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

How to draw HI Lewis Structure?

The hydrogen iodide or hydroiodic acid chemical formula is HI. Drawing HI Lewis Structure is very easy to by using the following method. Here in this post, we described step by step method to construct HI Lewis Structure. The iodine and hydrogen elements come as the member of the halogen and hydrogen family groups from the periodic table respectively. The valence electrons in iodine and hydrogen are seven and one respectively. Hydrogen iodide is used to make chemical reagents for organic chemical reactions as a catalyst or reagent in organic chemistry. More information about hydrogen iodide is reviewed in our previous post.

Key Points To Consider When Drawing The HI Electron Dot Structure

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

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

Finally, you must add their bond polarities to compute the strength of the one H-I single bonds (dipole moment properties of the HI molecule). The hydrogen-iodine bonds in hydrogen iodide(HI), for example, are polarised toward the more electronegative iodine in HI molecule, and because both bonds have the same size and are located around one hydrogen terminal of the tetrahedral or linear structure with three lone pairs (in total six electrons) on the iodine atom, their sum of dipole moment is nonzero due to the HI molecule’s bond dipole moment and less electron polarity to the hydrogen atoms. Because H-I bonds polarity is not canceled in the HI molecule due to the presence of three lone pairs of electrons in the tetrahedral structure. The hydrogen iodide(HI) molecule is classified as a polar molecule.

How to draw HI Lewis Structure?

The molecule of hydrogen iodide(with tetrahedral or linear-shaped molecular geometry) is tilted, the bond angles between iodine and hydrogen are 180 degrees. It has a difference in electronegativity values between iodine and hydrogen atoms, with central iodine’s pull being higher than terminal hydrogen’s in the HI molecule. But they not canceled each other due to the asymmetrical tetrahedral with three lone pairs in the molecular geometry of the HI molecule.

As a result, it has the nonzero dipole moment. The HI molecule has a nonzero dipole moment due to an unequal charge distribution of negative and positive charges. But both iodine and hydrogen atoms fall on the halogen and hydrogen family groups in the periodic table respectively. The iodine atom is a more electronegative value than hydrogen in the HI molecule. The HI molecule has the net dipole moment of 1.08D value in the ground state energy level.

HI molecule has two H-I single bonds. Its dipole moment in the ground state is totally different as compared with the excited state. If it absorbs light may be from visible or UV light. It undergoes pi to pi star and n to pi star transition from ground state energy level to excited state energy level. In the excited state energy level, the HI molecule shows a definite dipole moment. But it is very dynamic in nature.

Molecules can be classified as polar or nonpolar. The molecule polar behaves in a different manner as compared to nonpolar.

Overview: HI Lewis Structure

The central atom is iodine, which is bordered on two terminals with hydrogen atoms( in tetrahedral or linear geometry), and three lone pairs on the central iodine atom in the tetrahedral or linear molecular geometry. Iodine has seven outermost valence electrons, indicating that it possesses seven electrons in its outermost shell, whereas hydrogen also has one valence electron in its outermost shell. To complete the octet of the iodine atom requires one valence electron on each of their outermost shell.

One hydrogen atom establishes covalent connections with the central iodine atom as a result, leaving the iodine atom with three lone pairs. There are three lone pairs of electrons on the iodine central atom that resists the bond pairs of the H-I bond. According to VSEPR theory, the H-I bond pairs polarity lead the HI molecule to take on the linear or tetrahedral geometry structure.

The HI molecule’s two H-I bonds are arranged in asymmetrical polarity order around the linear or tetrahedral molecular geometry, giving rise to the HI molecular shape. The HI molecule has a tetrahedral or linear molecular geometry because there is an electrical repulsion between the lone pairs of electrons in iodine and one single bond pair(H-I) of the HI molecule.

Lewis structure of HI has dot electron representative structure. Valence electrons of atoms undergo orbitals mixing in the chemical reactions, gives new types of molecular species of HI. The molecule is nothing but a bundle of valence electrons from the atoms. But it is converted to bond pairs and lone pairs in the molecular structure.

Electronegative value Difference Calculation of HI Molecule:

Iodine and hydrogen Electronegative difference in HI:

The iodine atom has an electronegativity of 2.66, while hydrogen has an electronegativity of 2.2 in the HI molecule. The difference in electronegativity of iodine and hydrogen can be estimated using the method below.

