CH3OH is the molecular formula of methanol, also known as methyl alcohol, which is the simplest aliphatic alcohol. It is primary alkyl alcohol in which a methyl group is linked to a hydroxyl functional group.

It is a polar solvent due to the presence of a hydroxyl functional group. It has a molecular mass of 32.042g along with being colorless and having a pungent odor and volatile nature.

To know more about methanol, it is necessary to get acquainted with its bonding characteristics.

For this, it is required to understand in detail the Lewis structure, VSEPR theory, the Hybridization concept as well as its polar nature.

The Lewis Structure of a molecule gives the simplest representation of valence shell electrons around itself.

Here, the valence electrons are represented by small dots and since a single bond consists of two bonding electrons, the two dots between two atoms are represented by a line instead, which represents a bond between them.

A lewis structure gives us information about the existence of a bond between atoms, although it is not enough to tell us about the type of the bond.

For making the lewis structure of methanol, it is required to satisfy the octet rule first and then calculate the formal charge of every element in the molecule.

The octet rule states for every element to obtain a noble gas configuration i.e to have a total of 2 electrons (in the case of H and He) or a total of 8 electrons in their valence shells. This is because noble gases are considered to be stable in nature.

A formal charge is a theoretical charge present on elements of a molecule. The sum of these formal charges has to be equivalent to the net charge on the molecule. It is calculated by the following formula;

Formal charge of an element = (Valence electrons) – ((bonding electrons)/2) – (Non bonding electrons)

1) Calculation of total valence electrons in methanol

The total no. of valence electrons in a molecule is the sum of valence shell electrons of each element in the molecule.

Take a look at the following table for the same.

Now, there is 1 carbon atom, 1 oxygen atom, and 4 hydrogen atoms in one CH3OH molecule. Therefore the total valence electrons are:

(1*4) + (1*6) + (1*4) = 14 valence shell electrons

2) Sketching a rough diagram

For drawing a rough sketch, it is required to select the central atom of the molecule.

The central atom must not be strongly electronegative and must not be too small in size. Therefore in methanol, both C and O are appropriate central atoms.

However, the lewis structure of methanol can be drawn easily by considering either of the elements as the central atom.

Here, we will be moving further by taking carbon as the central atom. As we can observe, methanol has three C-H bonds, one C-O bond, and one O-H bond.

The rough sketch looks something like the following image.

3) Placing the valence electrons around individual atoms

The following shows how the valence electrons are placed around the respective atoms.

4) Satisfy the octet rule of each atom and sketch the final lewis structure

As there are a total of 14 valence shell electrons in methanol, therefore, there must be 7 electron pairs in the molecule.

We know that two bonding electrons together form a bond which is represented by a line.

The following image shows the lewis structure which has all the bonds and lone pairs visible along with the satisfied octet rule.

For completing the lewis diagram, the last step is to calculate the formal charge. As we discussed the formula to calculate the formal charge earlier, it can be done as follows;

Formal charge of an element = (Valence electrons) – ((bonding electrons)/2) – (Non bonding electrons)

Where;

Bonding electrons are the ones that are involved in bond formation.

Non-bonding electrons are lone pairs around each atom that are not involved in bonding.

The net charge on a molecule is the sum of the formal charges on each individual atom. Therefore, the net charge on methanol molecule is;

(1*0)C + (1*0)O + (4*0)H = 0

This means that methanol is a neutral compound that has 2 lone pairs situated on the oxygen atom and its final lewis structure will look something like as can be seen in the following image;

It is however agreeable that a Lewis structure gives a lot of information about a certain compound, although it is difficult to tell a compound’s geometry and hybridization exactly by this concept.

Therefore, it becomes necessary to know the VSEPR theory and hybridization concept.

The molecular geometry of a compound can be very well explained by the Valence Shell Electron Pair Repulsion (VSEPR) Theory which is based on the assumption that a molecule stabilizes itself in a particular shape in which it experiences a minimum amount of electron-electron repulsions.

The molecular shape of a molecule is not the same as its molecular geometry if the difference between the no. of bond pairs electrons and the steric number is not zero, i.e the molecule contains non-zero lone pairs in it.

The steric number is a certain value assigned to each type of geometry according to the VSEPR theory which can be calculated by the following formula:

Steric number (S) = ( X + M + |a| – b )/2

Where;

X = no. of valence electrons of the central atom

M = no. of monoatomic side atoms

a = negative charge on the molecule

b = positive charge on the molecule

The following table shows the assigned steric number to each geometry.

The steric number of CH3OH according to the VSEPR theory is calculated as follows:

1) Calculating the steric number of methanol with Carbon as the central atom.

● No. of valence electrons of the carbon atom = 4

● No. of monoatomic side atoms = 4

● As discovered by the formal charge calculation that methanol is neutral, therefore the values of a and b are 0.

Steric number of CH3OH wrt C = ( 4 + 4 ) / 2 = 4

Since the steric number of the methanol molecule is 4, this means that it has tetrahedral geometry with respect to the carbon atom.

2) Calculating the steric number of methanol with oxygen as the central atom.

● No. of valence electrons of the oxygen atom = 6

● No. of monoatomic side atoms = 2

● As discovered by the formal charge calculation that methanol is neutral, therefore the values of a and b are 0.

Steric number of CH3OH wrt O = ( 6 + 2 ) / 2 = 4

Since the steric number of the methanol molecule is 4, this means that it has tetrahedral geometry with respect to the oxygen atom as well.

A compound having tetrahedral geometry can have four types of molecular shapes depending on the no. of lone pairs present in the molecule.

The various shapes in a particular type of geometry happen due to the presence of lone pair-lone pair repulsions, lone pair-bond pair repulsions, and bond pair-bond pair repulsions in a molecule.

The no. of lone pairs in a compound can be calculated by subtracting the no. of bond pairs from the steric number.

● If a compound has 3 bond pairs and 1 lone pair, then it will have a pyramidal shape.

● If a compound has 2 bond pairs and 2 lone pairs, then it will have a bent shape or a linear shape.

● If a compound has 4 bond pairs and no lone pairs, then it will have the tetrahedral geometry.

Thus, in the case of methanol;

● Since the central carbon atom has 4 sigma bonds and no lone pairs, this means that methanol has a tetrahedral shape with respect to the carbon atom and equal bond angles of 109.5 degrees.

● Since the oxygen atom has 2 sigma bonds and 2 lone pairs, this means that methanol has a bent shape with respect to the oxygen atom and a bond angle of 104.5 degrees.

Hybridization is a concept that describes the formation of hybrid orbitals upon the mixing of pure atomic orbitals which are identical in both energy and shape.

It is however necessary that the no. of hybrid orbitals thus formed must always be equal to no. of atomic orbitals that were mixed together.

The hybridization concept helps in explaining the bonding in the geometrical shapes of compounds.

For instance, hybridization can be easily or directly calculated from the steric number of the compound itself.

The following table shows the respective hybridization of each steric number value.

By looking at the table, we can tell that the hybridization of methanol is sp3 wrt to both carbon and oxygen as the central atoms since both of them makes the steric number of the molecule equivalent to 4.

On the other hand, the detailed method to calculate the hybridization of a compound is as follows;

The electronic configuration of C in the ground state is 1s2 2s2 2p2 which is represented as

follows:

To combine with three hydrogen atoms and one hydroxyl group, carbon will be required to extend its octet by exciting one of its valence electrons in the 2s orbital to the 2p orbital.

The electronic configuration of carbon in an excited state is now 1s2 2s1 2p3 which is represented as

follows:

The four atomic orbitals; one 2s and three 2p orbitals mix with each other to form four sp3 hybrid orbitals which can be seen in the following image.

Therefore, we can say that the hybridization of CH3OH is sp3.

I have also written an article on ethanol. Go and check out the hybridization, geometry, and lewis structure of ethanol.

If the charge distribution between two atoms is unequal or there occurs an electronegativity difference between them, then the bond between the two atoms is said to be polar.

Another factor that determines the polarity of a compound is dipole moment, which is the magnitude of the product of the partial charge of atoms and the distance between them.

Any compound is said to be polar if the dipole moment (μ) is not zero i.e μ ≠ 0.

For a dipole moment to be present between two atoms, there needs to be an electronegativity difference between them.

We know that methanol has three C-H bonds, one C-O bond, and one O-H bond.

So we need to calculate the electronegativity difference between a C-H, a C-O, and an O-H bond.

The electronegativity of C is 2.55, O is 3.44 and H is 2.2.

● The electronegativity difference in the C-H bond is 2.55 – 2.2 = 0.35 which is almost negligible. So the C-H bond is nonpolar.

● The electronegativity difference in the C-O bond is 3.44 – 2.55 = 0.89 which cannot be neglected. So the C-O bond is polar.

● The electronegativity difference in the O-H bond is 3.44 – 2.2 = 1.24 which makes this bond polar.

Now, By looking at the image we can see that both the arrows depicting the direction of the dipole moment is pointed towards the oxygen atom.

This is because oxygen is comparatively more electronegative than carbon or hydrogen (as can be seen by the electronegativity values), so the electron density of the bond is shifted more towards oxygen and therefore the two dipole moments do not cancel each other out.

For more detailed information, you must check out the polarity of CH3OH.

This means that methanol is a polar molecule.

It is however a known fact that methanol is used in many chemical reactions as a polar solvent.

This happens due to the presence of the hydroxyl functional group which results in hydrogen bonding in the compound.

The presence of the hydroxyl group initiates the hydrogen bonding between the methanol molecule and the water molecules when dissolved in water.

This makes the compound soluble in water.

To summarise, CH3OH is a polar and a neutral compound that has a tetrahedral geometry and bent and tetrahedral shapes with respect to the oxygen and carbon atoms respectively.

It has three C-H sigma nonpolar bonds, one C-O sigma polar bond, and one O-H sigma polar bond along with the hybridization of sp3.