91Ó°ÊÓ

We need to determine the shape of$P{H}_3$.

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

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

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

The total number of valence electrons in the compound is:

$$\begin{gathered} totalVE=5+3×1 \\ =5+3 \\ =8 \end{gathered}$$

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

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

The central atom $P$is bonded to $3$atoms with $1$lone pair. So, based on VSEPR theory, the shape of the molecule is trigonal pyramidal. .

We need to determine the shape of$HCN$.

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.

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

The total number of valence electrons in the compound is:

$$\begin{gathered} totalVE=1+4+5 \\ =10 \end{gathered}$$

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

In order to complete the octets of the $CandN$atoms in the molecule, there should be a single bond between each $C-H$and triple bond $C≡N$in the molecule. Then, the Lewis structure of the molecule is:

The central atom $C$has $2$bonded atoms and $0$lone pairs.So, based on VSEPR theory, the shape of the molecule is linear.