I’m super excited to teach you the lewis structure of IBr4- ion in just 5 simple steps.
Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of IBr4- ion.
So, if you are ready to go with these 5 simple steps, then let’s dive right into it!
Lewis structure of IBr4- ion contains four single bonds between the Iodine (I) atom and each Bromine (Br) atom. The Iodine atom (I) is at the center and it is surrounded by 4 Bromine atoms (Br). The Iodine atom has 2 lone pairs and it also has -1 formal charge.
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 IBr4-).
5 Steps to Draw the Lewis Structure of IBr4-
Step #1: Calculate the total number of valence electrons
Here, the given ion is IBr4- ion. In order to draw the lewis structure of IBr4- ion, first of all you have to find the total number of valence electrons present in the IBr4- ion.
(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 IBr4-
- For Iodine:
Iodine is a group 17 element on the periodic table.
Hence, the valence electrons present in iodine is 7 (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 IBr4- ion,
Valence electrons given by Iodine (I) atom = 7
Valence electrons given by each Bromine (Br) atom = 7
Electron due to -1 charge, 1 more electron is added
So, total number of Valence electrons in IBr4- ion = 7 + 7(4) + 1 = 36
Step #2: Select the center atom
While selecting the center 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).
Here in the IBr4 molecule, if we compare the iodine atom (I) and bromine atom (Br), then iodine is less electronegative than bromine.
So, iodine 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 IBr4 molecule, put the two electrons (i.e electron pair) between each iodine atom and bromine atom to represent a chemical bond between them.
These pairs of electrons present between the Iodine (I) and Bromine (Br) atoms form a chemical bond, which bonds the iodine and bromine atoms with each other in a IBr4 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 IBr4, 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 IBr4- ion are used in the above structure.
But there are total 36 valence electrons in IBr4- ion (as calculated in step #1).
So the number of electrons left to be kept on the central atom = 36 – 32 = 4.
So let’s keep these four electrons (i.e 2 electron pairs) on the central atom.
Now, let’s move to the next step.
Step #5: 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 iodine atom (I) 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 Iodine:
Valence electrons = 7 (as it is in group 17)
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|
Let’s keep this charge on the iodine atom in the above lewis structure of IBr4 molecule.
As you can see in the above sketch, there is one -ve charge on the iodine atom, which indicates the -1 formal charge on the IBr4 molecule.
Hence, the above lewis structure of IBr4- ion is the stable lewis structure.
Each electron pair (:) in the lewis dot structure of IBr4- ion represents the single bond ( | ). So the above lewis dot structure of IBr4- ion can also be represented as shown below.