# How is "CO"_2 nonpolar?

Aug 14, 2017

By symmetry.

Polarity is a vector concept, i.e. a bond is polar because the electric dipole moment $\vec{\mu}$ points in a given direction ($x , y , z$) with a given magnitude $| \vec{\mu} |$, i.e.

$\vec{\mu} = \left\langle{\mu}_{x} , {\mu}_{y} , {\mu}_{z}\right\rangle$

As vectors of identical magnitudes in exactly opposite directions add to cancel out completely, a "perfectly symmetrical" compound must be nonpolar, so that

${\sum}_{i} {\vec{\mu}}_{i} = 0$

For ${\text{CO}}_{2}$, we would have...

So, what do we mean by "perfectly symmetrical"? We mean the parent geometries of each so-called VSEPR structure, i.e. the ones with no lone pairs of electrons:

• two-atom linear, e.g. ${\text{N}}_{2}$
• three-atom linear, e.g. ${\text{CO}}_{2}$
• trigonal planar, e.g. ${\text{BF}}_{3}$
• tetrahedral, e.g. ${\text{CCl}}_{4}$
• trigonal bipyramidal, e.g. ${\text{PF}}_{5}$
• octahedral, e.g. ${\text{SF}}_{6}$
• etc.

all of which were NONPOLAR as listed above.

The following, more usual examples, are POLAR:

• ${\text{NO}}^{+}$, i.e. :"N"-=stackrel((+))("O": ), two-atom linear
• $\text{N"_2"O}$, i.e. $: \stackrel{\left(-\right)}{\ddot{\text{N"=stackrel((+))"N"=ddot"O}}} :$, three-atom linear
• $\text{AlF"_2"Cl}$, trigonal planar
• $\text{CH"_3"Cl}$, tetrahedral
• ${\text{PF"_3"Cl}}_{2}$, trigonal bipyramidal
• $\text{SF"_5"Cl}$, octahedral