The oxidation state of underlined in compound `ulV_2O_(7)^(2-)`

Options

A. 0
B. `+8`
C. `+6`
D. 5

Answer

To find the oxidation state of vanadium (V) in the compound \( V_2O_7^{2-} \), we will follow these steps:

### Step 1: Identify the oxidation states of known elements
In the compound \( V_2O_7^{2-} \), we know that the oxidation state of oxygen (O) is typically -2.

### Step 2: Set up the equation
Let the oxidation state of vanadium (V) be \( x \). Since there are 2 vanadium atoms, their total contribution to the oxidation state will be \( 2x \). There are 7 oxygen atoms, contributing a total of \( 7 \times (-2) = -14 \).

### Step 3: Account for the overall charge of the compound
The overall charge of the compound \( V_2O_7^{2-} \) is -2. Therefore, we can set up the following equation:

\[ 2x + (-14) = -2 \]

### Step 4: Solve for \( x \)
Now, we simplify the equation:

\[ 2x - 14 = -2 \]

Adding 14 to both sides gives:

\[ 2x = -2 + 14 \]

\[ 2x = 12 \]

Now, divide both sides by 2:

\[ x = 6 \]

### Step 5: Conclusion
The oxidation state of vanadium (V) in the compound \( V_2O_7^{2-} \) is +6.

### Final Answer
The oxidation state of vanadium in \( V_2O_7^{2-} \) is +6.

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