Advanced Inorganic Chemistry (CHE 510) Final Examination December 16, 2005

Transcription

1 Advanced Inorganic Chemistry (CHE 510) Final Examination December 16, 2005 Name: Answer questions in the area provided, the back of an exam page, or on the clearly labeled spare sheet. Please show all of your work and explain all of your answers so that you may receive partial credit. 1. (32 pts) (a, 4 pts) Explain why all solids have some defects at temperatures above 0 K. All solids have some defects above 0 K because up to a certain concentration, defects lead to a reduction of free energy. Creating a defect results in an increase in ΔS. This term is large since there are a large number of positions to put the defect. Since free energy is given by ΔG = ΔH - TΔS, ΔG is negative (favorable) with respect to creating a defect. As the concentration of defects increases, the gain in ΔS is little because the crystal is already very disordered. Therefore, the energy (ΔH) to make more defects is relatively high. Thus, ΔG becomes positive and it doesn't happen. Graphically, this is expressed as shown below:!h E!G =!H - T!S "#!S [defects]

2 (b, 4 pts) How would the substitution of 0.5% of the cations in Na crystals with Mn 2+ affect the conductivity of the crystals? Explain why. Na Na Na Na Mn 2 + Na Na Na Na Na Na Na Na Schottky defects would be expected to predominate in Na crystals doped with Mn 2. Cation vacancies (Schottky defects) in the crystal lattice would result from the need to maintain charge balance in the crystal when a Mn 2+ ion replaces a Na + ion in the crystal lattice. As ionic conductivity in Na is primarily through a vacancy mechanism and we have increased the number of vacancies, we would expect that the cations would have higher mobility and thus display a greater conductivity. Remember, ionic conductivity can be expressed by the equation:! = n i e i u i σ = ionic conductivity n i = # of charge carriers e i = their charge µ i = ionic mobility (c, 8 pts) Discuss Schottky and Frenkel defects with examples. In each case, explain if any electrical conduction is possible and by what mechanism. Both Schottky and Frenkel defects are stoichiometric defects. Schottky defect involves a pair of vacant sites, anion and cation vacancy. Schottky defects occur most often in alkali halides such as sodium chloride: Na Na Na Na Na Na Na Na Na Na Na Na Na Na Na Na Na Frenkel defect involves the displacement of an atom off the lattice into an interstitial site that is normally empty. Taking silver chloride as an example, the ion can move from its normal octahedral site (in the rock salt lattice) into a tetrahedral site:

3 Frenkel defects are more frequent than Schottky defects because less energy is required for their formation. Schottky defects allow electrical conduction through a vacancy mechanism whereby an ion (usually the cation because it s smaller than the anion) migrates through the solid by moving into vacant sites. Frenkel defects allow electrical conduction through either interstitial or interstitialcy mechanism. The interstitial mechanism involves movement of a cation from interstitial site to interstitial site (below): In the interstitialcy mechanism, the interstitial moves onto a lattice site while knocking another atom into an interstitial position (below): (d, 4 pts) What effect would you expect to find on ionic conductivity of crystals if 0.5% of the silver ions were substituted with Cd 2+? Explain why. Substitution of Cd 2+ for + would create cation vacancies in the structure. These would provide attractive places for interstitial silver ions. Because the conductivity in occurs by an interstitial mechanism, this reduces the number of charge carriers and hence, the conductivity (see conductivity equation above)

4 (e, 4 pts) Consider a sample of pure silicon. Compare the conductivity of this material at 600 K to its conductivity at 4 K. Use appropriate drawings to support your explanation. Silicon is a semiconductor with a bandgap of 1.1eV. Above T = 0 K, the population of the MO energy levels is described by the Fermi-Dirac equation. The population decays exponentially at energies well above the Fermi level. T = 0 K T = > 0 K E! F E! F fraction populated fraction populated when electrons near the Fermi level are promoted to empty levels above the Fermi level, they can move freely through the solid and the substance is an electronic conductor. Hence Si is a better conductor at 600 K. (f, 4 pts) Which element would you add to germanium to make it a p- type semiconductor? Draw a diagram of its band structure and explain. A group III or 13 element such as boron or gallium would be suitable. If 0.01% B(3+) is doped for Ge. This creates localized atomic orbitals since the B atoms are so far apart there is little or no overlap between them. These orbitals lie just above the filled valence band and are called acceptor levels since they can accept electrons:

5 conduction band acceptor levels valence band e - This small gap means that thermal promotion to acceptor levels is possible. Electrons in acceptor bands now permits conductivity since holes are created in the valence band and they move towards a negative electrode. (g, 4 pts) When titanium(iv) oxide is heated in hydrogen, a blue color develops, indicating light absorption in the red. Does reduction of Ti(IV) to Ti(III) correspond to n-doping or p-doping? Explain. n-doping: reduction of Ti(IV) to Ti(III) puts electrons in the previously empty d-band (conduction band) formed by metal orbitals since a small deficit of O atoms results in the structure. The blue color is a result of internal d-d transitions within octahedrally coordinated Ti(III) ions. TiO 2 + H 2 Ti 2 O 3 + H 2 O 2. (28 pts) (a, 12 pts) Design a selective one-step synthesis for each of the three possible isomers of [() ( )]. Three isomers of [() ( )] are: H 3 N A B C Keeping in mind that (i) the relative order of the trans-directing ability of the ligands is: - > - > ~, and (ii) the - bond is labile due to the inherent mismatch between the soft (II) center and hard chloride ligand, we can devise selective procedures for

