Iodine pentafluoride (IF5) Lewis dot structure, molecular
geometry or shape, electron geometry, bond angle, formal
charge, hybridization
IF5 is the chemical formula for iodine pentafluoride. It exists as a colorless liquid. It is commonly used as a solvent and a fluorinating agent in the chemistry laboratory. In this article, we have summarized some important chemical properties of IF5.
You will learn through this article: how to draw the Lewis structure of IF5, what is its molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, etc. So, without any further delay, let’s begin reading!
So, without any further delay, let’s start reading!
The Lewis structure of iodine pentafluoride (IF5) consists of iodine (I) atom at the center. It is bonded to five atoms of fluorine (F) at the sides. There are a total of 6 electron pairs around the central iodine atom in the IF5 lewis dot structure. Out of these 6 electron pairs, there are 5 bond pairs and 1 lone pair of electrons.
You can easily draw the Lewis dot structure of IF5 using the simple step-by-step guide given below.
Steps for drawing the Lewis dot structure of IF5
1. Count the total valence electrons in IF5
The Lewis dot structure of a molecule is referred to as a simplified representation of all the valence electrons present in it. Therefore, the very first step while drawing the Lewis structure of IF5 is to count the total valence electrons present in the concerned elemental atoms.
There are 2 different elements present in IF5 i.e., iodine (I) and fluorine (F). When you search through the Periodic Table of elements, you will find that both I and F are halogens located in Group VII A of the Periodic Table.
Thus, both iodine and fluorine have a total of 7 valence electrons in each atom.
∴ The IF5 molecule consists of 1 iodine atom and 5 atoms of fluorine. Thus, the total valence electrons available for drawing the Lewis structure of IF5 = 7+5(7) = 42 valence electrons.
2. Find the least electronegative atom and place it at the center
The leastelectronegative atom is most likely to share its electrons with the atoms in its surroundings. Therefore, such an atom is placed at the center of the Lewis structure of a molecule. Fluorine (F) is the most electronegative atom in the Periodic Table.
Electronegativity decreases down the group. Thus, Iodine (I) is less electronegative than fluorine, and consequently, it is placed at the center of the IF5 Lewis structure. All the 5 F atoms are placed in its surroundings.
3. Connect outer atoms with the central atom
In this step, we join the outer F-atoms with the central I atom using single straight lines, as shown in the diagram below.
Each straight line represents a single covalent bond i.e., a bond pair containing 2 electrons. There are a total of 5 single bonds in the above diagram, so the total valence electrons used out of the 42 available are 5(2) = 10 valence electrons.
Total valence electrons available – electrons used till step 3 = 42-10 = 32 valence electrons.
This means 32 valence electrons are still available to be accommodated in the Lewis dot structure of IF5. Let’s see how we can do so.
4. Complete the octet of outer atoms
The fluorine (F) atoms are the outer atoms in the Lewis structure of IF5. Each F atom is bonded to the central I atom using single bonds. Thus, each F atom already has 2 valence electrons, and it requires 6 more valence electrons to achieve a stable octet electronic configuration.
Hence, 6 valence electrons are placed as 3 lone pairs around each F atom in the Lewis structure of IF5 as shown below.
5. Place the remaining electrons as lone pairs on the central atom
Total valence electrons used till step 4 = 5 single bonds + 5 (electrons placed around each F-atom, shown as dots) = 5(2) + 5(6) = 40 valence electrons.
Total valence electrons available – electrons used till step 4 = 42-40 = 2 valence electrons.
Thus, these 2 electrons are placed as a lone pair on the central iodine atom in the Lewis structure of IF5.
The central iodine (I) atom has a total of 12 valence electrons in the IF5 molecule. This means it has an expanded octet. It is due to the availability of the 4d subshell in the iodine atom that it can accommodate more the 8 electrons in its valence shell during chemical bonding. After completely filling the 5p atomic orbital, electrons start filling d orbitals.
As a final step, we need to check the stability of this Lewis structure. Let’s do that using the formal charge concept.
6. Check the stability of IF5 Lewis structure using the formal charge concept
The less the formal charge on the atoms of a molecule, the better the stability of its Lewis structure.
The formal charge can be calculated using the formula given below.
What are the electron and molecular geometry of IF5?
The ideal electron geometry of the iodine pentafluoride (IF5) molecule is octahedral. However, the molecular geometry or shape of IF5 is different from its electron geometry due to the presence of a lone pair on the central iodine atom. Thus, it occupies a square pyramidal shape.
Molecular geometry of IF5
The molecular geometry or shape of the IF5 molecule is square pyramidal. A lone pair of electrons present on the central iodine atom in IF5 leads to lone pair-bond pair repulsions.
Lone pair-bond pair repulsions are stronger than bond pair-bond pair repulsions which results in distorting the overall shape and geometry of the molecule.
Four F-atoms lie at the four corners of the square base, pushed away from the center of the molecule. Conversely, the fifth F atom and a lone pair occupy equatorial positions. Thus, a pyramid is formed at the top of the square base, as shown in the figure below.
