So you have seen the above image by now, right?
Let me explain the above image in short.
PI5 lewis structure has a Phosphorus atom (P) at the center which is surrounded by five Iodine atoms (I). There are 5 single bonds between the Phosphorus atom (P) and each Iodine atom (I).
If you haven’t understood anything from the above image of PI5 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of PI5.
So let’s move to the steps of drawing the lewis structure of PI5.
Steps of drawing PI5 lewis structure
Step 1: Find the total valence electrons in PI5 molecule
In order to find the total valence electrons in a PI5 molecule, first of all you should know the valence electrons present in phosphorus atom as well as iodine atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)
Here, I’ll tell you how you can easily find the valence electrons of phosphorus as well as iodine using a periodic table.
Total valence electrons in PI5 molecule
→ Valence electrons given by phosphorus atom:
Phosphorus is group 15 element on the periodic table. [1] Hence the valence electrons present in phosphorus is 5.
You can see the 5 valence electrons present in the phosphorus atom as shown in the above image.
→ Valence electrons given by iodine atom:
Iodine is a group 17 element on the periodic table. [2] Hence the valence electrons present in iodine is 7.
You can see the 7 valence electrons present in the iodine atom as shown in the above image.
Hence,
Total valence electrons in PI5 molecule = valence electrons given by 1 phosphorus atom + valence electrons given by 5 iodine atoms = 5 + 7(5) = 40.
Step 2: Select the central atom
For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.
Now here the given molecule is PI5 and it contains phosphorus atom (P) and iodine atoms (I).
You can see the electronegativity values of phosphorus atom (P) and iodine atom (I) in the above periodic table.
If we compare the electronegativity values of phosphorus (P) and iodine (I) then the phosphorus atom is less electronegative.
So here the phosphorus atom (P) is the center atom and the iodine atoms (I) are the outside atoms.
Step 3: Connect each atoms by putting an electron pair between them
Now in the PI5 molecule, you have to put the electron pairs between the phosphorus atom (P) and iodine atoms (I).
This indicates that the phosphorus (P) and iodine (I) are chemically bonded with each other in a PI5 molecule.
Step 4: Make the outer atoms stable
Now in this step, you have to check the stability of the outer atoms.
Here in the sketch of PI5 molecule, you can see that the outer atoms are iodine atoms.
These outer iodine atoms are forming an octet and hence they are stable.
Also, in step 1 we have calculated the total number of valence electrons present in the PI5 molecule.
The PI5 molecule has a total 40 valence electrons and all these valence electrons are used in the above sketch of PI5.
Hence there are no remaining electron pairs to be kept on the central atom.
So now let’s proceed to the next step.
Step 5: Check the stability of lewis structure
Now you have come to the final step in which you have to check the stability of lewis structure of PI5.
The stability of lewis structure can be checked by using a concept of formal charge.
In short, now you have to find the formal charge on phosphorus (P) atom as well as iodine (I) atoms present in the PI5 molecule.
For calculating the formal charge, you have to use the following formula;
Formal charge = Valence electrons – (Bonding electrons)/2 – Nonbonding electrons
You can see the number of bonding electrons and nonbonding electrons for each atom of PI5 molecule in the image given below.
For Phosphorus (P) atom:
Valence electrons = 5 (because phosphorus is in group 15)
Bonding electrons = 10
Nonbonding electrons = 0
For Iodine (I) atom:
Valence electrons = 7 (because iodine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6
Formal charge | = | Valence electrons | – | (Bonding electrons)/2 | – | Nonbonding electrons | ||
P | = | 5 | – | 10/2 | – | 0 | = | 0 |
I | = | 7 | – | 2/2 | – | 6 | = | 0 |
From the above calculations of formal charge, you can see that the phosphorus (P) atom as well as iodine (I) atom has a “zero” formal charge.
This indicates that the above lewis structure of PI5 is stable and there is no further change in the above structure of PI5.
In the above lewis dot structure of PI5, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of PI5.
I hope you have completely understood all the above steps.
For more practice and better understanding, you can try other lewis structures listed below.
Try (or at least See) these lewis structures for better understanding:
SO2F2 Lewis Structure | OPBr3 Lewis Structure |
PH2- Lewis Structure | NS2 Lewis Structure |
SiH2O Lewis Structure | OF3- Lewis Structure |
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. He is a founder of Pediabay and is passionate about helping students through his easily digestible explanations.
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