Single correct · +4 / −1

The oxidation state of iodine in HIO₄ (periodic acid) is:

  1. (a) +3
  2. (b) +5
  3. (c) +7
  4. (d) +1
Answer and solution

Answer (c) +7

Solution

Chemistry ma’am · 3 steps

Let H be +1 and each O be −2, then make the whole neutral molecule add up to zero.

  1. Fix the known ones. In HIO₄, hydrogen is +1 and each oxygen is -2. There is no peroxide link here, so oxygen keeps its usual value.
  2. Sum to zero. The molecule is neutral, so all the oxidation states add up to zero. Call the oxidation state of iodine x.

    (+1) + x + 4(-2) = 0 ⇒ x = +7

  3. So. Iodine is at +7 in periodic acid, which is option (c). That is its highest oxidation state, matching its 7 valence electrons.

The idea to keep. For any oxoacid, set H = +1 and O = −2, then balance to the overall charge. The prefix 'per-' tells you the central halogen is in its highest state, +7.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Redox Reactions

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