# What is the Lewis structure of OCS?

Aug 4, 2016

If you think about it, $\text{S}$ is right below $\text{O}$, so the number of valence electrons it contributes is the same.

Therefore, $\text{OCS}$ is isoelectronic with ${\text{CO}}_{2}$, having the same number of valence electrons but not identical atoms.

So, what you have in comparison is:

$\textcolor{b l u e}{: \stackrel{. .}{\text{O"="C"=stackrel(..)"S}} :}$ $\implies$ $\text{OCS}$

$: \stackrel{. .}{\text{O"="C"=stackrel(..)"O}} :$ $\implies$ ${\text{CO}}_{2}$

You can do the electron-counting method and get the same result.

$\text{O} : 6$
$\text{C} : 4$
$\text{S} : 6$

$6 + 4 + 6 = 16$ valence electrons

Distribute $8$ of them in two double bonds ($2$ per bond line), and you have $8$ remaining.

$\text{O"="C"="S}$

So, $4$ as lone pairs on $\text{O}$ and $4$ as lone pairs on $\text{S}$. This also gives you the above result.

$: \stackrel{. .}{\text{O"="C"=stackrel(..)"S}} :$