Geometry, Hybridization, and Polarity of ICl5 Lewis Structure

Iodine pentachloride (ICl5) has a molecular mass of 304.40 grammes per mol. It’s a helpful substance for conducting research in scientific laboratories. Unlike IF5, however, it is not a widely used chemical due to its instability.

The higher size of the chlorine atom in contrast to fluorine causes this instability. This is why, anytime ICl5 is needed in a reaction, it is freshly produced in labs.

ICl5 is an interhalogen molecule in which the iodine atom is bigger and more electropositive than the chlorine atom and also serves as the centre atom.

In layman’s terms, an interhalogen chemical is one in which two separate halogens react with one another. These are of the type ABx, with x being one of the following values: 1,3,5, and 7. Polyhalides are interhalogens with x values of 3,5 and 7.

As a result, ICl5 is classified as a polyhalide.

To learn more about this molecule, you must first understand its bonding properties as well as its polarity.

Understanding Lewis structure, VSEPR theory, and hybridization ideas is essential for this.

ICl5 Lewis Structure

The Lewis dot structure, based on the octet rule, gives a sense of bonding in the ions of a compound in terms of shared electron pairs, both bonding and non-bonding.

The following steps can be used to create the Lewis structure of ICl5:

Step 1: Determine the compound’s total valence electrons.

The electronic configuration of iodine and chlorine is necessary for the first step. Iodine has the electronic configuration [Kr]4d105s25p5.

The valence shell of iodine has seven electrons, as can be shown.

Second, chlorine has the electrical arrangement 1s22s22p63s23p5. Because chlorine and iodine belong to the same group, group 17, the valence shell of chlorine also has seven electrons.

The sum of the valence electrons of iodine and chlorine will equal the total number of valence electrons in the molecule.

It’s equal to ((71)+ (75), which equals 42 valence electrons in total.

Step 2: Make a rudimentary Lewis diagram

A skeletal diagram is made by simply putting the iodine and chlorine atoms in a way that five chlorine atoms circle the iodine atom for our own practicality.

The approximate Lewis diagram of ICl5 is depicted in the figure below.

Step 3: Finishing the Lewis structure by ensuring that every atom in the compound meets the octet rule criteria.

There are 8 valence electrons around each chlorine atom in the Lewis structure (a), and a total of 12 valence electrons surrounding the iodine atom (elements with a period >= 3 can expand their octet to more than 8 electrons).

Ten of the twelve electrons around iodine are involved in bonding (indicated by the single bond) in the Lewis structure (b), while one electron pair behaves as the lone pair.

When the formal charge on the compound is known, the lewis structure is complete.

It is necessary to know the formula for calculating the formal charge in order to do so. It can be calculated using the following formula:

V – B/2 – N is the formal charge.

Where;

V is the number of valence electrons.

B is the number of bonding electrons.

N is the number of non-bonding electrons in a system.

A compound’s formal charge is the sum of the formal charges of each bound atom in the complex. To put it another way, each atom’s formal charge must first be calculated individually.

I = 12 – 10/2 – 2 = 5 formal charge

Cl (a) formal charge = 8 – 2/2 – 6 = 1

Cl (b,c,d,e) = 1 has a similar formal charge.

Because chlorine has a higher electronegative value than iodine, it attracts the shared electron pair to itself, resulting in a negative charge.

As a result, each chlorine atom will have a formal charge of -1 rather than 1.

Now,

(formal charge of Iodine + formal charges of chlorine atoms) = (5 + (-1 + -1 + -1 + -1 + -1 + -1)

equals 0

As a result, ICl5 has a formal charge of 0, indicating that the chemical is neutral.

Although it is true that a Lewis structure provides a lot of information about a molecule, it is impossible to accurately predict a compound’s geometry and hybridization using this idea.

As a result, familiarity with the VSEPR theory and hybridization notion is required.

Molecular Geometry of ICl5

The VSEPR Theory, which is based on the assumption that a molecule takes the form that is the most stable and has the fewest electron-electron repulsions, can explain the geometry of a compound quite well.

The core atom of the compound is the least electronegative element when estimating the structure of a molecule.

According to the theory, a molecule’s shape is not the same as its geometry if the number of bond pairs electrons is not equal to the steric number, which is a numerical value ascribed to each type of geometry and can be computed using the formula:

(A + n + |x| – y)/2 = Steric number (S)

Where;

A represents the number of valence electrons in the centre atom.

n denotes the number of monoatomic side atoms.

x = molecule’s negative charge

y = molecule’s positive charge

The steric number allocated to each shape is listed in the table below.

