CH3I Molecular Geometry - Science Education and Tutorials

Q: What is the molecular notation for CH3I molecule?

CH3I molecular notation is AX4.

Iodomethane(CH3I) has the composition of one carbon, one iodine, and three hydrogen atoms. What is the molecular geometry of iodomethane?. Drawing and predicting the CH3I molecular geometry is very easy by following the given method. Here in this post, we described step by step to construct CH3I molecular geometry. Hydrogen, iodine, and carbon come from the1st, 17th, and 14th family groups in the periodic table. Hydrogen, iodine, and carbon have one, seven, and four valence electrons respectively.

Table of Contents

Key Points To Consider When drawing The CH3I Molecular Geometry

A three-step approach for drawing the CH3I molecular can be used. The first step is to sketch the molecular geometry of the CH3I molecule, to calculate the lone pairs of the electron in the central carbon, terminal iodine, and terminal hydrogen atom; the second step is to calculate the CH3I hybridization, and the third step is to give perfect notation for the CH3I molecular geometry.

The CH3I molecular geometry is a diagram that illustrates the number of valence electrons and bond electron pairs in the CH3I molecule in a specific geometric manner. The geometry of the CH3I molecule can then be predicted using the Valence Shell Electron Pair Repulsion Theory (VSEPR Theory) and molecular hybridization theory, which states that molecules will choose the CH3I geometrical shape in which the electrons have from one another in the specific molecular structure.

Finally, you must add their bond polarities characteristics to compute the strength of the C-I bond (dipole moment properties of the CH3I molecular geometry). The carbon-iodine bonds in the iodomethane molecule(CH3I), for example, are polarised toward the more electronegative value iodine atom, and because all (three C-H and one C-I) bonds have the same size and polarity, their sum is non zero due to the CH3I molecule’s bond dipole moment, and the CH3I molecule is classified as a polar molecule.

The molecule of iodomethane (with tetrahedral shape CH3I molecular geometry) is tilted at 111.5 degrees. It has a difference in electronegativity values between iodine and carbon atoms, with iodine’s pull the electron cloud being greater than carbon’s. As a result, it has a permanent dipole moment in its molecular structure. The CH3I molecule has a dipole moment due to an unequal charge distribution of negative and positive charges.

Overview: CH3I electron and molecular geometry

According to the VSEPR theory, CH3I possesses tetrahedral molecular geometry and CH4-like electron geometry. Because the center atom, carbon, has three C-H and C-I bonds with the hydrogen and iodine atoms surrounding it. The H-C-I bond generates 111.5 degrees in the tetrahedral molecular geometry. The CH3I molecule has a tetrahedral geometry shape because it contains one iodine and three hydrogen atoms.

There are three C-H and one C-I bond at the CH3I molecular geometry. After linking the three hydrogens and one iodine atom in the tetrahedral form, it maintains the tetrahedral-like structure. In the CH3I molecular geometry, the C-I and C-H bonds have stayed in the four terminals of the tetrahedral molecule.

The center carbon atom of CH3I has no lone pair of electrons, resulting in tetrahedral electron geometry. However, the molecular geometry of CH3I looks like a tetrahedral and has no lone pair out of the plane. It’s the CH3I molecule’s asymmetrical geometry. As a result, the CH3I molecule is polar.

How to find CH3I hybridization and molecular geometry

Calculating lone pairs of electron on carbon in the CH3I molecule:

1.Determine the number of lone pairs on the core carbon atom of the CH3I Lewis structure. Because the lone pairs on carbon are mostly responsible for the CH3I molecule geometry distortion, we need to calculate out how many there are on the central carbon atom of the Lewis structure.

Use the formula below to find the lone pair on the carbon atom of the CH3I molecule.
L.P(C) = V.E(C) – N.A(C-H and C-I)/2

Lone pair on the central carbon atom = L.P(C)
The core central carbon atom’s valence electron = V.E(C)
Number of C-H and C-I bonds = N.A (C-H and C-I)
calculation for carbon atom lone pair in CH3I molecule

For instance of CH3I, the central atom, carbon, has four electrons in its outermost valence shell, three C-H, and one C-I bond connection.

As a result of this, L.P(C) = (4 –4)/2=0

In the CH3I electron geometry structure, the lone pair on the central carbon atom is zero. It means there is no lone pair in the core carbon atom. These lone pair on the central carbon atom is responsible for the CH3I molecular geometry distortion.

If you imagine, there is no lone pair on the carbon atom of the CH3I molecule. Then, no electronic repulsion of C-I and C-H bonds pair in the CH3I. That gives stable tetrahedral geometry.

But in reality, the CH3I molecule undergoes distortion in its geometry due to the polarity of one iodine atom in the tetrahedral geometry. This leads to tetrahedral geometry for the CH3I molecule.

Calculating lone pairs of electron on iodine in the CH3I molecule:

1.Determine the number of lone pairs on the terminal iodine atom of the CH3I Lewis structure. Because the lone pairs on carbon are mostly responsible for the CH3I molecule geometry distortion, we need to calculate out how many there are on the terminal iodine atom of the Lewis structure. But we need to use different formulations for terminal atoms.

