Question

8. What are the oxidation states of manganese in the Ba1-x(H20)xMn5010 and how many of each are required for charge balance when x = 0? (There are multiple possible solutions, so choose the one that is the most viable solution.) 9. What are the oxidation states of manganese in the Ba1-x(H2O)xMng010 and how many of each are required for charge balance when x = 0.5? (Choose the most viable solution.)

Answer 1

8 We can calculate the average charge of Mn we would need in the extremes. If x is 0, the fomula would be BaMn₅O₁₀. Where there would be a charge of +2 from Ba and 10 (-2) charges from O. These two result in a charge of -18, which divided by 5 yields 3.6.

If x is 1, we would have (H₂O)Mn₅O₁₀ where there would be a charge of -20 from the O atoms, the H₂O fraction is neutral, so the oxidation state of Mn should be +4 in order to achieve neutrality.

These two values mean that the oxidation states of Mn are probably +2 and +4, and when x = 0, their proportion is given by (x is the fraction of atoms in the +2 state, while the rest, given by 1-x is in the +4 state):

$2x+4(1-x)=3.6$

$-2x+4=3.6$

$2x=0.4$

$x=0.2$

There should be 20% in the +2 state and 80% in the +4 state.

9. The possible values are the same. If x = 0.5, we would have: Ba_0.5(H₂O)Mn₅O₁₀

Which means that (x in the following is the average charge of Mn):

$0.5\cdot 2+5x+10\cdot (-2)=0$

$5x=20-1=19$

$x=19/5=3.8$

And now we can solve as we did before, where x is the fraction of atoms in the +2 state:

$2x+4(1-x)=3.8$

$-2x+4=3.8$

$2x=0.2$

$x=0.1$

10% of the Mn atoms should be in +2 state and 90% in +4 state to achieve this stoichiometry.