What is the formal charge on the phosphate ion in #PO_3^-#?
1 Answer
Assigning formal charges assumes
#"FC" = "valence e"^(-) - "owned e"^(-)# where:
- owned electrons are found by cleaving each bond homolytically so that one electron goes to each atom that was bonding.
- valence electrons are found from the group number of the main group elements.
Of course, this is meaningless for a molecule as a whole. Furthermore, the ion you've quoted does not exist.
Phosphate is actually
Assign the labels
#"O"_((1))# would bring in#bb6# valence electrons, and it owns#overbrace("2 lone pairs" xx "2 electrons each")^("lone pairs")# #+# #overbrace("2 bonds" xx "1 electron")^("bonding electrons") = bb6# of them.
Thus, its formal charge is
#0# .
#"O"_((2) - (4))# would each bring in#bb6# valence electrons, and they each own#overbrace("3 lone pairs" xx "2 electrons each")^("lone pairs")# #+# #overbrace("1 bond" xx "1 electron")^("bonding electrons") = bb7# of them.
Thus, their formal charges are all
#-1# .
- And lastly, the
#"P"# in the middle would bring in#5# valence electrons, but own#5 xx 1 = 5# of them, having a formal charge of#0# .
And we indeed still have a total charge of
#overbrace((0))^(O_((1))) + 3 xx overbrace((-1))^(O_((2) - (4))) + overbrace((0))^(P) = overbrace(3^-)^(PO_4^(3-))#