Triiodide [I3]- ion Lewis dot structure, molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, polar vs non-polar
The chemical formula I3– represents an anion composed of three iodine (I) atoms. This anion is known as the triiodide ion. The triiodide (I3–) ion is generated by the chemical reaction of the iodine (I2) molecule with the iodide (I–) ion.
In this article, we have discussed how to draw the Lewis dot structure of [I3]–. You will also learn interesting facts about the molecular geometry or shape of [I3]–, its electron geometry, bond angle, hybridization, formal charges, and polarity nature.
So, without any further delay, dive into the article and gain some valuable chemistry knowledge.
The Lewis structure of triiodide [I3]– consists of three identical iodine (I) atoms. One I atom acts as the central atom while the other two iodine atoms act as outer atoms.
There are a total of 5 electron density regions around the central I atom in the I3– Lewis structure. Out of the 5 electron density regions, there are 2 bond pairs and 3 lone pairs of electrons on the central I atom.
In the step-by-step guide given below, we will teach you how to draw the Lewis dot structure of the triiodide [I3]– ion. So grab a paper and pencil and draw this Lewis structure with us.
Steps for drawing the Lewis dot structure of [I3]–
1. Count the total valence electrons in [I3]–
The very first step while drawing the Lewis structure of [I3]– is to find the total valence electrons present in the concerned elemental atoms.
As the triiodide I3– ion is made up of three identical iodine (I) atoms so we just need to look for iodine in the Periodic Table. Iodine is a halogen that lies in Group VII A (or 17) of the Periodic Table. Thus it has a total of 7 valence electrons.
The [I3] – ion consists of 3 I-atoms, and a negative (-1) charge which means 1 extra valence electron.
∴ Therefore, the valence electrons available for the Lewis structure of [I3]– = 3(7) + 1 = 22 valence electrons.
2. Choose the central atom
In this second step, the least electronegative atom out of all the concerned atoms is chosen and placed as the central atom.
But in the case of I3–, the situation seems uncomplicated because it involves three identical Iodine atoms. Hence anyone I atom can be chosen as the central atom while the other two atoms are placed in its surroundings, as shown in the figure below.
3. Connect outer atoms with the central atom
In this step, the outer atoms are joined to the central atom using single straight lines. So, the outer I atoms are joined to the central I atom in the structure of I3– using straight lines.
Each straight line represents a single covalent bond i.e., a bonded electron pair containing 2 electrons. There are a total of 2 single bonds in the diagram shown above thus 2(2) = 4 valence electrons.
Total valence electrons available – electrons used till step 3 = 22 – 4 = 18 valence electrons.
This means 18 valence electrons are still available to be accommodated in the Lewis dot structure of I3–.
4. Complete the octet of outer atoms
In this step, we need to complete the octet of the two I-atoms bonded to the central iodine atom. Each I-atom requires a total of 8 valence electrons to achieve a stable octet electronic configuration.
Both the outer I-atoms are bonded to the central iodine atom using single bonds. That means each outer I-atom already has 2 valence electrons. It is short of 6 electrons that are required to complete its octet.
Therefore, 6 valence electrons are placed around both the outer I atoms as 3 lone pairs respectively. Refer to the figure below.
5. Complete the octet of the central atom
Total valence electrons used till step 4 = 2 single bonds + 2 (electrons placed around each outer I-atom, shown as dots) = 2(2) + 2(6) = 16 valence electrons.
Total valence electrons – electrons used till step 4 = 22 – 16 = 6 valence electrons.
There are 6 valence electrons still available thus these 6 electrons are placed as 3 lone pairs on the central iodine atom.
The figure above illustrates that both the outer I-atoms have a complete octet with 1 single bond and 3 lone pairs each, However, the central I-atom has a total of 10 valence electrons.
But there is nothing to get worried about. That is because in this case, the central I-atom has an expanded octet. It has 4d orbitals therefore iodine can accommodate more than 8 valence electrons during chemical bond formation.
