N: 5 electrons 8 electrons
3 O: 18 electrons 24 electrons
24 electrons 32 electrons
(32 – 24) electrons = 8 electrons = 4 bonds
Place N at the center:
(N and O both have a formal charge.)
They cannot be reduced because the N octet cannot be expanded. However, since
resonance will be present, a better version might be:
H 3 PO 4
has needs
3 H 3 electrons 3(2) = 6 electrons
P 5 electrons 8 electrons (at least)
4 O 24 electrons 32 electrons
32 electrons 46 electrons
(46 – 32) electrons = 14 electrons = 7 bonds (at least)
Place the P at the center, the four oxygens around it, and hydrogens on three of the
oxygens, with single bonds between them; this will use all seven bonds.
B.
