1988

JEE Main 1988 Chemistry question on Redox and Electrochemistry

The oxidation number of phosphorus in Ba(H2PO2)2\text{Ba}(\text{H}_2\text{PO}_2)_2 is
  1. A.
    +3
  2. B.
    +2
  3. C.
    +1
  4. D.
    -1
Show answer and solution

Correct option: C

Solution

Ba is +2, each H₂PO₂⁻ anion has charge −1. In H₂PO₂⁻: oxygen is −2 (×2 = −4), hydrogen is +1 (×2 = +2). So P + 2(+1) + 2(−2) = −1, giving P + 2 − 4 = −1, thus P = +1.

[1988]

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