Lewis Structure of TeO2 (With 6 Simple Steps to Draw!)

I’m super excited to teach you the lewis structure of TeO2 in just 6 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of TeO2 molecule.

So, if you are ready to go with these 6 simple steps, then let’s dive right into it!

Lewis structure of TeO2 contains two double bonds between the Tellurium (Te) atom and each Oxygen (O) atom. The Tellurium atom (Te) is at the center and it is surrounded by 2 Oxygen atoms (O). The Tellurium atom has 1 lone pair and both the Oxygen atoms have 2 lone pairs.

Let’s draw and understand this lewis dot structure step by step.

(Note: Take a pen and paper with you and try to draw this lewis structure along with me. I am sure you will definitely learn how to draw lewis structure of TeO2).

6 Steps to Draw the Lewis Structure of TeO2

Step #1: Calculate the total number of valence electrons

Here, the given molecule is TeO2. In order to draw the lewis structure of TeO2, first of all you have to find the total number of valence electrons present in the TeO2 molecule.
(Valence electrons are the number of electrons present in the outermost shell of an atom).

So, let’s calculate this first.

Calculation of valence electrons in TeO2

• For Tellurium: 

Tellurium is a group 16 element on the periodic table. ^[1]

Hence, the valence electrons present in tellurium is 6 (see below image).

• For Oxygen:

Oxygen is a group 16 element on the periodic table. ^[2]

Hence, the valence electron present in oxygen is 6 (see below image).

Hence in a TeO2 molecule, 

Valence electrons given by Tellurium (Te) atom = 6
Valence electrons given by each Oxygen (O) atom = 6
So, total number of Valence electrons in TeO2 molecule = 6 + 6(2) = 18

Step #2: Select the center atom

While selecting the atom, always put the least electronegative atom at the center. 

(Remember: Fluorine is the most electronegative element on the periodic table and the electronegativity decreases as we move right to left in the periodic table as well as top to bottom in the periodic table). ^[1]

Here in the TeO2 molecule, if we compare the tellurium atom (Te) and oxygen atom (O), then the tellurium is less electronegative than oxygen.

So, tellurium should be placed in the center and the remaining 2 oxygen atoms will surround it.

Step #3: Put two electrons between the atoms to represent a chemical bond

Now in the above sketch of TeO2 molecule, put the two electrons (i.e electron pair) between each tellurium atom and oxygen atom to represent a chemical bond between them.

These pairs of electrons present between the Tellurium (Te) and Oxygen (O) atoms form a chemical bond, which bonds the tellurium and oxygen atoms with each other in a TeO2 molecule.

Step #4: Complete the octet (or duplet) on outside atoms. If the valence electrons are left, then put the valence electrons pair on the central atom

Don’t worry, I’ll explain!

In the Lewis structure of TeO2, the outer atoms are oxygen atoms.

So now, you have to complete the octet on these oxygen atoms (because oxygen requires 8 electrons to have a complete outer shell).

Now, you can see in the above image that all the oxygen atoms form an octet.

Also, only 16 valence electrons of TeO2 molecule are used in the above structure.

But there are total 18 valence electrons in TeO2 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 18 – 16 = 2.

So let’s keep these two electrons (i.e electron pair) on the central atom.

Now, let’s move to the next step.

Step #5: Check whether the central atom has octet or not. If it does not have an octet, then move the electron pair from the outer atom to form a double bond or triple bond

In this step, we have to check whether the central atom (i.e tellurium) has an octet or not. 

In simple words, we have to check whether the central Tellurium (Te) atom has 8 electrons or not.

As you can see from the above image, the central atom (i.e tellurium) has only 6 electrons. So it does not fulfill the octet rule.

Now, in order to fulfill the octet of tellurium atom, we have to move the electron pair from the outer atom (i.e oxygen atom) to form a double bond.

Now you can see from the above image that the central atom (i.e tellurium), is having 8 electrons. So it fulfills the octet rule.

Step #6: Final step – Check the stability of lewis structure by calculating the formal charge on each atom

Now, you have come to the final step and here you have to check the formal charge on tellurium atom (Te) as well as each oxygen atom (O).

For that, you need to remember the formula of formal charge;

Formal charge = Valence electrons – Nonbonding electrons – (Bonding electrons)/2

• For Tellurium:
  Valence electrons = 6 (as it is in group 16)
  Nonbonding electrons = 2
  Bonding electrons = 6
• For 1st Oxygen:
  Valence electron = 6 (as it is in group 16)
  Nonbonding electrons = 4
  Bonding electrons = 4
• For 2nd Oxygen:
  Valence electron = 6 (as it is in group 16)
  Nonbonding electrons = 6
  Bonding electrons = 2

Formal charge	=	Valence electrons	–	Nonbonding electrons	–	(Bonding electrons)/2
Te	=	6	–	2	–	6/2	=	+1
1st O	=	6	–	4	–	4/2	=	0
2nd O	=	6	–	6	–	2/2	=	-1

So you can see above that the formal charges on tellurium is +1 and the formal charge on the 2nd oxygen atom is -1.

This indicates that the above lewis structure of TeO2 is not stable and so we have to minimize the charges to get a more stable lewis structure.

This can be done by shifting the lone pair from negatively charged oxygen atom to the positively charged tellurium atom to form a bond.

Now, in the above structure, you can see that the charges are minimized and the above lewis structure of TeO2 is the final stable structure.

Each electron pair (:) in the lewis dot structure of TeO2 represents the single bond ( | ). So the above lewis dot structure of TeO2 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of SbH3
Lewis Structure of KrCl4
Lewis Structure of PS3-
Lewis Structure of SOF2
Lewis Structure of SeBr4 

---

Article by;