Iodine trifluoride, or IF3, is a yellow solid classified as an interhalogen chemical. It is highly unstable and decomposes at temperatures higher than -28°C.
IF3 has a molar mass of 183.9 g/mol. Two ways can be used to make IF3:-
- F2 + I2 ——> IF3 in CCl3F at 45 °C.
- The fluorination reaction is utilised at low temperatures.
2IF3 + 3XeF2 ——> I2 + 3XeF2 ——> I2 + 3XeF2
Let’s get into the crucial components of IF3, such as Lewis structure, molecular geometry, hybridization, and so on.
Lewis Structure IF3
Let’s have a look at the Lewis structure of IF3 before we start learning how to draw it. Valence electrons are employed to draw the Lewis structures of any molecule, as you may know.
I is surrounded by two lone pairs of electrons in the Lewis structure of IF3 and makes three single bonds with each of the F atoms.
Let’s take a look at the stages involved in designing the Lewis structure:-
The first and most important step is to count the molecule’s total amount of valence electrons.
Step 2: Next, we must determine the molecule’s core atom.
Step 3: We’ll now begin drawing chemical linkages between the central atom and the atoms around it. Dots are used to depict these bonds.
Step 4: Now we fill in the remaining valence electrons to complete each atom’s octet, making the molecule stable overall.
Step 5. Depending on the molecule’s stability, transform lone pairs into double or triple bonds if possible.
As a final resort, we should always check each atom’s formal charge to ensure it is as low as feasible.
Each atom in the optimal Lewis structure for each molecule should have a formal charge of 0. The formal charge can be determined using the following formula:-
Drawing the Lewis Structure of IF3 in Steps
Step 1: To begin, count the total amount of valence electrons in the molecule IF3.
I and F are both in group 7, hence they have seven valence electrons apiece. However, because F has three atoms, it will have 733=21 valence electrons. As a result, the total number of valence electrons in the molecule IF3 is 21+7=28.
Step 2: The core atom is the one with the highest valence and the greatest number of bonding sites. As a result, I is the central atom in IF3.
Step 3: We’ll now arrange the electrons into a chemical bond, which is represented by two dots. Six valence electrons are used up because there are three atoms in F.
Step 4: Now we complete the octet by arranging the valence electrons around each atom.
Following the steps outlined above, we can see that 24 valence electrons have been used. There are four valence electrons left. On the core atom I, these valence electrons will function as lone pairs.
I now have a total of more than eight valence electrons. Because atoms below period 5 can have an enlarged octet and so carry more valence electrons, it can retain more than 8 valence electrons. The octet rule does not apply in this case.
Hence I have two lone pairs and have the capacity to hold more than eight valence electrons.
Step 5: Now that we’ve got our Lewis structure, let’s make sure it’s the best Lewis structure for IF3. When we look at the formal charge of each atom in the molecule, we notice that each atom has the lowest possible value of 0.
As a result, we don’t need to convert the lone pairs into double/triple bonds because we already know the molecule’s optimal Lewis structure, IF3.
With each atom having a formal charge of 0, this Lewis structure of IF3 with two lone pairs is the most stable.
Hybridization using IF3
Hybridization is useful in identifying the optimum, stable molecular shape of a molecule and in learning more about the mechanism of bonding in a molecule.
To obtain increased stability, molecules generate a combination of hybrid orbitals, as the term suggests. Sp3d is the hybridization of IF3.
The following approaches can be used to find hybridization of any molecule:
Method 1: The Addition Method consists of the following steps:
The total number of bonds produced by the central atom and the number of lone pairs on the central atom can be added to determine hybridization of any molecule.
The following factors influence the value of Hybridization (H):
If H=2 indicates sp hybridization, H=3 means sp2 hybridization, H=4 means sp3 hybridization, H=5 means sp3d hybridization, and H=6 means sp3d2 hybridization, then H=6 means sp3d2 hybridization.
The Iodine atom is known to be the core atom of IF3. It is bound to three F atoms and has two lone pairs.
When we multiply the number of bonded sites by the number of lone pairs, we get 2+3 = 5, indicating that IF3 is Sp3d hybridised.
Technique 2: The formula method consists of the following steps:
The following is the formula for calculating the Hybridization of any molecule:-
H stands for hybridization on the central atom, V for valence electrons on the central atom, M for monovalent atoms bound to the central atom, Cas for charge on the cation or more electropositive atom, and A for charge on the anion or more electropositive atom.
I is the core atom in IF3. Thus, V =7. (valence electrons of I). Because F is a monovalent atom with three F atoms coupled to I, M = 3.
Because the molecule’s overall charge is neutral, both C and A will be zero.
As a result of the formula,
H=5 This indicates that IF3 has been hybridised with Sp3d.
As a result, any molecule’s hybridization can be determined using these two methods.
Molecular Geometry IF3
The repulsion between the lone pairs and the bond pairs determines the molecular structure of any substance. Let me explain the distinction between molecule shape and molecular/electron geometry before we go any further.
The lone pairs are ignored by molecular shape, but they are taken into consideration by molecular geometry when establishing the geometry. IF3 has a trigonal bipyramidal electron shape.
IF3, on the other hand, has a T-shaped molecular form. The bond angles between each atom are around 90 degrees.
The molecular shape is determined using the symbol AXE.
A stands for the number of central atoms.
Atoms that are bonded to the centre atom are designated by the letter X.
The number of lone pairs on the centre atom is equal to E. (non-bonding).
Let’s use the aforementioned notation to establish the molecular structure of IF3.
A=1 (the core atom is Iodine).
3 = X (3 F atoms).
2 = E (2 lone pairs).
We get the formula AX3E2 for IF3 by plugging in the values for each notation.
We can observe that IF3 forms a T-shaped molecule by looking at this notation in the VSEPR graphic below.
ClF3 is a molecule that is similar to IF3 in terms of structure and hybridization. See the article ClF3 Lewis Structure, Hybridization, and Geometry for more information.
Polarity of IF3
The molecule IF3 is polar.
Determining the polarity of the molecule will be a piece of cake now that we know how to sketch the Lewis structure and the molecular shape of IF3.
The Lewis structure, or molecular shape, of IF3 reveals that it is a T-shaped molecule. The difference in electronegativity between I (2.5) and F (4.0) is now more than 0.5, causing polarity.
It is obvious from the structure of IF3 that the dipole moments do not cancel each other out because it is not symmetrical. Because of all of these variables, IF3 is a polar molecule.
The Lewis structure, hybridization, molecular shape, and polarity of IF3 were all discussed in this article. You should now be able to ace any IF3 questions that come your way. Please feel free to contact me if you have any questions about any of the issues addressed above.
Good luck with your studies!