The Lewis structure of MnO4– contains three double bonds and one single bond, with manganese in the center, and four oxygens on either side. The top oxygen atom, left oxygen atom, and right oxygen atom has two lone pairs. The bottom oxygen atom has three lone pairs, and the manganese atom does not have any lone pair.
Plus, there is a negative (-1) charge on the bottom oxygen atom.
Steps
By using the following steps, you can easily draw the Lewis structure of MnO4–.
#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Calculate formal charge and check stability (if octet is already completed on central atom)
#5 Convert lone pair and calculate formal charge again (if formal charges are not closer to zero)
Let’s one by one discuss each step in detail.
#1 Draw skeleton
In this step, first calculate the total number of valence electrons. And then, decide the central atom.
- Let’s calculate the total number of valence electrons
We know that… manganese is a transition metal that has an electron configuration [Ar] 3d54s2, so manganese has seven valence electrons. And oxygen is a group 16 element, so oxygen has six valence electrons.
Now MnO4– has one manganese atom and four oxygen atoms.
So the total number of valence electrons = valence electrons of manganese atom + (valence electrons of oxygen atom × 4)
And MnO4– has a negative (-1) charge, so we have to add one more electron.
Therefore, the total number of valence electrons = 7 + 24 + 1 = 32
- Now decide the central atom
The atom with the least electronegative value is placed at the center. By looking at the periodic table, we get the electronegativity values for manganese and oxygen as follows:
Electronegativity value of manganese = 1.55
Electronegativity value of oxygen = 3.44
Obviously, manganese is less electronegative than oxygen. Hence, assume that manganese is the central atom.
So now, put manganese in the center and oxygens on either side. And draw the rough skeleton structure for the Lewis structure of MnO4– something like this:
Also read: How to draw Lewis structure of XeCl2 (4 steps)
#2 Show chemical bond
Place two electrons between the atoms to show a chemical bond. Since manganese is surrounded by four oxygens, use eight electrons to show four chemical bonds as follows:
#3 Mark lone pairs
As calculated earlier, we have a total of 32 valence electrons. And in the above structure, we have already used eight valence electrons. Hence, twenty-four valence electrons are remaining.
Two valence electrons represent one lone pair. So twenty-four valence electrons = twelve lone pairs.
Note that manganese is a period 4 element, so it can keep more than 8 electrons in its last shell. And oxygen is a period 2 element, so it can not keep more than 8 electrons in its last shell.
Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.
The outside atoms are oxygens, so each oxygen will get three lone pairs. And the central atom (manganese) will not get any lone pair, because all twelve lone pairs are used.
So the Lewis structure of MnO4– looks something like this:
In the above structure, you can see that the octet is completed on the central atom (manganese), and also on the outside atoms. Therefore, the octet rule is satisfied.
Now calculate the formal charge and check the stability of the above structure.
Also read: How to draw Lewis structure of SeCl4 (4 steps)
#4 Calculate formal charge and check stability
The following formula is used to calculate the formal charges on atoms:
Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons
Collect the data from the above structure and then, write it down below as follows:
- For manganese atom
Valence electrons = 7
Nonbonding electrons = 0
Bonding electrons = 8
Formal charge = 7 – 0 – ½ (8) = +3
- For each oxygen atom
Valence electrons = 6
Nonbonding electrons = 6
Bonding electrons = 2
Formal charge = 6 – 6 – ½ (2) = -1
Mention the formal charges of atoms on the structure. So the Lewis structure of MnO4– looks something like this:
In the above structure, you can see that the formal charges of atoms are not closer to zero. Therefore, convert lone pair and calculate formal charge again.
Also read: How to draw Lewis structure of SeS2 (6 steps)
#5 Convert lone pair and calculate formal charge again
As mentioned earlier, manganese is a period 4 element, so it can keep more than 8 electrons in its last shell.
So convert one lone pair from the three oxygen atoms to make a new bond with the manganese atom. And then, the Lewis structure of MnO4– looks something like this:
Now one last thing we need to do is, calculate the formal charge again and check the stability of the above structure.
Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons
Collect the data from the above structure and then, write it down below as follows:
- For manganese atom
Valence electrons = 7
Nonbonding electrons = 0
Bonding electrons = 14
Formal charge = 7 – 0 – ½ (14) = 0
- For top oxygen, left oxygen, and right oxygen atom
Valence electrons = 6
Nonbonding electrons = 4
Bonding electrons = 4
Formal charge = 6 – 4 – ½ (4) = 0
- For bottom oxygen atom
Valence electrons = 6
Nonbonding electrons = 6
Bonding electrons = 2
Formal charge = 6 – 6 – ½ (2) = -1
Mention the formal charges of atoms on the structure. So the Lewis structure of MnO4– looks something like this:
In the above structure, you can see that the formal charges of atoms are closer to zero. Therefore, this is the most stable Lewis structure of MnO4–.
And each horizontal line drawn in the above structure represents a pair of bonding valence electrons.
Now MnO4– is an ion having a negative (-1) charge, so draw brackets around the above Lewis structure and mention that charge on the top right corner. And then, the Lewis structure of MnO4– looks something like this:
Related
- Lewis structure of XeCl2
- Lewis structure of SeCl4
- Lewis structure of SeS2
- Lewis structure of TeCl4
- Lewis structure of C2Br2
External links
- MnO4- (Permanganate) Ion Lewis Structure | Steps of Drawing – Chemistry School
- Potassium permanganate structurre – Chemistry Stack Exchange
- Permanganate – Wikipedia
- Draw the Lewis structure for the permanganate ion (MnO4-) – Quizlet
- Permanganate | MnO4- | CID 24401 – National Institutes of Health (.gov)
- Draw a Lewis structure with the lowest formal charges for MnO4- – Chegg
- Permanganate ion | MnO4 – ChemSpider
- The MnO4- ion does not contain any O – O bonds. Taking into account the 4s and 3d orbitals, draw a Lewis structure for the MnO4- ion with minimal formal charges – Pearson
- Write the Lewis structure for the permanganate ion with the best possible formal charge distribution. How many resonance structures must be drawn? – Numerade
- How many resonance structures are possible for permanganate ion? – Answers
- MnO4- (Permanganate Ion) Oxidation Number – ChemicalAid
Deep
Rootmemory.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.