Question

A non-stoichiometric compound is a compound that cannot be represented by a small whole-number ratio of atoms, usually because of point defects in the crystal lattice. What is the average oxidation state of vanadium in ? What is the average state of vanadium in VO 1.19? If each vanadium atom has either a 2 or 3 oxidation state in this compound, what percentage of the vanadium atoms are in the lower oxidation state?

Answer 1

The average oxidation state of vanadium in VO1.19 is +2.38, and 62% of the vanadium atoms are in the lower +2 oxidation state.

Step-by-step explanation:

The question concerns the calculation of the average oxidation state of vanadium in the nonstoichiometric compound VO1.19. If we assume that the oxygen is in its usual -2 oxidation state, the total oxidation state for the oxygen atoms in VO1.19 is -2 * 1.19 = -2.38. For the compound to be electrically neutral, the total positive oxidation state of the vanadium must balance this, which means the average oxidation state of vanadium must be +2.38.

If vanadium has either a +2 or +3 oxidation state, to find the percentage of vanadium atoms in the lower +2 state, we can set up an equation: x * (+2) + (1-x) * (+3) = +2.38, where x represents the fraction of vanadium atoms with a +2 oxidation state. Solving for x gives us x = 0.62. The percentage of vanadium atoms in the +2 oxidation state is therefore 62%.

Answer 2

let x be the percentage of the 2 oxidation state
2 oxidation state vanadium bonds with 1 oxygen
3 oxidation state bonds with 1.5 oxygen.
1*x+1.5*(1-x)=1.19
x+1.5-1.5x=1.19
0.31 = .5x
x = .62
hope it helps