Ready to learn how to draw the lewis structure of SI4?
Awesome!
Here, I have explained 5 simple steps to draw the lewis dot structure of SI4 (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 SI4 contains four single bonds between the Sulfur (S) atom and each Iodine (I) atom. The Sulfur atom (S) is at the center and it is surrounded by 4 Iodine atoms (I). The Sulfur atom has 1 lone pair while all the four iodine 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 SI4).
5 Steps to Draw the Lewis Structure of SI4
Step #1: Calculate the total number of valence electrons
Here, the given molecule is SI4. In order to draw the lewis structure of SI4, first of all you have to find the total number of valence electrons present in the SI4 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 SI4
- For Sulfur:
Sulfur is a group 16 element on the periodic table. [1]
Hence, the valence electrons present in sulfur is 6 (see below image).
- For Iodine:
Iodine is a group 17 element on the periodic table. [2]
Hence, the valence electrons present in iodine is 7 (see below image).
Hence in a SI4 molecule,
Valence electrons given by Sulfur (S) atom = 6
Valence electrons given by each Iodine (I) atom = 7
So, total number of Valence electrons in SI4 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: 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). [3]
Here in the SI4 molecule, if we compare the sulfur atom (S) and iodine atom (I), then the sulfur is less electronegative than iodine.
So, sulfur should be placed in the center and the remaining 4 iodine atoms will surround it.
Step #3: Put two electrons between the atoms to represent a chemical bond
Now in the above sketch of SI4 molecule, put the two electrons (i.e electron pair) between each sulfur atom and iodine atom to represent a chemical bond between them.
These pairs of electrons present between the Sulfur (S) and Iodine (I) atoms form a chemical bond, which bonds the sulfur and iodine atoms with each other in a SI4 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 SI4, the outer atoms are iodine atoms.
So now, you have to complete the octet on these iodine atoms (because iodine requires 8 electrons to have a complete outer shell).
Now, you can see in the above image that all the iodine atoms form an octet.
Also, only 32 valence electrons of SI4 molecule are used in the above structure.
But there are total 34 valence electrons in SI4 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 sulfur atom (S) as well as each iodine atom (I).
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 = 2
Bonding electrons = 8 - For Iodine:
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 | – | 2 | – | 8/2 | = | 0 |
I | = | 7 | – | 6 | – | 2/2 | = | 0 |
So you can see above that the formal charges on sulfur as well as iodine are “zero”.
Hence, there will not be any change in the above structure and the above lewis structure of SI4 is the final stable structure only.
Each electron pair (:) in the lewis dot structure of SI4 represents the single bond ( | ). So the above lewis dot structure of SI4 can also be represented as shown below.
Related lewis structures for your practice:
Lewis Structure of GaCl3
Lewis Structure of NSF
Lewis Structure of C2H4Br2
Lewis Structure of KrF4
Lewis Structure of TeCl4
Article by;
Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. With a desire to make learning accessible for everyone, he founded Knords Learning, an online learning platform that provides students with easily understandable explanations.
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