# TeO2 Lewis structure

TeO2 (tellurium dioxide) has one tellurium atom and two oxygen atoms.

In TeO2 Lewis structure, there are two double bonds around the tellurium atom, with two oxygen atoms attached to it. Each oxygen atom has two lone pairs, and the tellurium atom has one lone pair.

Contents

## Steps

Here’s how you can easily draw the TeO2 Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms
#4 Minimize formal charges by converting lone pairs of the atoms, and try to get a stable Lewis structure
#5 Repeat step 4 again if needed, until all charges are minimized

Now, let’s take a closer look at each step mentioned above.

### #1 Draw a rough skeleton structure

• 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 TeO2 has one tellurium atom and two oxygen atoms, so…

Valence electrons of one tellurium atom = 6 × 1 = 6
Valence electrons of two oxygen atoms = 6 × 2 = 12

And the total valence electrons = 6 + 12 = 18

• Second, find the total electron pairs

We have a total of 18 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 = 18 ÷ 2 = 9

• 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 Mention lone pairs on the atoms

Here, we have a total of 9 electron pairs. And two Te — O bonds are already marked. So we have to only mark the remaining seven 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 one lone pair.

Mark the lone pairs on the sketch as follows:

### #3 If needed, mention formal charges on the atoms

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 – 2 – ½ (4) = +2

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 Minimize formal charges by converting lone pairs of the atoms

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

### #5 Since there are charges on atoms, repeat step 4 again

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:

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 TeO2.

Next: SbH3 Lewis structure