SiCl2Br2 dibromo dichlorosilane is an inorganic compound composed of silicon and four halogen atoms. Silicon is a metalloid as it has properties of both metal and non-metal. Halogens are non metal. Out of four halogen atoms, two are chlorine, and the other two are bromine.
The molecular weight is 258.80 grams/mole. The melting point of SiCl2Br2 is -45.5°C. All four bonds are polar and covalent.
This article will discuss the concepts required to predict lewis structure, geometry, hybridization, and polarity of compounds.
Lewis structure is a 2D arrangement of electrons, bonds, and atoms in a compound. It does not depict an accurate picture of the compound always.
All atoms in a stable Lewis structure should obey the octet rule and satisfy formal charges.
This rule is applicable for main group elements.
All elements prefer to have their configuration similar to that of noble gas because noble gases are considered stable.
Noble gases are present in group 18 and have 8 electrons in their valence shell configuration. Thus, those atoms which have eight valence electrons are considered to be stable.
It is a theoretical concept.
A neutral compound does not ensure that all atoms are neutral.
Formal charge depicts the charge present on each atom in a compound, and it is developed when a charge is distributed on constituent atoms.
It is calculated using the formula given below.
Steps to draw the lewis structure of SiCl2Br2
Step 1: Count the total number of valence shell electrons on the compound.
This is done by adding the valence shell electrons of all the constituent atoms.
|Atom||Atomic Number||Group Number||Valence electrons according to group number||Electronic configuration (E.C.)||Valence shell from E.C.||Valence electrons from E.C.|
|One Si||14||14||4||1s2 2s2 2p6 3s2 3p2||n=3||4|
|Two Cl||17||17||7||1s2 2s2 2p6 3s2 3p5||n=3||7|
|Two Br||35||17||7||1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5||n=4||7|
|Total number of valence shell electrons= 4 + (7*2) +(7*2)=32|
The Lewis dot structure for Si, Cl, and Br are as follows-
Step 2: Choose a suitable central atom for the compound.
The central atom is supposed to be the least electronegative one out of the constituent atoms.
The central atom is supposed to share its electron density with all other atoms.
If the central atom is more electronegative than the side atom, the central atom will not share the electron density with side atoms.
Thus, Si is the central atom for this compound.
Step 3: Draw a skeletal diagram.
In this step, we have to arrange the side atoms and central atoms suitably.
Step 4: Arrange the valence electrons around the elemental symbols.
The total valence shell electrons (calculated in step 1) are placed according to bond formation.
Step 5: Complete the octet of atoms by forming bonds.
Each Cl and Br atom has seven valence electrons in the isolated state. They share one electron with Si to complete the octet.
Si has four valence electrons in the isolated state. It shares one electron from both Cl atoms and Br atoms to complete the octet.
Step 6: Calculate the formal charge on all atoms.
The net charge on this compound is zero. Therefore, the sum of formal charge on three atoms should come out to be zero.
|Atom||Total number of valence electrons in free atom||Total number of lone pairs||(Total number of bonding electrons)*0.5||Formal Charge|
Thus, the structure drawn in step 5 is the best Lewis structure for SiCl2Br2.
We cannot depend entirely on Lewis structure for various properties of a compound. For predicting the geometry of covalent compounds, we take the help of VSEPR theory.
Geometry is the 3D arrangement of atoms and bonds.
Some electron pairs do not participate in bond formation but affect the arrangement of atoms and bonds. Such electron pairs are lone pairs of electrons, and the distorted geometry is called shape.
VSEPR stands for valence shell electron pair repulsion theory.
According to VSEPR theory-
• The valence electron pairs repel each other, and this leads to instability.
• To make the arrangement of the electrons stable, the repulsions between them have to be decreased.
• As a result, electrons align themselves so that the repulsion is the least, and the distance between them is maximum.
• The stable arrangement of the valence electron pairs of atoms helps in determining the molecular geometry.
