Question

what is the oxidation number of each vanadium atom in v₂o₅?

Explanation:

Step1: Recall oxidation number of oxygen

In most compounds, oxygen has an oxidation number of -2.

Step2: Set up equation for compound

Let the oxidation number of vanadium be $x$. For $V_2O_5$, we have the equation $2x + 5\times(-2)=0$ (since the sum of oxidation numbers in a neutral compound is 0).

Step3: Solve for $x$

\[

$$\begin{align*} 2x-10&=0\\ 2x&=10\\ x& = 5 \end{align*}$$

\]

Answer:

5