Key idea behind the keyword question
The phrase “chromium iii oxide why is it cr2o3 and not cro3” is answered by one rule: a correct ionic formula must be electrically neutral, and the Roman numeral in the name fixes the metal’s oxidation state.
Step 1: Assign ionic charges
- Chromium(III) \(\rightarrow\) \(\mathrm{Cr^{3+}}\).
- Oxide \(\rightarrow\) \(\mathrm{O^{2-}}\).
Step 2: Enforce charge neutrality to get the subscripts
Let the formula be \(\mathrm{Cr_{x}O_{y}}\). Total charge must be zero:
\[ 3x - 2y = 0 \]The smallest whole-number solution comes from matching total positive and negative charge using the least common multiple of \(3\) and \(2\), which is \(6\):
- \(2\) chromium ions give \(2 \times (+3) = +6\).
- \(3\) oxide ions give \(3 \times (-2) = -6\).
Therefore, the neutral formula is:
\[ \mathrm{Cr_2O_3} \]A quick verification uses the charge sum:
\[ 2(+3) + 3(-2) = +6 - 6 = 0 \]Step 3: Why \(\mathrm{CrO_3}\) does not match chromium(III)
If the formula were \(\mathrm{CrO_3}\), the three oxygen atoms contribute a total oxidation number of \(3 \times (-2) = -6\). For a neutral compound, chromium would have to be \(+6\), not \(+3\):
\[ \text{In } \mathrm{CrO_3:}\quad x + 3(-2) = 0 \;\Rightarrow\; x = +6 \]Comparison table (same elements, different oxidation states)
| Name | Formula | Oxidation state of chromium | Charge-balance check |
|---|---|---|---|
| chromium(III) oxide | \(\mathrm{Cr_2O_3}\) | \(+3\) | \(2(+3) + 3(-2) = 0\) |
| chromium(VI) oxide (chromium trioxide) | \(\mathrm{CrO_3}\) | \(+6\) | \((+6) + 3(-2) = 0\) |
Visualization: ion ratio that makes a neutral formula unit
Final statement
Chromium(III) oxide is \(\mathrm{Cr_2O_3}\) because \(\mathrm{Cr^{3+}}\) and \(\mathrm{O^{2-}}\) must combine to give net charge zero; \(\mathrm{CrO_3}\) would instead require chromium in the \(+6\) oxidation state.