# How do you draw double and triple bonds in Lewis Symbols?

Dec 17, 2016

#### Answer:

Let's look at 2 small molecules: $\text{acetylene}$; and $\text{hydrogen cyanide}$.

#### Explanation:

Both molecules contain formal triple bonds.

$\text{Acetylene:}$ ${C}_{2} {H}_{2}$; there are $2 \times 4 \left(C\right) + 2 \left(H\right)$ $\text{valence electrons}$, thus $10$ valence electrons to distribute.

$H - C \equiv C - H$

$\text{Hydrogen cyanide:}$ $H C N$; there are $4 \left(C\right) + 5 \left(N\right) + 1 \left(H\right)$ $\text{valence electrons}$, thus $10$ valence electrons to distribute.

$H - C \equiv N :$

At the nitrogen centre 2 of the valence electrons are lone-pair, and thus the nitrogen is associated with 2("inner shell")+2("lone pair)"+3("C-N triple bond)= 7 electrons", whose charge is balanced by the 7 protons in the nitrogen nucleus. Are you happy with this?

Can you similarly account for the bonding in $\text{ethylene}$, ${H}_{2} C = C {H}_{2}$, with $\text{12 valence electrons}$?