How to draw CH3I Lewis Structure? - Science Education and Tutorials

The iodomethane chemical formula is CH3I. Drawing CH3I Lewis Structure is very easy to by using the following method. Here in this post, we described step by step method to construct CH3I Lewis Structure. The carbon, iodine, and hydrogen elements come as the member of the carbon, halogen, and hydrogen family groups from the periodic table respectively. The valence electrons in carbon, iodine, and hydrogen are four, seven, and one respectively. Iodomethane is used as an organic volatile reagent in methylation organic reactions.

Table of Contents

Key Points To Consider When Drawing The CH3I Structure

A three-step approach for drawing the CH3I Lewis structure can be used. The first step is to sketch the Lewis structure of the CH3I molecule, to add valence electron around the carbon atom; the second step is to add valence electrons to the one iodine and three hydrogen atoms, and the final step is to combine the step1 and step2 to get the CH3I Lewis Structure.

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

Finally, you must add their bond polarities to compute the strength of the C-I bond (dipole moment properties of the CH3I molecule). The carbon-iodine bonds in iodomethane(CH3I), for example, are polarised toward the more electronegative iodine and because both bonds have the same size and are located around four terminals with one iodine and three hydrogen atoms, their sum is non zero due to the CH3I molecule’s bond dipole moment and the lone pairs of electron on one iodine atoms. The CH3I molecule is classified as a polar molecule.

The molecule of iodomethane (with tetrahedral molecular geometry) is tilted, the bond angles between iodine, carbon, and hydrogen are 111.5 degrees. It has a difference in electronegativity values between carbon and iodine atoms, with carbon’s pull being less than iodine’s terminal in the CH3I molecule. As a result, it has the permanent dipole moment. The CH3I molecule has a permanent dipole moment due to an unequal charge distribution of negative and positive charges. The net dipole moment of the CH3I molecule is lower than other halogen methane derivatives such as CH3Cl, CH3Br, and CH3F.

CH3I Lewis Structure:

The central atom is carbon, which is bordered on four terminals with one iodine atom, three hydrogen atoms, and no lone pair on the carbon in the tetrahedral geometry. Carbon has four outermost valence electrons, indicating that it possesses four electrons in its outermost shell, whereas iodine only has seven valence electrons in its outermost shell. To complete the octet of the iodine atom, an iodine terminal atom requires one electron. If you’re interested in learning more about the iodine octet rule, please see in our previous post.

One iodine and three hydrogen atoms establish covalent connections with the central carbon atom as a result, leaving the carbon atom with no lone pair. There is no lone pair on the carbon central atom that resist the bond pairs of the one C-I and three C-H. According to VSEPR theory, no electronic repulsion of the lone pair and bond pair leads the CH3I molecule to take on a tetrahedral molecular geometry shape.

The CH3I molecule’s C-I bonds are arranged in asymmetrical order around the tetrahedral molecular geometry, giving rise to the CH3I molecular shape. The CH3I molecule has a tetrahedral molecular geometry because there is no electrical repulsion between lone pair and bond pairs of CH3I molecule.

Electronegative Difference Calculation of CH3I Molecule:

Calculating the electronegative difference in carbon and iodine in CH3I

The carbon atom has an electronegativity of 2.55, while iodine has an electronegativity of 2.96 in the CH3I molecule. The difference in electronegativity can be estimated using the method below.

The electronegative value difference between carbon and iodine in CH3I molecule
Electronegativity value of carbon in CH3I = 2.55
Electronegativity value of iodine in CH3I= 2.66
Difference of electronegativity value between carbon and iodine in CH3I molecule= 2.66– 2.55 = 0.11
Electronegativity difference between C-I bond calculation of CH3I molecule

The electronegative value difference between carbon and hydrogen in CH3I molecule
Electronegativity value of carbon in CH3I = 2.55
Electronegativity value of hydrogen in CH3I = 2.20
Difference of electronegativity value between carbon and hydrogen in CH3I = 2.55 – 2.20 =0.35
Electronegativity difference between C-H bond calculation of CH3I molecule

Due to the difference in electronegativity value of less than 0.5, the C-I bond of the CH3I molecule becomes nonpolar. Because of this difference in electronegativity, the CH3I molecule’s C-I bond becomes nonpolar. The electronegativity of an atom is the strength with which it may attract bound electron pairs to its side.

As a result, the C-I bond’s dipole moment is high due to the polarization of the bonds, and C-I bonds’ dipoles are arranged at the top of the tetrahedral molecular geometry. The CH3I molecule’s total dipole moment is predicted to be less than other halogen methane compounds. It has a partial negative charge for iodine atoms and a partial positive charge for the central carbon atom.

The electron dot structure of the CH3I molecule is also known as the CH3I Lewis structure. It determines the number of outermost valence electrons as well as the electrons engaged in the CH3I molecule’s bond formation. The outermost valence electrons of the CH3I molecule must be understood while considering the Lewis structure of the molecule.

