SiCl2Br2 Lewis Structure in 6 Steps (With Images)

So you have seen the above image by now, right?

Let me explain the above image in short.

SiCl2Br2 lewis structure has a Silicon atom (Si) at the center which is surrounded by two Chlorine atoms (Cl) and two Bromine atoms (Br). There are single bonds between the Silicon-Chlorine atoms and Silicon-Bromine atoms. There are 3 lone pairs on the Chlorine atoms (Cl) as well as Bromine atoms (Br).

If you haven’t understood anything from the above image of SiCl2Br2 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of SiCl2Br2.

So let’s move to the steps of drawing the lewis structure of SiCl2Br2.

Steps of drawing SiCl2Br2 lewis structure

Step 1: Find the total valence electrons in SiCl2Br2 molecule

In order to find the total valence electrons in a SiCl2Br2 molecule, first of all you should know the valence electrons present in silicon atom, chlorine atom as well as bromine 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 silicon, chlorine as well as bromine using a periodic table.

Total valence electrons in SiCl2Br2 molecule

→ Valence electrons given by silicon atom:

Silicon is a group 14 element on the periodic table. ^[1] Hence the valence electrons present in silicon is 4.

You can see the 4 valence electrons present in the silicon 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.

→ Valence electrons given by bromine atom:

Bromine is a group 17 element on the periodic table. ^[3] Hence the valence electrons present in bromine is 7.

You can see the 7 valence electrons present in the bromine atom as shown in the above image.

Hence, 

Total valence electrons in SiCl2Br2 molecule = valence electrons given by 1 silicon atom + valence electrons given by 2 chlorine atoms + valence electrons given by 2 bromine atoms = 4 + 7(2) + 7(2) = 32.

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 SiCl2Br2 and it contains silicon atom (Si), chlorine atoms (Cl) and bromine atoms (Br).

You can see the electronegativity values of silicon atom (Si) chlorine atom (Cl) and bromine atom (Br) in the above periodic table.

If we compare the electronegativity values of silicon (Si), chlorine (Cl) and bromine (Br) then the silicon atom is less electronegative.

So here the silicon atom (Si) is the center atom and the chlorine atoms (Cl) and bromine atoms (Br) are the outside atoms.

Step 3: Connect each atoms by putting an electron pair between them

Now in the SiCl2Br2 molecule, you have to put the electron pairs between the silicon atom (Si), chlorine atoms (Cl) and bromine atoms (Br).

This indicates that the silicon (Si), chlorine (Cl) and bromine (Br) are chemically bonded with each other in a SiCl2Br2 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 SiCl2Br2 molecule, you can see that the outer atoms are chlorine atoms and bromine atoms.

These outer chlorine and bromine 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 SiCl2Br2 molecule.

The SiCl2Br2 molecule has a total 32 valence electrons and all these valence electrons are used in the above sketch of SiCl2Br2.

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 octet on the central atom

In this step, you have to check whether the central silicon atom (Si) is stable or not.

In order to check the stability of the central silicon (Si) atom, we have to check whether it is forming an octet or not.

You can see from the above picture that the silicon atom is forming an octet. That means it has 8 electrons.

And hence the central silicon atom is stable.

Now let’s proceed to the final step to check whether the lewis structure of SiCl2Br2 is stable or not.

Step 6: 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 SiCl2Br2.

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 silicon (Si) atom, chlorine (Cl) atoms as well as bromine (Br) atoms present in the SiCl2Br2 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 SiCl2Br2 molecule in the image given below.

For Silicon (Si) atom:
Valence electrons = 4 (because silicon is in group 14)
Bonding electrons = 8
Nonbonding electrons = 0

For Chlorine (Cl) atom:
Valence electron = 7 (because chlorine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

For Bromine (Br) atom:
Valence electrons = 7 (because bromine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

From the above calculations of formal charge, you can see that the silicon (Si) atom, chlorine (Cl) atom as well as bromine (Br) atom has a “zero” formal charge.

This indicates that the above lewis structure of SiCl2Br2 is stable and there is no further change in the above structure of SiCl2Br2.

In the above lewis dot structure of SiCl2Br2, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of SiCl2Br2.

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:

About author

Jay Rana

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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