SiBr4 Lewis structure helps to determine the shape, formation and electronic facts of the compound. This article would be illustrated with several factors regarding the Lewis structure of SiBr4 that is Silicon tetra bromide or Tetrabromnosilane.
The facts about SiBr4 that would be described in this article are being listed below:
Drawing of SiBr4 Lewis structure
In order to draw the Lewis structure of any compound there are few general steps that need to be followed by maintaining the rules. These steps are the fundamental requirement to establish a true electronic stricture of a compound by showing the electron sharing aspects.
The very first step for initiating the drawing of Lewis structure of SiBr4 is to calculate the number of valance electrons possessed by the elements. Then the central; atom must be chosen with proper justification. Systematic formation of the compound is thoroughly discussed by the drawing rule given by Lewis.
In SiBr4, Si has four valance electrons. For getting ultimate stability like Neon, the element needs to gain for more electrons in its last energy level. As complementary, Br needs 1 electron to fill octet as it has 7 valance electrons. This would be supportive for the elements to fulfil the octet rule.
After calculating the valance electron in the elements, finding the central atom is the task before drawing the Lewis structure. In SiBr4 Silicon would be the central atom as it shares its 4 electrons with 4 bromine atoms. Each bromine atom also shares one electron with the Silicon atom.
Identification of the central atom fundamentally determines the shape of the compound. the Lewis structure gets an ultimate basic shape in this way.
SiBr4 Lewis structure shape
VSEPR (Valence Shell electron pair repulsion) theory brings forth the trick of identifying the shape of the compound. However, the geometry of the compound is the similar to its shape.
The total number of valance electrons is 32. Participated valance electrons are responsible for giving raise to the bonds in the compound. VSEPR theory represents that the bond pairs and lone pairs in the central atom of the compound assigns the shape of geometry of the compound is tetrahedral as the number of bind pairs is 4 and lone pairs is 0 in Silicon.
SiBr4 Lewis structure formal charges
It is significant to find the formal charges of the elements to understand internal electronic formation. If the compound shows neutral characteristics that does not say that the individual elements are also neutral by nature. The elements or atoms of the elements can hold charges.
There is a specific formula, which gives absolute measure of the formal charges in elements. The formula of finding the formal charges of atoms is (valence electrons – nonbonding electrons – ½ bonding electrons).
Formal charge of Silicon atom is 4-0- (1/2)*8)) = 0 and formal charge of each of the Bromine atom is 7-6-(1/2)*2)) =0.
SiBr4 Lewis structure lone pairs
The existence of lone pairs matters a lot in the formation of compound to measure the stability of the compound. VSEPR theory is the ideal theory, which describes the fact regarding the drawbacks of having lone pairs in the compounds.
The central atom that is Silicon does not have any lone pair. Each of the four Bromine atoms has 3 lone pairs. Therefore, it is obvious that the lone pairs of the Bromine atoms would show high repulsion.
According to VSEPR theory, the electron pairs or lone pairs exist in the bromine atoms repel each other therefore the compound becomes unstable. To keep the stability the electron pairs are noticed o align with highest distance from each other to avoid the repulsion and provide stability to the compound.
SiBr4 is a sp3-hybridised compound. However, this has been identified from the shape and the geometry of the polyatomic molecules present in Silicon tetra bromide. VSEPR theory is also responsible for calculating the hybridization of the compound.
In accordance with the VSEPR, theory it has been evaluated that the total electron pairs in SiBr4 is 4. The hybridised orbitals of the central atom in SiBr4 undergo promotion (from s orbital to p orbital) of sp3 orbital. Sigma bonds are formed by Bromine with sp3 hybrid orbitals.
SiBr4 Lewis structure resonance
Resonance refers to the oscillation process preceded by the electron system. It generally take place in the compounds with double bonds. The alterations in the bonds happen in the compounds with double bonds.
Resonance is not applicable for the tetrahedral shape of Silicon tetra bromide. The sigma bonds in the compound do not undergo resonance. The presence of lone pairs also is being a barrier in front of giving resonating structure to the compound. However, the bond angle in the compound is 109.5 °.
SiBr4 Lewis structure octet rule
The main aim of the periodic elements is to fulfil the octet rule. According to octet rule, it is needed for each of the element in periodic table to fill the last electronic shell like the nearest noble gas.
This is the main fact behind adopting of delivering electrons from the one atom to another atom in chemistry. Octet rule drives the elements in undergoing electron sharing process.
SiBr4 polar or nonpolar
Polarity of any compound depends on the total amount of dipole moment felt by the individual elements in the compound.
As Silicon is less electronegative than Bromine atoms, the compound must hold the polar characteristics but SiBr4 has been confirmed as a non-polar compound as the net dipole moment in the compound has been recognised to be cancelled out due to the symmetrical shape.
Frequently Asked Questions (FAQs)
Question 1: Calculate the total number of valance electrons present in SiBr4.
Answer: The total number of valance electron in Silicon is 4 and in the each of the Bromine atom it is 7 therefore, the total the number of valance electrons in four Bromine atom is (4*7 = 28). The sum of the number of valance electrons in the compound is (28+4 = 32).
Question 2: Is SiBr4 ionic or covalent compound?
Answer: SiBr4 is a covalent compound. As the elements shares their valance electrons with each other instead of donating the electrons from one to another, it is strongly considered as covalent compound.
Question 3: What is the reason behind not considering SiBr4 as a non-polar compound?
Answer: SiBr4 does not have dipole due to its clear symmetric structure. Cancellation of the dipole moment is the reason behind considering the compounds as non-polar. However, the compound has internal polarity due to the presence of lone pairs. It magnifies the intermolecular forces between the atoms.
Question 4: Write the facts about the solubility of SiBr4.
Answer: The compound is acidic and highly soluble in water. The crystalline structure of the compound makes it a reliable source for being compatible with Bromine atoms.