Question

Draw a Lewis structure for each of the following compounds:

(a) C${}^2$H${}^6$ \hspace{.2cm} (b) C${}^2$H${}^4$ \hspace{.2cm} (c) C${}^2$H${}^2$ \hspace{.2cm} (d) C${}^3$H${}^8$ \hspace{.2cm} (e) C${}^3$H${}^6$ \hspace{.2cm} (f) CH${}^3$OH

Answer 1

Solution

The task is to draw Lewis dot structures for the following compounds:

(a) C${}^2$H${}^6$ Math output error\hspace{.2cm} (b) C${}^2$H${}^4$ Math output error\hspace{.2cm} (c) C${}^2$H${}^2$ Math output error\hspace{.2cm} (d) C${}^3$H${}^8$ Math output error\hspace{.2cm} (e) C${}^3$H${}^6$ Math output error\hspace{.2cm} (f) CH${}^3$OH

 

When drawing Lewis structures, first determine the number of valence electrons for each atom. The number of valence electrons is indicated by the group number.

Therefore, carbon has 4 (one on each side) and hydrogen 1.

Then, connect atoms that can form more than one bond and place them in the centrum.

In the end, arrange hydrogens at terminals. Keep in mind the octet rule. The rule states that each atom tends to have 8 electrons in its valence shell (except hydrogen which tends to have a duet) in order to have a noble gas configuration and hence maximum stability.

 

(a) Since carbon can form more than one bond, connect the two carbons first and place them in the centrum. Three hydrogens go on each carbon and so the octet rule is satisfied.

(b) In this case, by the distribution of hydrogens, the octet is not satisfied. Therefore, the two remaining unpaired electrons are shared between two carbons to form a double bond.

 

(c) After the addition of hydrogens, each carbon is left with two unpaired electrons. So, four unpaired electrons are shared to form a triple bond so that each carbon would have an octet.

(d) First, connect the three carbon atoms since carbon can form multiple bonds. By the distribution of hydrogens, each carbon achieves an octet.

 

(e) A similar situation as in (b). There are not enough hydrogens for each carbon to have an octet, so two unpaired electrons are shared to form a double bond.

(f) Oxygen has 6 valence electrons. Four of them make up for two lone electron pairs (not shared), whereas two unpaired electrons are shared with other atoms. Place carbon and oxygen in the centrum and add hydrogens at terminals. Do not forget to place two lone pairs on oxygen.