Iodite (IO2-) lewis dot structure, molecular geometry or shape, electron geometry, bond angles, hybridization, polar or nonpolar
Iodite ion, also known as dioxidoiodate is represented by the chemical formula IO2–. It is an oxyanion of iodine produced by the deprotonation of iodous acid (HIO2).
IO2– is a relatively unstable and reactive chemical species used as an intermediate in iodine chemistry, redox reactions and titrimetry, specifically Iodometric titrations.
In this article, we will teach you how to draw the Lewis dot structure of IO2–, what is its molecular geometry or shape, electron geometry, bond angles, hybridization, formal charges, polarity, etc.
So if you want to learn all this and much more, let’s start reading!
The Lewis dot structure of the iodite (IO2–) ion comprises an iodine (I) atom at the center. It is surrounded by two oxygen (O) atoms, one on either side, via single and double covalent bonds, respectively.
There are 2 lone pairs of electrons on the central I-atom; conversely, the terminal O-atoms also carry 2 and 3 lone pairs in the IO2– Lewis structure, respectively.
You can easily learn to draw the Lewis dot structure of IO2– by following the simple steps given below.
Steps for drawing the Lewis dot structure of IO2–
1. Count the total valence electrons present in IO2–
The two distinct elements present in IO2– are iodine and oxygen.
Oxygen (O) is present in Group VI A (or 16) of the Periodic Table of Elements. Thus, it has a total of 6 valence electrons in each atom.
In contrast, iodine (I) is a halogen located in Group VII A (or 17), containing 7 valence electrons in each atom.
Total number of valence electrons in oxygen = 6
Total number of valence electrons in iodine = 7
The IO2– ion comprises 1 I-atom and 2 O-atoms.
An important point to remember is that the IO2– ion carries a negative (-1) charge, which means 1 extra valence electron is added in this Lewis structure.
∴ Therefore, the total valence electrons available for drawing the Lewis dot structure of IO2– = 1(7) + 2(6) = 19 +1 = 20 valence electrons.
2. Find the least electronegative atom and place it at the center
By convention, the least electronegative atom out of all those available is chosen as the central atom while drawing the Lewis structure of a molecule or molecular ion.
The least electronegative atom can easily form covalent bonds with other atoms by sharing its electrons.
Iodine (E.N = 2.66) is invariably less electronegative than oxygen (E.N = 3.44).
Therefore, an I-atom is placed as the central atom in IO2– Lewis structure while the two O-atoms occupy peripheral positions, as shown below.
3. Connect the outer atoms with the central atom
In this step, the outer atoms, i.e., 2 O-atoms, are joined to the central I-atom using single straight lines.
A straight line represents a single covalent bond, i.e., a bond pair containing 2 electrons.
In the above structure, there are a total of 2 single bonds, i.e., 2(2) = 4 valence electrons are already consumed out of the 20 initially available.
Now let’s see in the next steps where to place the remaining 16 valence electrons in the IO2– Lewis dot structure.
4. Complete the octet of the outer atoms
An O-atom needs a total of 8 valence electrons in order to achieve a stable octet electronic configuration.
An I-O bond represents 2 valence electrons already present around both oxygen atoms.
Therefore, to complete its octet, the remaining 6 valence electrons are placed as 3 lone pairs on each O-atom, as shown below.
5. Complete the octet of the central atom
Total valence electrons used till step 4 = 2 single bonds + 2(electrons placed around each O-atom, shown as dots) = 2(2) + 2(6) = 16 valence electrons.
Total valence electrons – electrons used till step 4 = 20 – 16 = 4 valence electrons.
Thus these 4 valence electrons are placed as 2 lone pairs on the central I-atom. This completes the octet of the central I-atom in the IO2– Lewis structure.
Now let’s check its stability by applying the formal charge concept.
6. Check the stability of Lewis’s structure using the formal charge concept
The less the formal charge on the atoms of a molecule or molecular ion, the better the stability of its Lewis structure.
