B2O3 (Boric anhydride) Lewis structure
Boric anhydride (B2O3) lewis structure contains two boron atoms and and three oxygen atoms. An oxygen atom has made single bonds with boron atoms. Each boron atom has made bonds with two oxygen atoms. There are no charges on atoms in boric anhydride lewis structure and you will learn how to draw the lewis structure of B2O3 .
B2O3 lewis structure
We will see what are the characteristics of B2O3 lewis structure.
- Both boron atoms have made single bonds with one oxygen atom. Also, that oxygen atom has two lone pairs on its last shell.
- As well as, each boron atom has made a double bond with an oxygen atom. So, there are three bonds around a boron atom.
- Oxygen atoms which are located outside the molecule has two lone pairs in their valence shell in the lewis structure of boric anhydride.
Steps of drawing lewis structure of B2O3
When we draw a lewis structure, guidelines are given. Number of steps you need to draw the lewis structure, can be changed according the complexity of the molecule or ion. Those all steps are mentioned and explained in detail in this tutorial for your knowledge improvement.
- Find total number of electrons of the valance shells of boron and oxygen atoms
- Find total number of electrons pairs existing as lone pairs and bonds
- Determine center atom
- Mark lone pairs on atoms
- Mark charges on atoms if there are.
- Check the stability and minimize charges on atoms by converting lone pairs to bonds to obtain best lewis structure.
Total number of electrons of the valance shells of B2O3
There are only two elements in B2O3 molecule. Oxygen, a VIA group element has six electrons in its last shell. Boron is a group IIIA element in the periodic table and contains three electrons in its last shell. Now, we know how many electrons are there in valence shells of boron and oxygen atoms.
- valence electrons given by boron atoms = 3 * 2 = 6
- valence electrons given by oxygen atoms = 6 * 3 = 18
- Because, there is no overall charge in molecule, no additions or deductions to total number of valence electrons
- Total number of valence electrons = 6 + 18
- Total number of valence electrons = 24
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, boric anhydride, total number of pairs of electrons are twelve in their valence shells.
Selection of center atom of B2O3
- Fact 1: Because, boron atom can show higher valence (3) than oxygen atom (2), boron atom has the higher priority to be the center atom.
- Fact 2: As well as, because boron is more electropositive than oxygen, again boron should be the center atom.
- From above two facts, it can be decided that boron should be the center atom in B2O3.
- But, there are two boron atoms and we will face to a problem of how to select one boron atom as center atom.
- We can draw several skeletal and can build lewis structure for each skeletal.
- From each drawn structure, you can decide which structure is the most stable structure.
- By doing so, we have found that following skeletal is the most suitable one to draw the most stable lewis structure of boric anhydride.
Mark lone pairs on atoms
After determining the center atom and skeletal of B2O3 molecule, we can start to mark lone pairs on atoms. Remember that, there are total of twelve electron pairs.
- There are already four bonds in the above drawn skeletal. So, there are eight remaining lone pairs to mark on atoms.
- Start to mark remaining lone pairs on outside atoms (in this case: oxygen atoms). Each oxygen atom will take three lone pairs. So, six electron pairs are marked on outside two oxygen atoms as lone pairs. Now, two more lone pairs are remaining.
- Then, mark remaining two more valence electrons pairs are marked on center oxygen atom.
Mark charges on atoms
There are charges above drawn structure as below (each oxygen atom has -1 charge and boron atom has +1 charge).
Check the stability and minimize charges on atoms by converting lone pairs to bonds
There are charges on four atoms in above structure. If we can reduce charges furthermore by converting lone pairs to bonds to get more stable structure, try that step.
Therefore, convert a lone pair on oxygen atom to form a double bond between one boron atom and one outside oxygen atom as following image. Then, charges of atoms will be reduced.
We can convert one more lone pair of other outside oxygen atom to form another double bond between other boron atom and oxygen atom as below.
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