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.

Alternative method: Lewis structure of TeO₂

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 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

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

• 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

Lone pair

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:

Formal charge

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.

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

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:

Final structure

The final structure of TeO₂ consists of a central tellurium atom linked to two oxygen atoms through double covalent bonds. In this arrangement, the tellurium atom utilizes an expanded octet to accommodate ten valence electrons, comprising four bonding pairs and one lone pair. Each oxygen atom satisfies the octet rule by retaining two lone pairs alongside its double bond. This specific configuration is preferred because it results in formal charges of zero for every atom in the molecule, representing the most energetically favorable state for the molecule. Consequently, this electronic pattern serves as the definitive and most accurate Lewis representation for tellurium dioxide.

Next: SbH₃ Lewis structure

External links

• https://www.numerade.com/ask/question/i-find-the-total-number-of-valence-electrons-ii-draw-the-lewis-structure-if-the-species-has-resonance-draw-additional-lewis-structures-iii-for-any-bonds-that-are-polar-covalent-draw-the-corr-49963/
• https://www.chegg.com/homework-help/questions-and-answers/draw-lewis-dot-structure-tellurium-dioxide-teo2-bonding-best-described-q49983229
• https://oneclass.com/homework-help/chemistry/7032628-teo2-lewis-structure.en.html