Q.6.59 - An Introduction to General, Organic, and Biological Chemistry

Question

Draw the Lewis structure for each of the following molecules.

a. $\mathrm{HF}$

b. ${\mathrm{SF}}_2$

c. ${\mathrm{NBr}}_3$

d. ${\mathrm{ClON}}_2$

Step-by-Step Solution

Answer

The Lewis structures can be given as:

a. 

b. 

c.

d.

Step 1: Part (a) Given Information

To draw the Lewis structure of $\mathrm{HF}$.

Step 2: Part (a) Explanation

The electrons that are present in an outermost shell of an atom are known as Valence electrons.

In $\mathrm{HF}$, there are two atoms sulfur and hydrogen. The number of valence electrons present in fluorine is 7 while for hydrogen it is one.

Thus, the total number of valence electrons in $\mathrm{HF}$ is as follows:

$$\begin{gathered} \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{valence} \mathrm{electrons}= \mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{F} + \mathrm{Valence} {\mathrm{e}}^{- }\mathrm{of} \mathrm{H} \\ = 7 {\mathrm{e}}^{-}+1 {\mathrm{e}}^{-} \\ =8 {\mathrm{e}}^{-} \end{gathered}$$

Hence, the total number of valence electrons in $\mathrm{HF}$ are 8.

Now, determining Lewis structure, in order to obtain octet, one valence electron of a hydrogen atom is shared with one valence electron of a fluorine atom. The remaining six electrons are used as lone pairs that are placed around the $\mathrm{F}$ atom leading to octets at all the atoms.

Therefore, the Lewis structure of $\mathrm{HF}$ is shown below:

Step 1: Part (b) Given Information

To determine the Lewis structure of ${\mathrm{SF}}_2$.

Step 2: Part (b) Explanation

In ${\mathrm{SF}}_2$, there are three atoms, one sulfur and two fluorine atoms. The number of valence electrons present in sulfur is 6 while for fluorine it is seven.

Thus, the total number of valence electrons in ${\mathrm{SF}}_2$ is as follows:

$$\begin{gathered} \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{valence} \mathrm{electrons}= \mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{S} + 2\times \mathrm{Valence} {\mathrm{e}}^{- }\mathrm{of} \mathrm{F} \\ = 6 {\mathrm{e}}^{-}+(7\times 2) {\mathrm{e}}^{-} \\ =6+14=20 {\mathrm{e}}^{-} \end{gathered}$$

Hence, the total number of valence electrons in ${\mathrm{SF}}_2$ are 20.

Now, determining Lewis structure, four electrons $\left(2\times 2{\mathrm{e}}^{-}\right)$ are used to bond the Fluorine atoms with the central atom i.e., sulfur. Thus, the remaining electrons will be 16 electrons.

Thus, these 16 remaining electrons are used to complete the octets. These electrons are used as lone pairs that are placed around sulfur and fluorine atoms so that all atoms have octets.

Therefore, the Lewis structure of ${\mathrm{SF}}_2$ is shown below:

Step 1: Part (a) Given Information

To determine the Lewis structure of ${\mathrm{NBr}}_3$.

Step 2: Part (c) Explanation

In ${\mathrm{NBr}}_3$ , the central atom is nitrogen, and the number of valence electrons present in nitrogen is 5 while the three bromine atoms have seven valence electrons each.

Thus, the total number of valence electrons in ${\mathrm{NBr}}_3$ is as follows:

$$\begin{gathered} \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{valence} \mathrm{electrons}= \mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{N} + 3\times \mathrm{Valence} {\mathrm{e}}^{- }\mathrm{of} Br \\ = 5 {\mathrm{e}}^{-}+(3\times 7) {\mathrm{e}}^{-} \\ =5+21=26 {\mathrm{e}}^{-} \end{gathered}$$

Hence, the total number of valence electrons in  are 26.

Now, determining Lewis structure, three electrons $\left(2\times 3{\mathrm{e}}^{-}\right)$ are used to bond the bromine atoms with the central atom i.e., nitrogen. Thus, the remaining electrons will be 20 electrons.

Thus, these 20 remaining electrons are used to complete the octets. These electrons are used as lone pairs that are placed around nitrogen and bromine atoms so that all atoms have octets.

Therefore, the Lewis structure of  ${\mathrm{NBr}}_3$ is shown below:

Step 1: Part (d) Given Information

To determine the Lewis structure of ${\mathrm{ClNO}}_2$

Step 2: Part (d) Explanation

In ${\mathrm{ClNO}}_2$ , the central atom is nitrogen, and the number of valence electrons present in nitrogen is 5 while chlorine has 7 valence electrons and  each oxygen has six valence electrons.

Thus, the total number of valence electrons in ${\mathrm{ClNO}}_2$  is as follows:

$$\begin{gathered} \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{valence} \mathrm{electrons}= \mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{N} +\mathrm{Valence} {\mathrm{e}}^{- }\mathrm{of} \mathrm{Cl}+ 2\times \mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{O} \\ = 5 {\mathrm{e}}^{-}+7 {\mathrm{e}}^{-}+\left(2\times 6\right) {\mathrm{e}}^{-} \\ =5+7+12=24 {\mathrm{e}}^{-} \end{gathered}$$

Hence, the total number of valence electrons in  ${\mathrm{ClNO}}_2$ are 24.

Now, determining Lewis structure, three electrons $\left(2\times 3{\mathrm{e}}^{-}\right)$ are used to bond the chlorine and two oxygen atoms with the central atom i.e., nitrogen. Thus, the remaining electrons will be 16 electrons.

Thus, these 16 remaining electrons are used to complete the octets. These electrons are used as lone pairs that are placed around nitrogen, chlorine and oxygen atoms so that all atoms have octets.

Therefore, the Lewis structure of  ${\mathrm{ClNO}}_2$ is shown below: