Anon-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?

Average oxidation state = VO1.19
Oxygen is-2. Then 1.19 (-2) = -2.38
Average oxidation state of V is +2.38 

Consider 100 formula units of VO1.19
There would be 119 Oxide ions = Each oxide is -2. Total charge = -2(119) = -238 
The total charge of all the vanadium ions would be +238. 
Let x = number of of V+2 
Then 100 – x = number of V+3 
X(+2) + 100-x(+3) = +238 
2x + 300 – 3x = 238 
-x = 238-300 = -62 
x = 62
 
Thus 62/100 are V+2 
62/100 * 100 = 62%

62 % is the percentage of the vanadium atoms are in the lower oxidation state. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.