairs as lone pairs on the sketch.

Also remember that tellurium is a period 5 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 tellurium, there is zero lone pair because all nine electron pairs are over.

Mark the lone pairs on the sketch as follows:

### #3 Calculate and mark formal charges on the atoms, if required

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For tellurium atom, formal charge = 6 – 0 – ½ (6) = +3

For each oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

Here, both tellurium and oxygen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because both tellurium and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

### #4 Convert lone pairs of the atoms, and minimize formal charges

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

### #5 Repeating step 4 to get a stable Lewis structure

Since there are charges on tellurium and oxygen atoms, again convert a lone pair of the oxygen atom to make a new Te — O bond with the tellurium atom as follows:

### #6 Minimize charges again

There are still charges on tellurium and oxygen atoms, so again convert a lone pair of the oxygen atom to make a new Te — O bond with the tellurium atom as follows:

In the above structure, you can see that the central atom (tellurium) forms an octet. And the outside atoms (oxygens) also form an octet. Hence, the octet rule is satisfied.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the stable Lewis structure of TeO3.

Next: TeO2 Lewis structure