Chemistry 104B - Spring 2003 - JohnArnold - Midterm 2

Chemistry 104B - Spring 2003 - JohnArnold - Midterm 2 - 1....

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1. (12 points) Complete the following: He Ne Ar Kr Xe Rn At I Br Cl F O S S e T e P o B i S b A s P N C S i G e S n P b Tl In Ga Al B Cd Hg Au Cu Pd Pt Co Fe Os Tc C r M o W T a N b T i L a S c C a M g B e S r B a R a F r C s R b K N a L i H Periodic Table V Mn Y Zr Hf Ru Ni Zn Re Rh Ir Ag 2. (6 points) Put in order of increasing rate of substitution by H 2 O the complexes: (a) [Rh(NH 3 ) 6 ] 3+ (b) [Co(NH 3 ) 6 ] 3+ (c) [Ni(H 2 O) 6 ] 2+ (d) [Mn(H 2 O) 6 ] 2+ (e) [Ir(NH 3 ) 6 ] 3+ Explain your reasoning. The overall order is (e) < (a) < (b) < (c) < (d). Co 3+ , Rh 3+ , and Ir 3+ are going to be low spin, d 6 , and thus relatively inert. Since metal-ligand bond strengths increase as one goes down a column in the transition metal series, expect the order of rates to be Co 3+ > Rh 3+ > Ir 3+ within that series. Mn 2+ is HS, d 5 , while Ni 2+ is HS, d 8 , both of which will be labile due to population of eg set orbitals by two electrons. For the d 5 case, there will be no LFSE during ligand exchange, but for d 8 LFSE will be negative; thus (d) is faster than (c). 3. (6 points) For each of the following pairs of electronic transitions, state which would be expected to be more intense and why. i) 3 A 2g 3 T 2g in [NiCl 6 ] 4- OR 3 T 1 3 T 2 in [NiCl 4 ] 2- Laporte allowed, there is no inversion center in T d complex. ii) 4 A 2g 2 E g in [Cr(H 2 O) 6 ] 3+ OR 4 A 2g 2 E g in [W(H 2 O) 6 ] 3+ Coupling of spin and orbital angular momentum can relax the spin selection rule, therefore, for heavier elements (greater spin-orbit coupling) more intense bands are observed. iii) 1 A 1 (metal) 1 E (metal) OR 1 A 1 (metal) 1 E (ligand) Symmetry allowed, there is a change in parity from metal to ligand.
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Chemistry 104B - Spring 2003 - JohnArnold - Midterm 2 - 1....

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