Step 1: Understanding the Concept:
Stock notation uses Roman numerals placed in parentheses immediately after the name of the metal to indicate its oxidation state in a compound. To solve this, we must systematically determine the oxidation state of the metal in each formula.
Step 2: Key Formula or Approach:
Apply the rule that the sum of the oxidation states of all atoms in a neutral chemical formula must be zero. Assume the standard oxidation state of oxygen is -2. Let the metal's oxidation state be \(x\).
Step 3: Detailed Explanation:
Let's evaluate each compound:
a) \(\text{MnO}\):
Let the oxidation state of Mn be \(x\).
\(x + (-2) = 0\)
\(x = +2\)
Stock notation corresponds to (II). Therefore, a matches with ii.
b) \(\text{Mn}_2\text{O}_3\):
Let the oxidation state of Mn be \(x\).
\(2x + 3(-2) = 0\)
\(2x - 6 = 0\)
\(2x = 6 \implies x = +3\)
Stock notation corresponds to (III). Therefore, b matches with i.
c) \(\text{MnO}_2\):
Let the oxidation state of Mn be \(x\).
\(x + 2(-2) = 0\)
\(x - 4 = 0 \implies x = +4\)
Stock notation corresponds to (IV). Therefore, c matches with iii.
d) \(\text{Fe}_2\text{O}_3\):
Let the oxidation state of Fe be \(x\).
\(2x + 3(-2) = 0 \implies x = +3\)
The correct notation should be Iron(III) oxide. However, the only remaining option in the right column is 'iv) I'. This suggests an error in the question's premise for item 'd' (it likely meant to be a Copper(I) compound like \(\text{Cu}_2\text{O}\)).
Despite this error, we can look at the options to see which matches our first three definitive calculations:
a \(\rightarrow\) ii
b \(\rightarrow\) i
c \(\rightarrow\) iii
Only option C provides this exact combination: a-ii, b-i, c-iii, d-iv. By elimination, this must be the intended correct answer.
Step 4: Final Answer:
Following the correct matches for the first three compounds, Option C is the right choice.