Download Assignment #9 KEY 1. Most ligands are written as abbreviations

Assignment #9 KEY
1. Most ligands are written as abbreviations. Write out the full name and draw the structure of
the following ligands: en, THF, phen, py, acac1-, ox2-
2. Name, draw, give the denticity, and identify the donor atoms of the following ligands: en, EDTA4, DMSO, dien, bpy
En: bidentate, both nitrogen atoms are donors
DMSO: monodentate, donor is either the sulfur or the oxygen
EDTA: hexadentate, donor atoms are the nitrogen atoms (neutral) and the oxygen atoms (-1 charge
for each)
Dien: tridentate, donor atoms are the three nitrogen atoms
Bipy: bidentate, donor atoms are the two nitrogen atoms
3. Draw the cis and trans isomers of:
a) PtCl2(NH3)2
b) Co(en)2Cl2
4. Draw the meridinal/facial isomers of:
a) CrCl3(OH2)3
b) Cr(terpy)Cl3
Same for the terpy with the tridentate terpy in place of water.
5. Draw the d-orbital splitting of the following geometries:
a) Octahedral
b) Tetrahedral
c) Square Planar
Octahedral (left): lower set is t2g and upper set is eg*
Tetrahedral (middle): lower set is e and upper set is t2
Square planar: remember the upper level is anti bonding
6. Assign the oxidation state and give the d-electron count for the transition metal in the following
complexes:
a) CuCl4
b) PtCl42c) Co(OH2)63+
d) NiCl2(PPh3)2
e) PdCl2(PPh3)2
a) CuIV ; d7
b) PtII ; d8
c) CoIII ; d6
d) NiII ; d8
e) PdII ; d8
7. Draw the ∆ and Λ isomers of Co(en)3.
∆ is the on the left, Λ is on the right.
8. Classify the following ligands are high field or low field: CO, CN, Cl, bipy, OH2, PR3
High field ligands are pi acceptors: CO, CN, bipy (less than CO ad CN)
Low field ligands are pi donors: Cl
In between: OH2, PR3
9. Explain the following observation: NiCl2(PPh3)2 is paramagnetic and PdCl2(PPh3)2 is diamagnetic.
The oxidation state of nickel here is plus two and the d-electron count here is eight. The same is true
for the palladium complex. The nickel complex will be tetrahedral and will have two unpaired
electrons (nickel is the only d8 metal that will adopt a tetrahedral configuration – this happens about
50% of the time). Since palladium will be square planar, there are no unpaired electrons.
10. Explain the following observation: KMnO4 is an intense purple colour and Co(NH3)62+ is a pale
yellow-straw colour.
Permanganate has ligand to metal charge transfer from the oxygen atoms to the manganese metal.
Since charge transfer are allowed transitions the value for the epsilon will be high and the colour will
be intense. In the cobalt (III) hexaammonia species, there are only d-to-do transitions which are
Laporte forbidden and spin allowed which leads to low values of epsilon (<100) which leads to pale
colours.