Iodine trifluoride (IF3) Lewis dot structure, molecular geometry or shape, electron geometry, bond angle, formal charge, hybridization
IF3 is the chemical formula for an unstable interhalogen chemical compound i.e., trifluoro iodine or iodine trifluoride. It appears as a yellow solid that readily decomposes at temperatures above -28°C. It is because of the vulnerability of IF3 that it is relatively unfamiliar and un-talked about in the science world.
Keeping all that in mind, this article introduces you to the basics of IF3 including how to draw its Lewis structure, molecular geometry, or shape of IF3, electron geometry, bond angle, hybridization, polar or non-polar nature, and/or any formal charges present on IF3 atoms.
So, without any further delay, let’s start reading!
|Name of Molecule||Iodine trifluoride|
|Molecular geometry of IF3||T-shaped|
|Electron geometry of IF3||Trigonal bipyramidal|
|Bond angle (F-I-F)||88.5°|
|Total Valence electron in IF3||28|
|Overall Formal charge in IF3||Zero|
How to draw lewis structure of IF3?
The Lewis structure of IF3 consists of an iodine (I) atom at the center. It is bonded to 3 atoms of fluorine (F), one on each side of the molecule. There are a total of 5 electron pairs around the central Iodine atom in the IF3 lewis dot structure. Out of these 5 electron pairs, there are 3 bond pairs and 2 lone pairs.
Drawing the Lewis structure of IF3 is quite an easy task. Let’s see how we can do so by following the simple steps given below.
Steps for drawing the Lewis dot structure of IF3
1. Count the total valence electrons in IF3
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 IF3 is to count the total valence electrons present in the concerned elemental atoms.
The valence electrons present in the atom of an element can be determined by identifying that element in the Periodic Table.
The two different elements present in IF3 are iodine (I) and fluorine (F). Both of these elements are halogens present in Group VII A of the Periodic Table. So, both I and F atoms contain 7 valence electrons.
- Total number of valence electrons in iodine = 7
- Total number of valence electrons in fluorine = 7
∴ The IF3 molecule consists of 1 iodine atom and 3 atoms of fluorine. Thus, the total valence electrons available for drawing the Lewis structure of IF3 = 7+7(3) = 28 valence electrons.
2. Find the least electronegative atom and place it at the center
The least electronegative atom is most likely to share its electrons with other atoms. 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 IF3 Lewis structure. All the 3 F atoms are placed in its surroundings.
3. Connect outer atoms with the central atom
The iodine atom is the central atom in the IF3 Lewis structure while the fluorine atoms are the outer atoms. So, in this step, we connect the three F atoms with the central I atom using single straight lines.
Each straight line represents a single covalent bond containing a bonded pair of electrons i.e., 2 electrons. There are a total of 3 single bonds in the IF3 molecule, so the total valence electrons used so far are 3(2) = 6 valence electrons.
- Total valence electrons available – electrons used till step 3 = 28-6 = 22 valence electrons.
- This means we still have 22 valence electrons to be accommodated in the Lewis structure of IF3.
4. Complete the octet of outer atoms
The F atoms are the outer atoms in the Lewis structure of IF3. Each F atom requires a total of 8 valence electrons in order to achieve a stable octet electronic configuration.
There are three I-F bonds in the IF3 Lewis diagram drawn till now. This means each F atom already has 2 valence electrons, so it needs 6 more electrons to acquire a complete octet.
Thus, 6 valence electrons are placed as 3 lone pairs around each F atom in the Lewis structure, as shown below.
5. Place the remaining electrons as lone pairs on the central atom
- Total valence electrons used till step 4 = 3 single bonds + 3 (electrons placed around each F atom, shown as dots) = 3(2) + 3(6) = 24 valence electrons.
- Total valence electrons available – electrons used till step 4 = 28 – 24 = 4 valence electrons.
Thus, these 4 valence electrons are placed as 2 lone pairs around the central I atom in the IF3 Lewis structure.
In this way, the central atom iodine in IF3 Lewis structure has a total of 10 valence electrons. It falls under the expanded octet rule. Iodine (I) having d-subshells available can accommodate more than 8 valence electrons during chemical bonding.
As a final step, we just need to check the stability of the above Lewis structure and we can do so by using the formal charge concept.
6. Check the stability of IF3 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.
- Formal charge = [ valence electrons – nonbonding electrons- ½ (bonding electrons)]
Now let us count the formal charges present on IF3 atoms using this formula and the Lewis structure obtained in step 5.
For fluorine atom
- Valence electrons of fluorine = 7
- Bonding electrons = 1 single bond = 2 electrons
- Non-bonding electrons = 3 lone pairs = 6 electrons
- Formal charge = 7 – 6 – 2/2 = 7 – 6 – 1 = 7 – 7 = 0
For iodine atom
- Valence electrons of iodine = 7
- Bonding electrons = 3 single bonds = 3(2) = 6 electrons
- Non-bonding electrons = 2 lone pairs = 2(2) = 4 electrons
- Formal charge = 7 – 4 – 6/2 = 7 – 4 – 3 = 7 – 7 = 0
Zero formal charges present on all the bonded atoms in the IF3 molecule mark the stability of its Lewis structure.
Also check –
What are the electron and molecular geometry of IF3?
