Determine the oxidation number of V in the V_4O_12^4− ion given that the oxidation number of oxygen is −2
[V4O12]^4-
Remember the V and O must add to -4.
4*V + 12*-2 = -4
4V - 24 = -4
4V = +20
V = ?
To determine the oxidation number (or oxidation state) of V in the V4O12^4- ion, we can use the following steps:
1. Identify the oxidation number of oxygen: In this case, the oxidation number of oxygen is -2, as given in the question.
2. Assign a variable (x) to the oxidation number of V: Let x be the oxidation number of V.
3. Set up an equation: The sum of the oxidation numbers in the V4O12^4- ion should equal the charge on the ion, which is -4.
So, we have:
4x + 12(-2) = -4
4. Solve the equation for x:
Simplifying the equation, we have:
4x - 24 = -4
Adding 24 to both sides, we get:
4x = 20
Dividing both sides by 4, we find:
x = 5
Therefore, the oxidation number of V in the V4O12^4- ion is +5.
To determine the oxidation number of V in the V4O124− ion, we need to consider the fact that the sum of the oxidation numbers of all the atoms in a compound or ion must add up to the overall charge.
In this case, we have the ion V4O124−, which has a charge of −4. From the given information, we know that the oxidation number of oxygen is −2.
Let's assign the oxidation number of V as x. Since there are four V atoms in the ion, the total oxidation number contributed by V will be 4x. Each oxygen atom contributes an oxidation number of −2, and since there are twelve oxygen atoms, the total oxidation number contribution by oxygen will be 12(−2) = −24.
Now, we can write the equation to represent the sum of the oxidation numbers:
4x + (-24) = -4
Simplifying the equation:
4x - 24 = -4
4x = -4 + 24
4x = 20
x = 20/4
x = +5
Therefore, the oxidation number of V in the V4O124− ion is +5.