The final step is to check the stability of the Lewis structure of I3– obtained in this step. Let us do that using the formal charge concept.
6. Check the stability of the I3– 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.
The above calculation shows that a zero formal charge is present on each outer I-atom in the I3– Lewis structure. However, a -1 formal charge is present on the central I-atom which is equivalent to the charge present on the ion overall.
Therefore, the above Lewis structure is enclosed in square brackets and a -1 charge is placed at the top right corner, as shown below. The final Lewis structure obtained above is thus the correct and most stable Lewis representation of the triiodide (I3–) ion.
What are the electron and molecular geometry of I3-?
The triiodide [I3]– ion has a trigonal bipyramidal electron geometry while the molecular geometry or shape of the triiodide ion is linear. It is due to the 3 lone pairs of electrons present on the central I-atom in I3– that the molecular ion adopts a different shape from its ideal electron pair geometry.
Molecular geometry of [I3]–
The triiodide [I3]– ion has a linear shape or molecular geometry.
It is due to 3 lone pairs of electrons on the central I-atom in [I3]– that a strong lone pair-lone pair and lone pair-bond pair repulsive effect exists in the molecular ion, in addition to the bond pair-bond pair electronic repulsions between the bonded iodine atoms.
The valence bond theory (VBT) states that lone pair-lone pair repulsions > lone pair-bond pair repulsions > bond pair-bond pair repulsions.
This strong repulsive effect distorts the shape and geometry of I3–. The iodine atoms are pushed farthest away from the 3 lone pairs above and below the molecular ion. The two outer I-atoms consequently occupy terminal positions and I3– adopts a linear shape, as shown below.
An important point to remember is that the molecular geometry or shape of a molecule or molecular ion is strongly influenced by the different number of lone pairs and bond pairs present around the central atom.
Contrarily, the ideal electronic geometry only depends on the total number of electron density regions around the central atom no matter whether it’s a bond pair or a lone pair.
Now, let us see how this concept applies to the electronic geometry of [I3]–
Electron geometry of [I3]–
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule or a molecular ion containing a total of 5 electron density regions around the central atom is trigonal bipyramidal.
In I3– ion, there are 2 single bonds and 3 lone pairs around the central iodine atom which makes a total of 2+3 = 5 electron density regions. Thus, its electron geometry is trigonal bipyramidal.
An ideal trigonal bipyramidal molecule consists of a triangular base while two pyramids are formed, one on top and the other at the bottom, as shown below.
A shortcut to finding the electron and the molecular geometry of a molecule or a molecular ion is by 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 [I3]– molecular ion
A in the AXN formula represents the central atom. In the [I3]– ion, iodine is present at the center so A = I.
X denotes the atoms bonded to the central atom. In [I3]–, two iodine (I) atoms are bonded to the central Iodine so X = 2.
N stands for the lone pairs present on the central atom. As per the Lewis structure of [I3]– there are three lone pairs of electrons on the central Iodine so N = 3.
As a result, the AXN generic formula for the [I3]– ion is AX2N3.
Now, you may have a look at the VSEPR chart below.
The VSEPR chart above shows that the ideal electron geometry of a molecule or molecular ion with AX2N3 generic formula is trigonal bipyramidal while its molecular geometry or shape is linear, as we already noted down for the triiodide [I3]– ion.
During chemical bonding, one electron pair from the 5p orbitals get unpaired, and an electron shift to the empty 5d atomic orbital of iodine. The 5s orbital, three half-filled 5p orbitals, and one 5d orbital of the central iodine atom consequently mix to produce five sp3d hybrid orbitals.
Out of these five sp3d hybrid orbitals, three sp3d hybrid orbitals contain paired electrons which are situated as three lone pairs on the central I-atom in I3–.
The other two sp3d hybrid orbitals contain a single electron each which they use for I-I sigma (σ) bond formation by sp3d-p orbital overlap.
