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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
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
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
Problem :
Combining what you know about resonance and hybridization, predict the hybridization of each oxygen in the acetate ion--CH3CO2 -.
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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
Solution for Problem 1 >>
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.
Close
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
Solution for Problem 2 >>
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.
Close
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
Solution for Problem 3 >>
a. Tetrahedral--sp3
b. Linear--sp
c. Octahedral--d2sp3
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d. Trigonal Bipyramidal--dsp3
e. Trigonal Planar--sp2
Close
Problem :
Combining what you know about resonance and hybridization, predict the hybridization of each oxygen in the acetate ion--CH3CO2 -.
Solution for Problem 4 >>
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.
