1) The correct arrangement of the following complex ions interms of increasing crystal field splitting energy (D) is:
[CrI6]3-Â Â Â Â Â Â Â Â Â Â [CrF4]-Â Â Â Â Â Â Â Â Â Â [Cr(en)3]3+Â Â Â Â Â Â Â Â Â Â [W(en)3]3+
a) Smallest [CrI6]3- <[Cr(en)3]3+ <[CrF4]- < [W(en)3]3+Largest
b) Smallest [CrI6]3- <[W(en)3]3+ < [CrF4]-< [Cr(en)3]3+ Largest  Â
c) Smallest [CrF4]- <[CrI6]3- <[Cr(en)3]3+ <[W(en)3]3+ Largest
d) Smallest [CrF4]- <[W(en)3]3+ <[CrI6]3- <[Cr(en)3]3+ Largest
e) Smallest [W(en)3]3+ <[Cr(en)3]3+ <[CrF4]- < [CrI6]3-Largest
2) Determine the number of unpaired electrons for each of thefollowing coordination complexes:
[Ni(H2O)6]2+ | [CoCl6]2– | Cr(CO)6 | [Pt(CN)4]2– |
Arrange these complexes in order of increasing number ofunpaired electrons (in the answers below an = sign is used if twocomplexes have the same number of electrons).
a) Fewest [Pt(CN)4]2– <[Ni(H2O)6]2+ 6 < [CoCl6]2–Greatest
b) Fewest [Pt(CN)4]2– = Cr(CO)6< [CoCl6]2– <[Ni(H2O)6]2+ Greatest
c) Fewest [Pt(CN)4]2– 6 < [CoCl6]2– <[Ni(H2O)6]2+  Greatest
d) Fewest [Pt(CN)4]2– 6 <[Ni(H2O)6]2+ <[CoCl6]2– Greatest
e) Fewest [Pt(CN)4]2– = Cr(CO)6< [Ni(H2O)6]2+ <[CoCl6]2– Greatest
