# Understanding the Lewis Structure, Geometry, Hybridization, and Polarity of CH2Br2

By brominating methane, a covalent molecule with the formula CH2Br2 is produced. Dibromomethane is an odourless, colourless liquid. It is a key ingredient in the production of a wide variety of pharmaceuticals, perfumes, agricultural chemicals, and other similar products.

Prediction of Lewis structure, geometry, hybridization, and polarity are among the foundational ideas we’ll cover in this essay.

## Instance of Lewis’s Structure

An illustration of a molecule in two dimensions. Because only electrons in the outermost valence shell are involved in chemical bonds, this model dictates that only these electrons should be shown in the lewis structure of an atom. Lewis structures are a very early method of attempting to explain chemical bonds.

The Lewis structure of a chemical may take on more than one possible form. For the most stable Lewis structure, we need to ensure that the octet rule is met as well as formal charges.

## Rule of Eight

To achieve stability, every atom forms bonds with others. Stable noble gases (with the exception of Helium) have eight valence electrons.

In order to get a configuration similar to that of a noble gas, atoms must either lose or gain electrons until they have eight electrons in their valence shell.

The term “octet rule” describes this inclination for clusters of eight electrons.

## Statement of Formal Charge

This is merely a theoretical concept at this point. It does this by contrasting the number of electrons in a single, neutral atom with the number of electrons on the atom in the compound. The correct formal charge formula is

Formula for formal charge: (total number of electrons in the atom’s valence shell minus the number of unpaired electrons) / (0.5 *). (number of electrons involved in bond formation)

When there are more electrons on an atom in a compound than there are on an isolated atom, the formal charge is negative, and when there are less electrons, the charge is positive.

A bond forms between atoms with complementary charges when those charges are opposite one another.

Both formal charges, positive and negative, are more stable on electronegative than electropositive atoms.

## How to Represent the CH2Br2 Lewis Structure

First, determine how many electrons are in each atom’s valence shell.

Knowing the total number of electrons in all of the atoms’ valence shells is necessary before sketching the Lewis structure.

Atomic Number Group Number Atomic Mass

A breakdown of valence electrons by atomic number

Electrical and electronic set-up (E.C.)

An atomic valence shell originating from E.C.

Electrons with a high valence come from the element helium.

One C6144122222222n=2

4

H 1 1 1 1s1 n=1 1

Two Br 35 17 7

In a mathematical progression going from 1s2 to 2p6 to 3s2 to 3d10 to 4s2 to 4p5, n=4

7

The sum of the electrons in the outer, conduction, and valence shells is twenty.

The second step is to sketch an elemental lewis dot diagram.

To see an element’s Lewis structure, we arrange the electrons in its valence shell around its chemical symbol.

The third step is to select an appropriate central atom for the complex.

Careful consideration must be given to which atom will serve as the nucleus. The least electronegative of the component atoms is required (C, H, Br).

If the central atom is more electronegative than the side atoms, it will attract electrons to itself.

Among the component atoms, bromine has the highest electronegativity and hence cannot serve as the centre atom. Since hydrogen makes only one link, it can never serve as the centre atom. Therefore, carbon (C) is the key element here.

A skeletal schematic, as the fourth step, is drawn.

The next step is to put the central atoms and the ones on the sides in the right places.

The fifth step is to position the valence electrons around the elemental signs.

Predicted bond formation is used to assign the total number of electrons in the valence shell (determined in step 1).

Step 6: Attempt to complete the octet for each atom by forming bonds with its neighbours.

In its ground state, carbon possesses four valence electrons. Because of its complete valence shell arrangement, it shares an electron with both H and Br.

One valence electron is shared between the hydrogens and the carbon in this fully filled structure.

In its ground state, each bromine atom possesses 7 electrons in its valence shell, and it shares one of those electrons with carbon to achieve a complete electronic configuration.

In dibromomethane, every atom has an octet.

Identify the formal charge of each atom in Step 7.

This compound has no overall net charge.

As a result, the total formal charge of C+H+H+Br+Br+Br+Br+Br2 is zero.

Atom

According to the periodic table, the valence electron count is the number of electrons in an atom’s outermost shell.

Number of elements in the periodic table,

meaning of electrons as

Number of bonded electrons * 0.5 = Formal Charge on a lone pair

In the subatomic realm

C 4 0 8*0.5=4

4-0-4=0

H1 1 0 2*0.5=1

1-0-1=0

H2 1 0 2*0.5=1

1-0-1=0

Br1 7 6 2*0.5=1

7-6-1=0

Br2 7 6 2*0.5=1

7-6-1=0

The Lewis structure for CH2Br2 is hence the one depicted in step 6.

## Molecular Geometry of CH2Br2

Molecular geometry is the study of the spatial relationships between a molecule’s atoms in three dimensions. Since Lewis structure cannot be used to predict a compound’s shape, an alternative theory was required. The geometry and form of a compound can be calculated with the assistance of VSEPR theory.

VSEPR theory refers to the repulsion between electron pairs in a valence shell.

The theory relies on the idea that the electrons in the valence shells of the atoms in a molecule will naturally cluster in a way that minimises the attractive forces between them. Geometry refers to the consistent pattern.

The distinction between geometry and shape is often blurred. The geometry of an atom is defined by the configuration of its bonding pairs of electrons, whereas the geometry of an atom’s form is defined by the configuration of its bonding and lone pairs of electrons.

## VSEPR Studies on the Geometry of CH2Br2

First, let’s assume the number of electrons in the valence shell of the core atom is A. (arbitrary variable)

The carbon in CH2Br2 serves as the core atom. The valence electron count for carbon is 4. (Demonstrated at the beginning of the lewis-structure-drawing process)

A=4

It’s as simple as counting the atoms on the sides and setting that number equal to B. (arbitrary variable). Due to the presence of four adjacent atoms (two hydrogen and two bromine), the value of B in CH2Br2 is 4.

If the compound has a net positive charge, subtract the charge from B, and if the compound has a net negative charge, add the charge to B. For neutral chemicals, this isn’t necessary.

No net charge is contributed to CH2Br2, and B=4 alone.

Fourth, incorporate the atomic and electric contributions from the sides (A+B) into the total (A+C).

When applied to CH2Br2, A+B = 8

Fifth, to obtain the total number of electron pairs that contribute to the shape, divide A+B by 2.

This molecule of CH2Br2 has four valence electron pairs.

Six, classify all electron pairs into bonding and non-bonding categories. The number of side atoms is the same as the number of bonding electron pairs.

In the case of CH2Br2, there are four atoms on the periphery. Consequently, there are four pairs of electrons involved in covalent bonds and no unpaired electrons.

With this data at hand, one can quickly and readily determine the geometry and shape of dibromo methane.

Tetrahedral geometry and form characterise electrons. To put it another way, compounds with no lone pairs have the same geometry and structure.

## Incorporating CH2Br2 into a hybrid system

Pauling established the concept of hybridization as a means of explaining bond formation.

By combining orbitals from different atoms, hybrid orbitals are created that have the same energy, shape, and size as the original orbitals. Bonds are formed by hybridised orbitals, and their overlap is more extensive than that of unhybridized ones.

Only orbitals with similar energy, shape, and size are capable of hybridization.

For instance, four sp3 hybrid orbitals of the same size, shape, and energy can be formed by combining one 2s and three 2p, but two 2s and six 6d cannot.

They are not literally mixed together. Only wavefunction mixing takes place.

The carbon atom is the most important one in CH2Br2. Here, we are just concerned with how the core atom hybridises. The ground state configuration of carbon comprises two electrons that are unpaired. Only two bonds can be formed.

All four valence electrons are unpaired as a result of the electron promotion process.

There are 4 electrons total, and they’re all spread out in different orbitals. Hybridization occurs between all four orbitals, resulting in sp3 hybrid orbitals.

In this approach, we count the number of electron pairs on the centre atom, just as we did in the VSEPR method.

Electron pair count for dibromo methane is 4.

## The steric numbers can be calculated directly using our formula.

To calculate the steric number, add all the sigma bonds and lone pairs on the centre atom.

## Polarity of CH2Br2

To define polarity, we look at whether or not the net dipole moment is positive or negative for a given chemical. The dipole moment is measured in vector space.

To what extent-

An individual bond’s dipole moment

â€¢ The electronegativity disparity between the atoms that make up the bonds

To what extent does the geometry and symmetry of a compound

It’s important to note that CH2Br2 has both carbon-hydrogen and carbon-bromine bonds.

The electronegativity scale developed by Linus Pauling-

It has an electronegativity of 2.55 for carbon (C).

Hydrogen has an electronegativity of 2.2.

Electronegative value of bromine is 2.96.

Differences between C-H and C-Br are calculated to be 0.35 and 0.41, respectively. Since the bond dipole moment is not zero, we conclude that the bonds are polar.

There is no certainty that a molecule will be polar just because its links are polar. You can think of it as a tetrahedral composite. Different substituents cause the vector sum of the dipole moments of the four bonds to be positive rather than zero.

Thus, CH2Br2 is an example of a polar molecule.

## Conclusion

Dibromomethane is an organic molecule with many applications.

Because it complies with the octet rule and the formal charges, the Lewis structure depicted above is the best possible one.

Dibromomethane is found to be tetrahedral in both geometry and form (CH2Br2).

The hybridization of the central atom in CH2Br2 is sp3.

An example of a polar molecule.

Misha Khatri
Misha Khatri is an emeritus professor in the University of Notre Dame's Department of Chemistry and Biochemistry. He graduated from Northern Illinois University with a BSc in Chemistry and Mathematics and a PhD in Physical Analytical Chemistry from the University of Utah.