Electron Configuration and Oxidation Numbers?

Question

Rank these transition metal ions in order of decreasing number of unpaired electrons.

Fe^3+
Ni^2+
Cu^1+
V^3+
Mn^4+

Consider the following vanadium species. Arrange them in order of strongest to weakest oxidizing agent.

[V(H2O)6]^3+
[V[H2O)4]^2+
[V(H2O)6]^2+
VO4^3-


And any explanations of how you know would be greatly appreciated!

Answer 1

1) The first questions deal with electrons occupying the d-orbitals. There are 5 d-orbitals altogether, and the following calculation is done with the assumption that the metal complexes are in high-spin configuration (i.e all orbitals are filled first before pairing take place).

Fe3+ have 5 electrons in the d-orbitals => each orbital contains 1 unpaired electron. (Total unpaired electron = 5)

Ni2+ have 8 electrons in the d orbitals => 3 fully filled orbitals + 2 singly filled orbitals. (Total unpaired electron = 2)

Cu1+ have 10 electrons which occupy all the d orbitals (Total unpaired electron = 0)

V3+ have 2 electrons, so only give 2 singly filled orbitals (Total unpaired electron = 2)

Mn4+ have 3 electrons, and give 3 singly filled orbitals. (Total unpaired electron = 3)

Fe3+ > Mn4+ > Ni2+ = V3+ > Cu+

2) The strongest oxidising agent would have the highest oxidation state.

[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+
(VO4)3- : Oxidation state on V = 5+

I do not think number of coordinated water will affect the reaction strength of the oxidising agent as the oxidising property should be determined by the metal itself.

So from the strongest oxidising agent: (VO4)3- > [V(H2O)6]3+ > [V(H2O)6]2+ = [V(H2O)4]2+

Answer 2

V3 is equal to Ni

Answer 3

something in group a million has a million valence electron, by way of fact the electron ability stages fill up with a million left over. the table is prepared that way. group 2, 2 valence electrons group 14, 4 group 15, 5 group sixteen, 6 group 17, 7 group 18, 8, finished and stable oxidation numbers: group a million: +a million group 2: +2 communities 3-12: you're given the ox. #, often by utilising roman numerals group13: +3 group 14 +/- 4, often -4 group 15: -3 group sixteen: -2 group 17: -a million group 18 is in no way used by way of fact the climate are already stable.

Answer 4

Fe3+ > Mn4+ > V3+ > Ni2+ > Cu+

VO4(3-)>[V(H2O)6]^3+>[V[H2O)4]^2+>[V(H2O)6]^2+

Answer 5

This is the right answer for the second part:
VO4^3- > [V[H2O)4O]^2+ > [V(H2O)6]^3+ > [V[H2O)6]^2+
O.No. +5 +4 +3 +2