ClO
2- Total number of electrons 7 + 2 x 6 + 1 = 20 = 10 pairs
Identify the central atom and connect the peripheral atoms with a pair of electrons as two dots and count the remaining electron pairs
![](../../../../images/module1/lec1/eqn2.png)
20-4 = 16 electrons left (8 pairs)
Complete the octet of oxygen atoms and count the remaining electron pairs
![](../../../../images/module1/lec1/eqn3.png)
16-12 = 4 electrons
Complete the octet of chlorine atom,
![](../../../../images/module1/lec1/eqn4.png)
Since all the electrons are utilized and octet is satisfied, no need of any multiple bonds.
The structure is
![](../../../../images/module1/lec1/eqn5.png)
or
![](../../../../images/module1/lec1/eqn6.png)
or
![](../../../../images/module1/lec1/eqn7.png)
CO (carbond monoxide) Total number of electrons 4 + 6 = 10
Connect the two atoms with a pair of electrons as two dots and count the remaining electron pairs
![](../../../../images/module1/lec1/eqn8.png)
10-2 = 8 electrons left
Complete the octet of oxygen atoms and count the remaining electron pairs
![](../../../../images/module1/lec1/eqn9.png)
8-6 = 2 electrons
Place the remaining electron pair on carbon atom,
![](../../../../images/module1/lec1/eqn10.png)
All the electrons are utilized and octet is not satisfied for carbon as there is a shortage of four more electrons.
![](../../../../images/module1/lec1/eqn11.png)
Drag two electron pairs on oxygen atom in between carbon and oxygen to establish two more bonds so that a triple bond exists between C and O which satisfies the octet of both C and O.
The structure is
![](../../../../images/module1/lec1/eqn12.png)
or
![](../../../../images/module1/lec1/eqn13.png)
or
![](../../../../images/module1/lec1/eqn14.png)