test5 - (50 points Give the following information for each...

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Unformatted text preview: (50 points) Give the following information for each of the following compounds: a) its name, b) the structure of the complex portion of each species, and c) the hydridization of the central transition metal in each species. If stereoisomers (geometrical or optical) are possible, give the structures of all stereoisomers. a) Na3[Fe(C2O4)3]H2O name:sodium trioxalatoferrate(III) mono hydrate Oxalate is a bidentate ligand. Three oxalates yield a coordination number of 6. Octahedral coordination results. The complex anion lacks a center and planes of symmetry so optical isomers exist, i.e. non-superimposable mirror images. hybridization of the Fe: d2sp3 structures of the isomers b) K2[CoBr2I2] name:potassium dibromodiiodocobaltate(II) Cobalt(II) is a d7 case and all of the four monodentate ligands are halides so a 4 coordinate, tetrahedral structure is expected. No isomers. hybridization of the Co: sp3 structure of the single species c) Pd[(NH3)2(CN)(NO2)] (Pd, palladium, atomic number 46) name: diamminecyanonitropalladium(II) Palladium(II) is a d8 case so a square-planar geometry with cis and trans isomers is expected. hybridization of the Pd: dsp2 structures of the isomers (50 points) There are striking differences in the colors of titanium compounds. Solid anhydrous (i.e. no water) titanium(II) chloride is very dark and blue whereas solid anhydrous titanium(IV) oxide is colorless and is a common pigment in white paint. Explain why one compound is colored and the other colorless. Ti(IV) is a d0 case and there are no d electrons to excite. Excitation of the core electrons requires high-energy UV photons. Ti(II), on the other hand, is a d2 case and transitions between the d orbitals which lie in the visible are possible. What structural inferences can be drawn from the spectroscopic properties of titanium(II) chloride? The dark color indicates a high molar extinction coefficient and a lack of a center of symmetry. This rules out linear, square planar, and octahedral coordination of the Ti(II) cations by the chloride anions. Tetrahedral coordination is likely. Predict the crystal structure of titanium(IV) oxide which is known to geologists as the mineral rutile. Include the coordination numbers of the titanium(IV) cation and the oxide anion in your description. The ratio of the ionic radii of the Ti(IV) and oxide ions is 0.49 which lies in the range for packing of ions in octahedral holes. The stoichiometry is TiO2 or TiO0.5, i.e. N/2 Ti's per N O's. Hence a likely structure is one in which Ti(IV) ions fill half of the octahedral holes in a cp lattice of oxide anions. Octahedral packing means a coordination number of 6 for the Ti(IV). The oxide anions are outnumbered by a factor of 2:1 so the coordination number of oxide is (6)/(2) =3. Titanium(IV) chloride, a volatile compound of titanium, is used in sky writing. i) Explain why the titanium(IV) chloride is volatile. The titanium has a high oxidation number and covalent bonding rather than ionic bonding is likely. The high volatility indicates weak interactions between the TiCl4 molecules which is possible with the molecular solid class of materials. ii) When the titanium(IV) tetrachloride vapors are released from the aircraft, they react with the moisture in the air. Write a balanced chemical equation for the reaction. Species with a high oxidation number in water form tetrahedrally coordinated oxyanions. TiCl4(l) + 4 H2O(l) = TiO4-4(aq) + 4 Cl-(aq) + 8 H+(aq) Alternatively, titanium dioxide is not very soluble and the formation of this solid is likely. The solid titanium dioxide is responsible for the white plume in sky writing. TiCl4(l) + 2 H2O(l) = TiO2(s) + 4 Cl-(aq) + 4 H+(aq) (12 points) The degreaser dichloroethane, ClCH2CH2Cl, has a low solubility in water, 8 g/liter, whereas ethylene glycol, HOCH2CH2OH, which is used in antifreeze is soluble in water in all proportions. Provide a molecular explanation for the unusually high solubility of ethylene glycol. Ethylene glycol has two hydroxyl (-OH) groups which can participate in hydrogen bonding with water. (14 points) Borazon, a cubic form of BN, is one of the hardest substances known. Provide a molecular explanation for its extreme hardness. What can be concluded about its solubility in most solvents? B and N are non-metals so covalent bonding is expected. The hardness indicates a network solid in which there is an infinite network of covalent bonds throughout the sample. The material is expected to be insoluble in virtually all solvents. (24 points ) 0.001 mole of iron(III) nitrate is dissolved in 1.0 liter of 1.0 M aqueous NaCN. The following questions relate to the iron(III) complex which is present in the solution. (MO = molecular orbital) Cyanide is a strong ligand and is present in large excess. An octahedral cyanide complex of iron(III) is expected. Give the formula of the complex ion: Fe(CN)6-3 Give the number of unpaired electrons: 1 Iron(III) is d5. Cyanide is a very strong-field ligand so the five 3d electrons go into the three t2g non-bonding orbitals. Only one of the three MO's is half-filled and hence only one unpaired electron. No electrons are left for the eg* antibonding MO's. Give the number of electrons in bonding MO's: 12 Each of the 6 cyanide ligands contributes a pair of electrons to be placed into the bonding MO's. Give the number of electrons in anti-bonding MO's: 0 Give the number of electrons in non-bonding MO's: 5 Give the color of the solution: at the red end of the spectrum Cyanide is a strong-field ligand so the gap between the anti-bonding and non-bonding MO's is large. Absorption occurs at the blue end of the spectrum and transmission at the red end. ...
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