MCQ
The oxidation number of $\text{Cr}$ in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is :
- A$+3$
- B$−3$
- ✓$+6$
- D$−6$
✓
Answer
Correct option: C.
$+6$
The sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion.
The oxidation number of simple ions is equal to the charge on the ion.
The metals in Group $\text{IA (K)}$ atom has an oxidation number of $+1$.
Oxygen usually has an oxidation number of $−2.$
Oxidation no of $\text{Cr}$ in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
$=2\times 1+2\times \text{x}+7\times (−2)=0$
$\text{x}=\frac{14-2}{2}=+6$
The oxidation number of simple ions is equal to the charge on the ion.
The metals in Group $\text{IA (K)}$ atom has an oxidation number of $+1$.
Oxygen usually has an oxidation number of $−2.$
Oxidation no of $\text{Cr}$ in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
$=2\times 1+2\times \text{x}+7\times (−2)=0$
$\text{x}=\frac{14-2}{2}=+6$
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