The ideal electron geometry of the iodine trifluoride (IF3) molecule is trigonal bipyramidal. But it is due to the 2 lone pairs of electrons present on the central iodine atom that the molecule adopts a different molecular geometry or shape from its electron geometry i.e., T-shaped.
Molecular geometry of IF3
IF3 is a T-shaped molecule. Two lone pairs situated on the central iodine atom in the IF3 molecule set up a lone pair-lone pair and lone pair-bond pair repulsive effect which distorts the symmetry of the molecule.
Consequently, the molecule (IF3) adopts an asymmetric T-shape. The 2 lone pairs occupy the equatorial positions while the 3 F-atoms occupy the two axial positions as well as one equatorial position in order to minimize the repulsive effect.
Electron geometry of IF3
There are a total of 5 electron density regions around the central I atom in the IF3 molecule. According to the VSEPR concept, its ideal electron geometry is trigonal bipyramidal.
The electron geometry depends on the total number of electron density regions around the central atom i.e., total electron pairs in this case. It does not take into account whether an electron pair is a bond pair or a lone pair unlike that important for molecular geometry.
A quick and more straightforward way of finding the electron and molecular geometry or shape of a molecule such as IF3 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 the IF3 molecule
- A in the AXN formula represents the central atom. In IF3, iodine (I) acts as the central atom so A= I.
- X denotes the atoms bonded to the central atom. 3 F atoms are bonded to the central I atom in the IF3 molecule thus X=3.
- N stands for the lone pairs present on the central atom. As 2 lone pairs of electrons are present on central iodine in IF3 thus N=2.
So, the AXN generic formula for IF3 molecule is AX3N2.
Now have a quick look at the VSEPR chart given below to identify where you find AX3N2.
The VSEPR chart confirms that molecules with an AX3N2 generic formula have a T-shape while their ideal electron geometry is trigonal bipyramidal, as we already noted down for the IF3 molecule.
Hybridization of IF3
The central iodine (I) atom is sp3d hybridized in the IF3 molecule.
The electronic configuration of iodine is [Kr] 4d10 5s2 5p5.
During chemical bonding, one 5p electron of iodine gets excited and shifts to the empty 5d atomic orbital. The 5s, three 5p, and one 5d atomic orbitals of iodine hybridize to yield five sp3d hybrid orbitals.
Two of these five hybrid orbitals contain paired electrons. These paired electrons are situated as lone pairs on the central I atom in IF3. The other three sp3d hybrid orbitals contain a single electron each which they use for sigma bond formation with the p orbitals of fluorine atoms in IF3, 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 IF3 is 5 so it has sp3d hybridization.
The IF3 bond angle
The ideal F-I-F bond angles are 180° and 90° but it is due to the electronic repulsions and distortion present in the molecule that the F-I-F bond angle decreases. Thus, the F-I-F bond angle in IF3 is approx. 88.5°.
There are two different bond lengths in iodine difluoride i.e., a short I-F bond length of 191 pm and a longer I-F bond length of 202 pm. Refer to the figure below.
Also check:- How to find bond angle?
Is IF3 polar or nonpolar?
Fluorine (F) is a highly electronegative element. 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 in the IF3 molecule. Thus, each I-F bond is polar.
The asymmetric T-shape of the IF3 molecule maintains the polarity effect. The dipole moments of I-F bonds do not get canceled. Rather, their effect adds up. In conclusion, the IF3 molecule is polar with net dipole moment μ=0.632 D.
Read in detail–
How many lone pairs and bond pairs are in the IF3 lewis structure?
The iodine trifluoride (IF3) molecule has a total of 11 lone pairs. 3 lone pairs are present on each fluorine (F) atom while 2 lone pairs are situated on the central iodine (I) atom, as displayed in the Lewis structure of IF3 using dots (see the figure below).
Why is the shape of IF3 different from its electron pair geometry?
The ideal electron pair geometry of a molecule depends on the total electron pairs around the central atom, which are 5 in the case of IF3, so its electron pair geometry is trigonal bipyramidal.
The shape of IF3 is different from its electron geometry because of 2 lone pairs present on the central I atom.
The lone pair-lone pair and lone pair-bond pair repulsive effect make the molecule adopt a distorted T-shape.
Why is IF3 not stable?
One reason for the instability of IF3 is its distorted shape and molecular geometry. The lone pair-lone pair and lone pair-bond pair repulsions present account for steric crowding in the molecule and thus its instability.
All the chemical compounds want to achieve greater stability. In the same way, IF3 undergoes decomposition to transform into more stable products.
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- The total valence electrons available for drawing the IF3 Lewis structure are 28.
- The IF3 molecule has a T-shape and molecular geometry.
- The ideal electron geometry of IF3 is trigonal bipyramidal.
- It is due to the presence of 2 lone pairs on the central I atom in the IF3 molecule that it adopts a distorted shape, different from its ideal electron pair geometry.
- The central I atom is sp3d hybridized in IF3.
- The F-I-F bond angle is 88.5° in the IF3 molecule while the I-F bond lengths are 191 pm and 202 pm respectively.
- IF3 is a polar molecule in nature (net μ=0.632 D).
- Zero formal charges are present on both I and F atoms in the IF3 molecule which makes sure the stability of its Lewis structure.
About the author
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/