The formal charges can be calculated using the formula given below.
Formal charge = [valence electrons- nonbonding electrons- ½ (bonding electrons)].
Now let us use this formula and the Lewis structure obtained in step 5 to determine the formal charges present on the IO2– bonded atoms.
As per the above calculation, a +1 formal charge is present on the central I-atom, while each O-atom carries a formal charge of -1.
But as already told, the fewer the formal charges, the greater the stability of a Lewis structure. So let’s see how we can reduce these formal charges in the next step.
7. Minimize the formal charges by converting a lone pair into a covalent bond and again check the stability of Lewis’s structure
A lone pair present on a terminal O-atom is converted into an additional covalent bond between the central I-atom and the corresponding O-atom.
This makes a total of 10 valence electrons surrounding the central I-atom in the IO2– Lewis dot structure. This situation falls under the expanded octet rule.
The iodine atom can accommodate more than 8 valence electrons due to the availability of d-atomic orbitals.
Now let us again check the stability of the Lewis structure obtained above.
For iodine atom
Valence electrons of iodine = 7
Bonding electrons = 1 single bond + 1 double bond = 2 + 4 = 6 electrons
The above calculation shows that the formal charges present on iodine and one of the two oxygen atoms are reduced to zero. However, a -1 formal charge is present on the other O-atom, which is also the charge present on IO2– ion overall.
Thus we have obtained the stable and correct Lewis representation of the iodite (IO2–) ion.
However, you may note that the following two resonance forms are possible for representing the Lewis structure of the iodite (IO2–) ion.
The pi-bonded electrons and lone pairs keep revolving from one position to another on the molecular ion, resulting in multiple Lewis representations.
The actual IO2– structure is a hybrid of both the above resonance forms, known as the resonance hybrid.
Now let us discuss its electron and molecular geometry or shape.
What are the electron and molecular geometry of IO2-?
The molecular geometry or shape of the iodite (IO2–) ion w.r.t the central I-atom is bent, angular or V-shape. In contrast, its ideal electronic geometry is tetrahedral.
The presence of 2 lone pairs of electrons on the central I-atom leads to strong lone pair-lone pair and lone pair-bond pair electronic repulsions, thus distorting the overall molecular shape of IO2–.
Molecular geometry of IO2–
The molecular geometry or shape of the iodite (IO2–) ion is bent, angular or V-shaped.
The presence of 2 lone pairs of electrons on the central I-atom in IO2– leads to strong lone pair-lone pair and lone pair-bond pair electronic repulsions. This strong repulsive effect tilts the bonded O-atoms away from the center. Hence the iodite ion occupies a bent, angular or inverted V-shape, as shown below.
Electron geometry of IO2–
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule or molecular ion containing a total of 4 electron density regions around the central atom is tetrahedral.
In IO2–, the I-atom at the center is surrounded by 2 bond pairs and 2 lone pairs of electrons, making a total of 4 electron density regions. Hence, the ideal electron pair geometry of the IO2– ion is tetrahedral.
An easy trick to finding a molecule’s electron and molecular geometry is using the AXN method.
AXN is a simple formula representing the number of bonded atoms and lone pairs present on the central atom.
It is used to predict the shape and geometry of a molecule or molecular ion using the VSEPR concept.
AXN notation for IO2–
A in the AXN formula represents the central atom. In IO2–, an iodine (I) atom is present at the center, so A = I.
X denotes the atoms bonded to the central atom. In IO2–, 2 O-atoms are directly bonded to the central I-atom. So X = 2 for IO2–.
N stands for the lone pairs present on the central atom. As per the Lewis structure of IO2–, the central I-atom has 2 lone pairs of electrons. Thus, N = 2 for IO2–.
As a result, the AXN generic formula for IO2– is AX2N2.
Now, you may have a look at the VSEPR chart below.
The VSEPR chart confirms that the molecular geometry or shape of a molecule or molecular ion with an AX2N2 generic formula is bent, angular or V-shaped while its electron geometry is tetrahedral, as we already noted down for the iodite (IO2–) ion.
