Q. 6.139 - An Introduction to General, Organic, and Biological Chemistry

Question

Use the Lewis structure to determine the shape of the following: $(6.6 ,6.8)$

a. $CB{r}_4$

b. $H_{2}O$

Step-by-Step Solution

Answer

a.  The shape of $CB{r}_4$ is tetrahedral.

b.  The shape of $H_{2}O$ is bent $\left(109°\right)$.

Part (a) step 1: Given Information

We need to determine the shape of $CB{r}_4$.

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.

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

The total number of valence electrons in the compound is: 

$$\begin{gathered} total VE=4+4\times 7 \\ =4+32 \\ =32 \end{gathered}$$

So, the total number of valence electrons is $32$.

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

The central atom  $C$ is bonded to $4$  atoms with  $0$lone pair. So, based on VSEPR theory, the shape of the molecule is tetrahedral.

Part (b) step 1: Given Information

We need to determine the shape of $H_{2}O$.

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.

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

The total number of valence electrons in the compound is: 

$$\begin{gathered} total VE=1\times 2+6 \\ =2+6 \\ =8 \end{gathered}$$

So, the total number of valence electrons is $8$.

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

The central atom  $O$ has $2$ bonded atoms with  and $2$ lone pairs. So, based on VSEPR theory, the shape of the molecule is bent $\left(109°\right)$.