91Ó°ÊÓ

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

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:

$$\begin{gathered} totalVE=4+4×7 \\ =4+32 \\ =32 \end{gathered}$$

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.

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

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:

$$\begin{gathered} totalVE=1×2+6 \\ =2+6 \\ =8 \end{gathered}$$

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 $\left(109°\right)$.