What is the oxidation state of V in VO2?
Oxygen is -2; two of them makes a total of -4. Since compounds are zero, V must be +4 to balance the -4 of the oxygen atoms.
+4
To determine the oxidation state of the element vanadium (V) in the compound VO2, we need to consider the overall charge of the compound and the known oxidation states of other elements involved.
In this case, oxygen (O) typically has an oxidation state of -2, since it's more electronegative than most other elements. The compound VO2 consists of two oxygen atoms, so their total charge is -4 (-2 × 2).
To find the oxidation state of vanadium (V), we can assign x as the oxidation state. Therefore, the algebraic expression for the compound VO2 is:
(V^x)(O^-2)2 = 0
Since the compound is neutral (oxidation states must balance out to zero), the sum of the oxidation states should be zero.
Let's solve it step-by-step:
1. Multiply the oxidation state of vanadium by one V atom: x * 1
2. Multiply the oxidation state of oxygen by two O atoms: -2 * 2
(x) + (-2 * 2) = 0
Simplifying the equation:
x - 4 = 0
x = 4
Therefore, the oxidation state of vanadium (V) in VO2 is +4.
To determine the oxidation state of an element in a compound, you can follow these steps:
Step 1: Determine the oxidation state of the other elements in the compound.
In the compound VO2, oxygen (O) almost always has an oxidation state of -2. Therefore, in this case, the total oxidation state contributed by the oxygen atoms would be -4 (since there are two oxygen atoms).
Step 2: Use the rule that the sum of the oxidation states in a compound is equal to the overall charge of the compound.
In this case, we know that the compound is neutral, meaning it has an overall charge of 0. Therefore, the sum of the oxidation states of the elements in the compound should be 0.
Step 3: Use algebra to find the oxidation state of the element of interest.
Let's assume that the oxidation state of vanadium (V) is x. Since there is only one vanadium atom in VO2, the oxidation state of vanadium can be represented as x.
Now, using the information from steps 1 and 2, we can set up an equation:
x + (-4) = 0
Simplifying the equation:
x - 4 = 0
Adding 4 to both sides:
x = 4
Therefore, the oxidation state of vanadium (V) in VO2 is +4.