Question

Question: Draw one valid Lewis structure for each compound. Assume the atoms are arranged as drawn

a,

b.

c.

d.

Short answer

Answer

a,

b.

c.

d.

Step-by-step solution

Step-by-Step SolutionStep 1:Meaning of Lewis structures

Another name for Lewis structures is “electron dot structures.” The idea of bonding is obtained with the help of Lewis structures

Formation of Lewis structure

The primary step involved in the formation of Lewis structure is the arrangement of atoms near each other. After this arrangement, the valence electrons are counted.

Then the bonds are placed between different atoms, and the formal charge is provided to the atoms.

 Step 3: Lewis structure of compounds

a. The valence electrons of carbon, hydrogen, and nitrogen are four, one, and five, respectively. The electrons are represented as bonds between the atoms, and the lone pair that is present in nitrogen is also indicated in the Lewis structures.

$\begin{array}{rcl}\mathrm{Number}\mathrm{of}\mathrm{valence}{\mathrm{e}}^{-}& =& 1\mathrm{C}\times 4{\mathrm{e}}^{-}+2\mathrm{H}\times 1{\mathrm{e}}^{-}+2\mathrm{N}\times 5{\mathrm{e}}^{-}\\ & =& 4{\mathrm{e}}^{-}+2{\mathrm{e}}^{-}+10{\mathrm{e}}^{-}\\ & =& 16{\mathrm{e}}^{-}\end{array}$

Representation of Lewis structure of a

b. The valence electrons of carbon, hydrogen, nitrogen, and oxygen are four, one, five, and six, respectively. Bonds are utilized to indicate the electrons between the atoms. The oxygen consists of a lone pair, and this is also indicated in the Lewis structure.

$\begin{array}{rcl}\mathrm{Number}\mathrm{of}\mathrm{valence}{\mathrm{e}}^{-}& =& 1\mathrm{C}\times 4{\mathrm{e}}^{-}+3\mathrm{H}\times 1{\mathrm{e}}^{-}+1\mathrm{N}\times 5{\mathrm{e}}^{-}+2\mathrm{O}\times 6{\mathrm{e}}^{-}\\ & =& 4{\mathrm{e}}^{-}+3{\mathrm{e}}^{-}+5{\mathrm{e}}^{-}+12{\mathrm{e}}^{-}\\ & =& 24{\mathrm{e}}^{-}\end{array}$

Representation of Lewis structure of b

c.The valence electrons of carbon, hydrogen, and nitrogen, and oxygen are four, one, five, and six, respectively. The total count of valence electrons is identified.

$\begin{array}{rcl}\mathrm{Number}\mathrm{of}\mathrm{valence}{\mathrm{e}}^{-}& =& 2\mathrm{C}\times 4{\mathrm{e}}^{-}+3\mathrm{H}\times 1{\mathrm{e}}^{-}+1\mathrm{N}\times 5{\mathrm{e}}^{-}+1\mathrm{O}\times 6{\mathrm{e}}^{-}\\ & =& 8{\mathrm{e}}^{-}+3{\mathrm{e}}^{-}+5{\mathrm{e}}^{-}+6{\mathrm{e}}^{-}\\ & =& 22\mathrm{e}\end{array}$

Representation of Lewis structure of c

d. The valence electrons of carbon, hydrogen, and nitrogen are four, one, and five, respectively. The valence electrons of each atom are added to obtain the total number.

$\begin{array}{rcl}\mathrm{Number}\mathrm{of}\mathrm{valence}{\mathrm{e}}^{-}& =& 2\mathrm{C}\times 4{\mathrm{e}}^{-}+2\mathrm{H}\times 1{\mathrm{e}}^{-}+1\mathrm{N}\times 5{\mathrm{e}}^{-}+1{\mathrm{e}}^{-}\\ & =& 8{\mathrm{e}}^{-}+2{\mathrm{e}}^{-}+5{\mathrm{e}}^{-}+1{\mathrm{e}}^{-}\\ & =& 16{\mathrm{e}}^{-}\end{array}$

Representation of Lewis structure of d