How is #"CO"_2# nonpolar?
1 Answer
Aug 14, 2017
By symmetry.
Polarity is a vector concept, i.e. a bond is polar because the electric dipole moment
#vecmu = << mu_x, mu_y, mu_z >>#
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 vecmu_i = 0#
For
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.
#"N"_2# - three-atom linear, e.g.
#"CO"_2# - trigonal planar, e.g.
#"BF"_3# - tetrahedral, e.g.
#"CCl"_4# - trigonal bipyramidal, e.g.
#"PF"_5# - octahedral, e.g.
#"SF"_6# - etc.
all of which were NONPOLAR as listed above.
The following, more usual examples, are POLAR:
#"NO"^(+)# , i.e.#:"N"-=stackrel((+))("O": )# , two-atom linear#"N"_2"O"# , i.e.#:stackrel((-))ddot"N"=stackrel((+))"N"=ddot"O":# , three-atom linear#"AlF"_2"Cl"# , trigonal planar#"CH"_3"Cl"# , tetrahedral#"PF"_3"Cl"_2# , trigonal bipyramidal#"SF"_5"Cl"# , octahedral
