How to draw IBr3 Lewis Structure? - Science Education and Tutorials

The iodine tribromide chemical formula is IBr3. Drawing IBr3 Lewis Structure is very easy to by using the following method. Here in this post, we described step by step method to construct IBr3 Lewis Structure. The iodine and bromine elements come as the member of the halogen family group from the periodic table. The valence electrons in iodine and bromine are seven. The branch of halogen chemistry is used to make chemicals reagents for fluorination reactions.

Table of Contents

Key Points To Consider When Drawing The IBr3 Electron Dot Structure

A three-step approach for drawing the IBr3 Lewis structure can be used. The first step is to sketch the Lewis structure of the IBr3 molecule, to add valence electrons around the iodine atom; the second step is to add valence electrons to the three bromine atoms, and the final step is to combine the step1 and step2 to get the IBr3 Lewis Structure.

The IBr3 Lewis structure is a diagram that illustrates the number of valence electrons and bond electron pairs in the IBr3 molecule. The geometry of the IBr3 molecule can then be predicted using the Valence Shell Electron Pair Repulsion Theory (VSEPR Theory), which states that molecules will choose the IBr3 geometrical shape in which the electrons have from one another.

Finally, you must add their bond polarities to compute the strength of the three I-Br bonds (dipole moment properties of the IBr3 molecule). The iodine-bromine bonds in iodine tribromide(IBr3), for example, are polarised toward the more electronegative bromine in IBr3 molecule, and because both bonds have the same size and are located around three bromine terminals of the trigonal bipyramidal with two lone pairs (in total four electron) on the iodine atom, their sum of dipole moment is nonzero due to the IBr3 molecule’s bond dipole moment and more electron polarity to the bromine atoms. Because each three I-Br bonds polarity not canceled each other in the IBr3 molecule. The trifluoro iodine or iodine tribromide(IBr3) molecule is classified as a polar molecule.

The molecule of iodine tribromide (with trigonal bipyramidal molecular geometry) is tilted, the bond angles between iodine and bromine are slightly smaller than 90 degrees. It has a difference in electronegativity values between iodine and bromine atoms, with central iodine’s pull being less than terminal bromine’s in the IBr3 molecule. But they canceled each other due to the asymmetrical molecular geometry of the IBr3 molecule.

As a result, it has the nonzero dipole moment. The IBr3 molecule has a nonzero dipole moment due to an unequal charge distribution of negative and positive charges. But both these atoms fall on the halogen family group. The bromine atom is a more electronegative value than iodine in the IBr3 molecule. The net dipole moment of the IBr3 molecule is smaller than the IF3 molecule.

Molecules can be classified as polar or nonpolar. The molecule polar behaves in a different manner as compared to nonpolar.

Overview: IBr3 Lewis Structure

The central atom is iodine, which is bordered on three terminals with bromine atoms( in distorted T-shaped or trigonal bipyramidal), and two lone pairs on the central in the trigonal bipyramidal geometry. Iodine has seven outermost valence electrons, indicating that it possesses seven electrons in its outermost shell, whereas bromine also has seven valence electrons in its outermost shell. To complete the octet of the iodine and bromine atoms requires one valence electron on each of their outermost shell.

Three bromine atoms establish covalent connections with the central iodine atom as a result, leaving the iodine atom with two lone pairs. There are two lone pairs of electrons on the iodine central atom that resists the bond pairs of the three I-Br. According to VSEPR theory, the I-Br bond pairs polarity lead the IBr3 molecule to take on the distorted T-Shape or trigonal bipyramidal geometry structure.

The IBr3 molecule’s three I-Br bonds are arranged in unsymmetrical polarity order around the trigonal bipyramidal molecular geometry, giving rise to the IBr3 molecular shape. The IBr3 molecule has a distorted T- shaped or trigonal bipyramidal molecular geometry because there is electrical repulsion between the lone pairs of electrons in bromine and three bond pairs(I-Br) of the IBr3 molecule.

Lewis structure of IBr3 has dot electron representative structure. Valence electrons of atoms undergo orbitals mixing in the chemical reactions, gives new types of molecular species. The molecule is nothing but a bundle of valence electrons from the atoms. But it is converted to bond pairs and lone pairs in the molecular structure.

Electronegative value Difference Calculation of IBr3 Molecule:

Iodine and bromine Electronegative difference in IBr3:

The iodine atom has an electronegativity of 2.66, while bromine has an electronegativity of 2.96 in the IBr3 molecule. The difference in electronegativity of iodine and bromine can be estimated using the method below.

The electronegative value difference between iodine and bromine
Electronegativity value of iodine = 2.66
Electronegativity value of bromine= 2.96
Difference of electronegativity value between iodine and bromine= 2.96 – 2.66=0.3
Electronegativity difference between I-Br bond calculation of IBr3 molecule

The electronegative difference between iodine and bromine is greater than 0.5. This indicated the bond polarity moves near to polar nature. I-Br bond polarity in the IBr3 molecule is polar.

Because of this difference in electronegativity of iodine and bromine atoms, the IBr3 molecule’s I-Br bond becomes polar. The total net dipole moment of the IBr3 molecule is nonzero due to the no cancellation of the bond dipole moment in the distorted T-shaped or trigonal bipyramidal geometry. The electronegativity of an atom is the strength with which it may attract bound electron pairs to its side. The polarity of IBr3 is discussed in our previous post.

As a result, the I-Br bond’s dipole moment is high due to the polarization of the bonds and two lone pairs of electrons on iodine, and all I-Br bonds’ dipoles are arranged in the asymmetrical IBr3 molecular geometry. The IBr3 molecule’s total dipole moment is predicted to be lesser than the BrF3 molecule.

The electron dot structure of the IBr3 molecule is also known as the IBr3 Lewis structure. It determines the number of outermost valence electrons as well as the electrons engaged in the IBr3 molecule’s bond formation. The outermost valence electrons of the IBr3 molecule must be understood while considering the Lewis structure of the molecule.

The iodine atom is the middle element in IBr3 molecular geometry, with seven electrons in its outermost valence electron shell, whereas the bromine atom has seven electrons in its outermost valence electron shell. The bromine atom has seven valence electrons.

The IBr3 has a total of 28 valence electrons as a result of the foregoing above-said reasoning. With the core central iodine atom, the three terminals with three bromine atoms form covalent bonds, leaving the iodine atom with two lone pairs in the middle of distorted T-shaped or trigonal bipyramidal geometry.

Because lone pair on the terminal bromine atom creates interaction with I-Br bond pairs(but it is negligible). The bond angle of the Br-I-Br bond in the distorted T- shaped or trigonal bipyramidal molecular geometry is approximately smaller than 90 degrees. This angle is smaller than the CH4 molecule bond angle. The I-Br bond length of the IBr3 molecule is longer than the I-F bond length in the IF3 molecule.

To sketch the IBr3 Lewis structure by following these instructions:

Step-1: IBr3 Lewis dot Structure by counting valence electrons on the iodine atom

To calculate the valence electron of each atom in IBr3, look for its periodic group from the periodic table. The halogen group families, which are the 17th group in the periodic table, are both made up of iodine and bromine atoms. In their outermost shells, bromine and iodine have seven valence electrons.

Calculate the total number of valence electrons in the IBr3 molecule’s outermost valence shell. The first step is to determine how many electrons are in the IBr3 Lewis structure’s outermost valence shell. An electron in an atom’s outermost shell is known as a valence electron. It is represented by dots in the IBr3 Lewis diagram. The IBr3 molecule’s core iodine atom can be represented as follows:

Total outermost valence shell electron of iodine atom in IBr3= 7
Total outermost valence shell electron of bromine atom in IBr3= 7
The molecule has one central iodine and three bromine atoms. Then the total outermost valence shell electrons can be calculated as follows
∴ Total outermost valence shell electrons available for IBr3 Lewis structure( dot structure) = 7+3*7=28 valence electrons in IBr3.
calculation of total valence electron of IBr3 molecule

Choose the atom with the least electronegative value atom and insert it in the center of the molecular geometry of IBr3. We’ll choose the least electronegative value atom in the IBr3 molecule to place in the center of the IBr3 Lewis structure diagram in this phase. The electronegativity value in periodic groups grows from left to right in the periodic table and drops from top to bottom.

The first step is to put seven valence electrons around the iodine atom as given in the figure.

Step-2: Lewis Structure of IBr3 for counting valence electrons around the terminal bromine atoms

As a result, iodine is the fourth atom in the periodic table’s halogen family group. Bromine is the third member of the halogen family. The electronegative value of the bromine atom is higher than that of the iodine atom in the IBr3 molecule. Furthermore, bromine has a seven electrons limit since iodine is the less electronegative element in the IBr3 molecule.

In the IBr3 Lewis structure diagram, the iodine atom can be the center atom of the molecule. As a result, central iodine in the IBr3 Lewis structure, with all three bromine atoms arranged in distorted T-shaped or trigonal bipyramidal geometry.

Add valence electrons around the bromine atom, as given in the figure.

Step-3: Lewis dot Structure for IBr3 generated from step-1 and step-2

Connect the exterior and core central atom of the IBr3 molecule with three single I-Br bonds. In this stage, use three bromine atoms on the outside of the IBr3 molecule to the central iodine atom in the middle.

Count how many electrons from the outermost valence shell have been used in the IBr3 structure so far. Each I-Br bond carries two electrons because each iodine atom is connected to three bromine atoms by three I-Br bonds. Bond pairings of I-Br are what they’re called.

So, out of the total of 28 valence electrons available for the IBr3 Lewis structure, we used 6 electrons for the IBr3 molecule’s three I-Br bonds. The IBr3 molecule has two lone pairs of electrons in the central iodine atom.

Place the valence electrons in the I-Br bond pairs starting with the core iodine, three bromine atoms in the IBr3 molecule. In the IBr3 Lewis structure diagram, we always begin by introducing valence electrons from the central iodine atom(in step1). As a result, wrap around the central iodine atom’s bond pair valence electrons first (see figure for step1).

