TeBr2 Lewis Structure in 6 Steps (With Images)

So you have seen the above image by now, right?

Let me explain the above image in short.

TeBr2 lewis structure has a Tellurium atom (Te) at the center which is surrounded by two Bromine atoms (Br). There are 2 single bonds between the Tellurium atom (Te) and each Bromine atom (Br). There are 2 lone pairs on the Tellurium atom (Te) and 3 lone pairs on both the Bromine atoms (Br).

If you haven’t understood anything from the above image of TeBr2 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of TeBr2.

So let’s move to the steps of drawing the lewis structure of TeBr2.

Steps of drawing TeBr2 lewis structure

Step 1: Find the total valence electrons in TeBr2 molecule

In order to find the total valence electrons in a TeBr2 molecule, first of all you should know the valence electrons present in tellurium atom as well as bromine atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)

Here, I’ll tell you how you can easily find the valence electrons of tellurium as well as bromine using a periodic table.

Total valence electrons in TeBr2 molecule

→ Valence electrons given by tellurium atom:

Tellurium is a group 16 element on the periodic table. ^[1] Hence the valence electrons present in tellurium is 6.

You can see the 6 valence electrons present in the tellurium atom as shown in the above image.

→ Valence electrons given by bromine atom:

Bromine is a group 17 element on the periodic table. ^[2] Hence the valence electrons present in bromine is 7.

You can see the 7 valence electrons present in the bromine atom as shown in the above image.

Hence, 

Total valence electrons in TeBr2 molecule = valence electrons given by 1 tellurium atom + valence electrons given by 2 bromine atoms = 6 + 7(2) = 20.

Step 2: Select the central atom

For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.

Now here the given molecule is TeBr2 and it contains tellurium atom (Te) and bromine atoms (Br).

You can see the electronegativity values of tellurium atom (Te) and bromine atom (Br) in the above periodic table.

If we compare the electronegativity values of tellurium (Te) and bromine (Br) then the tellurium atom is less electronegative.

So here the tellurium atom (Te) is the center atom and the bromine atoms (Br) are the outside atoms.

Step 3: Connect each atoms by putting an electron pair between them

Now in the TeBr2 molecule, you have to put the electron pairs between the tellurium atom (Te) and bromine atoms (Br).

This indicates that the tellurium (Te) and bromine (Br) are chemically bonded with each other in a TeBr2 molecule.

Step 4: Make the outer atoms stable. Place the remaining valence electrons pair on the central atom.

Now in this step, you have to check the stability of the outer atoms.

Here in the sketch of TeBr2 molecule, you can see that the outer atoms are bromine atoms.

These outer bromine atoms are forming an octet and hence they are stable.

Also, in step 1 we have calculated the total number of valence electrons present in the TeBr2 molecule.

The TeBr2 molecule has a total 20 valence electrons and out of these, only 16 valence electrons are used in the above sketch.

So the number of electrons which are left = 20 – 16 = 4.

You have to put these 4 electrons on the central tellurium atom in the above sketch of TeBr2 molecule.

Now let’s proceed to the next step.

Step 5: Check the octet on the central atom

In this step, you have to check whether the central tellurium atom (Te) is stable or not.

In order to check the stability of the central bromine (Br) atom, we have to check whether it is forming an octet or not.

You can see from the above picture that the tellurium atom is forming an octet. That means it has 8 electrons.

And hence the central tellurium atom is stable.

Now let’s proceed to the final step to check whether the lewis structure of TeBr2 is stable or not.

Step 6: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of lewis structure of TeBr2.

The stability of lewis structure can be checked by using a concept of formal charge.

In short, now you have to find the formal charge on tellurium (Te) atom as well as bromine (Br) atoms present in the TeBr2 molecule.

For calculating the formal charge, you have to use the following formula;

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

You can see the number of bonding electrons and nonbonding electrons for each atom of TeBr2 molecule in the image given below.

For Tellurium (Te) atom:
Valence electrons = 6 (because tellurium is in group 16)
Bonding electrons = 4
Nonbonding electrons = 4

For Bromine (Br) atom:
Valence electron = 7 (because bromine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

Formal charge	=	Valence electrons	–	(Bonding electrons)/2	–	Nonbonding electrons
Te	=	6	–	4/2	–	4	=	0
Br	=	7	–	2/2	–	6	=	0

From the above calculations of formal charge, you can see that the tellurium (Te) atom as well as bromine (Br) atom has a “zero” formal charge.

This indicates that the above lewis structure of TeBr2 is stable and there is no further change in the above structure of TeBr2.

In the above lewis dot structure of TeBr2, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of TeBr2.

I hope you have completely understood all the above steps.

For more practice and better understanding, you can try other lewis structures listed below.

Try (or at least See) these lewis structures for better understanding:

AsF5 Lewis Structure	HI Lewis Structure
PO3- Lewis Structure	BBr3 Lewis Structure
IF2- Lewis Structure	BrF2- Lewis Structure