To determine the oxidation number of chromium in the dichromate ion \(\text{Cr}_2\text{O}_7^{2-}\) and in \(\text{Cr}^{3+}\), we will analyze each case separately.
For \(\text{Cr}_2\text{O}_7^{2-}\):
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Assign known oxidation states:
- Oxygen (O) typically has an oxidation state of \(-2\).
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Calculate the total contribution of oxygen:
- In \(\text{Cr}_2\text{O}_7^{2-}\), there are 7 oxygen atoms: \[ \text{Total oxidation state from O} = 7 \times (-2) = -14. \]
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Set up an equation:
- Let the oxidation state of chromium (Cr) be \(x\). Since there are 2 chromium atoms, their total contribution will be \(2x\).
- The overall charge of the dichromate ion is \(-2\), therefore the equation is: \[ 2x + (-14) = -2. \]
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Solve for \(x\): \[ 2x - 14 = -2 \ 2x = 12 \ x = 6. \] Hence, the oxidation number of each chromium atom in \(\text{Cr}_2\text{O}_7^{2-}\) is \(+6\).
For \(\text{Cr}^{3+}\):
- Define the oxidation state:
- The notation \(\text{Cr}^{3+}\) indicates that chromium is in the \(3+\) oxidation state.
Conclusion:
- In \(\text{Cr}_2\text{O}_7^{2-}\), the oxidation number of Cr is \(+6\).
- In \(\text{Cr}^{3+}\), the oxidation number of Cr is \(+3\).