(b) Calculate the Do and CFSE in kJ/mol for this complex.
(Recall the relationship between wavelength, nm, and wavenumber, cm-1 and that
1.1962 x 10-2 x E (cm-1)= E (kJ/mol).
1. The point symmetry of [Ni(CN)5]3- is C4v. What is the geometry of the ion?
This symmetry only fits a square pyramidal geometry.
2. Consider the reactions below to answer questions (a) and (b).
(i) 3 phenanthroline + [Ru(H2O)6]2+ ---> 6 H2O + [Ru(phen)3] 2+
(ii) 3 oxalate + [Ru(H2O)6]2+ ---> 6 H2O + [Ru(oxalate)3]4-
(a) Which reaction, (i) or (ii), is predicted to have larger value for its Kf and why?
Reaction (i) will have a larger Kf. While both reactions use bidentate ligands and so benefit from the chelate effect, the first reaction uses ligands with N-atom donors. Since Ru(2+) is a soft ion it will prefer teh N atoms of phen ligands to the harder O atoms of oxalate.
(b) What is the point symmetry of the Ru product?
D3
3. (a) Create five complexes having symmetries of these point groups:
D4h, C3v, C2v, D3, Cs
by exchanging water ligands (i.e., making ligand substitutions) on [Ni(H2O)6]2+
using any of the following ligands:
ethylenediamine (en), chloride, acetylacetonate (acac), thiocyanate, ammonia.
Draw structures to illustrate the symmetries of your complexes.
4. Square planar Pt(NH3)2Cl2 can be made in both the cis and the trans form, but only the cis form is active as an the anti-cancer drug .
(a) Describe an experiment that can distinguish between two isomers.
IR spectroscopy will distinguish between
cis and trans isomers of square planar complexes. Cis-complexes will have both
IR-acive symmetric and asymmetric stretches (for example of Pt-Cl bonds)
and trans complexes will only display one antisymmetric stretch; the symmetric
one is IR inactive for the trans isomer.
(b) Use CFT to explain why all Pt(II) complexes are square planar.
CFT theory applied to a sqaure planar geometry, D4h symmmetry, places the d x2-y2 orbital very high in energy, the degenerate d xz and yz orbitals lowest and the d xy and d z2 slightly about these. For a d8 M ion, the 8 e- nicely fill the lowest energy four d orbitals leaving the high energy one vacant. For 2nd and 3rd row metals, since the CF is exceptionally large, this favors low spin electron configurations and the diamagnetic square planar geometry for all complexes regardless of the field strength of their ligands in the spectrochemical series.
5. Do you think it is ever possible for the 5-fold degenerate d-orbitals in a spherical field to adopt a symmetry that would cause the five-fold degeneracy to split so that all five orbitals are non-degenerate, i.e., each is stabilized or destabilized by a different energy? How might this happen? Can you suggest specific situation that would make a metal’s d-orbitals all non-degenerate?
It's all about SYMMMETRY! We have seen that, when the symmetry is lowered, degenerate orbitals split, or become non-degenerate. So, to make a complex with each d orbital unique—non-degenerate— put different ligands around the M. Six different L around M would necessitate that all teh d orbitals are different, i.e. non-degenerate.
6. A new protein has been isolated from a squishy pink marine animal. A
conspicuous feature of the new protein is an amino acid sequence corresponding
to a metal binding site. The residues of this binding site are two cysteines,
one histidine and one methionine. It is known that this metalloprotein
is involved in one-electron redox reactions and it is speculated that the metal
is the catalytic site of the redox reaction. Which first row transition
metal ion(s) could be the metal used by this marine protein?
The characteristic residues of amino acids cysteine, methionine and histidine are teh -SH, the S-CH3 and the imidazole groups. THese are potential S atom and N atom donors to metals. These atoms are all soft and so will prefer to bind a soft metal. Of teh first row metals, Cu(+) is the softest. Cu is also able to do 1 electron redox reaction, easily moving from Cu(I) to Cu(II).
7. Use the molecular orbital pictures included in the MO diagram below for ammonia to identify the label the MO’s in the diagram (i.e. say what types of MO’s the picture depict). Then, compare the bonding description of this MO diagram with the bonding description for ammonia given by VSEPR.
The assignments are, readaing from the
bottom goign to the top: sigma binding, two degenerate sigma bonding, sigma
non-bonding, then degenerate sigma antibonding and the highest energy orbital
is sigma antibonding. VSEPR predicts 3 N-H bonds, all identical. MOT predicts
3 N-H bonds, but of different types. A lone pair from VSEPR is matched by
a non-bonding orbital localized in a N orbital. MOT includes antibonding
orbitals that are absent in VSEPR bonding descriptions.
8.
The lambda max = 450 nm for [Cr(NH3)6]3+.
The lambda max = 420 nm for [Co(NH3)6]3+.
Assume P= 18,000 cm-1 for both complexes.
(a) Draw CFT diagrams for each complex.
(b) Calculate the Do and CFSE in kJ/mol for this complex.
(Recall the relationship between wavelength, nm, and wavenumber, cm-1 and
that
1.1962 x 10-2 x E (cm-1)= E (kJ/mol).
9. a) The hybrid orbital hybrid(a) is composed of 60% 3dx2-y2 and
40% 4py.
[i.e. : hybrid(a)
= c1 . 3dx2-y2 + c2 . 4py
]
b) Draw the hybrid orbital hybrid(a) . click for my
drawings
c) What are the values of c1 and c2?
c1 = 0.77 and c2 = 0.63
d) Is the hybrid hybrid(a) normalized? Support
your answer with an argument.
Yes it is normalized. This can be determined because (a) the sum of % composition =100% and/or (b) (0.77)62 + (0.63)^2 = 1
e) Draw the hybrid orbital, hybrid(b), that
is a partner to Fa and write its mathematical expression.
hybrid(b) = 0.63 3dx2-y2 - 0.77 4py click for my drawings
f) Are the two hybrids, hybrid(a) and hybrid(b),
orthogonal? Support your answer with an argument.
Yes because integrating the product {[0.77 dx2-y2 + 0.63 py] x [0.63 dx2-y2 - 0.77 py]} =0.
10. Using the table of Valence Orbital Ionization Energies below, predict
which member of
Group 16 will make the most covalent bonds to Zn in a compound ZnX.
Which atomic orbital(s) of the chosen Group 16 element will mainly be
involved in bonding
to Zn?
principle quantum # orbital energies (eV)
O 2 -32.3 -15.8
S 3 -20.7 -11.6
Se 4 -20.8 -10.8
The most covalent bond will involve orbitals close in energy on each atom of the bond. Using the Zn 4s orbital at -9.4 eV, the orbital closest in energy is the Se 4p at -10.8 eV. ZnSe is the most covalent compound.
11. The hybrid orbital expressions are given below for the central atom (S) in SF2.
(a) Which hybrid orbitals will be used for S-F bonds? hybrid (a) and hybrid (b)
(b) Which hybrid orbitals
will be used for non-bonding electron. hybrid (c) and hybrid
(d)
(c) Predict the molecular geometry of SF2 Bent
hybrid(a)
= 0.33 3s + 0.63 3pz + 0.71 3py
hybrid hybrid(b)
= 0.33 3s + 0.63 3pz - 0.71 3py
orbitals hybrid(c)
= 0.90 3s - 0.44 3pz
on sulfur: hybrid(d)
= 1.0 3px