Lewis Structure of PI5 (With 5 Simple Steps to Draw!)

I’m super excited to teach you the lewis structure of PI5 in just 5 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of PI5 molecule.

So, if you are ready to go with these 5 simple steps, then let’s dive right into it!

Lewis structure of PI5 contains five single bonds between the Phosphorus (P) atom and each Iodine (I) atom. The Phosphorus atom (P) is at the center and it is surrounded by 5 Iodine atoms (I). The Phosphorus atom does not have a lone pair while all the 5 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 PI5).

5 Steps to Draw the Lewis Structure of PI5

Step #1: Calculate the total number of valence electrons

Here, the given molecule is PI5. In order to draw the lewis structure of PI5, first of all you have to find the total number of valence electrons present in the PI5 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 PI5

• For Phosphorus: 

Phosphorus is a group 15 element on the periodic table.

Hence, the valence electrons present in phosphorus is 5 (see below image).

• For Iodine:

Iodine is a group 17 element on the periodic table.

Hence, the valence electrons present in iodine is 7 (see below image).

Hence in a PI5 molecule, 

Valence electrons given by Phosphorus (P) atom = 5
Valence electrons given by each Iodine (I) atom = 7
So, total number of Valence electrons in PI5 molecule = 5 + 7(5) = 40

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 PI5 molecule, if we compare the phosphorus atom (P) and iodine atom (I), then the phosphorus is less electronegative than iodine.

So, phosphorus should be placed in the center and the remaining 5 iodine atoms will surround it.

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

Now in the above sketch of PI5 molecule, put the two electrons (i.e electron pair) between each phosphorus atom and iodine atom to represent a chemical bond between them.

These pairs of electrons present between the Phosphorus (P) and Iodine (I) atoms form a chemical bond, which bonds the phosphorus and iodine atoms with each other in a PI5 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 PI5, 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, all the 40 valence electrons of PI5 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 PI5.

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 phosphorus atom (P) 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 Phosphorus:
  Valence electrons = 5 (as it is in group 15)
  Nonbonding electrons = 0
  Bonding electrons = 10
• 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
P	=	5	–	0	–	10/2	=	0
I	=	7	–	6	–	2/2	=	0

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

Hence, there will not be any change in the above structure and the above lewis structure of PI5 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of PI5 represents the single bond ( | ). So the above lewis dot structure of PI5 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of SO2F2
Lewis Structure of OPBr3
Lewis Structure of PH2-
Lewis Structure of NS2
Lewis Structure of SiH2O 

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