Lewis structure of TeBr4

The information on this page is ✔ fact-checked.

The Lewis structure of TeBr₄ illustrates the molecular arrangement of tellurium tetrabromide, a compound comprising one tellurium and four bromine atoms. Within this structure, the tellurium atom serves as the central atom, forming single bonds with four bromine atoms. One lone pair is present on the tellurium atom, while each bromine atom carries three lone pairs.

To accurately draw the Lewis structure of TeBr₄, begin by creating a rough diagram of the structure and then depict the chemical bonds between each atom. Include any lone pairs present in the structure. Ensure that the octet rule is satisfied for both the central atom and the surrounding atoms. Finally, check the stability of the structure by calculating the formal charges of each atom and aim for a stable Lewis structure with the fewest formal charges possible.

Steps

Draw skeleton

In this step, first calculate the total number of valence electrons. And then, decide the central atom.

• Let’s calculate the total number of valence electrons

We know that… tellurium is a group 16 element and bromine is a group 17 element. Hence, tellurium has six valence electrons and bromine has seven valence electrons.

Now TeBr₄ has one tellurium atom and four bromine atoms.

So the total number of valence electrons = valence electrons of tellurium atom + (valence electrons of bromine atom × 4)

Therefore, the total number of valence electrons = 6 + 28 = 34

• Now decide the central atom

The atom with the least electronegative value is placed at the center. By looking at the periodic table, we get the electronegativity values for tellurium and bromine as follows:

Electronegativity value of tellurium = 2.1
Electronegativity value of bromine = 2.96

Obviously, tellurium is less electronegative than bromine. Hence, assume that tellurium is the central atom.

So now, put tellurium in the center and bromines on either side. And draw the rough skeleton structure for the Lewis structure of TeBr₄ something like this:

Also read: How to draw Lewis structure of GeH₄ (3 steps)

Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since tellurium is surrounded by four bromines, use eight electrons to show four chemical bonds as follows:

Also read: How to draw Lewis structure of HOF (4 steps)

Mark lone pairs

As calculated earlier, we have a total of 34 valence electrons. And in the above structure, we have already used eight valence electrons. Hence, twenty-six valence electrons are remaining.

Two valence electrons represent one lone pair. So twenty-six valence electrons = thirteen lone pairs.

Note that tellurium is a period 5 element, so it can keep more than 8 electrons in its last shell. And bromine is a period 4 element, so it can keep more than 8 electrons in its last shell.

Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.

The outside atoms are bromines, so each bromine will get three lone pairs. And the central atom (tellurium) will get one lone pair.

So the Lewis structure of TeBr₄ looks something like this:

In the above structure, you can see that the octet is completed on the central atom (tellurium), and also on the outside atoms. Therefore, the octet rule is satisfied.

After completing the octet, one last thing we need to do is, calculate the formal charge and check the stability of the above structure.

Also read: How to draw Lewis structure of XeCl₄ (4 steps)

Calculate formal charge and check stability

The following formula is used to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

Collect the data from the above structure and then, write it down below as follows:

• For tellurium atom

Valence electrons = 6
Nonbonding electrons = 2
Bonding electrons = 8

Formal charge = 6 – 2 – ½ (8) = 0

• For each bromine atom

Valence electrons = 7
Nonbonding electrons = 6
Bonding electrons = 2

Formal charge = 7 – 6 – ½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of TeBr₄ looks something like this:

In the above structure, you can see that the formal charges of both (tellurium and bromine) are zero. Therefore, this is the stable Lewis structure of TeBr₄.

And each horizontal line drawn in the above structure represents a pair of bonding valence electrons.

📝 Your feedback matters. Visit our contact page.

Related

• Lewis structure of GeH₄
• Lewis structure of HOF
• Lewis structure of XeCl₄
• CH₃⁺ Lewis structure
• Lewis structure of CH₃NCO

External links

• TeBr4 Lewis Structure in 5 Steps (With Images) – Pediabay
• Tellurium tetrabromide | TeBr4 | CID 82311 – National Institutes of Health (.gov)
• Draw the best Lewis dot structure for TeBr4 – Chegg
• What is the molecular geometry of TeBr4? – Quora
• Tellurium tetrabromide | Br4Te – ChemSpider
• Draw the Lewis structure for the tellurium tetrabromide (TeBr4) molecule – Brainly
• Draw the best Lewis dot structure for TeBr4 – Numerade
• Draw the Lewis structure of TeBr4? – Quizlet
• What is the lewis structure of tellurium tetrabromide – OneClass
• Tellurium Bromide TeBr4 – American Elements
• Which of the following best describes the VSEPR structure of Br03- and TeBr4 – Bartleby