SiBr4 Lewis Structure, Geometry, Hybridization, and Polarity

SiBr4 Lewis Structure

Silicon tetrabromide or Tetrabromosilane is an inorganic compound in which one Si-atom and four Br-atoms are bonded through covalent bonds.

It is synthesized from silicon and vapors of Bromine at high temperatures (T>600°C).

SiBr4 is a colorless liquid with a very suffocating odor. This characteristic odor is because of its ability to hydrolyze and release HBr gas (an irritating odor) as shown in the equation.

SiBr4  +  2H2O  →  SiO2  +  4 HBr

The molar mass is 347.701 grams/ mole, and density is 2.79 g cm-3. The melting and boiling points are 15°C and 153°C, respectively.

It is a reactive compound. It is used in the synthesis of Si and its compounds, SnO2 whiskers, Si3N4, etc. Si3N4 is further used in making coatings, ceramics, cutting tools, etc.

It is corrosive, and irritant and therefore, caution should be maintained while using it.

In this article, we will closely look at some of the characteristics of SiBr4 through various concepts.

First, we will understand the bonding through the Lewis concept, and then we will study the molecular geometry by VSEPR theory and finally discuss hybridization and polarity of SiBr4.


SiBr4 Lewis Structure

Octet Rule– Main group elements have a preference for eight electrons in their outermost shell. If the electrons are more or less than 8, the element is considered to be unstable.

To gain stability, these unstable elements form bonds either by transferring electrons or sharing electrons.

The completion of the octet rule is the driving force for bond formation in main group elements.

Lewis structure cannot explain the molecule completely but provides valuable information regarding formation and properties.

It provides a 2-D diagrammatic representation of bonding through the arrangement of valence electrons and follows the octet rule.

Electron-dot notation is used for drawing the lewis structure. In this, electrons are represented as dots around the symbol of the element.


Steps to draw the lewis structure of SiBr4

Step 1: Count the number of valence shell electrons on each atom of the molecule to get the total valence electron count.

SiBr4 has two elements ie; Si and Br.

Si belongs to group 14 and has the atomic number 14.

For group 14, the valence electron is 4. Also, the electronic configuration of Si is 1s22s22p63s23p2. The valence shell is n=3, and it has 4 electrons, 2 in 3s subshell and 2 in 3p subshell.

The Lewis dot structure for Si is shown.

Silicon lewis structure

Br belongs to group 17 and has the atomic number 35. For group 17, the valence electron is 7. Also, the electronic configuration of Br is

1s22s22p63s23p63d104s24p5. The valence shell is n=4 and has 7 electrons, 2 in 4s subshell and 5 in 4p subshell.

The Lewis dot structure for Bromine is shown

Bromine lewis structure

Total valence electrons= 1*(valence shell electron in Si) +4*(valence shell electron in Br )= (1*4) +(4*7)= 4+28= 32

Step 2. Select the central atom.

Si atom is chosen as the central atom to provide stability to the molecule and facilitate a better spread of electron density.

The central atom is supposed to share the electron density with surrounding atoms. Thus, a more electronegative atom (Br in this case) is not suitable as a central atom because it tends to keep the electron density towards itself.

Step 3: Draw a skeletal diagram

Sketch a skeletal diagram keeping the central atom and surrounding atoms in mind.

SiBr4 atoms

Step 4: Place the valence electrons around each atom using electron-dot notation.

The total valence shell electrons are arranged according to bond formation.

SiBr4 valence electrons


Step 5: Complete the octet of atoms by forming bonds. A single bond consists of two electrons.

SiBr4 lewis structure

Each Br atom has seven valence electrons. It shares one electron with Si to complete the octet.

Si has four valence electrons. It shares one electron with all four Br atoms to complete the octet.

Step 6: Formal Charge

For polyatomic ions, a net charge is present on the whole molecule.

For SiBr4, the net charge is zero.

If a molecule as a whole is neutral, it does not mean all atoms are neutral. It is better to assign each element with a formal charge.

formal charge formulae

Formal charge on Si= 4-0- (0.5*8) = 0

Formal charge on Br= 7-6-(0.5*2) =0

The formal charge on all atoms comes out to be zero. Hence the lewis structure drawn in step 5 is accurate.

To get a visual understanding of the Lewis structure of SiBr4, you can check this video.



SiBr4 Geometry

Molecular geometry tells the arrangement of all the atoms and bonds in a molecule.

Lewis Structure has many drawbacks, and one of them is that it cannot explain the geometry of all molecules.

