Ready to learn how to draw the lewis structure of TeBr4?
Here, I have explained 5 simple steps to draw the lewis dot structure of TeBr4 (along with images).
So, if you are ready to go with these 5 simple steps, then let’s dive right into it!
Lewis structure of TeBr4 contains four single bonds between the Tellurium (Te) atom and each Bromine (Br) atom. The Tellurium atom (Te) is at the center and it is surrounded by 4 Bromine atoms (Br). The Tellurium atom has 1 lone pair while all the four bromine atoms have 3 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 TeBr4).
5 Steps to Draw the Lewis Structure of TeBr4
Step #1: Calculate the total number of valence electrons
Here, the given molecule is TeBr4. In order to draw the lewis structure of TeBr4, first of all you have to find the total number of valence electrons present in the TeBr4 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 TeBr4
- For Tellurium:
Tellurium is a group 16 element on the periodic table.
Hence, the valence electrons present in tellurium is 6 (see below image).
- For Bromine:
Bromine is a group 17 element on the periodic table.
Hence, the valence electrons present in bromine is 7 (see below image).
Hence in a TeBr4 molecule,
Valence electrons given by Tellurium (Te) atom = 6
Valence electrons given by each Bromine (Br) atom = 7
So, total number of Valence electrons in TeBr4 molecule = 6 + 7(4) = 34
Step #2: Select the center atom
While selecting the atom, always put the least electronegative atom at the center.
(Remember: Bromine 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).
Here in the TeBr4 molecule, if we compare the tellurium atom (Te) and bromine atom (Br), then the tellurium is less electronegative than bromine.
So, tellurium should be placed in the center and the remaining 4 bromine atoms will surround it.
Step #3: Put two electrons between the atoms to represent a chemical bond
Now in the above sketch of TeBr4 molecule, put the two electrons (i.e electron pair) between each tellurium atom and bromine atom to represent a chemical bond between them.
These pairs of electrons present between the Tellurium (Te) and Bromine (Br) atoms form a chemical bond, which bonds the tellurium and bromine atoms with each other in a TeBr4 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 TeBr4, the outer atoms are bromine atoms.
So now, you have to complete the octet on these bromine atoms (because bromine requires 8 electrons to have a complete outer shell).
Now, you can see in the above image that all the bromine atoms form an octet.
Also, only 32 valence electrons of TeBr4 molecule are used in the above structure.
But there are total 34 valence electrons in TeBr4 molecule (as calculated in step #1).
So the number of electrons left to be kept on the central atom = 34 – 32 = 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: 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 bromine atom (Br).
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 = 8
- For Bromine:
Valence electron = 7 (as it is in group 17)
Nonbonding electrons = 6
Bonding electrons = 2
|Formal charge||=||Valence electrons||–||Nonbonding electrons||–||(Bonding electrons)/2|
So you can see above that the formal charges on tellurium as well as bromine are “zero”.
Hence, there will not be any change in the above structure and the above lewis structure of TeBr4 is the final stable structure only.
Each electron pair (:) in the lewis dot structure of TeBr4 represents the single bond ( | ). So the above lewis dot structure of TeBr4 can also be represented as shown below.