Chapter U2.32, Problem 8E

Interpretation Introduction

Interpretation: Lewis structure for ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ and ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ should be drawn.

Concept Introduction :

• Lewis structures are the diagrams that show the bonding between the atoms of the molecules and existing lone pairs of electrons.
• Bonding electrons are those electrons which are shared between the atoms resulting in the formation of bond.
• Non-bonding electrons are the valence electrons of the atom which are not shared with another atom.

Answer to Problem 8E

Lewis structure for ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ and ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is as follows:

  

Explanation of Solution

For drawing the Lewis structure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ :

• Calculate the total number of valence electrons in molecule ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$:

Total number of valence electrons = 2(valence electrons of C) + 4(valence electrons of H)

  $=2(4)+4(1)={\text{12 e}}^{-}$

• Arrange the atoms in such a way that the least electronegative atom is at the center. Then put the valence electrons around them such that each atom contributes at least 1 electron to single bond and the octet rule for each atom is followed.
• To complete the octet each hydrogen, H atoms forms single bond with carbon, C and the two carbon will form double bond between them. The rest number of electrons are present as lone pair, the number of lone pair of electrons = Total number of valence electrons − Total number of electrons involved in bond formation.

The number of lone pair of electrons = 12 − 12 = 0 electrons

The Lewis structure for ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ is:

  

For drawing the Lewis structure of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ :

• Calculate the total number of valence electrons in molecule ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$:

Total number of valence electrons = 2(valence electrons of N) + 4(valence electrons of H)

  $=2(5)+4(1)={\text{14 e}}^{-}$

• Arrange the atoms in such a way that the least electronegative atom is at the center. Then put the valence electrons around them such that each atom contributes at least 1 electron to single bond and the octet rule for each atom is followed.
• To complete the octet each hydrogen, H atoms forms single bond with nitrogen, N and the two nitrogen will form single bond between them. The rest number of electrons are present as lone pair on nitrogen atom, the number of lone pair of electrons = Total number of valence electrons − Total number of electrons involved in bond formation.

The number of lone pair of electrons = 14 − 10 = 4 electrons

The Lewis structure for ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is: