CHEM231 - Problem Set #3
(Level 1 problems)
1. A metal complex of general formula ML6 can have two geometries: octahedral
and trigonal prismatic.
These are shown below, octahedral on the left and trigonal prismatic on the right.
a) What are the point groups of each of these geometries?
octahedral: Oh (of course!)
trigonal prismatic: D3h
b) Find the character tables for the two point groups you identified above in (a). Determine
the symmetry of the d- orbitals of the central metal in each of these point groups.
The d- orbitals are d(z2), d(x2-y2), d(xy), d(xz) and d(yz).
The d-orbtials in Oh transform as: Eg [d(z2), d(x2-y2)] and T2g [d(xy), d(xz), d(yz)] .
The d-orbtials in D3h transform
as: A1'
[d(z2)] , E' [d(xy), d(x2-y2)] and E" [d(xz),
d(yz)] .
c) The sigma bonds (MO’s) in an octahedral ML6 complex have symmetries a1, eg and t1u. Based
on this information which d- orbitals participate in sigma bonds and which
d-orbitals are non-bonding?
The doubly degenerate set
Eg [d(z2), d(x2-y2)] will participate in sigma bondign while the triply degerate
T2g set [d(xy), d(xz), d(yz)] will be non-bonding.
d) Will the trigonal prismatic complex have any triply degenerate sigma orbitals of any type? Why,
or why not?
No, because there are no triply degenerate symmetry types (irreducible representations) in D3h.
2. For the nitrate ion, NO3 _ :
a) Draw its structure as predicted by VSEPR theory.
Align the molecule in the usual way.
Hint: it will be easier if one oxygen atom is put on the negative x-axis.
b) What valence atomic orbitals should be used as a basis set for making a MO diagram for this ion?
c) Do the valence atomic orbitals of nitrogen have the same symmetry in this molecule?
Provide specific information to convince me?
(d) Using the equations for
the calculated molecular orbitals below (i.e., Y#1-#4) draw pictures
of these four molecular orbitals in NO3 _. Bold-type
numbers are coefficients on atomic orbitals.
(Note that these four, Y#1-#4, represent only a few of the many MO’s
in this ion).
#1 = - 0.76 . 2pz (N) + 0.37 . 2pz(Oa) + 0.37 . 2pz(Ob) + 0.37 . 2pz (Oc)
#2 = 0.47 . 2pz (N) + 0.51 . 2pz (Oa) + 0.51 . 2pz (Ob) + 0.51 . 2pz (Oc)
#3 = 0.40 . 2pz (Oa) + 0.40 . 2pz (Ob) - 0.83 . 2pz (Oc)
#4 = 0.71 . 2pz (Oa) - 0.71 . 2pz (Ob)
e) Is there any significance that there are four orbitals of this type?
f) Provide complete MO labels for the four orbitals, #1-#4
Click here for all answers to this problem
1. Using the valence atomic orbitals as a basis set, develop the MO diagram for the nitrite ion, NO2-. Include in your answer (a) linear combination of atomic orbitals for each MO (coefficients do not need to be specified) and (b) sketches of the MO’s (as best as you can manage) . Attempt an assignment of bonding character (anti-, non-, etc) to the MO’s.
Click here for the A1 molecular orbitals
Click here for the MO diagram (color coded to MO picture borders)
