Hybridization, Polarity, and the MO Diagram in the I3 Lewis Structure

The triiodide ion, often known as I3-, is a polyatomic molecule with a net negative charge of -1. It is a charged molecule.

I2 + I- —-> I3- (repeatedly)

Ion formation occurs in this “exergonic equilibrium,” where a positive flow of energy is generated from the system to its environment.

This ion’s non-reactivity with starch results in a distinctive blue-black hue that is frequently utilised for identifying purposes.

Many salts contain it in significant amounts.

Now that we have a better understanding of the interior chemistry and its bonding nature, we can better understand its properties.

Lewis’s I3 Structure

Anyone who wants to learn more about a molecule’s Lewis structure must do so. Why is it so important to have an understanding of this subject?

Chemical bonding within a molecule can be represented diagrammatically using something called a Lewis Structure.

The octet rule of valence electrons provides the basis for the Lewis electron dot formations.

Understanding a molecule’s properties and hybridization is critical since it aids in the creation of a chemical structure.

Let’s go through the following approach step by step to figure out what the Lewis structure of I3 is.

The First Step

We must first determine the total amount of valence electrons.

The ‘+’ or positive sign and the ‘-‘ or negative sign must be dealt with in this process.

When electrons are lost, the molecule’s negative charge decreases. When electrons are gained, the molecule’s charge increases.

The next step is to

Following the measurement of valence electrons, we must determine which atom is the core.

The atom with the highest valence, or least electronegative, must be checked. While the other elements have a smaller number of bonding sites, this centre atom has more of them.

The Third Step

Sketching the molecule’s skeletal diagram is the third stage. In this way, single bonds can be seen in their actual locations.

Step Four

Many people have heard of the ‘octet rule’. It’s a simple rule that stipulates that each atom in a molecule must have eight valence electrons (electron count in the outermost shell).

As a result, it is our responsibility to complete the octet. If possible, start with the more electropositive ones.

This is the fifth and last step.

Once the octets of atoms around other atoms have been finished, we must check to see whether any more bonds are required.

This results in the formation of double bonds or, if necessary, triple bonds.

The sixth and last step is here.

The formal charge notion is the final step in our study of bond creation.

Our Lewis structure is complete if we can verify that the formal charges of all the atoms within a given molecule remain at their lowest possible value.

This formula is all that is needed to calculate the formal charge.

With regards to I3 (Triiodide ion)

I3 is composed of three iodine atoms. The outermost electron shell of this atom has seven electrons, giving it an atomic number of 53.

The I3 molecule has an overall negative charge because of an extra electron.

Emitters: 3 * 7 + 1

A total of twenty-one plus one

=22

iodine is used as the centre atom since it is the only iodine atom in the molecule.

The skeletal structure of I3 with a negative charge has been depicted in the diagram above.

The octet rule is being satisfied in this stage.

According to the octet rule, the highest multiple of 8 under the total number of valence electrons (step 1) must be taken into account.. The value 16 is used here. In this case, we have 22-16=6.

Iodine, which is in the middle of two other atoms, has three pairs of lone electrons with itself.

Three single bonds are formed with the two additional IODs, revealing that the central iodine has two bond pairs and three single bonds.

The negative charge should be placed outside the circuit, as shown in the diagram.

Now the lewis structure is complete.

With the help of I3,

I3 (Triiodide ion) hybridization is sp3d.

Two-dimensional drawings of chemical compounds on paper are the only way to represent their structure.

Understanding chemical reactions and physical qualities can be improved by visualising them in a 3-D platform.

The hybridization notion is the basis for this representation of the atoms’ orientation in any simple or complicated chemical.

Checking the hybridization number is as simple as using the following formula:

H=0.5(V+M-C+A)

Charges of cations and anions are represented by C, whereas valence electrons are represented by A, and monovalent atoms are represented by M. The final hybridization value is represented by H.

[I3]- has two monovalent atoms, M=2, seven valence electrons, V=7, A=1, and C = 0 in this case.

In this case, H = 5.

Since the sp hybridization value is 2, we can conclude that the type of sp3d is sp3d for H=5.

I3’s Molecular Geometry

We can plainly see that the core iodine has three lone pairs in the Lewis Structure of the ion I3-.

Due to repulsion, the lone pairs are placed equatorially.

Thus, the other two iodine atoms are axially positioned and the bond angle is 1800.

The donor is the core atom, which has a negative charge, according to the formal charge theory.

As a result, the I3- ion has a straight and symmetrical form.

Shape of triiodide ions

Polarity

When it comes to chemical bonding, the molecule I3- is both fascinating and challenging.

The electrical geometry is trigonal bipyramidal, despite the fact that the molecular geometry is linear, as previously stated.

The centre atom of triiodide ions has the ‘-‘ charge because of the ionic structure.

Let’s begin by defining polarity.

When we talk about polarity in chemistry, we imply something that has a non-zero dipole moment.

The electronegativity difference within a molecule causes the dipole moment, which is a measure of polarity.

The separation of positive and negative charges is usually the cause, and the distance between them is what determines how often it occurs.

If we look at I3-, we see that it has a negative charge and that its molecular geometry is symmetrical, meaning that it has no dipole moment. Instead, as previously discovered, the molecule’s overall charge is negative.

If we look at the solubility factor, we may conclude that the triiodide ion, which is polar, is soluble in water ( I3- is soluble in water).

Diagram of the I3 MO (Molecular Orbital)

An excellent illustration of 3-center-4-electron chemical bonding is the triiodide ion.

In light of this, what does this symbolise?

MO or Molecular Orbital theory is laid out in front of you first.

Molecular orbitals, as most of us know, are a tool for analysing bonding in terms of its wave-like nature as well as its position.

The maximum number of electrons in a MO is two. An MO (atomic orbital) is formed when two AOs of the same energy overlap.

The number of valence electrons in I3- is 22. The valence orbitals of an iodine atom are arranged in the following manner:

5s, 5px, 5py, and 5pz.

Because they have low energies and are weak, 5s and 5px, 5py do not form bonds. In spite of the fact that 5pz remains, it possesses both bonding and a non-bonding lone pair on each iodine, each of which has an official charge of -0.5.

As previously computed, the ten non-bonding pairs and one single bond provide a total of 22 valence electrons.

Conclusion

In terms of chemical bonding, the triiodide ion stands out as a very useful topic.

In this case, we were given a taste of the hybridization (Lewis Structure), polarity, shape, and orbital notion.

You’ll have a better grasp of the physicochemical properties of molecules containing I3- after reading this.

Read more: What is the polarity of OCl2?

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.

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