IF5 Molecular Geometry - Science Education and Tutorials

Pentafluoro Iodine or Iodine pentafluoride(IF5) has the composition of one Iodine and five fluorine atoms. What is the molecular geometry of Iodine pentafluoride?. Drawing and predicting the IF5 molecular geometry is very easy by following the given method. Here in this post, we described step by step to construct IF5 molecular geometry. Iodine and fluorine come from the 17th family group in the periodic table. Iodine or fluorine has seven valence electrons.

Table of Contents

Key Points To Consider When drawing The IF5 Molecular Geometry

A three-step approach for drawing the IF5 molecular can be used. The first step is to sketch the molecular geometry of the IF5 molecule, to calculate the lone pairs of the electron in the central Iodineatom; the second step is to calculate the IF5 hybridization, and the third step is to give perfect notation for the IF5 molecular geometry.

The IF5 molecular geometry is a diagram that illustrates the number of valence electrons and bond electron pairs in the IF5 molecule in a specific geometric manner. The geometry of the IF5 molecule ion 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 IF5 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 five I-F bonds (dipole moment properties of the IF5 molecular geometry). Five Iodine-fluorine bonds in the Iodine pentafluoride(IF5), for example, are polarised toward the more electronegative value fluorine atoms, and because all five (I-F) bonds have the same size and polarity, their sum is nonzero due to the IF5 molecule’s bond dipole moment due to pulling the electron cloud to the downside in the square pyramidal geometry, and the IF5 molecule is classified as a polar molecule.

The molecule of Iodine pentafluoride(with square pyramidal shape IF5 molecular geometry) is tilted at 82 degrees bond angle of I-F. It has a difference in electronegativity values between Iodine and fluorine atoms, with fluorine’s pull the electron cloud being greater than Iodine’s. But bond polarity of I-F is not canceled to each other in the square pyramidal geometry. As a result, it has a permanent dipole moment in its molecular structure. The IF5 molecule has a dipole moment due to an unequal charge distribution of negative and positive charges.

Overview: IF5 electron and molecular geometry

According to the VSEPR theory, the IF5 molecule ion possesses square pyramidal molecular geometry. Because the center atom, Iodine, has five I-F bonds with the five fluorine atoms surrounding it. The I-F bond angle is 82 degrees in the square pyramidal IF5 molecular geometry. The IF5 molecule has a square planar geometry shape because it contains four fluorine atoms in the plan and one fluorine lies out of the plan.

There are five I-F bonds at the IF5 molecular geometry. After linking the five fluorine atoms and one lone pair of electrons in the square pyramidal form, it maintains the square bipyramidal-shaped structure. In the IF5 molecular geometry, the I-F bonds have stayed in the five terminals and one lone pair of electrons on the Iodine atom of the square pyramidal molecule.

The center Iodine atom of IF5 has one lone pair of electrons, resulting in square pyramidal IF5 electron geometry. However, the molecular geometry of IF5 looks square pyramidal-shaped and one lone pair of electrons on the Iodine of the IF5 geometry. It’s the IF5molecule’s asymmetrical geometry. As a result, the IF5 molecule is polar.

How to find IF5 hybridization and molecular geometry

Calculating lone pairs of electrons on Iodine in the IF5 geometry:

Determine the number of lone pairs of electrons in the core atom of the IF5 Lewis structure. Because the lone pairs of electrons on the Iodine atom are mostly responsible for the IF5 molecule geometry distortion, we need to calculate out how many there are on the central Iodine atom of the IF5 Lewis structure.

Use the formula below to find the lone pair on the Iodine atom of the IF5 molecule.
L.P(I) = V.E(I) – N.A(I-F)/2

Lone pair on the central Iodine atom in IF5= L.P(I)
The core central Iodine atom’s valence electron in IF5 = V.E(I)
Number of I-F bonds = N.A (I-F)
calculation for Iodine atom lone pair in IF5 molecule.

For instance of IF5, the central atom, Iodine, has seven electrons in its outermost valence shell, five I-F bond connections. This gives a total of five connections.

As a result of this, L.P(I) = (7 –5)/2=1

The lone pairs of electrons in the Iodine atom of the IF5 molecule are one.

Calculating lone pairs of electrons on fluorine in the IF5 geometry:

Finding lone pairs of electrons for the terminal atom is not similar to the central Iodine atom. We use the following formula as given below

Use the formula below to find the lone pair on the fluorine atom of the IF5 molecule.
L.P(F) = V.E(F) – N.A(I-F)

Lone pair on the terminal fluorine atom in IF5 = L.P(F)
Terminal fluorine atom’s valence electron in IF5 = V.E(F)
Number of I-F bonds = N.A ( I-F)
calculation for fluorine atom lone pair in IF5 molecule.

