how do we find the oxidation state of N and P in ZnNH4PO4 since both N and P have more than one oxidation states?
Asked by SIMRAN | 10th Jan, 2014, 07:38: PM
This compound is a mixed salt of orthophosphoric acid, zinc and ammonium cation.
NH4+ is a derivative of ammonia NH3. Oxidation number of hydrogen is always 1 in compounds with nitrogen. Based on the fact that the molecule is electrically neutral, we get,
X + 3 = 0
X = -3
X - Oxidation number of N
The same principle defines oxidation number of Zn = 2, O = -2
-3 + 4 × 1 + 2 + 4 × (-2) + x = 0
X = 5
Answer: Oxidation number of N = -3, P= +5
This compound is a mixed salt of orthophosphoric acid, zinc and ammonium cation.
NH4+ is a derivative of ammonia NH3. Oxidation number of hydrogen is always 1 in compounds with nitrogen. Based on the fact that the molecule is electrically neutral, we get,
X + 3 = 0
X = -3
X - Oxidation number of N
The same principle defines oxidation number of Zn = 2, O = -2
-3 + 4 × 1 + 2 + 4 × (-2) + x = 0
X = 5
Answer: Oxidation number of N = -3, P= +5
Answered by Hanisha Vyas | 15th Jan, 2014, 11:45: AM
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