Hybridization of IO2–
The central I-atom is sp3 hybridized in IO2–.
The electronic configuration of iodine is [Kr] 4d10 5s2 5p5.
During chemical bonding in IO2–, one 5p electron of iodine shifts to an empty d atomic orbital.
Consequently, the 5s atomic orbitals of iodine hybridize with its 5p orbitals to produce four sp3 hybrid orbitals.
Each sp3 hybrid orbital possesses a 25 % s-character and a 75% p-character.
However, these sp3 hybrid orbitals are not all equivalent, as two of these possess paired electrons which are situated as 2 lone pairs on the central I-atom.
In contrast, the other two sp3 hybrid orbitals of iodine-containing unpaired electrons form I-O sigma bonds by overlapping with the half-filled p orbitals of adjacent oxygen atoms.
Conversely, the unhybridized d orbital of iodine overlaps with the unhybridized p orbital of the adjacent oxygen atom to form an I=O pi bond.
Refer to the figure drawn below.
Another shortcut to finding the hybridization present in a molecule or molecular ion is using its steric number against the table below.
The steric number of the I-atom in IO2– is 4, so it has sp3 hybridization.
Steric number
Hybridization
2
sp
3
sp2
4
sp3
5
sp3d
6
sp3d2
The bond angle of IO2–
The ideal bond angle in a tetrahedral molecule is 109.5°. However, it is due to the distortion present in the molecular shape of the iodite ion that the O-I=O bond angle reduces to about 105°.
As per Pauling’s electronegativity scale, a polar covalent bond is formed between two dissimilar atoms with an electronegativity difference between 0.4 and 1.6 units.
A high electronegativity difference of 0.78 units is present between the covalently bonded iodine (E.N = 2.66) and oxygen (E.N = 3.44) atoms in each of the I-O and I=O bonds.
Therefore, both the bonds present in IO2– are strongly polar in nature.
Each O-atom strongly attracts the shared electron cloud largely towards itself. Each terminal O-atom thus gains a partial negative (δ–) charge while the central I-atom obtains a partial positive (δ+) in IO2–.
It is due to the asymmetric bent, angular, or V-shape of IO2– that the I-O dipole moments stay uncancelled.
This results in an overall polar iodite ion with a non-uniformly distributed electron cloud spread over it (net µ> 0).
The Lewis dot structure of the iodite (IO2–) ion displays a total of 20 valence electrons, i.e., 20/2 = 10 electron pairs.
Out of the 10 electron pairs, there are 3 bond pairs and 7 lone pairs of electrons.
An I-atom at the center is double-covalently bonded to an O-atom on one side and single-covalently bonded to another O-atom on the other side.
There are 2 lone pairs of electrons on the central I-atom in the IO2– Lewis structure, while 2 and 3 lone pairs are present on the terminal O-atoms.
How many lone pairs are present on the central I-atom in IO2– Lewis structure?
The central I-atom has 3 bond pairs and 2 lone pairs in IO2– Lewis structure.
What is the VSEPR shape of IO2–?
The VSEPR shape of IO2– w.r.t the central iodine atom is bent, angular or V-shaped.
Why is the molecular shape of IO2– different from that of its electronic geometry?
The presence of 2 lone pairs of electrons on the central I-atom leads to strong lone pair-lone pair and lone pair-bond pair electronic repulsions in IO2–.
The covalently bonded O-atoms tilt away from the center to overcome this strong repulsive effect. Hence IO2– occupies a bent shape, different from its electronic geometry, i.e., tetrahedral.
How is the molecular shape of IO3– different from that of IO2–?
In the iodate (IO3–) ion, an iodine atom at the center is directly bonded to three oxygen atoms at the sides. There is 1 lone pair of electrons on the central I-atom.
Thus the shape of IO3– w.r.t the central I-atom is trigonal pyramidal, different from the bent shape of iodite (IO2–).
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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