##CO_2## has 2 .
First, start with the molecule’s , which allows you to determine the of each atom.
We can see that ##C## has two regions of electron around it, which means it has a steric number equal to 2. This implies that it is ##sp## hybridized, therefore has 2 unhybridized p-orbitals with which it can form with the oxygen atoms.
On the other hand, each ##O## atom has three regions of electron density around it, which means it is ##sp^2## hybridized. This allows each ##O## atoms to have 1 unhybridized p-orbital with which to form a pi bond.
The in the ##CO_2## molecule looks like this:
##C##’s sp hybridized orbitals are shown in yellow and its two unhybridized p-orbitals are shown in blue. ##O##’s ##sp^2## hybridized orbitals are shown in green, while its remaining unhybridized p-orbital is shown in blue.
So, ##C## uses its two sp hybridized orbitals to form with the two ##O## atoms (each ##O## atom uses an ##sp^2## orbital for this).
Let’s say ##C## has its ##p_z## and ##p_x## orbitals left unhybridized. Its ##p_x## orbital will form a with the ##O## atom that has its ##p_x## orbital unhybridized, while its ##p_z## orbital will for a with the other ##O## atom’s ##p_z## orbital.
Therefore, ##CO_2## has 2 pi bonds and 2 .
A faster way to determine how many pi bonds the molecule has is to know that a double bond is comprised of 1 sigma and 1 pi bond. Since ##CO_2## has 2 double bonds, it will have 2 pi bonds.
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