# Molecular Orbitals

## Contents

#### Problems and Solutions

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. B2
b. C2
c. O2
d. NO
e. CO

a. B2 is paramagnetic because it has two unpaired electrons, one in each of its p orbitals.
b. C2 is diamagnetic because all of its electrons are paired.
c. O2 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. B2
b. C2
c. O2
d. NO
e. CO

a. B2 has 2 bonding pairs and 1 antibonding pair so it has a bond order of 1.
b. C2 has 3 bonding pairs and 1 antibonding pair so it has a bond order of 2.
c. O2 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. CH4
b. HCCCH
c. IF4 -
d. IF 4 +
e. CH2O

a. Tetrahedral--sp3
b. Linear--sp
c. Octahedral--d2sp3
d. Trigonal Bipyramidal--dsp3
e. Trigonal Planar--sp2

Problem :

Combining what you know about resonance and hybridization, predict the hybridization of each oxygen in the acetate ion--CH3CO2 -.

Because each oxygen is equivalent, their hybridizations must also be equivalent. Therefore, we must choose between sp2 and sp3 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 sp2 for both oxygens.