The oxidation number of iodine in the iodate ion (IO -3) is

  • A -5
  • B -1
  • C +1
  • D +5
waec 2006chemistry

The correct answer is D. +5

1O3

[oxidation number of 1] + [oxidation number of 0] x 3 = -1

x + -2 x 3 = 1

x - 6 = - 1

x = - 1 + 6 = +5

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