## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

# Chapter 3 - Chemical Reactions - Exercise 3.13 - Assigning Oxidation Numbers - Page 118: c

#### Answer

The first $N$ has an oxidation number equal to $(-3)$ Each $H$ has that equal to $(+1)$ The other nitrogen has an oxidation number equal to $(+5)$ Each $O$ has an oxidation number equal to $(-2)$

#### Work Step by Step

We can separate $NH_4NO_3$ in 2 ions: $NH{_4}^+$ and $N{O_3}^-$. 1. $N{H_4}^+$. According to rule 6(d), the oxidation number for hydrogen is equal to $(+1)$ Using rule 4, and calling the oxidation number for $N$ "x": $(x) * 1 + (+1)*4 = +1$ x + 4 = + 1 x = 1 - 4 x = -3 The oxidation number for the nitrogen in $N{H_4}^+$ is $(-3)$ 2. $N{O_3}^-$. According to Rule 7(a), each oxygen has an oxidation number of $(-2)$. Using rule 4 and calling the oxidation number for $N$ "y": $(y) * 1 + (-2) * 3 = -1$ y - 6 = -1 y = + 5

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