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

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.

Answer 2

Part A:

Average oxidation state of vanadium is 2.38.

Part B:

Percentage of the vanadium atoms are in the lower oxidation state=62%

Step-by-step explanation:

Part A:

1 atom vanadium(V) combines with 1.19 atoms of O to form
`VO_{1.19`.

Formula:

`\sum(Oxidation\ number\ of\ atoms\ in\ molecule* Number\ of\ atoms)=Oxidation\ number\of\ molecule`

In our Case:

Note: Oxidation number of O is always -2

`(Oxidation\ number\ of\ V* Number\ of\ atoms\ of\ vanadium)+(Oxidation\ number\ of\ O* Number\ of\ atoms\ of\ Oxygen)=Oxidation\ Number\ of\ VO_(1.19)\\\\(Oxidation\ number\ of\ V* 1)+(-2* 1.19)=Oxidation\ Number\ of\ VO_(1.19)\\\\\\Since, Oxidation\ Number\ of\ VO_(1.19)\ is\ 0\ as\ it\ is\ neutral\ molecule.\\(Oxidation\ number\ of\ V* 1)+(-2* 1.19)=0\\Oxidation\ number\ of\ V=2.38`

Average oxidation state of vanadium is 2.38.

Part B:

Let's say we have total 100 atoms of vanadium, a will show +2 oxidation state while (100-a) will show +3 oxidation state.

Now:

`Total\ Atoms* Average\ oxidation\ number\ of\ Vanadium=(Oxidation\ number* Number\ of\ atoms\ of\ with\ +2\ oxidation\ number )+(Oxidation\ number* Number\ of\ atoms\ of\ +3\ oxidation\ number)\\\\100*\ Average\ oxidation\ number\ of\ Vanadium\ = \ (+2*a)+[+3*(100-a)]\\100*2.38=2a+300-3a\\\\a= 62 atoms (+2\ oxidation\ state)`

`Percentage=(62)/(100)*100\ =62\%`

Percentage of the vanadium atoms are in the lower oxidation state=62%