5 Steps to Draw Cbr4 Lewis Structure,Hybridization (Solved!)

Carbon tetrabromide (CBr4) has a central carbon (C) atom with 4 valence electrons, bonded to four bromine (Br) atoms, each contributing 7 valence electrons. The Lewis structure shows four single C-Br bonds, with 8 bonding electrons and no lone pairs on carbon. The molecule adopts a tetrahedral geometry with bond angles of about 109.5°. Despite the polar C-Br bonds due to electronegativity differences (C: 2.55, Br: 2.96), CBr4 is nonpolar overall due to its symmetrical structure. This symmetry influences its physical properties and chemical behavior.

Facts on CBr4                      

Molecular Name	Tetrabromomethane Carbon Tetrabromide Chemical FormulaCBr4Central atom of the lewis structureCarbonlone pairs present on the central atom 0Molecular Geometry of CBr4TetrahedralElectron Geometry of CBr4TetrahedralBond Angle Br – C – Br109.5 degreeNo. of valence electron for CBr4	32
The formal charge of CBr4 molecule	0
Solubility	Insoluble in water Soluble in ether,chloroform,C2H5OH
Acidic or basic nature	CBr4 is neither acidic nor basic

How to draw Lewis dot structure for CBr4?

Following are the steps to follow to draw the Lewis structure of CBr4 molecule

 Step 1: Count total valence electrons present in CBr4 molecule

First step is to find the no. of valence electrons present in CBr4 molecule.

Carbon belongs to group 14th and Bromine belongs to group 17th of the periodic table.

Valence electron of Carbon = 4

Valence electron of Bromine = 7

	Valence electrons	Atom In CBr4	Total Electrons
C	4	1	4*1= 4
Br	7	4	7*4=28
			32

Step 2: Find least electronegative element in CBr4

Electronegativity increases in a period from left to right so Carbon is least electronegative than Bromine. Keep the least electronegative atom in the centre.

Now draw the skeletal structure of CBr4 molecule

Step 3: Put two electrons or a bond between C and Br atom

Step 4: Complete octet of the Outer atoms

Main group elements have tendency to attain octet configuration of their nearest noble gas element present in the 18 group. This is known as Octet Rule. Except hydrogen which completes duplet configuration like He

Complete the octet on outside atoms then move on to central atom

Step 5: Complete octet of central atom or make covalent bonds if necessary

There is no need of this step for CBr4 as the octet of central atom is already complete.

                         

                                                 

Step 6: Check formal charge value

Before we confirm that this is a perfect Lewis structure, we need to check the formal charge values

Formal charge = Valence electrons – (1/2) * bonding electrons – non-          bonding electrons

Valence electrons of Carbon = 4

Non-bonding electrons of carbon = 0

Bonding electrons of carbon = 8 (4 pairs)

Formal charge for carbon atom = 4 – 1/2*8 – 0 = 0

Valence electrons of Bromine = 7

Non-bonding electrons of Bromine = 6 (3 lone pairs)

Bonding electrons of Bromine = 2

Formal charge for Bromine atom = 7 – 1/2*2 – 6 = 0

As the C and Br elements have least possible formal charges, we have obtained suitable Lewis structure

                           

What is the formal charge in CBr4 and how it is calculated?

We can calculate formal charge by using the formula –

Formal charge = Valence electrons – (1/2) * bonding electrons – non-          bonding electrons

Valence electrons of Carbon = 4

Non-bonding electrons of Carbon = 0

Bonding electrons of carbon = 8 (4 pairs)

Formal charge for Carbon atom = 4 – 1/2*8 – 0 = 0

Valence electrons of Bromine = 7

Non-bonding electrons of Bromine = 6 (3 lone pairs)

Bonding electrons of fluorine = 2

Formal charge for Bromine atom =7 – 1/2*2 – 6 = 0

Formal charge on carbon and bromine atom in CBr4 Lewis dot structure is zero.

Does CBr4 molecule follow the octet rule?

CBr4 contains 32 valence electrons and all the atoms in CBr4 complete their octet as can be seen from the Lewis structure of CBr4 molecule. This proves that octet rule is followed by the CBr4 molecule.

How to find the molecular shape and electron geometry of CBr4 molecule?

In order to find the geometry (molecular/electron) of OF2 molecule, the three steps have to be followed. They are –

1)Count the lone pairs present on the central atom in the CBr4 Lewis structure

Or

Number of lone pairs can also be found by using the formula

Lone Pair =1/2*(V.E. –N.A.)

V.E.= valence electron on the central atom

N.A.= Number of atoms which are attached to the central atom

In CBr4 molecule, the valence electrons of central Carbon atom are 4 and four outer atoms are attached to it.