The carbon atom is the middle element in CH3I molecular geometry, with four electrons in its outermost valence electron shell, whereas the iodine atom has seven electrons in its outermost valence electron shell.

The CH3I molecule has a total of 14 valence electrons as a result of the foregoing above said reasoning. With the core central carbon atom, the four terminal with one iodine and three hydrogen atoms form covalent bonds, leaving the carbon atom with no lone pairs on it.

The tetrahedral molecular geometry and structure of the CH3I molecules are similar to that of the methane (CH4) molecule. Because no lone pair of a central carbon atom create interaction with C-I bond pairs. The bond angle of the I-C-H bond in the tetrahedral molecular geometry is approximately 111.5 degrees. The C-I and C-H bond lengths are 214 and 111 pm(picometer) respectively.

To sketch the Lewis structure of CH3I by following these instructions:

Step-1: CH3I Lewis dot Structure by counting valence electrons on carbon atom

To calculate the valence electron of each atom in CH3I, look for its periodic group from the periodic table. The carbon and halogen families, which are the 14th and 17th groups in the periodic table, are both made up of carbon and iodine atoms. In their outermost shells, carbon and iodine have four and seven valence electrons, respectively.

Because carbon and iodine are members of the periodic table’s carbon and halogen family groups, their valence electrons are four and seven, respectively.

Calculate the total number of valence electrons in the CH3I molecule’s outermost valence shell. The first step is to determine how many electrons are in the CH3I Lewis structure’s outermost valence shell. An electron in an atom’s outermost shell is known as a valence electron. It is represented by dots in the CH3I Lewis diagram. The CH3I molecule’s core carbon atom can be represented as follows:

Total outermost valence shell electron of carbon atom in CH3I= 4
Total outermost valence shell electron of the iodine atom in CH3I = 7
The CH3I molecule has one central carbon, three hydrogen, and one iodine atoms. Then the total outermost valence shell electrons can be calculated as follows
∴ Total outermost valence shell electrons available for CH3ILewis structure( dot structure) = 4 +1*7+ 3*1= 14 valence electrons in CH3I
calculation of total valence electron of CH3I molecule

Choose the atom with the least electronegative value atom and insert it in the center of the molecular geometry of CH3I. We’ll choose the least electronegative value atom in the CH3I molecule to place in the center of the CH3I Lewis structure diagram in this phase. The electronegativity value in periodic groups grows from left to right in the periodic table and drops from top to bottom.

Step-2: Lewis Structure of CH3I for constructing around the terminal hydrogen and iodine atoms

As a result, carbon is the first atom in the periodic table’s carbon family group. Iodine is the fourth member of the halogen family. The electronegative value of the carbon atom is lower than that of the iodine atom in the CH3I molecule. Furthermore, carbon has a four electrons limit since iodine is the most electronegative element in the CH3I molecule.

In the CH3I Lewis structure diagram, the carbon atom can be the center atom of the molecule. As a result, central carbon in the CH3I Lewis structure, with all one iodine and three hydrogens arranged in the tetrahedral geometry.

Add valence electrons around the iodine atom and add valence hydrogen atom, as given in the figure.

Step-3: Lewis dot Structure for CH3I generated from step-1 and step-2

Connect the exterior and core central atom of the CH3I molecule with four single bonds (one C-I and three C-H). In this stage, use four single bonds to connect all one iodine and three hydrogen atoms on the outside of the CH3I molecule to the central carbon atom in the middle.

Count how many electrons from the outermost valence shell have been used in the CH3I structure so far. Each C-I bond carries two electrons because each carbon atom is connected to one iodine and three hydrogen atoms by one C-I and three C-H bonds. Bond pairings of C-I and C-H are what they’re called.

So, out of the total of 14 valence electrons available for the CH3I Lewis structure, we used 8 for the CH3I molecule’s one C-I and three C-H bonds. The CH3I molecule has no lone pair electron in the center of carbon. We need to put no extra electrons in the molecular geometry of CH3I.

Place the valence electrons in the C-H and C-I bond pairs starting with the core carbon, three hydrogen, and one iodine atoms in the CH3I molecule. In the CH3I Lewis structure diagram, we always begin by introducing valence electrons from the central carbon atom(in step1). As a result, wrap around the central carbon atom’s bond pair valence electrons first (see figure for step1).

Carbon requires 8 electrons in its outermost valence shell to complete the molecular stability, 8 electrons bond pairs in C-H and C-I bonds. Then place no electrons as a lone pair of electrons on the carbon atom of the CH3I molecule. Carbon already shares 8 electrons to the four single C-I and C-H bonds. Then place the valence electron in the iodine atom, it placed around seven electrons(step-2). Totally, 6 valence electrons were placed on one iodine atom of the CH3I molecule.

We’ve positioned 8 electrons around the central carbon atom(step-3), which is represented by a dot, in the CH3I molecular structure above. The carbon atom completes its molecular stability in the CH3I molecule because it possesses 8 electrons in its bond pairs with one iodine and three hydrogens in the outermost valence shell.