According to the VSEPR theory, the steric number of ICl5 is computed as follows: A is the number of valence electrons in iodine, which is equal to 7.

n is the number of chlorine atoms, which is 5.

Because the ICl5 molecule is neutral (as previously established during the Lewis concept), x or y will be equal to 0.

Henceforth,

6 = S = (7 + 5) / 2

As a result, S = 6 and ICl5 has an octahedral geometry.

Subtracting the number of bond pairs from the steric number yields the number of lone pairs in a compound.

Compounds having octahedral geometry can adopt three different geometries depending on the number of bond pairs and lone pairs;

1) The shape is square planar when the number of bond pairs equals 4 and the number of lone pairs equals 2.

2) The shape is square pyramidal when the number of bond pairs equals 5 and the number of lone pairs equals 1.

3) When the number of bond pairs equals the steric number, which is 6, the shape is the same as the octahedral geometry of the compound.

Because ICl5 has one lone pair and five bond pairs, it must have a square pyramidal form.

Although the compound exhibits high-intensity lone pair – bond pair repulsions due to the existence of a lone pair.

The compound tends to stabilise itself by assuming a stable shape in order to decrease these repulsions. It achieves this by slightly distorting its structure from the square pyramidal shape.

As a result, ICl5 has a deformed square pyramidal shape.

IF5, a molecule with a comparable geometry and structure, is one example. You should also look into IF5’s Lewis structure.

Hybridization using ICl5

Hybridization is the process of creating hybrid orbitals by combining pure atomic orbitals that are identical in energy and structure.

It’s crucial to remember that the number of hybrid orbitals created must always equal the number of atomic orbitals combined.

While VSEPR provides a mechanism for almost accurately predicting the geometry of a molecule, the concept of hybridization aids in understanding the bonding that underpins these geometrical shapes.

Although the notion of orbital mixing is the same for all types of compounds, hybridization differs because each compound arranges itself in the most stable structure to minimise electron-electron repulsions.

Let us now examine the hybridization in ICl5.

I has the electrical configuration [Kr]4d105s25p5 in the ground state (note that we always consider the valence shells electrons only).

The ground state of the iodine atom is shown below.

Iodine must extend its octet by stimulating two of its valence electrons in the 5p orbital to the 5d orbital in order to join with five chlorine atoms.

In the excited state, iodine has the following electronic configuration:

Six sp3d2 hybrid orbitals are formed when one 5s, three 5p, and two 5d atomic orbitals are mixed together. As a result, ICl5 hybridization is sp3d2.

The electron pair that started in the 5s orbital and ended up in one hybrid orbital operates as the compound’s lone pair.

The five remaining hybrid orbitals form sigma bonds with the five chlorine atoms, as shown in the picture below.

Hybridization, on the other hand, can be determined directly from the steric number (explained in the VSEPR theory).

The hybridizations for various steric numbers are shown in the table below.

To determine the hybridization of a chemical, one can utilise one of the two procedures at their leisure.

Polarity of ICl5

If a compound has a dipole moment, it is considered to be polar in nature.

If there is an electronegativity difference between two linked atoms, they have a dipole moment. Iodine has an electronegativity of 2.66, while chlorine has a value of 3.16. As a result, there is an electronegativity difference between I and Cl.

It is possible to compute a compound’s dipole moment from its geometrical shape, which in the case of ICl5 is square pyramidal.

Chlorine is more electronegative than Iodine, indicating that the dipole moment is pointing in the other direction.

A negative sign is produced by a dipole moment with the same value in the opposite direction.

As seen in the diagram, the four chlorine atoms (b.c.d.e) situated in the same square plane in the geometry cancel out each other’s dipole moment.

In the diagram, the arrows represent the direction of the dipole moment.

There is one chlorine atom (a) that does not lie in the square plane. The dipole moment of ICl5 is caused by the link between this chlorine atom (a) and iodine. Because ICl5 has a polar character, it is polar in nature.

Conclusion

To summarise, in ICl5, I and Cl are joined by five sigma bonds and the molecule has a single lone pair. ICl5 is a neutrally charged chemical with a distorted square pyramidal form and an octahedral geometry.

It has a steric number of 6 and is sp3d2 hybridised. The compound has a non-zero dipole moment and is hence polar in nature.

Read more: Does Propane Have a Shelf Life?

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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