VSEPR theory (Valence Shell electron pair repulsion theory) overcomes this drawback. 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.

Electrons are negatively charged subatomic particles.

Valence shell electrons that are involved in bonding are known as bonding pair of electrons (bp), and those valence shell electrons which are not involved in bonding are termed as lone pairs of electrons (lp).

The repulsion order between lone pair and bond pair of electrons follows the order➔ lp – lp > lp – bp > bp –bp .

The trick for predicting geometry through VSEPR

• Count the number of valence shell electrons (N) in a molecule (as calculated in step 1 of drawing lewis’s structure).

For SiBr4, the total number of valence shell electrons is 32.

• if N>8, divide total valence shell electrons by 8 else divide by 2.

For SiBr4, since 32>8, we divide 32 by 8. The quotient comes out to be 4.

• If it is exactly divisible by 8 and the quotient equals the number of surrounding atoms, it implies that we don’t have any lone pairs on the molecule.

In this case, molecular geometry is the same as the shape of the molecule.


This implies SiBr4 does not have lone pairs on the central atom

• If it is not exactly divisible by 8 or quotient > the number of surrounding atoms, then lone pairs are present on the central atom, and geometry is not the same as the shape.

• The number of lone pairs can be found out either by dividing the remainder of (N/8) by 2 if N is not exactly divisible by 8 or by subtracting the number of surrounding atoms from the quotient if quotient> surrounding atoms.

• We can predict the shape using the following table by finding out the total number of lone pairs and bond pairs.

Here total domain = lone pair+ bond pair. The generic formula can be matched for further confirmation.

For SiBr4, the total number of electron pairs is 4.

Thus, the geometry and shape of SiBr4 are tetrahedral.

VSEPR chart



The concept of hybridization explains the geometrical shape and bonding in polyatomic molecules.

An orbital is a 3D region around the nucleus where the probability of finding an electron is maximum.

Hybridization can be defined as the mixing of pure atomic orbitals to form hybrid atomic orbitals. This mixing is feasible if the pure atomic orbitals are of similar shape and energy.

The number of pure orbitals combined for hybridization = the number of hybrid orbitals.

For instance, 2s and one 2p can form sp hybrid orbitals, but 2s and 5d cannot. Also, in C (carbon atom), one 2s and three 2p orbitals are mixed to form 4 equivalent sp3 hybrid orbitals.

SiBr4 hybridization

The type of hybridization also indicates the molecule’s geometry as the orbitals arrange themselves in such a way that repulsion between electrons in them is minimum.


SiBr4 Hybridization

In the SiBr4 molecule, the central atom is Si, as it is the least electronegative atom in the molecule.

The electronic configuration of Si is 1s22s22p63s23p2 . Only valence orbitals are used in hybridization.

Si orbital state

In the excited state, one electron of 3s gets promoted to 3p orbital. However, it is not always that promotion of electrons takes place.

Now, these 4 orbitals (one 3s and three 3p) undergo hybridization to form four sp3 orbitals, which will form bonds with the surrounding atoms.

In SiBr4, there are 4 surrounding atoms. Each atom forms bond with one sp3 hybrid orbital.

Br hybridized state

All Br atoms form a sigma bond with sp3 hybrid orbitals. Thus, SiBr4 has sp3 hybridization.

The trick for calculating the type of hybridization

In VSEPR theory, we calculated the total electron pairs in the last step. For SiBr4, it came out to be 4.

From the table below, we can predict hybridization using total electron pairs or steric numbers.  The bond angle in the SiBr4 molecule is around 109.5°, which can also be seen in one of the above images.

Hybridization Table


SiBr4 Polarity

A compound can be categorized as polar or non-polar depending on the net dipole moment.

Polarity also depends on the difference in electronegativity of elements, charge separation, and molecular geometry.

The electronegativity of Si is 1.9, and that of Br is 2.96. The difference is 1.06, so the bonds can be termed as polar and covalent.

The net polarity of the molecule is decided by its geometry.

SiBr4 is tetrahedral and tetrahedral is symmetrical shape. The dipole moment vectors cancel each other out and the compound as a whole turns out to be non-polar.

The compound would have been polar if the shape was not symmetrical.

SiBr4 Polarity



Si forms one covalent bond with each Br. The formal charge on all atoms is zero. The molecular geometry and shape of SiBr4 is tetrahedral.

The hybridization of SiBr4 is sp3. It is a non-polar molecule.

I hope you enjoyed the chemistry of SiBr4.

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