Question

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−

Answer 1

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+