Lewis Structure of SBr6 (With 5 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of SBr6?

Awesome!

Here, I have explained 5 simple steps to draw the lewis dot structure of SBr6 (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 SBr6 contains six single bonds between the Sulfur (S) atom and each Bromine (Br) atom. The Sulfur atom (S) is at the center and it is surrounded by 6 Bromine atoms (Br). The Sulfur atom does not have a lone pair while all the 6 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 SBr6).

5 Steps to Draw the Lewis Structure of SBr6

Step #1: Calculate the total number of valence electrons

Here, the given molecule is SBr6. In order to draw the lewis structure of SBr6, first of all you have to find the total number of valence electrons present in the SBr6 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 SBr6

• For Sulfur:

Sulfur is a group 16 element on the periodic table.

Hence, the valence electrons present in sulfur 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 SBr6 molecule, 

Valence electrons given by Sulfur (S) atom = 6
Valence electrons given by each Bromine (Br) atom = 7
So, total number of Valence electrons in SBr6 molecule = 6 + 7(6) = 48

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 SBr6 molecule, if we compare the sulfur atom (S) and bromine atom (Br), then the sulfur is less electronegative than bromine.

So, sulfur should be placed in the center and the remaining 6 bromine atoms will surround it.

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

Now in the above sketch of SBr6 molecule, put the two electrons (i.e electron pair) between each sulfur atom and bromine atom to represent a chemical bond between them.

These pairs of electrons present between the Sulfur (S) and Bromine (Br) atoms form a chemical bond, which bonds the sulfur and bromine atoms with each other in a SBr6 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 SBr6, 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, all the 48 valence electrons of SBr6 molecule (as calculated in step #1) are used in the above structure. So there are no remaining electron pairs.

Hence there is no change in the above sketch of SBr6.

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 sulfur atom (S) 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 Sulfur:
  Valence electrons = 6 (as it is in group 16)
  Nonbonding electrons = 0
  Bonding electrons = 12
• 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
S	=	6	–	0	–	12/2	=	0
Br	=	7	–	6	–	2/2	=	0

So you can see above that the formal charges on sulfur as well as bromine are “zero”.

Hence, there will not be any change in the above structure and the above lewis structure of SBr6 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of SBr6 represents the single bond ( | ). So the above lewis dot structure of SBr6 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of IO3-
Lewis Structure of HOFO
Lewis Structure of BrF
Lewis Structure of AlH3
Lewis Structure of MgF2 

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