Lone Pairs = 1/2* (4 – 4) = 0

2) Find the hybridization number of CBr4 molecule

Hybridization of central atom can be found out by using the formula.

Hybridization number = N.A. + L.P

Where-

N.A.- is the number of atoms attached to the central atom

L.P. – is the number of lone pairs on the central atom

There are 4 atoms attached to the Carbon with no lone pairs present on it.

Hybridization number = 4 + 0 = 4

Hybridization number of CBr4 molecule is 4

So, hybridization of CBr4 molecule is Sp3

3) Use VSEPR Theory to determine CBr4 molecular/electron geometry

We have determined hybridization of CBr4 as SP3 and no lone pairs present on the central atom.

Now according to AXnEx notation of VSEPR theory we are going to find VSEPR notation for CBr4 molecule

AXnEx notation

Where-

A – is the central atom

X – is the number of atoms attached to the central atom

E – is the number of lone pairs of electrons on the central atom

According to the Lewis dot structure of Carbon Tetrabromide, central atom is Carbon which has 4 fluorine atoms are connected to it and no lone pairs present on it.

So CBr4 formula becomes AX4

According to VSEPR chart, the molecule with AX4 formula has molecular shape as tetrahedral and electron geometry tetrahedral.

Total Domains	General Formula	Bonded atoms	Lone Pairs	Molecular Shape	Electron Geometry
1	AX	1	0	Linear	Linear
2	AX2	2	0	Linear	Linear
	AXE	1	1	Linear	Linear
3	AX3	3	0	Trigonal Planar	Trigonal Planar
	AX2E	2	1	Bent	Trigonal Planar
	AXE2	1	2	Linear	Trigonal Planar
4	AX4	4	0	Tetrahedral	Tetrahedral
	AX3E	3	1	Trigonal Pyramid	Tetrahedral
	AX2E2	2	2	Bent	Tetrahedral
	AXE3	1	3	Linear	Tetrahedral

VSEPR CHART

                       

Is CBr4 molecule polar or non-polar?

CBr4 is a non-polar molecule is confirmed by three factors –

1)Electronegativity:

Bromine (electronegativity 2.96) is more electronegative than Carbon (electronegativity 2.55) Being more electronegative than Carbon, Bromine attracts electrons towards it more strongly.

The molecule is polar when the difference in electronegativity between the atoms in the molecule is greater than 0.4

The electronegativity difference between Carbon and Bromine is 0.41

2)Geometrical/Molecular Shape: 

Geometrical Structure of CBr4 molecule is tetrahedral.

The four dipoles of four C-Br bond are in opposite direction. Now, due to the tetrahedral geometry (which is symmetrical) all the dipoles get cancelled by each other.

3)Dipole Moment:

Due to the symmetric geometry of the CBr4 (tetrahedral) all the 4 dipoles get cancelled by each other resulting in overall 0 dipole moment for CBr4.

All these factors show that CBr4 is a non-polar molecule.

What is the Hybridisation for CBr4 molecule?

The central atom carbon atom shares four electrons with four Bromine atoms. Hence in the Lewis structure there is a presence of four bonding pairs of electrons on the carbon atom in Lewis structure of CBr4 molecule

To share the electrons with the Bromine atoms, the orbitals of Carbon atom undergo hybridization to accommodate the electrons.

The electronic configuration of Carbon and Bromine in ground state is –

Carbon in ground state 1s2 2s2 2p2

Bromine in ground state 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5

Electronic configuration of Carbon after gaining four electrons to fulfil the octet by forming bonds with Bromine-

Carbon 1S2 2S2 2Px2 2Py2 2Pz2

Steric Number = Number of atoms bonded to central atom + Number of lone pairs of electrons attached to central atom

Steric Number = 4 + 0 = 4

So, hybridization for the central atom oxygen in CBr4 is Sp3

Does CBr4 molecule shows resonance?                                      

A molecule shows resonance when –

• The molecule has alternate double and single bonds.
• The molecule has presence of lone pair in conjugation with the double bond.

Carbon Tetrafluoride does not have a double bond so CBr4 does not show resonance.

Is CBr4 ionic or a covalent compound?

CBr4 is a covalent compound as the bond between Carbon and Bromine is formed by sharing of electrons.

Conclusion:

To summarize what we have discussed in this article –

The hybridisation of CBr4 molecule is Sp3. Both the molecular geometry and electron geometry of CBr4 is tetrahedral. The bond angle Br-C- Br is 109.5 degrees. The total valence electrons present in CBr4 molecule is 32.

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