The electronegative value difference between iodine and hydrogen in HI molecule

Electronegativity value of iodine= 2.66

Electronegativity value of hydrogen= 2.20

Difference of electronegativity value between iodine and hydrogen in HI molecule = 2.66 – 2.20 = 0.46

Electronegativity difference between H-I bond calculation of HI molecule

The electronegative difference between iodine and hydrogen is greater than 0.5. This indicated the bond polarity moves polar nature. H-I bond polarity in the HI molecule is polar.

Because of this difference in electronegativity of iodine and hydrogen atoms, the HI molecule’s H-I bond becomes polar in nature. The total net dipole moment of the HI molecule is nonzero due to the noncancellation of the bond dipole moment in the linear or tetrahedral geometry due to the presence of three lone pairs of electrons. The electronegativity of an atom is the strength with which it may attract bound electron pairs to its side. The polarity of HI is discussed in our previous post.

As a result, the H-I bond’s dipole moment is higher due to the polarization of the bonds and three lone pairs of electrons on iodine, and all H-I bonds’ dipoles are arranged in the asymmetrical HI molecular geometry. The HI molecule has a nonzero net dipole moment.

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

The iodine atom is the middle element in HI molecular geometry, with seven electrons in its outermost valence electron shell, whereas the hydrogen atom has one electron in its outermost valence electron shell. The hydrogen atom has one valence electron.

The HI has a total of 8 valence electrons as a result of the foregoing above-said reasoning. With the core central iodine atom, the one terminals with one hydrogen atom form covalent bonds, leaving the iodine atom with three lone pairs in the middle of linear or tetrahedral geometry.

Because three lone pairs on the terminal iodine atoms create interaction with H-I bond pairs(but it is negligible in the ground state of the HI molecule). The bond angle of the H-I bond in the linear or tetrahedral molecular geometry is approximately 180 degrees. This angle is greater than the CH4 molecule bond angle. The H-I bond length is 161pm(picometer).

To sketch the HI Lewis structure by following these instructions:

Step-1: HI Lewis dot Structure by counting valence electrons on the iodine atom

To calculate the valence electron of each atom in HI, look for its periodic group from the periodic table. The halogen and hydrogen group families, which are the 17th and 1st groups in the periodic table, are both made up of iodine and hydrogen atoms respectively. In their outermost shells, hydrogen and iodine have one and seven valence electrons respectively.

Calculate the total number of valence electrons in the HI molecule’s outermost valence shell. The first step is to determine how many electrons are in the HI 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 HI Lewis diagram. The HI molecule’s core iodine atom can be represented as follows:

Total outermost valence shell electron of iodine atom in HI= 7

Total outermost valence shell electron of the hydrogen atom in HI= 1

The HI molecule has one central iodine and one hydrogen atom. Then the total outermost valence shell electrons can be calculated as follows

∴ Total outermost valence shell electrons available for HI Lewis structure( dot structure) = 7+1*1= 8 valence electrons  in HI.  

calculation of total valence electron of HI molecule

Choose the atom with the least electronegative value atom and insert it in the center of the molecular geometry of HI. We’ll choose the least electronegative value atom in the HI molecule to place in the center of the HI Lewis structure diagram in this phase.

But in this case, hydrogen is the least electronegative than iodine. Hydrogen takes a maximum of two-electron in its orbital. This gives hydride ion(H-). So that iodine stays in the central molecular structure. The electronegativity value in periodic groups grows from left to right in the periodic table and drops from top to bottom.

How to draw HI Lewis Structure?

The first step is to put seven valence electrons around the iodine atom as given in the figure.

Step-2: Lewis Structure of HI for counting valence electrons around the terminal hydrogen atom

As a result, iodine is the first atom in the periodic table’s halogen family group. Hydrogen is the first member of the hydrogen family. It is the first element in the periodic table. The electronegative value of the iodine atom is higher than that of the hydrogen atom in the HI molecule. Furthermore, hydrogen has a one-electron limit since it is the less electronegative element in the HI molecule.

In the HI Lewis structure diagram, the iodine atom can be the center atom of the molecule. As a result, central iodine in the HI Lewis structure, with one hydrogen atom arranged in a linear or tetrahedral geometry.

How to draw HI Lewis Structure?

Add valence electron around the hydrogen atom, as given in the figure.

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

Connect the exterior and core central atom of the HI molecule with one single H-I bond. In this stage, use one hydrogen atom on the outside of the HI molecule to the central iodine atom in the middle.

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

So, out of the total of 8 valence electrons available for the HI Lewis structure, we used four electrons for the HI molecule’s one H-I single bond. The HI molecule has three lone pairs of electrons in the central iodine atom.

