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Question

During the disproportionation of lodline to iodide and iodate ions, the ratio of iodate and iodide ions formed in alkaline medium is:


A
1:5
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B
5:1
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C
3:1
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D
1:3
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Solution

The correct option is A 1:5
$$3{I}_{2} + 3{H}_{2}O \rightarrow 5{I}^{-} + {I{O}_{3}}^{-} + 3{H}^{+}$$
Ratio of iodate and iodide ions is $$1:5$$

Chemistry

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