In: Chemistry
Part A:
Rank these transition metal ions in order of decreasing number of unpaired electrons. If two ions have the same number of unpaired electrons, overlap them such that the two appear in a single column:
V3+, Mn4+, Ni2+, Fe3+, Cu+ |
Consider the following vanadium species. Arrange them in order of strongest to weakest oxidizing agent:
[V(H2O)6]2+, [V(H2O)6]3+, [V(H2O)4O]2+, VO43− |
The decreasing order of unpaired electrons is Fe3+, Mn4+, (V3+ = Ni2+), Cu+
V3+ is 1s22s22p63s23p63d2and has 2 unpaired electrons
Mn4+ is 1s22s22p63s23p63d3and has 3 unpaired electrons
Ni2+ is 1s22s22p63s23p63d8 and has 2 unpaired electrons
Fe3+ is 1s22s22p63s23p63d5 and has 5 unpaired electrons
Cu+ is 1s22s22p63s23p63d10 and has 0 unpaired electrons
The strongest to weakest oxidizing agents are VO43- , [V(H2O)6]3+ , [V(H2O)6]2+ = [V(H2O)4]2+
[V(H2O)6]3+ Oxidation state on V = 3+
[V(H2O)4]2+ Oxidation state on V = 2+
[V(H2O)6]2+ Oxidation state on V = 2+
VO43- Oxidation state on V = 5+