Q. 6.138 - An Introduction to General, Organic, and Biological Chemistry

Question

For each of the following, draw the Lewis structure and determine the shape:  $(6.6, 6.8)$

a. $COC{l}_2$($C$ is the central atom)

b. $HCCH$

c. $S{O}_2$

Step-by-Step Solution

Answer

a. The shape is trigonal planar and the Lewis structure is .

b. The shape is linear and the Lewis structure is .

c. The shape is bent $\left(120°\right)$ and the Lewis structure is  .

Part (a) step 1: Given Information

We need to draw the Lewis structure and determine the shape of $COC{l}_2$.

Part (a) step 2: Simplify

Now, count the number of the total valence electrons of the compound. 

$C$ is in group $4A$, it means it has $4$ valence electrons.

$O$ is in group  $6A$, it means it has $6$  valence electrons.

$Cl$ is in group $7A$, it means it has $7$  valence electrons.

Here, the total number of valence electrons in the compound is:

$$\begin{gathered} total VE=4+6+2\times 7 \\ =10+14 \\ =24 \end{gathered}$$

Then, the total number of valence electrons is  $24$.

In order to complete the octets of the $C, O, Cl$ atoms in the molecule, there should be a single bond between each  $C-Cl$ and a double bond between $C=O$in the molecule. So, the Lewis structure of the molecule is:

Based on the VSEPR theory, the theory states that the molecular geometry of a molecule with the following bonded atoms and lone pair is below. 

Mol Geom	Bonded	Lone pair
Linear	$2$	$0$
Trigonal Planar	$3$	$0$
Bent  $(120 °)$	$2$	$1$
Tetrahedral	$4$	$0$
Trigonal Pyramidal	$3$	$1$
Bent $\left(109°\right)$	$2$	$2$

The central atom $C$  is bonded to $3$atom with $0$lone pair. So, based on VSEPR theory, the shape of the molecule is trigonal planar.

Part (b) step 1: Given Information

We need to draw the Lewis structure and determine the shape of  $HCCH$.

Part (b) step 2: Simplify

Now, count the number of the total valence electrons of the compound. 

$H$ is in group $1A$, it means it has  $1$valence electrons.

$C$ is in group  $4A$ , it means it has $4$ valence electrons.

Here, the total number of valence electrons in the compound is:

$$\begin{gathered} total VE=2\times 1+2\times 4 \\ =2+8 \\ =10 \end{gathered}$$

Then, the total number of valence electrons is $10$.

In order to complete the octets of the  $C$ atoms in the molecule, there should be a triple bond between the $C\equiv C$ in the molecule. So, the Lewis structure of the molecule is:

The central atom $C$ are bonded to $2$ atom with $0$ lone pairs. So, based on VSEPR theory, the shape of the molecule is linear.

Part (c) step 1: Given Information

We need to draw the Lewis structure and determine the shape of $S{O}_2$

Part (c) step 2: Simplify

Now, count the number of the total valence electrons of the compound. 

$S$ is in group $6A$, it means it has $6$ valence electrons.

$C$ is in group $6A$ , it means it has  $6$  valence electrons.

Here, the total number of valence electrons in the compound is:

$$\begin{gathered} total VE=6+2\times 6 \\ =6+12 \\ =18 \end{gathered}$$

Then, the total number of valence electrons is  $18$.

In order to complete the octets of the $O$ atoms in the molecule, there should be a double bond between the $S=O$ in the molecule. So, the Lewis structure of the molecule is:

The central atom  $S$ is bonded to  $2$ atom and with $1$ lone pairs. So, based on VSEPR theory, the shape of the molecule is Bent $\left(120°\right)$.