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

The iodine atom in the molecule gets only 8 electrons around its molecular structure. This central iodine atom is octet stable. But it has three lone pairs. Iodine gas(Cl2) is a sublimating brown color solid. Iodine is very oxidative in nature. Iodination reactions in organic chemistry are carried out with help of an iodine reagent.

Hydrogen molecule(H2) is in the gaseous state at normal temperature and pressure. It is used as a hydrogenating agent in the field of organic chemistry. It is highly flammable in nature. It is applied in fuel cells. During the combustion, hydrogen gas gives the stream as the final product. This reduces environmental pollution.

Iodine requires 8 electrons in its outermost valence shell to complete the molecular octet stability, two electrons bond pairs in one H-I single bond, and three lone pairs in the central iodine atom. No lone pairs of electrons on the hydrogen atoms of the HI molecule are placed in a linear or tetrahedral geometry. Iodine already shares 8 electrons to the one H-I single bonds. Then place the valence electron in the hydrogen atoms, it placed around one electron on each atom(step-2). There are no valence electrons placed around one hydrogen atom as lone pair of electrons.

We’ve positioned 8 electrons around the one-terminal hydrogen atoms(step-3), which is represented by a dot, in the HI molecular structure above. The iodine atom completes its molecular octet stability in the HI molecule because it possesses 2 electrons in its (one H-I single bonds) bond pairs with one hydrogen in the outermost valence shell.

How to draw HI Lewis Structure?

Count how many outermost valence shell electrons have been used so far using the HI Lewis structure. One electron bond pairs are shown as dots in the HI chemical structure, whereas one single bond contains two electrons. The outermost valence shell electrons of the HI molecule(bond pairs) are 2 as a result of the calculation. The total valence electron in an iodine atom is 8.

So far, we’ve used 8 of the HI Lewis structure’s total 8 outermost valence shell electrons. Three lone pairs of electrons on the iodine atom in the linear or tetrahedral geometry of the HI molecule.

Complete the middle HI atom stability and, if necessary, apply a covalent bond. The central iodine atom undergoes octet stability(due to one single bond pair of electrons).

The core atom in the HI Lewis structure is iodine, which is bonded to the one hydrogen atom by single bonds (one H-I). With the help of one single bond, it already shares 8 electrons. As a result, the iodine follows the octet rule and has 8 electrons surrounding it on the one terminal of the HI molecule’s linear or tetrahedral geometry.

How to calculate the formal charge on iodine and hydrogen atoms in HI Lewis Structure?

Calculating formal charge on the iodine of HI molecule:

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

To calculate the formal charge on the central iodine atom of the HI molecule by using the following formula:

The formal charge on the iodine atom of HI molecule= (V. E(I)– L.E(I) – 1/2(B.E))

V.E (I) = Valence electron in an iodine atom of HI molecule

L.E(I) = Lone pairs of an electron in the iodine atom of the HI molecule.

B.E = Bond pair electron in F atom of HI molecule

calculation of formal charge on iodine atom in HI molecule

The iodine core atom (one single bond connected to one hydrogen atom) of the HI molecule has seven valence electrons, three lone pairs of electrons(six electrons), and 2 bonding pairing valence electrons. Put these values for the iodine atom in the formula above.

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

In the Lewis structure of HI, the formal charge on the central iodine atom is zero.

Calculating formal charge on the hydrogen atom of HI molecule:

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

To calculate the formal charge on the terminal hydrogen atom of the HI molecule by using the following formula:

The formal charge on the hydrogen atom of HI molecule= (V. E(H)– L.E(H) – 1/2(B.E))

V.E (H) = Valence electron in a hydrogen atom of HI molecule

L.E(H) = Lone pairs of an electron in the hydrogen atom of the HI molecule.

B.E = Bond pair electron in H atom of HI molecule

calculation of formal charge on hydrogen atom in HI molecule

The hydrogen terminal atoms of the HI molecule have one valence electron, no lone pairs of electrons(zero electrons), and two bonding pairing valence electrons(single bond). Put these values for the hydrogen atom in the formula above.

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

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

Summary:

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

What is the HI Lewis structure?

HI Lewis structure is dot representation

What is the formal charge on the HI Lewis structure?

Zero charges on the HI molecular structure

The polarity of the molecules

The polarity of the molecules are listed as follows

Lewis Structure and Molecular Geometry

Lewis structure and molecular geometry of molecules are listed below

External Reference:

Information on hydrogen iodide(HI) molecule

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