Question

Determine the formal charges on all the atoms in the following Lewis diagrams.

Which one would best represent bonding in the $C{l}_{2}O$

Short answer

The second option is more favorable because the formal charges have lower absolute values.

Step-by-step solution

Step 1

Formal charges are used as guidelines for Lewis representation. Each atom in a molecule can be assigned a formal charge in the molecule:

Now, the formal charge of $X=groupofX-\left(loneelectronsonX+\frac{1}{2}electronsinbondstoX\right)$

So, in the first option:

Terminal $Cl$in the $7^{th}$group, $6$lone electrons, $1$bond of electrons

Formal charge of terminal

$$\begin{gathered} Cl=7-\left(6+\frac{1}{2}·2\right) \\ =7-7=0 \end{gathered}$$

Then, Central $Cl$in the$7^{th}$group , lone electrons, $1$ bond of $2$electrons

So, the formal charge of central $Cl$:

$$\begin{gathered} Cl=7-\left(4+\frac{1}{2}·2·2\right) \\ =7-6=1 \end{gathered}$$

And, then $O$in the $6^{th}$group, $6$lone electrons, $1$
bond of electrons

Thus, the formal charge of central $O$:

localid="1663678834507" $$\begin{gathered} O=6-\left(6+\frac{1}{2}·1·2\right) \\ =6-7=\left(-1\right) \end{gathered}$$

Step 2

Now, in the second option:

Terminal $Cl$ in the$7^{th}$group, $6$lone electrons, $1$bond of $2$electrons

Formal charge of terminal :

$$\begin{gathered} Cl=7-\left(6+\frac{1}{2}·2\right) \\ =7-7=0 \end{gathered}$$

And, then $O$in the $6^{th}$group, $6$lone electrons, $1$bond of $2$electrons

Thus, the formal charge of central :

$$\begin{gathered} O=6-\left(6+\frac{1}{2}·1·2\right) \\ =6-7=\left(-1\right) \end{gathered}$$

Thus, in the second option the formal charges are all $0$whereas in the first option, oxygen and central chlorine have formal charges of $-1$and $+1$respectively.

Therefore, the second option is more favorable because the formal charges have lower absolute values.

Hence, the Lewis structures in which atoms have smaller formal charges are considered better representations.