Use the formula below to find the lone pair on the iodine atom of the CH3I molecule.
L.P(Br) = V.E(Br) – N.A(C-I)

Lone pair on the terminal iodine atom in CH3I = L.P(Br)
The core terminal iodine atom’s valence electron = V.E(Br)
Number of C-I bonds = N.A (C-I)
calculation for iodine atom lone pair in CH3I molecule

For instance of CH3I, the terminal atom, iodine, has seven electrons in its outermost valence shell, only one C-I connection.

As a result of this, L.P(Br) = (7 –1)=6

In the CH3I electron geometry structure, the lone pair on the terminal iodine atom is six. It means there are six lone pairs in the terminal iodine atom. These lone pair on the terminal iodine atom is responsible for the CH3I molecular interaction.

Calculating lone pairs of electron on hydrogen in the CH3I molecular geometry:

1.Determine the number of lone pairs on the terminal hydrogen atom of the CH3I Lewis structure. we need to calculate out how many there are on the terminal hydrogen atom of the Lewis structure. As we know clearly, the hydrogen atom has only s orbital in the ground state. It can take a maximum of two electrons.

Hydrogen does not show up any lone pairs of electrons. But we are trying to define this hypothesis in a mathematical manner.

Use the formula below to find the lone pair on the hydrogen atom of the CH3I molecule.
L.P(H) = V.E(H) – N.A(C-H )

Lone pair on the terminal hydrogen atom = L.P(H)
The core terminal hydrogen atom’s valence electron = V.E(H)
Number of C-H bonds = N.A (C-H )
calculation for hydrogen atom lone pair in CH3I molecule

For instance of CH3I, the terminal atom, hydrogen, has one electron in its outermost valence shell, only one C-H connection.

As a result of this, L.P(H) = (1 –1)=0

In the CH3I electron geometry structure, the lone pair on the terminal hydrogen atom is zero. It means there is no lone pair in the terminal hydrogen atoms. This simple mathematical illustration gives mathematical importance to chemical molecular structures.

Calculate the number of molecular hybridizations of CH3I molecule

What is CH3I hybrizidation? This is a very fundamental question in the field of molecular chemistry. All the molecules made by atoms. In chemistry, atoms are the fundamental particles. There are four different types of orbitals in chemistry. They are named as s, p, d, and f orbitals.

The entire periodic table arrangement is based on these orbital theories. Atoms in the periodic table are classified as follows:

s- block elements
p- block elements
d-block elements
f-block elements
Atoms are classified in the periodic table

CH3I molecule is made of one carbon, three hydrogens, and one iodine atom. The iodine and carbon atoms have s and p orbital. hydrogen comes as the first element in the periodic table. The hydrogen atom has s orbital.

When these atoms combine to form the CH3I molecule, its orbitals are mixed and form unique molecular orbitals due to hybridization.

How do you find the CH3I molecule’s hybridization? We must now determine the molecular hybridization number of CH3I.

The formula of CH3I molecular hybridization is as follows:
No. Hyb of CH3I= N.A(C-H and C-I bonds) + L.P(C)
No. Hy of CH3I= the number of hybridizations of CH3I
Number of C-H and C-I bonds = N.A (C-I and C-H bonds)
Lone pair on the central carbon atom = L.P(C)
Calculation for hybridization number for CH3I molecule

In the CH3I molecule, carbon is a core atom with one iodine and three hydrogen atoms connected to it and no lone pair. The number of CH3I hybridizations (No. Hyb of CH3I) can then be estimated using the formula below.

No. Hyb of CH3I = 4+0 =4

The CH3I molecule hybridization is four. The sp3 hybridization is formed when one S orbital and three p orbital join together to form a molecular orbital.

Molecular Geometry Notation for CH3I Molecule :

Determine the form of CH3I molecular geometry using VSEPR theory. The AXN technique is commonly used when the VSEPR theory is used to calculate the shape of the CH3I molecule.

The AXN notation of CH3I molecule is as follows:
The center carbon atom in the CH3I molecule is denoted by the letter A.
The bound pairs (C-H and C-I) of electrons to the core carbon atom are represented by X.
The lone pairs of electrons on the center carbon atom are denoted by the letter N.
Notation for CH3I molecular geometry

We know that carbon is the core atom, with four electron pairs bound (three C-H and one C-I) and no lone pair. The general molecular geometry formula for CH3I is AX4.

According to the VSEPR theory, if the CH3I molecule has an AX4 generic formula, the molecular geometry and electron geometry will both be tetrahedral forms.

Name of Molecule	Iodomethane
Chemical molecular formula	CH3I
Molecular geometry of CH3I	Tetrahedral
Electron geometry of CH3I	Tetrahedral
Hybridization of CH3I	sp3
Bond angle (H-C-I)	111.5º degree
Total Valence electron for CH3I	14
The formal charge of CH3I on carbon	0

Summary:

In this post, we discussed the method to construct CH3I molecular geometry, the method to find the lone pairs of electrons in the central carbon atom, CH3I hybridization, and CH3I molecular notation. Need to remember that, if you follow the above-said method, you can construct the CH3I molecular structure very easily.