IO₄^- (Periodate ion) Lewis Structure and Resonance Structures

Periodate ion (IO₄^-) is an oxianion of iodine and its lewis structure is drawn step by step in this tutorial. Resonance structures of IO₄^- is drawn by drawn lewis structure of IO₄^-. Stability of those resonance structures are tested to select best resonance structures. It is similar to drawing lewis structure of ClO₄^- ion and resonance structures of ClO₄^- ion

In this tutorial, we are going to learn followings.

1. Lewis structure of IO₄^- ion
2. Resonance structures of IO₄^- ion
3. Steps of drawing lewis structures
4. Steps of drawing resonance structures

Lewis Structure of IO₄^- ion

There are four oxygen atoms and iodine atom in periodate ion. Iodine atom is located at the center and there is a -1 charge on one oxygen atom.

Resonance structures of IO₄^- ion

We can draw four resonance structures for IO₄^- ion and they are shown below. These resonance structures are built from the lewis structure drawn for IO₄^- ion.

Now, we are going to learn, how to draw the lewis structure of ClO₄^- ion step by step. Later in this tutorial, we will see how to draw resonance structures of IO₄^- ion.

Steps of drawing IO₄^- lewis structure

Following steps are required to draw IO₄^- lewis structure and they are explained in detail in this tutorial.

1. How to find total number of electrons of the valance shells of chlorine and oxygen atoms and including charge of the anion
2. How many electrons pairs in valence shells
3. Determine center atom from chlorine and oxygen atom
4. Put lone pairs on atoms
5. Stability of lewis structure - Check the stability and minimize charges on atoms by converting lone pairs to bonds to obtain the best lewis structure.

Drawing correct lewis structure is important to draw resonance structures of IO₄^- ion.

Total number of electrons of the valance shells of iodine and oxygen atoms and charge of the anion

There are one iodine atom and four oxygen atoms in the periodate ion. Also there is a -1 overall charge on the IO₄^- ion.

Iodine and oxygen are located at 7 and 6 groups respectively in the periodic table. So iodine has seven electrons in its valence shell and for oxygen atom, there are six electrons in its valence shell.

• Total valence electrons given by iodine atoms = 7*1 = 7
• Total valence electrons given by oxygen atoms = 6 * 4 = 24
• Due to -1 charge, received electrons to valence electrons= 1

• Total valence electrons = 7 + 24 + 1 = 32

Total valence electrons pairs

Total valance electrons pairs = σ bonds + π bonds + lone pairs at valence shells

Total electron pairs are determined by dividing the number total valence electrons by two. For, IO₄^- there are 32 valence electrons pairs, so total pairs of electrons are 16. In next steps, we are going to mark those 16 lone pairs on oxygen atoms and chlorine atoms as bonds and lone pairs.

Center atom of IO₄^- ion

To be the center atom, ability of having greater valance is important. Iodine can show valence upto 7. But, oxygen's maximum valence is 2. Therefore iodine has more chance to be the center atom (See the figure). So, now we can build a sketch of IO₄^- ion.

Sketch of IO₄^- ion

Lone pairs on atoms

• There are already four I-O bonds in the sketch. Therefore only twelve valence electrons pairs are remaining to draw the rest of ion.
• Next step is, marking those twelve valence electrons pairs on outside atoms (in this case, oxygen atoms) as lone pairs. One oxygen atom will take three lone pairs following the octal rule (oxygen atoms cannot keep more than eight electrons in their valence shells). Therefore, twelve electrons pairs are marked on four oxygen atoms. Now, all electrons pairs are finished due to marking on oxygen atoms.
• So, there is no valence electrons pair to mark on center atom; iodine.

Mark charges on atoms after marking valence electrons

If charges exist on atoms, they should be marked on now. Marking charges of atoms is important because it is used to determine the best lewis structure.

Each oxygen atom has a -1 charge and iodine has a +3 charge. Having so many charges on atoms on a structure is not good to be the lewis structure.

Check the stability of drawn structure of IO₄^- ion and reduce charges on atoms by converting lone pairs to bonds

Above drawn structure is not a stable one because there are charges on every atoms. Therfore, we should try to convert lone pairs to bonds to reduce charges on atoms (if possible).

The drawn structure for IO₄^- is still not a stable structure because there are charges on three oxygen atoms and iodine atom. Also, when charge of iodine atom (+2) is large, that structure is unstable and cannot be a good lewis structure.

Now, we should try to minimize charges furthermore by converting one more lone pair which exist on oxygen atom to a I-O bond. So we convert one lone pair of one oxygen atom as a I-O bond as in the following figure.

Now charges are reduced. But, if possible, we can try to reduce charges furthermore to obtain the best lewis structure. yes,we can reduce one more electron pair (lone pair) of another oxygen atom (which holds a negative charge) to a I-O bond.

Now, there are fourteen electrons around iodine atom. This is acceptable because iodine can keep more than eight electrons because iodine has unfilled 3d orbits.

In new structure, charges of atoms are reduced furthermore. Now you understand this structure of IO₄^- is more stable than previous structure due to less charges on atoms.

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