Electron geometry of IF5
According to the VSEPR concept, the IF5 molecule has an octahedral electron geometry as there are a total of 6 electron density regions around the central I atom in the IF5 molecule.
The ideal electron geometry is based on the total electron density regions around the central atom in a molecule. Each electron pair whether it’s a bond pair or a lone pair is considered one region of electron density.
A quick and more straightforward way of finding the electron and molecular geometry or shape of a molecule such as IF5 is using the AXN method.
AXN is a simple formula to represent the number of atoms bonded to the central atom in a molecule and the number of lone pairs present on it.
It is used to predict the geometry or shape of a molecule using the VSEPR concept.
AXN notation for IF5 molecule
A in the AXN formula represents the atom present at the center of a molecule. In IF5, iodine (I) is present at the center so A=I.
X denotes the number of atoms directly bonded to the central atom. In IF5, 5 fluorine (F) atoms are bonded to central I thus X=5.
N stands for the number of lone pairs present on the central atom. In IF5, 1 lone pair is present on the central I hence N=1.
This shows that the AXN generic formula for the IF5 molecule is AX5N1.
Now use the VSEPR chart shown below to identify the electron and molecular geometry, or shape assigned next to AX5N1.
The VSEPR chart indicates that molecules with AX5N1 generic formula have a square pyramidal shape or molecular geometry while their electron geometry is octahedral, as we already noted down for the IF5 molecule.
Hybridization of IF5
The central iodine (I) atom is sp3d2 hybridized in the IF5 molecule.
The electronic configuration of iodine is [Kr] 4d10 5s2 5p5.
During chemical bonding, two 5p electrons of iodine get excited and shift to two separate empty 5d atomic orbitals. The 5s, three 5p, and two 5d atomic orbitals of iodine hybridize to yield six sp3d2 hybrid orbitals.
One of these six hybrid orbitals contains paired electrons. These paired electrons are situated as a lone pair on the central I atom in IF5. The remaining five sp3d2 hybrid orbitals contain a single electron each which they use for sigma bond formation with the p orbitals of fluorine atoms in IF5, one on each side of the molecule, as shown below.
A short trick for finding the hybridization present in a molecule is to memorize the table given below. You can calculate the steric number of a molecule and use that against this table to determine its hybridization.
The steric number of central I in IF5 is 6 so it has sp3d2 hybridization.
Steric number
Hybridization
2
sp
3
sp2
4
sp3
5
sp3d
6
sp3d2
The IF5 bond angle
The F-I-F bond angle in the IF5 molecule is 81.9°. Electronic repulsions present in the molecule distort the geometry or shape of the molecule and decrease the bond angle.
There are two main I-F bond lengths present in IF5 i.e., 184.4 pm at the axial position and a slightly longer 186.9 pm at the equatorial position, as shown below.
Iodine pentafluoride (IF5) is a polar molecule. Fluorine (F) is highly electronegative in nature. An electronegativity difference of 1.32 units exists between the bonded fluorine (E.N = 3.98) and iodine (E.N =2.66) atoms in each I-F bond. Thus, all the 5 I-F bonds present in the IF5 molecule are polar and possess a specific dipole moment value (symbol μ).
The dipole moments do not get canceled in the molecule overall due to its asymmetric shape. The electron cloud stays non-uniformly distributed throughout thus IF5 is polar (net μ=4.81 Debye).
How many lone pairs and bond pairs are in the IF5 lewis structure?
There are a total of 42 valence electrons i.e., 42/2 = 21 electron pairs in the Lewis structure of IF5.
Out of these 21 electron pairs, there are 5 bond pairs and 16 lone pairs.
Out of the 16 lone pairs, 1 lone pair of electrons is present at the central iodine atom while 3 lone pairs are present on each of the 5 F-atoms.
Is the electron and molecular geometry or shape of IF5 the same or different?
The IF5 molecule has a different electron and molecular geometry or shape i.e., an octahedral electron geometry while a square pyramidal molecular geometry/shape.
A lone pair of electrons situated on the central iodine atom accounts for this difference in shape and electron geometry of IF5.
Why are bonds in IF5 not equal in length?
The unequal I-F bond lengths in IF5 are accredited to the distortion present in its geometry. Lone pair-bond pair repulsions push F atoms away from the center.
Thus, the fluorine atoms occupy two different positions in the molecule i.e., axial and equatorial consequently two different bond lengths i.e., 184.4 pm and 186.9 pm are present in the molecule respectively.
Vishal Goyal is the founder of Topblogtenz, a comprehensive resource for students seeking guidance and support in their chemistry studies. He holds a degree in B.Tech (Chemical Engineering) and has four years of experience as a chemistry tutor. The team at Topblogtenz includes experts like experienced researchers, professors, and educators, with the goal of making complex subjects like chemistry accessible and understandable for all. A passion for sharing knowledge and a love for chemistry and science drives the team behind the website. Let's connect through LinkedIn: https://www.linkedin.com/in/vishal-goyal-2926a122b/
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