Q. 6.137 - An Introduction to General, Organic, and Biological Chemistry

Question

For each of the following, draw the Lewis structure and determine the shape: $(6.6 , 6.8)$

a. $N{F}_3$

b. $SiB{r}_4$

c. $CS{e}_2$

Step-by-Step Solution

Answer

a. The shape is trigonal pyramidal and the Lewis structure is .

b. The shape is tetrahedral and the Lewis structure is  .

c. The shape is linear and the Lewis structure is .

Part (a) step 1: Given Information

We need to draw the Lewis structure and determine the shape of $N{F}_3$.

Part (a) step 2: Simplify

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

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

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

Here, the total number of valence electrons in the compound is:

$$\begin{gathered} total VE=5+3\times 7 \\ =5+21 \\ =26 \end{gathered}$$

Then, the total number of valence electrons is $26$.

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

Based on the VSEPR theory, the theory states that the molecular geometry of a molecule with the following bonded atoms and lone pair is below.

Geometery	Bonded	Lone Pairs
Linear	$2$	$0$
Trigonal Planar	$3$	$0$
Bent $\left(120°\right)$	$2$	$1$
Tetrahedral	$4$	$0$
Trigonal Pyramidal	$3$	$1$
Bent$\left(109°\right)$	$2$	$2$

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

Part (b) step 1: Given Information

We need to draw the Lewis structure and determine the shape of $SiB{r}_4$.

Part (b) step 2: Simplify

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

$Si$ is in group $4A$, it means it has $4$ valence electrons.

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

Here, the total number of valence electrons in the compound is:

$$\begin{gathered} total VE=5+3\times 7 \\ =5+21 \\ =26 \end{gathered}$$

Then, the total number of valence electrons is  $26$.

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

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

Part (c) step 1: Given Information

We need to draw the Lewis structure and determine the shape of $Cs{e}_2$.

Part (c) 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.

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

Here, the total number of valence electrons in the compound is:

$$\begin{gathered} total VE=4+2\times 6 \\ =4+12 \\ =16 \end{gathered}$$

Then, the total number of valence electrons is $16$.

In order to complete the octets of the  $C$and  $Se$ atoms in the molecule, there should be a double bond between each  $C=Se$  in the molecule. So, the Lewis structure of the molecule is:

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