Q.6.60 - An Introduction to General, Organic, and Biological Chemistry

Question

Draw the Lewis structure for each of the following molecules.

a. $\mathrm{HCCH}$

b. ${\mathrm{CS}}_2$

c. ${\mathrm{H}}_2\mathrm{CO}$

d. ${\mathrm{SiF}}_4$

Step-by-Step Solution

Answer

The Lewis structures are given as follows:

a.

b. 

c.

d.

Step 1: Part (a) Given Information

Determine the Lewis structure of $\mathrm{HCCH}$.

Step 2: Part (a) Explanation

In $\mathrm{HCCH}$, there are two atoms of carbon and hydrogen each. The number of valence electrons present in hydrogen is 1 while for the carbon it is four.

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

$$\begin{gathered} \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{valence} \mathrm{electrons}=2\times \mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{H}+2\times \mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{C} \\ =\left(2\times 1\right)+\left(2\times 4\right) \\ =2+8=10 {\mathrm{e}}^{-} \end{gathered}$$

Hence, the total number of valence electrons in $\mathrm{HCCH}$ are 10.

Now, determining Lewis structure, in order to obtain an octet, one valence electron of a hydrogen atom is shared with one valence electron of a carbon atom for both the atoms, while the remaining valence electrons of carbon atoms are shared between each other forming a double bond.

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

Step 1: part (b) Given Information

To determine Lewis structure of ${\mathrm{CS}}_2$

Step 2: part (b) Explanation

In ${\mathrm{CS}}_2$ , there are three atoms, two sulfur, and one carbon atom. The number of valence electrons present in sulfur is 6 while for the carbon it is four.

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

$$\begin{gathered} \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{valence} \mathrm{electrons}=2\times \mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{S}+\mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{C} \\ =\left(2\times 6\right)+4 \\ =12+4=16 {\mathrm{e}}^{-} \end{gathered}$$

Hence, the total number of valence electrons in ${\mathrm{CS}}_2$ are 16.

Now, determining Lewis structure, four electrons $(2\times 2{\mathrm{e}}^{-})$ are used to bond the Sulfur atoms with the central atom i.e., carbon Thus, the remaining electrons will be 12 electrons.

Thus, these 12 remaining electrons are used to complete the octets. These electrons are used as lone pairs that are placed around sulfur atoms in order to complete octets. 

 The carbon atom forms double bonds with both the sulfur atoms.

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

Step 1: Part (c) Given information

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

Step 2: Part (c) Explanation

In ${\mathrm{H}}_2\mathrm{CO}$, there are two hydrogen atoms, one carbon, and one oxygen atom present. the number of valence electrons of hydrogen is 1, for carbon, it is 4 while for oxygen it is 6.

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

$$\begin{gathered} \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{valence} \mathrm{electrons}=2\times \mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{H}+\mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{C}+\mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{O} \\ =\left(2\times 1\right)+4+6 \\ =2+4+6=12 {\mathrm{e}}^{-} \end{gathered}$$

Hence, the total number of valence electrons in ${\mathrm{H}}_2\mathrm{CO}$ are 12.

Now, determining Lewis structure, four electrons $(4\times 1{\mathrm{e}}^{-})$ are used to bond the oxygen and hydrogen atoms with the central atom i.e., carbon. Thus, the remaining electrons will be 8 electrons.

Thus, these 8 remaining electrons are used to complete the octets. These electrons are used as lone pairs that are placed around oxygen atoms in order to complete octets. 

 The carbon atom forms double bonds with the oxygen atom and a single bond with two hydrogen atoms.

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

Step 1: Part (d) Given Information

To determine the Lewis structure of ${\mathrm{SiF}}_4$.

Step 2: Part (d) Explanation

In ${\mathrm{SiF}}_4$, there are one silicon atom and four fluorine atoms present. The number of valence electrons of silicon is 4 while for fluorine it is 7.

Thus, the total number of valence electrons in ${\mathrm{SiF}}_4$ are as follows:

$$\begin{gathered} \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{valence} \mathrm{electrons}=\mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{Si}+2\times \mathrm{Valence} {\mathrm{e}}^{-} \mathrm{of} \mathrm{F} \\ =4+\left(2\times 7\right) \\ =4+14=18 {\mathrm{e}}^{-} \end{gathered}$$

Hence, the total number of valence electrons in ${\mathrm{SiF}}_4$ is 18.

Now, determining Lewis structure, four electrons $(4\times 1{\mathrm{e}}^{-})$ are used to bond the fluorine atoms with the central atom i.e., silicon. Thus, the remaining electrons will be 14 electrons.

Thus, these 14 remaining electrons are used to complete the octets. These electrons are used as lone pairs that are placed around fluorine atoms in order to complete octets. 

Therefore, the Lewis structure of ${\mathrm{SiF}}_4$ is shown below: