Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 3 - Sections 3.1-3.12 - Exercises - Problems by Topic: 45b


Its formula is $IrBr_{3}.4H_{2}O$

Work Step by Step

The cation of iridium(III) bromide tetrahydrate has a charge of (III), i.e., it is $Ir^{3+}$. The anion is bromide, i.e $Br^{-}$. To balance the charges, 3 $Br^{-}$ combines with 1 $Ir^{3+}$. Then it also has 4 (tetra) water molecules of hydration. So, its formula is $IrBr_{3}.4H_{2}O$
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