What is the average oxidation state of vanadium in VO 1.14?

Average oxidation state: VO1.14
Oxygen is usually -2. Then 1.14(-2) = -2.28
Average oxidation state of V is 2.28

Consider 100 formula units of VO1.14
There would be 114 Oxide ions = Each oxide is -2. Total charge = -2(114) = -228
The total charge of all the vanadium ions would be 228.
Let x = number of of V 2
Then 100 – x = number of V 3
X( 2) 100-x( 3) = 228
2x 300 – 3x = 228
-x = 228-300 = -72
x = 72
Thus 72/100 are V 2
72/100 * 100 = 72%

When the oxidation state of V is +2 :
The formula of the oxide is VO.
The mole ratio V : O = 1 : 1

When the oxidation state of V is +3 :
The formula of the oxide is V₂O₃.
The mole ratio V : O = 2 : 3 = 1 : 1.5

Let a be the number of moles of V in VO, and b be that in V₂O₃.
Then number of moles of V in VO and V₂O₃ = (a + b) mol
and number of moles of V in VO and V₂O₃ = (a + 1.5b) mol

Overall mole ratio V : O is
(a + b) : (a + 1.5b) = 1 : 1.14
a + 1.5b = 1.14 × (a + b)
a + 1.5b = 1.14a + 1.14b
0.14a = 0.36 b
a/b = 18/7
a : b = 18 : 7

Percentage of V atoms in lower oxidation state (+2)
= [a/(a + b)] × 100%
= [18/(18 + 7)] × 100%
= 72%