Count how many outermost valence shell electrons have been used so far using the CH3I Lewis structure. Four electron bond pairs are shown as dots in the CH3I chemical structure, whereas four single bonds each contain two electrons. The outermost valence shell electrons of the CH3I molecule are 8 +6 = 14 as a result of the calculation.

So far, we’ve used 14 of the CH3I Lewis structure’s total 14 outermost valence shell electrons. No lone pair of electrons on the carbon atom in the tetrahedral geometry of the CH3I molecule.

Complete the middle carbon atom stability and, if necessary, apply a covalent bond. The central carbon atom undergoes octet stability. Because it has a total of eight electrons in the outermost valence shell.

The core atom in the CH3I Lewis structure is carbon, which is bonded to one iodine and three hydrogen atoms by four single bonds (C-I and C-H). With the help of four single bonds, it already shares eight electrons. As a result, iodine follows the octet rule and has eight electrons surrounding it on the one terminal of the CH3I molecule’s tetrahedral geometry.

How to calculate the formal charge on a carbon, hydrogen, and iodine atoms in CH3I molecule?

Calculating formal charge on carbon in CH3I:

The formal charge on the CH3I molecule’s carbon central atom often corresponds to the actual charge on that carbon central atom. In the following computation, the formal charge will be calculated on the central carbon atom of the CH3I Lewis dot structure.

To calculate the formal charge on the central carbon atom of the CH3I molecule by using the following formula:
The formal charge on the carbon atom of CH3I molecule= (V. E(C)– L.E(C) – 1/2(B.E))
V.E (C) = Valence electron in a carbon atom of CH3I molecule
L.E(C) = Lone pairs of an electron in the carbon atom of the CH3I molecule.
B.E = Bond pair electron in C atom of CH3I molecule
calculation of formal charge on carbon atom in CH3I molecule

The carbon core atom (four single bonds connected to one iodine and three hydrogen atoms ) of the CH3I molecule has four valence electrons, no lone pair of electrons, and eight bonding electrons. Put these values for the carbon atom in the formula above.

Formal charge on carbon atom of CH3I molecule = (4- 0-(8/2)) =0

In the Lewis structure of CH3I, the formal charge on the central carbon atom is zero.

Calculating formal charge on iodine in CH3I:

The formal charge on the CH3I molecule’s iodine terminal atom often corresponds to the actual charge on that iodine terminal atom. In the following computation, the formal charge will be calculated on the terminal iodine atom of the CH3I Lewis dot structure.

To calculate the formal charge on the terminal iodine atom of the CH3I molecule by using the following formula:
The formal charge on the iodine atom of CH3I molecule= (V. E(Br)– L.E(Br) – 1/2(B.E))
V.E (Br) = Valence electron in a iodine atom of CH3I molecule
L.E(Br) = Lone pairs of an electron in the iodine atom of the CH3I molecule.
B.E = Bond pair electron F atom of CH3I molecule
calculation of formal charge on iodine atom in CH3I molecule

The iodine core terminal (one single bond connected with the central carbon atom) of the CH3I molecule has seven valence electrons, six lone pairs of electrons, and two bonding electrons. Put these values for the carbon atom in the formula above.

Formal charge on iodine atom of CH3I molecule = (7- 6-(2/2)) =0

In the Lewis structure of CH3I, the formal charge on the terminal iodine atom is zero.

Calculating formal charge on hydrogen in CH3I:

The formal charge on the CH3I molecule’s hydrogen terminal atom often corresponds to the actual charge on that hydrogen terminal atom. In the following computation, the formal charge will be calculated on the terminal hydrogen atom of the CH3I Lewis dot structure.

To calculate the formal charge on the terminal hydrogen atom of the CH3I molecule by using the following formula:
The formal charge on the terminal hydrogen atom of CH3I molecule= (V. E(H)– L.E(H) – 1/2(B.E))
V.E (H) = Valence electron in the hydrogen atom of CH3I molecule
L.E(H) = Lone pairs of an electron in the hydrogen atom of the CH3I molecule.
B.E = Bond pair electron in H atom of CH3I molecule
calculation of formal charge on hydrogen atom in CH3I molecule

The hydrogen terminal atom ( three hydrogen atoms ) of the CH3I molecule has one valence electron, no lone pair of electrons, and two bonding electrons. Put these values for the carbon atom in the formula above.

Formal charge on hydrogen atom of CH3I molecule = (1- 0-(2/2)) =0

In the Lewis structure of CH3I, the formal charge on the terminal hydrogen atom is zero.

Summary:

In this post, we discussed the method to construct the CH3I Lewis structure. First, the valence electrons are placed around the carbon atom. Second, place the valence electron on the iodine and hydrogen atoms. Finally, when we combined the first and second steps. It gives CH3I Lewis structure. Need to remember that, if you follow the above-said method, you can construct molecular dot structure very easily.