Question

What is the oxidation number of V in $V_2O_7^{4-}$ ion?

• A. $+4$
• B. $+3$
• C. $+6$
• D. $+5$

Hint:

For polyatomic ions, the sum of oxidation numbers of all atoms equals the charge on the ion.

Answer 1

The Correct Option is D

Step 1: Assign oxidation state of oxygen.
Oxidation state of oxygen is $-2$.
Step 2: Let oxidation state of vanadium be $x$.
There are 2 vanadium atoms and 7 oxygen atoms in the ion.
Step 3: Write the oxidation number equation.
\[ 2x + 7(-2) = -4 \]
\[ 2x - 14 = -4 \]
\[ 2x = 10 \]
\[ x = +5 \]
Step 4: Conclusion.
The oxidation number of vanadium in $V_2O_7^{4-}$ ion is $+5$.