Steps to predict Geometry of SiCl2Br2 using VSEPR
Step 1: Count the number of valence shell electrons on the central atom and let it be equal to A (arbitrary variable).
In the case of SiCl2Br2, the central atom is Si. S has 4 valence electrons. (Shown in step1 of drawing lewis structure)
Step 2: Count the number of side atoms and let it be equal to B (arbitrary variable).
In SiCl2Br2, there are four side atoms (2 chlorine and 2 bromine) and B=4
Step 3: If the compound is charged, subtract the charge from B for the positively charged compound and add the charge to B for the negatively charged compound.
This step can be avoided for neutral compounds.
In SiCl2Br2, there is no contribution of charge and B=6 only.
Step 4: Add the contribution of side atoms and charge to the contribution of the central atom, i.e., A+B
For SiCl2Br2, A+B=8
Step 5: Divide A+B by 2 to find total electron pairs affecting the shape.
For SiCl2Br2, there are 4 electron pairs.
Step 6: Divide the total electron pairs as bonding and non-bonding. The bonding electron pair is equal to the number of side atoms.
For SiCl2Br2, there are four side atoms. Thus, there are four bonding pairs of electrons and 0 non-bonding pairs of electrons.
Using this information, we can predict geometry and shape using the following table.
Electron geometry and shape are tetrahedral. Here, geometry and shape are the same due to the absence of lone pair of electrons.
The geometry and bonding of some polyatomic covalent compounds are explained using a unique concept called hybridization.
Hybridization can be defined as the mixing of atomic orbitals to form equivalent hybrid orbitals. It involves the redistribution of energy.
All atomic orbitals cannot be mixed. Only those orbitals that are similar in shape, size, and energy can mix and form hybrid orbitals.
For example, one 3s and two 3p orbitals can be mixed to form three sp2 orbitals, but 1s and 5d cannot because the energy difference is very high.
There is no literal mixing of orbitals; the operations are just carried on the wavefunction of orbitals. All compounds do not undergo hybridization like PH3.
In SiCl2Br2, the central atom is Si. Here, only the central atom is hybridized.
The electronic configuration of Si in ground state is 1s2 2s2 2p6 3s2 3p2.
The electronic configuration of Si in excited state is 1s2 2s2 2p6 3s1 3p3.
In an excited state, all 4 electrons are unpaired.
The halogens form a sigma bond with the Si atom. But the strength of the sigma bond is different when the bonds are formed by overlap with 3s and 3p.
All the halogens atom wants an equally good overlap.
Thus, to form a stable compound, all four orbitals of silicon are mixed to form four equivalent orbitals called sp3 orbitals.
The trick for calculating hybridization
We can also predict hybridization by calculating the total electron pairs, which we calculated in the prediction of geometry using VSEPR.
For SiCl2Br2, the total domain comes out to be 4. From the table, we see that for steric number 4, hybridization is sp3.
SiCl2Br2 is a polar compound.
The polarity of a compound depends on the presence or absence of net dipole moment.
The net dipole moment, in turn, depends on various factors like-
• The dipole moment of individual bond
• Difference in electronegativities
There are two types of bonds in this compound, S-Br, and S-Cl.
The electronegativity of Si, Br, and Cl are 1.9, 2.96, and 3.16.
For the Si-Br bond, the difference in electronegativity is 1.06, and for the Si-Cl bond, the difference in electronegativity is 1.26. Due to this difference in electronegativity, the bonds develop a polar character.
Polar bonds do not guarantee polar compounds. The shape is tetrahedral. If all side atoms were the same, dipole moments would have canceled each other.
In this compound, dipole moments are not canceled because there are two types of side atoms, or we can say that it is an unsymmetrical compound.
Therefore, the net dipole moment of SiCl2Br2 is not zero and it is a polar compound.
SiCl2Br2 is a covalent compound.
The Lewis structure predicted is the most stable one as all atoms obey the octet rule and satisfy the formal charge of the compound.
The geometry and shape come out to be tetrahedral.
Hybridization of the central atom is sp3.
It is a polar compound.