NEETChemistryRedox Reactions
White phosphorus reacts with aqueous alkali as shown in the following equation: P ₄ + 3 OH ^- + 3 H ₂ O PH ₃ + 3 H ₂ PO ₂^- What are the changes in the oxidation number of phosphorus in this disproportionation reaction?
Options
- A0 to -3 and 0 to +1
- B0 to +3 and 0 to -1
- C0 to -3 and 0 to +3
- D-3 to 0 and +1 to 0
Correct answer
A. 0 to -3 and 0 to +1
Step-by-step solution
In elemental white phosphorus ( P ₄ ), the oxidation number of phosphorus is 0 . In phosphine ( PH ₃ ), let the oxidation number of P be x . The oxidation number of H is +1 . x + 3(+1) = 0 x = -3 In the hypophosphite ion ( H ₂ PO ₂^- ), let the oxidation number of P be y . The oxidation number of H is +1 and O is -2 . y + 2(+1) + 2(-2) = -1 y + 2 - 4 = -1 y - 2 = -1 y = +1 Thus, the oxidation number of phosphorus changes from 0 to -3 and from 0 to +1 . Answer: 0 to -3 and 0 to +1