What is the oxidation state of the vanadium atom in VO4−
rule: O is -2, making V+4
To determine the oxidation state of the vanadium (V) atom in the VO4^- ion, we can use the known charges of the other atoms involved.
In this case, oxygen typically has an oxidation state of -2, since there are four oxygen atoms in the VO4^- ion, their total charge is -8 (-2 x 4 = -8).
The overall charge of the VO4^- ion is -1, meaning the total charge of the vanadium and oxygen atoms combined is -1.
From this information, we can set up the equation:
(V) + (-8) = -1
By solving this equation, we find that the oxidation state of vanadium (V) in VO4^- is +5.
To determine the oxidation state of the vanadium atom in VO4^-, we need to consider the overall charge of the ion and the known oxidation states of the other elements involved.
Here's a step-by-step process to find the oxidation state of vanadium:
Step 1: Determine the overall charge of the ion
The ion VO4^- has a charge of -1, indicated by the superscript '-'. This means that the total sum of the charges of all the atoms in the ion adds up to -1.
Step 2: Assign the known oxidation states of the other elements
In the oxyanion VO4^-, oxygen (O) usually has an oxidation state of -2. Since there are four oxygen atoms in VO4^-, their total charge is -8 (4 x -2 = -8).
Step 3: Calculate the oxidation state of vanadium
As the total charge of the VO4^- ion is -1, and the oxygen atoms contribute -8, the remaining charge of -1 must be balanced by the vanadium atom.
Let's assume that the oxidation state of vanadium is 'x'. Since each oxygen atom has an oxidation state of -2, and there are four oxygen atoms, the overall charge contributed by oxygen is -8.
To balance this, we can set up the equation:
x + (-8) = -1
Simplifying the equation, we get:
x - 8 = -1
Adding 8 to both sides:
x = 7
Hence, the oxidation state of the vanadium atom in VO4^- is +7.