The iodine atom in the molecule gets only six electrons around its molecular structure. This central iodine atom is extra octet stable. But it has two lone pairs. Iodine (I2) is a brownish vaporize solid in nature. when iodine acted as an iodinating agent, it is used as a chemical reagent. But IBr3 is used as a fluorinating agent in organic chemistry (this decreases the chemical reactivity).

Iodine requires 10 electrons in its outermost valence shell to complete the molecular extra octet stability, 6 electrons bond pairs in three I-Br bonds. Then lone pairs of electrons on the bromine atoms of the IBr3 molecule are placed in a distorted T-shaped or trigonal bipyramidal geometry. Iodine already shares six electrons to the three I-Br bonds. Then place the valence electron in the bromine atoms, it placed around seven electrons on each atom(step-2). 18 valence electrons were placed around bromine atoms as lone pairs of electrons.

We’ve positioned 18 electrons around the terminal bromine atoms(step-3), which is represented by a dot, in the IBr3 molecular structure above. The iodine atom completes its molecular extra octet stability in the IBr3 molecule because it possesses six electrons in its (three I-Br) bond pairs with three bromine in the outermost valence shell.

Count how many outermost valence shell electrons have been used so far using the IBr3 Lewis structure. Three electron bond pairs are shown as dots in the IBr3 chemical structure, whereas three single bonds each contain two electrons. The outermost valence shell electrons of the IBr3 molecule(bond pairs) are six as a result of the calculation.

So far, we’ve used 28 of the IBr3 Lewis structure’s total 28 outermost valence shell electrons. Two lone pairs of electrons on the iodine atom in the distorted T structure or trigonal bipyramidal of the IBr3 molecule.

Complete the middle iodine atom stability and, if necessary, apply a covalent bond. The central iodine atom undergoes extra octet stability(due to two lone pairs of electrons).

The core atom in the IBr3 Lewis structure is iodine, which is bonded to the three bromine atoms by single bonds (three I-Br). With the help of three single bonds, it already shares six electrons. As a result, the iodine follows the extra octet rule and has six electrons surrounding it on the three terminals of the IBr3 molecule’s trigonal bipyramidal geometry.

How to calculate the formal charge on iodine and bromine atoms in IBr3 Lewis Structure?

Calculating formal charge on the iodine of IBr3 molecule:

The formal charge on the IBr3 molecule’s iodine central atom often corresponds to the actual charge on that iodine central atom. In the following computation, the formal charge will be calculated on the central iodine atom of the IBr3 Lewis dot structure.

To calculate the formal charge on the central iodine atom of the IBr3 molecule by using the following formula:
The formal charge on the iodine atom of IBr3 molecule= (V. E(I)– L.E(I) – 1/2(B.E))
V.E (I) = Valence electron in a iodine atom of IBr3 molecule
L.E(I) = Lone pairs of an electron in the iodine atom of the IBr3 molecule.
B.E = Bond pair electron in I atom of IBr3 molecule
calculation of formal charge on iodine atom in IBr3 molecule

The iodine core atom (three single bonds connected to three bromine atoms ) of the IBr3 molecule has seven valence electrons, two lone pairs of electrons(four electrons), and six bonding pairing valence electrons. Put these values for the iodine atom in the formula above.

Formal charge on iodine atom of IBr3 molecule = (7- 4-(6/2)) =0

In the Lewis structure of IBr3, the formal charge on the central iodine atom is zero.

Calculating formal charge on the bromine of IBr3 molecule:

The formal charge on the IBr3 molecule’s bromine terminal atom often corresponds to the actual charge on that bromine terminal atom. In the following computation, the formal charge will be calculated on the terminal bromine atom of the IBr3 Lewis dot structure.

To calculate the formal charge on the terminal bromine atom of the IBr3 molecule by using the following formula:
The formal charge on the bromine atom of IBr3 molecule= (V. E(Br)– L.E(Br) – 1/2(B.E))
V.E (Br) = Valence electron in a bromine atom of IBr3 molecule
L.E(Br) = Lone pairs of an electron in the bromine atom of the IBr3 molecule.
B.E = Bond pair electron in Br atom of IBr3 molecule
calculation of formal charge on bromine atom in IBr3 molecule

The bromine terminal atom of the IBr3 molecule has seven valence electrons, three lone pairs of electrons(six electrons), and two bonding pairing valence electrons. Put these values for the bromine atom in the formula above.

Formal charge on bromine atom of IBr3 molecule = (7- 6-(2/2)) =0

In the Lewis structure of IBr3, the formal charge on the terminal bromine atom is zero.

Summary:

In this post, we discussed the method to construct the IBr3 Lewis structure. First, the valence electrons are placed around the iodine atom. Second, place the valence electron on the bromine atoms. Finally, when we combined the first and second steps. It gives IBr3 Lewis structure. Need to remember that, if you follow above said method, you can construct molecular dot structure very easily.