Lewis dot structure of carbonic acid H2CO3

A simple method for drawing Lewis structures: Carbonic Acid

A simple method for writing Lewis structures (electron Lewis dot structures) was given in a previous post entitled “Lewis Structures and the Octet Rule”.

Several worked examples relevant to this procedure were given in previous posts please see the post entitled "Lewis Structures & the Octet Rule - Theory & Examples"

Let us examine the case of carbonic acid H₂CO₃. How can we draw the carbonic acid Lewis dot structure?

Step 1: Connect the atoms with single bonds. 

 

Fig. 1 : Connect the atoms of carbonic acid with single bonds

Step 2: Calculate the # of electrons in π bonds (multiple bonds) using  formula (1):

Where n in this case is 4 since H₂CO₃ consists of six atoms but two of them are H atoms.

Where V = (1 + 6 + 4 + 6 + 6 + 1) = 24 

Therefore, P = 6n + 2 – V = 6 * 4 + 2 – 24 = 2      So  there are 2 π electrons in H₂CO₃and therefore 1 double bond must be added to the structure of Step 1.

Step 3 & 4: The Lewis structure of H₂CO₃  is derived below:

Fig. 2: Lewis structure of carbonic acid H2CO3

For a detailed description of the above method please see the following video:

Lewis electron dot structures of carbonic acid

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Relevant Posts

Lewis Structures and the Octet Rule - A simple method to write Lewis Structures

Simple method for writing Lewis Structures – Ozone O₃ and carbonate CO₃^-2 ion

Lewis dot structures - Electron dot structure of CO

Simple Method for drawing Lewis structures (Video): H2CO3

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References

1. G.N. Lewis, J.A.C.S, 38, 762-785, (1916)
2.  E. C. McGoran, J. Chem. Educ., 68, 19-23 (1991)
3. A.B.P. Lever, J. Chem. Educ., 49, 819-821, (1972)

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Key Terms

electron dot Lewis structures, Lewis Structures and the Octet Rule, how can we draw the carbonic acid Lewis dot structure

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Electron Lewis Dot Structure of chlorosulphonic acid HSO3Cl

A simple procedure for writing electron Lewis dot structures was given in a previous post entitled “Lewis Structures and the Octet Rule”.

Several worked examples relevant to this procedure were given in previous posts please see the Sitemap - Table of Contents (Lewis Electron Dot Structures).

Let us examine the case of chlorosulphonic acid HSO3Cl . Chlorosulphonic acid is used to prepare alkyl sulfates, which are useful as detergents and as chemical intermediates. How can we construct the chlorosulphonic acid Lewis dot structure?

Step 1: Connect the atoms with single bonds. 

Fig. 1 : Connect the atoms of chlorosulphonic acid with single bonds.

 Step 2:  Calculate the # of electrons in π bonds (pi bonds, multiple bonds) using  formula (1): 

Where n in this case is 5 since  HSO₃Cl  consists of six atoms but one of them is a H atom.

Where V = (1 + 6 + 6 * 3 + 7) = 26 

Therefore, P = 6n + 2 – V = 6 * 5 + 2 – 32 = 0     So  there are no  π electrons (pi electrons) in HSO₃Cl  

Step 3 & 4: The Lewis structures of  HSO₃Cl are derived below:

Fig. 2 : Plausible Lewis structure of the chlorosulphonic acid.

Construct the electron dot structure of acetamide

A simple procedure for writing Lewis dot structures was given in a previous post entitled “Lewis Structures and the Octet Rule”.

Several worked examples relevant to this procedure were given in previous posts please see the Sitemap - Table of Contents (Lewis Electron Dot Structures).

Let us consider the case of acetamide CH₃CONH₂

Acetamide has many uses and applications such as:

• As a general solvent (molten acetamide is excellent solvent for many organic and inorganic compounds)

•  in biocides

• in preparation of cosmetics and hypnotics, 

• in various organic and inorganic syntheses, 

• as a drug intermediate in the manufacture of ampicilline, cephaclor, cephalexin, cephradine, enalapril

• as an antacid in lacquers and cosmetics

• as a plasticiser in leather and coatings

Step 1: Connect the atoms with single bonds.

Fig. 1 : Connect the atoms of acetamide with single bonds.

Step 2: Calculate the # of electrons in π bonds (pi bonds, multiple bonds)  using  formula (1): 


 

Where n in this case is 4 since CH₃CONH₂  consists of nine atoms but five of them is H.

Where V = (1*3 + 4 + 4 + 6 + 5 + 1*2) = 24 

Therefore, P = 6n + 2 – V = 6 * 4 + 2 – 24 = 2       So there are 2 π electrons (pi electrons) in CH₃CONH₂ and therefore1 double bond must be added to the structure of Step 1.

Step 3 & 4: One double bond must therefore be placed in the structure shown in Fig 1:

Figure 2: Lewis structures of acetamide