Question

What is the oxidation number of N or P in following compound :- ZnNH4PO4

Answer 1

What is the oxidation number of N or P in following compound: $\mathsf{ZnNH}_4\mathsf{PO}_4$

Solution

This compound is the mixed salt of orthophosphoric acid, Zinc and ammonium cation.

$$\begin{array}{c c c} H _ {4} ^ {4} N ^ {3} & 0 ^ {- 2} \\ Z n ^ {* 2} & 0 ^ {- 2} & P ^ {* 5} = 0 ^ {- 2} \\ & 0 ^ {- 2} \end{array}$$

$NH_{4}^{+}$ is a derivative of ammonia $NH_{3}$. Hydrogen oxidation number is always 1 in compounds with nitrogen. Based on the fact that the molecule is electrically neutral, you get

$$\begin{array}{l} X + 3 = 0 \\ X = - 3 \\ X - \text{Oxidation number of N} \\ \end{array}$$

The same principle defines oxidation number of P. Oxidation number of Zn = 2, O = -2

$$\begin{array}{l} - 3 + 4 \cdot 1 + 2 + 4 \cdot (- 2) + x = 0 \\ X = 5 \\ \end{array}$$

Answer: oxidation number of: $N = -3$ , $P = +5$