To determine the oxidation state of chromium in chromium(III) sulfate and its chemical formula, let's analyze the details step-by-step:
1. Identify the Polyatomic Ion and Its Charge:
- The sulfate ion ([tex]\( \text{SO}_4^{2-} \)[/tex]) has a charge of -2.
2. Given Data about Chromium:
- Chromium(III) indicates that each chromium ion has a charge of [tex]\(+3\)[/tex].
3. Balancing the Charges:
- We need to balance the overall charge of the compound to be neutral.
- Since we have three sulfate ions (each with a charge of -2), the total negative charge is [tex]\( 3 \times (-2) = -6 \)[/tex].
4. Calculate the Number of Chromium Ions:
- To balance the total charge of -6 from the sulfate ions, we need a total positive charge of +6 from chromium ions.
- Given that each chromium ion has a charge of [tex]\(+3\)[/tex], we need [tex]\( \frac{+6}{+3} = 2 \)[/tex] chromium ions.
5. Chemical Formula:
- Thus, with 2 chromium ions ([tex]\( \text{Cr}^{3+} \)[/tex]) and 3 sulfate ions ([tex]\( \text{SO}_4^{2-} \)[/tex]), the chemical formula is [tex]\( \text{Cr}_2(\text{SO}_4)_3 \)[/tex].
6. Oxidation State of Chromium:
- Each chromium ion in this compound has an oxidation state of [tex]\( +3 \)[/tex].
Thus, the oxidation state of chromium in chromium(III) sulfate is [tex]\( \boxed{+3} \)[/tex], and the chemical formula of the compound is [tex]\( \boxed{\text{Cr}_2(\text{SO}_4)_3} \)[/tex].
