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Chem 416, Problem Set 1 Answer Key
Problem 1 (10 points).
A)
Order: B > A > C
B)
Order: A > D > C > B
C)
Order: C > A > B
D) Cyanate is the anion with the [OCN]- arrangement—two of its resonance structures are quite good
and we therefore expect the compound to be stable. Fulminate is the anion with the [CNO]arrangement; none of its resonance structures are great, but at (1B) is at the very least acceptable, and
we thus expect it to be obtainable, however unstable. The arrangement [CON]- gives no good resonance
structure, and its salts are expected to be unknown.
E) Linear
Problem 2+3 (32 points).
(A) Cr(NH3)3Cl3 = Cr3+ (d3) + 3 NH3 + 3 ClCr(NH3)3Cl3 is a 15-electron complex: 3 d electrons + 6 two-electron ligands.
(B) Ni(H2O)62+ = Ni2+ (d8) + 6 H2O
Ni(H2O)62+ is a 20-electron complex: 8 d electrons + 6 two-electron ligands.
(C) Ru(py)62+ = Ru2+ (d6) + 6 py
Ru(py)62+ is an 18-electron complex: 6 d electrons + 6 two-electron ligands.
(D) Ir(CO)(Cl)(PPh3)2 = Ir+ (d8) + CO + Cl– + 2 PPh3
(PPh3, triphenylphosphine, is a neutral ligand)
Ir(CO)(Cl)(PPh3)2 is a 16-electron complex: 8 d electrons + 4 two-electron ligands. This is a famous
compound, called “Vaska’s complex”
(E) Fe(CO)5 = Fe0 (d8) + 5 CO
Fe(CO)5 is an 18-electron complex: 8 d electrons + 5 two-electron ligands. (Yes, I know that gas phase
iron atoms are 4s23d6, but all metal complexes are s0dn, so always count the number of valence
electrons and put them all in the d shell.)
(F) Hf(NMe2)4 = Hf4+ (d0) + 4 NMe2–
Hf(NMe2)4 is an 8-electron complex: no d electrons + 4 two-electron ligands
(except that each NMe2– ligand has two lone pairs on N and so can make a  bond with the Hf4+ center,
in addition to the  bonds we’re counting here).
(G)
Cu2+ (d9) + 2 H2O + 2
This is a 21-electron complex: 9 d electrons + 6 two-electron donor ligands.
The dots indicate the lone pairs that bind to the metal, not all the lone pairs in the ligand.
(H)
Co3+ (d6) + 3
Ethylene diamine ligands are neutral (like two ammonias attached
together.
This is an 18-electron complex: 6 from the metal + 6 two-electron donor amine ligands
Problem 4 (20 points).
A)
C3v
D3h
D∞h
C3v
C2v
C3v
C2v
C2v and C2v
D3h
C2v
C2v
B)
H3BNH3
H2
30.87 g/mol
(2.02 g/mol)3 eq. = 6.06 g/mol
(6.06/30.87)100% = 19.6 wt%
C)
C2v
3 kinds of Mn
D4h
1 kind of Mn
Problem 5 (28 points).
A)
PbO2 + Pb + 2 H2SO4 - 2 PbSO4 + 2 H2O
Oxidation states:
PbO2
SUM
(+4) + (-4) = 0
Pb
O
4+
2(2-)
Pb
0
0
H
S
O
4(1+)
2(6+)
8(2-)
(+4) + (+12) + (-16) = 0
Pb
S
O
2(2+)
2(6+)
8(2-)
(+4) + (+12) + (-16) = 0
O
H
2(2-)
4(1+)
(-4) + (+4) = 0
Pb
H2SO4
PbSO4
H2O
Oxidant: PbIVO2
Reductant: Pb(0)
-----------------------------------------------------------------------------4 FeS2 + 11 O2  2 Fe2O3 + 8 SO2
Oxidation states:
FeS2
SUM
(+8) + (-8) = 0
Fe
S
4(2+)
8(1-)
O
11(0)
0
Fe
O
4(3+)
6(2-)
(+12) + (-12) = 0
S
O
8(4+)
16(2-)
(+32) + (-32) = 0
O2
Fe2O3
SO2
Oxidant: O2
Reductant: Fe, S
B)
Problem 6 (10 points).
B) A redox noninnocent ligand is a ligand that can be oxidized or reduced.
C) “The structure of (smif){Li(smif-smif)}Ti eshibits three anionic ligand or ligand fragments: the smif
anion, the reduced pyridine-imine considered as a radical anion, and the anionic aziridine of the coupled
Li(smif-smif) ligand. From the structure and its diamagnetism, it is proposed that 1 contains a d1 Ti(III)
center antiferromagnetically coupled to the pyridine-imine radical anion.”
D) It looks like a smiley face! 