**Problem : **

Paramagnetic materials, those with unpaired electrons, are attracted by magnetic fields whereas diamagnetic materials, those with no unpaired electrons, are weakly repelled by such fields. By constructing a molecular orbital picture for each of the following molecules, determine whether it is paramagnetic or diamagnetic.

a. B_{2}

b. C_{2}

c. O_{2}

d. NO

e. CO

a. B_{2} is paramagnetic because it has two unpaired electrons, one
in each of its p orbitals.

b. C_{2} is diamagnetic because all of its electrons are paired.

c. O_{2} is paramagnetic because it has two unpaired electrons, one
in each of its p* orbitals.

d. NO has an odd number of electrons and, therefore, must be paramagnetic.

e. CO is diamagnetic because all of its electrons are paired.

**Problem : **

Calculate the bond order for each of the following molecules (Hint: first draw a correlation diagram for each).

a. B_{2}

b. C_{2}

c. O_{2}

d. NO

e. CO

a. B_{2} has 2 bonding pairs and 1 antibonding pair so it has a
bond order of 1.

b. C_{2} has 3 bonding pairs and 1 antibonding pair so it has a
bond order of 2.

c. O_{2} has 4 bonding pairs and 2 antibonding pairs so it has a
bond order of 2.

d. NO has 4 bonding pairs and 1.5 antibonding pairs so it has a bond order
of 2.5.

e. CO has 4 bonding pairs and 1 antibonding pair so it has a bond order of 3.

**Problem : **

Predict the hybridization of the central atom in each of the following molecules.

a. CH_{4}

b. HCCCH

c. IF_{4}
^{-}

d. IF _{4}
^{+}

e. CH_{2}O

a. Tetrahedral--sp^{3}

b. Linear--sp

c. Octahedral--d^{2}sp^{3}

d. Trigonal Bipyramidal--dsp^{3}

e. Trigonal Planar--sp^{2}

**Problem : **

Combining what you know about resonance and hybridization, predict the
hybridization of each
oxygen in the acetate ion--CH_{3}CO_{2}
^{-}.

Because each oxygen is equivalent, their hybridizations must also be
equivalent. Therefore, we must
choose between sp^{2} and sp^{3} hybridization. For best
overlap, it makes
sense to assume that the resonating lone pair is in a p-orbital so that it
is properly positioned to make
a p bond to the central carbon. Because we must
have that p-orbital free,
i.e. not hybridized, the hybridization must be sp^{2} for both oxygens.