6 synthesizing each of the three isomers of [() ( )]. Thus, one such approach is: + A H 3 N + H 3 N B + - C (b, 16 pts) Design a selective two-step synthesis for cis- and trans- [( )(PMe 3 ) 2 ] starting with To prepare cis-[(pme 3 )( ) 2 ], we first react [ 4 ] 2- with ammonia, because it is exercises a weak trans effect. Next, we react the complex obtained with PMe 3 to obtain the desired complex PMe PMe 3 To prepare trans-[(pme 3 )( ) 2 ], we react [ 4 ] 2- first with the strongly trans directing PMe 3 ligand. Subsequent reaction with ammonia produces the desired product. 2- PMe 3 - PMe 3 + PMe NH 3 - -

7 3. (25 pts) (a, 6 pts) [Cu( ) 4 ] + is completely colorless while [Cu( ) 4 ] 2+ is intensely blue. Explain this observation using appropriate drawings. [Cu( ) 4 ] + [Cu( ) 4 ] 2+ t 2 t 2! O! O e Both compounds are tetrahedral. Cu + is a d 10 ion while Cu 2+ is a d 9 ion. Consequently, the d orbitals in [Cu( ) 4 ] + are fully occupied and there is no possibility of electronic transition between the e g and t 2g levels. In the case of [Cu( ) 4 ] 2+, a d-d transition is possible and is spin allowed. Furthermore, it s Laporte allowed since the compound is noncentrosymmetric. e (b, 10 pts) For octahedral complexes, we discussed d n electron configurations for which a Jahn-Teller distortion can be expected. On the basis of our discussions, predict which d n configurations would not show Jahn-Teller splitting for the tetrahedral case (ignore possible low-spin cases). Explain. Any nonlinear molecule containing an orbitally degenerate ground electronic configuration is intrinsically unstable. Hence a distortion must occur to lower the symmetry, remove the degeneracy, and lower the energy of the system. The splitting of the d orbitals in a tetrahedral field is shown below: T d t 2! O e early, d 2, d 5, d 7, and d 10 ions will experience no Jahn-Teller distortion.

8 (c, 9 pts) Given that the maximum absorption in the d-d peak for [Ti(H 2 O) 6 ] 3+ occurs at 20,300 cm -1, predict where the d-d peaks will occur for [Ti(CN) 6 ] 3- and [Ti() 6 ] 3-. Explain your reasoning. These are all Ti 3+ complexes and the relative field strength for the ligands follows the order: CN - > H 2 O > -. Hence, we can safely predict that the absorption maximum will occur at greater than 20,300 cm -1 for [Ti(CN) 6 ] 3-. In contrast, [Ti() 6 ] 3- will show its absorption maximum at lower than for 20,300 cm -1. This is because the stronger the ligand field, the greater the splitting between the t 2g and e g orbital sets of octahedral complexes. 4. (15 pts) Both dissociative and associative (solvent attack) ratedetermining steps have been suggested for the mechanism of cistrans isomerization of square planar (II) complexes. Given the data below, suggest the most likely mechanism for isomerization of [(PEt 3 ) 2 (Mes)Br] (Mes = 2,4,6-Me 3 C 6 H 2 ) in CH 3 OH. Explain your reasoning. Reaction k 1 x 10 4 H S isomerization [(PEt 3 ) 2 (Mes)Br] + I The negative value of S for the isomerization rules out a dissociative pathway. We would expect the entro to increase in the transition state if a dissociative mechanism were involved since the number of species present would increase. The reaction between [(PEt 3 ) 2 (2,4,6- Me 3 C 6 H 2 )Br] and I - clearly proceeds via associative mechanism ( S < -10 eu). It s reasonable to suggest that isomerization via solvent attack proceeds by a similar mechanism.

9 5. (25 pts) The following data refer to the gas-phase substitution reaction: [M(CO) 6 ] (g) + *CO (g) [M(CO) 5 (*CO)] (g) + CO (g) M H kj/mol S J/mol K Cr Mo W In decalin solution, the H values are: Cr, kj/mol; Mo, kj/mol; W, kj/mol. (a, 5 pts) What is the significance of the agreement between gasphase and solution activation enthalpies? Explain. The fact that the activation enthalpies are nearly the same suggest that the reaction mechanism in both the gas phase and in decalin must be the same. (b, 10 pts) Based on the above data and an observed rate dependence on *CO at low concentration, draw a mechanism for the reaction and discuss it. The large positive activation entropies clearly imply a dissociative mechanism. Thus, CO dissociation is the first step. Presumably, the five coordinate intermediate generated is then trapped by *CO. The observed rate dependence on *CO at low concentration suggests CO dissociation is reversible. [M(CO) 6 ] k 1 M(CO) 5 + CO k -1 k 2 *CO [M(CO) 5 (*CO)] + CO

10 (c, 5 pts) What is the form of the reaction rate law at low [*CO]? rate = -d[m(co) 6 ]/dt = k 1 k 2 [M(CO) 6 ][*CO] k -1 [CO] + k 2 [*CO] (d, 5 pts) What is the form of the reaction rate law at high [*CO]? At high *CO concentration (or if k 2 >> k -1 ), the expression above reduces to: rate = k 1 [M(CO) 6 ].