For instance of IF5, five terminal atoms, fluorine, have seven electrons in its outermost valence shell, one I-F bond connection. This gives a total of five IF bond connections. But we are considering only one connection for the calculation.

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

The lone pairs of electrons in the fluorine atom of the IF5 molecule are six. Five fluorine atoms are connected with the central Iodine atom.

In the IF5 electron geometry structure, the lone pair on the central Iodine atom is one, lone pairs of electrons in the fluorine atom have six. Five fluorine atoms have 30 lone pairs of electrons.

It means there are one lone pair of electrons in the core Iodine atom. One lone pair of electrons on the central Iodine atom is responsible for the square pyramidal nature of IF5 molecular geometry. But in the structure fluorine atoms are polarised sidewise in their geometry.

The one lone pair of electrons are placed at another side of the IF5 geometry. Because the Iodine atom is a lower electronegative value as compared with other atoms in the IF5 molecule. Five fluorine atoms are polarized towards the sidewise in the IF5 structure.

But in reality, the IF5 has one lone pair of electrons in its structure. This makes the IF5 more irregular structure of the molecule. Because there is electric repulsion between bond pairs and lone pairs. But some sort of interaction is there between fluorine lone pairs and bond pairs. But it is negligible.

Calculate the number of molecular hybridizations of the IF5 molecule

What is IF5 hybridization? This is a very fundamental question in the field of molecular chemistry. All the molecules are made of atoms. In chemistry, atoms are the fundamental particles. There are four different types of orbitals in chemistry. They are named 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

IF5 molecule is made of one Iodine, five fluorine atoms. The fluorine and Iodine atoms have s and p orbitals. Fluorine comes as the first element from the halogen family in the periodic table. The Iodine atom also belongs to the same family group. But it falls as the fourth element in the periodic table.

When these atoms combine to form the IF5 molecule, its atomic orbitals are mixed and form unique molecular orbitals due to hybridization.

How do you find the IF5 molecule’s hybridization? We must now determine the molecular hybridization number of IF5.

The formula of IF5 molecular hybridization is as follows:
No. Hyb of IF5= N.A(I-F bonds) + L.P(Br)
No. Hy of IF5 = the number of hybridizations of IF5
Number of I-F bonds = N.A (I-F bonds)
Lone pair on the central Iodine atom = L.P(I)
Calculation for hybridization number for IF5 molecule

In the IF5 molecule, Iodine is a core central atom with five fluorine atoms connected to it. It has one lone pair of electrons on Iodine. The number of IF5 hybridizations (No. Hyb of IF5) can then be estimated using the formula below.

No. Hyb of IF5= 5+1=6

The IF5 molecule ion hybridization is six. The Iodine and fluorine atoms have s,p, and d orbitals. The sp3d2 hybridization of the IF5 molecule is formed when one S orbital, three p orbitals, and two d orbitals join together to form the IF5 molecular orbital.

Molecular Geometry Notation for IF5 Molecule :

Determine the form of IF5 molecular geometry using VSEPR theory. The AXN technique is commonly used when the VSEPR theory is used to calculate the shape of the IF5 molecule.

The AXN notation of IF5 molecule is as follows:
The central Iodine atom in the IF5 molecule is denoted by the letter A.
The bound pairs (five I-F bonds) of electrons to the core Iodine atom are represented by X.
The lone pairs of electrons on the central Iodine atom are denoted by the letter N.
Notation for IF5 molecular geometry

We know that Iodine is the core atom, with five electron pairs bound (five I-F) and one lone pair of electrons. The general molecular geometry formula for IF5 is AX5N1.

According to the VSEPR theory, if the IF5 molecule ion has an AX5N1 generic formula, the molecular geometry and electron geometry will both square pyramidal forms.

Name of Molecule	Iodine pentafluoride
Chemical molecular formula	IF5
Molecular geometry of IF5	Square pyramidal
Electron geometry of IF5	Square pyramidal
Hybridization of IF5	sp3d2
Bond angle (I-F)	82º degree
Total Valence electron for IF5	42
The formal charge of IF5 on Iodine	0

Summary:

In this post, we discussed the method to construct IF5 molecular geometry, the method to find the lone pairs of electrons in the central Iodine atom, IF5 hybridization, and IF5 molecular notation. Need to remember that, if you follow the above-said method, you can construct the IF5 molecular structure very easily.