The oxidation state of iodine in is: | Redox Reactions and Volumetric Analysis — Chem-Mantra

Redox Reactions and Volumetric AnalysiseasyMCQ SINGLE

Question

The oxidation state of iodine in $H_4IO_6^-$ is:

Answer: B

💡 Solution & Explanation

Calculation: $4(+1) + x + 6(-2) = -1 \Rightarrow 4 + x - 12 = -1 \Rightarrow x - 8 = -1 \Rightarrow x = +7$. [1, 3]

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