Predict the shape of each of the following molecules: $(6.8)$

a. $N{H}_{2}Cl(Nisthecentralatom)$

b.$Te{O}_2$

We need to find the shape of$N{H}_{2}Cl$.

Here, identify the total number of valence electrons in each molecule.

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

$N$is in group $5A$, it means it has $5$valence.

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

The total number of valence electrons.

localid="1652896514744" $$\begin{gathered} totalVE=2\times 1+5+7 \\ =2+12 \\ =14 \end{gathered}$$

the total valence electrons is localid="1652896519767" $14$.

the central atom is localid="1652896528093" $N$.

In order to complete the octet of the localid="1652896533146" $NandCl$atoms, there should be a single bond between the localid="1652896541264" $NandCl$atoms and single bonds in localid="1652896545863" $NandH$atoms.

Therefore, the Lewis structure of the molecule should be:

Based on VSEPR theory:

The molecule that its central atom $N$has $3$bonded atoms and $1$lone pair. So, the corresponding shape according to the VSEPR theory is trigonal pyramidal.

We need to find the shape of$Te{O}_2$.

Here, identify the total number of valence electrons in each molecule.

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

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

The total number of valence electrons.

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

the total valence electrons is $18$.

the central atom is $Te$because there is only one of it in the molecule.

In order to complete the octet of the $O$, there should be a double bond between the $Te=O$bond. So, the lewis structure of the molecule should be:

The molecule, the central atom $Te$has $2$bonded atom and $1$lone pair. So, the corresponding shape according to the VSERP theory is Bent$(120°)$.