Manganese heptoxide (Mn2O7) Lewis Structure

Lewis structure of Manganese heptoxide (Mn2O7) ion is drawn step by step in this tutorial. Total valence electrons of two manganese atoms and seven oxygen atoms are considered to draw the Mn2O7 Lewis structure. There are double bonds and single bonds in the lewis structure of Mn2O7.

Written by: Damsarani Pothuvillamulla, (undergraduate), Civil Engineering, University of Peradeniya,

Lewis Structure of Mn2O7

There are seven oxygen atoms and two chromium atoms in Manganese heptoxide. Also, there are 6 (Mn=O) double bonds and two Mn-O bonds.

Manganese heptoxide Mn2O7 lewis structure

Now, we are going to learn, how to draw the Lewis structure of Mn2O7 step by step. You will learn all basic steps and rules of Lewis structure drawing.

Basic steps of drawing Mn2O7 Lewis structure

There are defined steps in lewis structure drawings and they are required to draw Mn2O7 lewis structure.

  • How to find total number of electrons of the valance shells of manganese and oxygen atoms.
  • How many electrons pairs in valence shells.
  • Determine center atom from manganese and oxygen atom.
  • Put lone pairs on atoms.
  • 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.

Total number of electrons of the valance shells of manganese and oxygen atoms

There are two manganese atoms and seven oxygen atoms in the Manganese heptoxide.

Manganese and oxygen are located at 7 and 16 groups respectively in the periodic table. So, both manganese atom has 7 electrons and oxygen atom has six electrons in their valence shells.

Therefore to find total number of valence electrons given by oxygen, and manganese atoms multiply relevant atom's valence electrons by number of relevant atoms atoms. See followings to understand.

  • Total valence electrons given by oxygen atoms = 6 * 7 = 42
  • Total valence electrons given by manganese atoms = 7 * 2 = 14

  • Total number of valence electrons = 42 + 14 = 56

Total valence electrons pairs

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

Total electron pairs are determined by dividing the number total valence electrons by two. For, Mn2O7 there are 56 valence electrons, so total pairs of electrons are 28. In next steps, we are going to mark those 28 lone pairs on oxygen atoms and manganese atom as bonds and lone pairs.

Center atom and skeletal structure of Mn2O7

To be the center atom, ability of having greater valance is important. Manganese can show maximum valence upto 7. But oxygen's maximum valence is 2. Therefore, manganese has more chance to be the center atom (See the figure). So, now we can build a sketch of Mn2O7.

skeletal structure of Mn2O7

Mark lone pairs on manganese and oxygen atoms

There are already eight Mn-O bonds in the sketch. Therefore, only twenty valence electrons pairs are remaining to mark as lone. pairs on manganese and oxygen atoms.

Next step is, marking those twenty valence electrons pairs on outside atoms (in this case, oxygen atoms) as lone pairs. One oxygen atom (the oxygen atom in middle will only take two lone pairs) will take three lone pairs following the octal rule (oxygen atoms cannot keep more than eight electrons in their valence shells). Therefore, all twenty electrons pairs are marked on seven oxygen atoms. Now, all electrons pairs are finished due to marking on oxygen atoms.

So, there is no valence electrons pair to mark on manganese atom.

marking lone pairs on oxygen and manganese atoms in Mn2O7

Mark charges on atoms

Check charges on atoms and mark them as below figure. You can observe both manganese and six oxygen atoms have charges. Marking charges is important to decide the best structure of the ion since in the best structure, charges on atoms should be minimized.

Marking charges on atoms in Mn2O7

Each oxygen atom except center oxygen atom has -1 charge and both manganese atoms have +3 charge.

Check the stability of drawn structure of Mn2O7 and reduce charges on atoms by converting lone pairs to bonds

The above drawn structure for Mn2O7 is not a stable structure because six oxygen atoms and both manganese atoms have charges. Also, when charge of an atom (in manganese atom, there is a +3 charge) is large, that structure become more unstable and cannot be a good lewis structure. When an ion or molecule has so many charges on atoms, that structure is not stable. Now, we are going to find a more stable structure.

Now, we should try to minimize charges by converting lone pair or pairs which exist on oxygen atoms to bonds. So, we convert one lone pair of one oxygen atom as a Mn-O bond as in the following figure.

Reducing charges on the atoms in Mn2O7

Now there is a double bond between one chromium and one oxygen atom. There are also seven single bonds (Mn-O) with manganese atoms and other six oxygen atoms. (The oxygen atom in middle already completed the valence shell.)

But there are still lot of charges on atoms in above structure and given structure is not stable yet. If possible, we should reduce charges furthermore. Another lone pair on another oxygen atom is transferred as a Mn-O bond to reduce charge as below.

Reducing charges on the atoms furthermore in Mn2O7

You can see charges are furthermore decreased. But, not all. So, we can continue this step until charges on atoms are minimized.

We can do the same procedure to the other chromium atom too and will get the lewis structure of Manganese heptoxide. In the end of this process, both manganese atoms and seven oxygen atoms should not have charges..

Lewis Structures of Molecules