Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 16 - Oxidation and Reduction - Exercises - Problems - Page 603: 51

Answer

a) In $CO_{3}^{2-}$ ion, oxidation state of $C = +4, O=-2$. b) In $OH^{-}$ ion, oxidation state of $O = -2, H = +1$ c) In $NO_{3}^{-}$ ion, oxidation state of $N = +5, O = -2$ d) In $NO_{2}^{-}$ ion, oxidation state of $ N = +3, O = -2$

Work Step by Step

For a polyatomic ion, algebraic sum of oxidation state will be equal to the charge of the ion. a) In $CO_{3}^{2-}$ , oxidation state of O = -2. Let's say oxidation state of $C = x$ Hence, $x + (-2)\times3 = -2$ by solving we get $x = +4$ So, oxidation state of $C = +4$ b) In $OH^{-}$, oxidation state of O = -2 and H = +1 so that algebraic sum is equal to -1. c) In $NO_{3}^{-}$, oxidation state of O = -2, let's say oxidation of N = $x$ Hence, $x + (-2)\times3 = -1$, by solving we get $x = +5$ So, oxidation state of $N = +5$ d) In $NO_{2}^{-}$, oxidation state of O = -2, let's say oxidation of N = $x$ Hence, $x + (-2)\times2= -1$, by solving we get $x = +3$ So, oxidation state of $N = +3$
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