Using stock notation represent the following compound:
A. $Sn{O_2}$
B. $C{r_2}{O_3}$

Hint: As we know that the stock notation is basically the method of representing the oxidation number of any metal using roman numerals such as I, II, III and so on within the parenthesis. We also know that the oxidation number of any metal can be identified by adding the charge on all the atoms present in a compound.

Complete step by step solution:
As we already know that stock notation is the representation of the oxidation number of any metal with the help of roman numbers within parenthesis. So we will calculate the oxidation number of the metals present in the given compounds. Let us talk about these compounds one by one.

A. The first one is stannic dioxide $Sn{O_2}$. Let us assume the charge present on tin in this compound is $x$ and we know that the charge on oxygen is generally $ - 2$. So, the oxidation number of stannic oxide would be:
$x + ( - 4) = 0$
$x = + 4$
So we can represent the calculated oxidation number in stock notation as: Tin(IV)dioxide or Stannic(IV)dioxide.

B. Similarly, we can identify the oxidation number of chromium oxide, $C{r_2}{O_3}$. So, let us assume that the charge on chromium is $x$ and the oxygen possesses a charge of $ - 2$. Thus the oxidation number of $C{r_2}{O_3}$ would be:
$2x + ( - 6) = 0$
$x = + 3$
Hence, in stock notation chromium oxide will be represented as $Chromium{\text{ }}(III){\text{ }}oxide$.

Note: Always remember that the charge on any metal or element usually represents its oxidation state and the oxidation state or number represented as roman numbers inside parentheses is stock notation representation of compounds. Also, stannic dioxide in mineral form is commonly called cassiterite and it is the main ore of tin.