Q. 6.140 - 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. $P{H}_3$

b. $HCN$

Step-by-Step Solution

Answer

a. The shape of $P{H}_3$ is trigonal pyramidal.

b. The shape of $HCN$ is linear.

Part (a) step 1: Given Information

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

Part (a) step 2: Simplify

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:  

$$\begin{gathered} total VE=5+3\times 1 \\ =5+3 \\ =8 \end{gathered}$$

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. .

Part (b) step 1: Given Information

We need to determine the shape of $HCN$.

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. 

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

The total number of valence electrons in the compound is:   

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

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

In order to complete the octets of the  $C and N$ atoms in the molecule, there should be a single bond between each $C-H$ and triple bond $C\equiv N$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.