You may also keep in mind that the eighth electron in the sp3d hybrid orbital of central iodine is the extra electron gained while I3– formation.
A shortcut to finding the hybridization present in a molecule or a molecular ion is by using its steric number against the table given below. The steric number of central I in [I3]– is 5 so it has sp3d hybridization.
Steric number
Hybridization
2
sp
3
sp2
4
sp3
5
sp3d
6
sp3d2
The I3– bond angle
The three iodine atoms present in the triiodide [I3]– ion lie in a planar arrangement, on a straight line, and form a mutual bond angle of 180°.
According to Pauling’s electronegativity scale, a covalent chemical bond is considered polar if the bonded atoms have an electronegativity difference between 0.5 to 1.6 units.
As we have already seen that the I3– ion consists of three identical iodine atoms. The electronegativity value of each iodine atom is 2.66 units. Therefore, a zero electronegativity difference exists between the bonded atoms in an I-I bond.
Also I3– has a planar linear shape. Thus, by definition, the triiodide [I3]– ion should be non-polar.
But some chemists report that it is due to the negative charge present on the ion overall and an odd number of lone pairs on the central I-atom in I3– that its symmetry gets disturbed and it exhibits some polar characteristics.
A particular evidence of this argument is that I3– is water soluble. As like dissolves like and H2O is definitely a polar molecule therefore the I3– ion may also possess some polar character (net dipole moment µ >0).
In conclusion, I3– is considered neither polar nor non-polar rather, there is always some room for debate on this topic.
How many bond pairs and lone pairs are there in the Lewis structure of I3–?
The Lewis structure of the triiodide [I3]– ion displays a total of 22 valence electrons i.e., 22/2 = 11 electron pairs.
Out of the 11 electron pairs, there are 2 bond pairs and 9 lone pairs.
Out of the 9 lone pairs, 3 lone pairs of electrons are present on the central iodine atom while the remaining 6 lone pairs are present on the terminal iodine atoms, as 3 lone pairs each.
What is the shape of the triiodide I3– ion?
The triiodide I3– ion has a linear shape or molecular geometry. However, its ideal electron pair geometry is trigonal bipyramidal.
The three lone pairs of electrons present on the central iodine atom in I3– distort the geometry of the molecular ion and make it adopt a different shape from its electronic geometry.
How is the shape of I3+ different from I3–?
The I3+ ion has a bent or V-shape as opposed to the linear shape of the I3– ion.
There are 2 lone pairs of electrons on the central iodine atom in I3+. The AXN generic formula for this ion is AX2N2. So according to the VSEPR concept, the cation adopts a bent shape.
The bent shape is attributed to the lone pair-lone pair and lone pair-bond pair repulsions in the molecular ion. This repulsive effect distorts the shape of the molecule and makes it adopt a different shape from its ideal electron pair geometry i.e., tetrahedral.
Contrarily, there are 3 lone pairs of electrons on the central iodine atom in I3–. Its AXN generic formula is AX2N3.
The lone pair repulsive effect makes the ion adopt a linear shape, different from the ideal trigonal bipyramidal electron geometry.
The total number of valence electrons available for drawing triiodide [I3]– ion Lewis structure is 22.
The negative 1 charge present on the ion accounts for 1 extra electron added in its Lewis structure.
The [I3]– ion has a linear shape or molecular geometry.
The ideal electron pair geometry of [I3]– is trigonal bipyramidal.
The I3– ion has sp3d hybridization.
The polarity of I3– ion is a debatable topic. It is considered non-polar owing to its identical atoms and linear shape. However, the negative charge present on the ion and its water solubility accounts for some definite polar characteristics in I3–.
-1 formal charge is present on the central iodine atom while zero formal charges are present on both the outer iodine atoms in the I3– Lewis structure
This accounts for an overall negative charge on the monovalent poly halogen anion i.e., the triiodide ion.
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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