Lewis Dot Structures|How to draw Lewis Structure – Lewis structures of Cyanamide NCN-2 | Chemistry Net

Lewis Dot Structures|How to draw Lewis Structure – Lewis structures of Cyanamide NCN-2

Lewis Dot Structures|How to draw Lewis Structure – NCN-2

Lewis Dot Structures of the Cyanamide ion NCN-2

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 the NCN-2 ion and draw the Lewis dot structures:

Step 1: Connect the atoms with single bonds.

Fig. 1: Connecting the atoms of the NCN-2 with single bonds according to Step 1 of the method

 

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

Where n in this case is 3 since NCN-2consists of  3 atoms.

Where V = 5 + 4 + 5 – (-2) = 16

Therefore, P = 6n + 2 – V = 6 *3 + 2 – 16 = 4   Therefore, there are either 2 double bonds or 1 triple  bond.

Step 3 & 4: The Lewis structures of of NCN-2 are as follows:

Fig. 2: Lewis resonance structures of the cyanamide NCN-2 ion

 

The ion represented by the electron dot structure #1 is the most stable comparing to the #2 and #3 because of less charge separation. The cyanamide ion is isoelectronic and hence isostructural with the CO2 molecule. 


 

Relevant Posts

Lewis Structures|Octet Rule: A Simple Method to write Lewis Structures

Canonical Structures - Lewis Dot Structure of OCN-

 


 

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)

 

Key Terms

resonance structures of cyanate ion NCN-2, Lewis electron structures of cyanamide ion, chemical formula of the cyanamide, simple procedure for drawing Lewis structures of cyanamide ion,

 


 

1 comment:

  1. I'm confused, I thought there would be 18 electrons not 16, there is a negative charge of 2, thus there are 2 extra electrons? In which case there would be a single bond and a double bond, along with a lone pair. No?

    ReplyDelete