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Isomers and Ligands

4. Circle the letter corresponding to the correct answer for the following.
For which of the following complexes can a tetralredral coordination geometry be unequivocally
excluded based upon its magnetic properties?
a) Cu(PPh3)3Cl (diamagnetic) b) P(PPhr3)2Cl2 (diamagnetic)
c) Ni(PPh3)2Br2 (paramagnetic) d) Co(PPh3)2Cl2 (paramagnetic)
Which ion will undergo the greatest change in bond distance upon one-electron oxidation? (1.s. =
low spin; h.s. : high sprn)
a) h.s. [Cr(OH2)6]2+ b) [V(OH2)6]2+ c) l.s. [Ru(OH2)6]2+ d) l.s. [Ir(OH2)6]3+*
For which of the following compounds is the absorptivity of a 0.1 M solution expected to be lowest?
a) [Mn(OH2)6]3+ b) [Fe(OH2)6]3+ c) [Mn(CN)6]3- d) [Fe(CN)6]3-
Which of the following reactions will have the largest equilibrium constant? dien =
a) [Ni(OH2)6]2+ + 6 NH3 = [Ni(NH3)6]'2+ + 6H2O
b) [Ni(OH2)6]2+ + 2 dien = [Ni(dien)2]2+ + 6H2O
c) [Ni(NH3)6]2+ + 2 dien = [Ni(dien)2]2+ + 6NH3
d) b & c are the same since the chelate effect is operative in both

The complex cations [Co(NH3)5(NCS)]2+, and [Co(NH3)5(SCN)]2+ are what type of isomers?
a) diastereomers b) structural c) linkage d) ionization e) b & c
As a ligand to transition metal ions, CP- (oxide ion) is
a) an omega donor only
b) an omega donor and pi donor
c) an omega donor and pi acceptor
d) uses vacant p orbitals for its pi acceptor interactions
e) uses filled p orbitals for its pi donor interactions
f) c&d
g) b&e

Which of the following complexes contain d2 metal ions (circle all correct answers)?
a) [Re2Cl4(dppe)2] (dppe is Me2PCH2CH2PMe2 a bidentate ligand, which, in this case connects
the two metals across the metal-metal bond)
b) [V(O)(acac)4]
c) [Os(O)2Cl4]2-
d) [Ti(OH2)4Cl2]
e) [Re2Cl8]3-


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Please see the revised file (attached).


(b) Pt(PPh3)2Cl2
Cu+: [Ar] 3d10, Pt2+: [Xe] 4f145d8, Ni2+: [Ar] 4s23d6, Co2+: [Ar] 4s23d5
Diamagnetic means no unpaired electron, paramagnetic indicates presence unpaired electron. As all given compounds are 4 coordinated, from the given magnetic properties and keeping in mind the crystal field splitting for tetrahedral (remember all tetrahedral complexes are high spin, as the splitting energy is quite low in comparison to octahedral field) and square planar complexes, it is only the Pt2+ complex that can not be tetrahedral.

For example, a d10 system with no unpaired electron (diamagnetic), can be tetrahedral or square planar. A d6 system with unpaired electron (paramagnetic) has to be tetrahedral, not square planar. A d5 system with unpaired electron (paramagnetic), can be tetrahedral or square planar. But a d8 system with no unpaired electron (diamagnetic), has to be square planar. ...

Solution Summary

The solution discusses isomers and ligands