So you have seen the above image by now, right?
Let me explain the above image in short.
XeCl4 lewis structure has Xenon atom (Xe) at the center which is surrounded by four Chlorine atoms (Cl). There are 4 single bonds between the Xenon atom (Xe) and each Chlorine atom (Cl). There are 2 lone pairs on the Xenon atom (Xe) and 3 lone pairs on all the four Chlorine atoms (Cl).
If you haven’t understood anything from the above image of XeCl4 (xenon tetrachloride) lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of XeCl4.
So let’s move to the steps of drawing the lewis structure of XeCl4.
Steps of drawing XeCl4 lewis structure
Step 1: Find the total valence electrons in XeCl4 molecule
In order to find the total valence electrons in XeCl4 (xenon tetrachloride) molecule, first of all you should know the valence electrons present in xenon atom as well as chlorine 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 xenon as well as chlorine using a periodic table.
Total valence electrons in XeCl4 molecule
→ Valence electrons given by xenon atom:
Xenon is a group 18 element on the periodic table. [1] Hence the valence electrons present in xenon is 8.
You can see the 8 valence electrons present in the xenon atom as shown in the above image.
→ Valence electrons given by chlorine atom:
Chlorine is group 17 element on the periodic table. [2] Hence the valence electrons present in chlorine is 7.
You can see the 7 valence electrons present in the chlorine atom as shown in the above image.
Hence,
Total valence electrons in XeCl4 molecule = valence electrons given by 1 xenon atom + valence electrons given by 4 chlorine atoms = 8 + 7(4) = 36.
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 XeCl4 (xenon tetrachloride) and it contains xenon atom (Xe) and chlorine atoms (Cl).
You can see the electronegativity values of xenon atom (Xe) and chlorine atom (Cl) in the above periodic table.
If we compare the electronegativity values of xenon (Xe) and chlorine (Cl) then the xenon atom is less electronegative.
So here the xenon atom (Xe) is the center atom and the chlorine atoms (Cl) are the outside atoms.
Step 3: Connect each atoms by putting an electron pair between them
Now in the XeCl4 molecule, you have to put the electron pairs between the xenon atom (Xe) and chlorine atoms (Cl).
This indicates that the xenon (Xe) and chlorine (Cl) are chemically bonded with each other in a XeCl4 molecule.
Step 4: Make the outer atoms stable. Place the remaining valence electrons pair on the central atom.
Now in this step, you have to check the stability of the outer atoms.
Here in the sketch of XeCl4 molecule, you can see that the outer atoms are chlorine atoms.
These outer chlorine 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 XeCl4 molecule.
The XeCl4 molecule has a total 36 valence electrons and out of these, only 32 valence electrons are used in the above sketch.
So the number of electrons which are left = 36 – 32 = 4.
You have to put these 4 electrons on the central xenon atom in the above sketch of XeCl4 molecule.
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 XeCl4.
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 xenon (Xe) atom as well as chlorine (Cl) atoms present in the XeCl4 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 XeCl4 molecule in the image given below.
For Xenon (Xe) atom:
Valence electrons = 8 (because xenon is in group 18)
Bonding electrons = 8
Nonbonding electrons = 4
For Chlorine (Cl) atom:
Valence electron = 7 (because chlorine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6
Formal charge | = | Valence electrons | – | (Bonding electrons)/2 | – | Nonbonding electrons | ||
Xe | = | 8 | – | 8/2 | – | 4 | = | 0 |
Cl | = | 7 | – | 2/2 | – | 6 | = | 0 |
From the above calculations of formal charge, you can see that the xenon (Xe) atom as well as chlorine (Cl) atom has a “zero” formal charge.
This indicates that the above lewis structure of XeCl4 is stable and there is no further change in the above structure of XeCl4.
In the above lewis dot structure of XeCl4, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of XeCl4.
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:
AlF3 Lewis Structure | IBr Lewis Structure |
SeCl4 Lewis Structure | HOF Lewis Structure |
XeO2F2 Lewis Structure | XeH4 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.
Read more about our Editorial process.