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When a basic solution of Cr(OH)4-ion is slowly acidified, solidCr(OH)5precipitates and then redissolves in excess acid. Ifrole="math" localid="1663334451033" width="69" height="29">Cr(OH)4- exists asCr2H2O)2(OH)4-, write equations that represent these two reactions.

Short Answer

Expert verified

The solution of given statement is

CrOH4aq-+Haq+→CrOH3s+H2OlCrH2O2OH4aq-+Haq+→CrOH3s+3H2Ol

Step by step solution

01

Definition of Coordination Compound

A coordination complex consists of a central atom or ion, which is typically metallic and is named the coordination centre, and a surrounding array of bound molecules or ions, that are successively called ligands or complexing agents.

02

Solve the given statements

Basic Cr(OH)4-turns into solidCr(OH)3when slowly acidified. One of the OH - groups of the complex ion reacts with H+ in an acid-base neutralization reaction to form water:

role="math" localid="1663334602733" CrOH4aq-+Haq+→CrOH3s+H2Ol

The same happens when CrOH4H2O2-is given. The complex still turns into solidCr(OH)3when slowly acidified; aside from releasing one of the OH groups it also releases two moles of water into the solution:

CrH2O2OH4aq-+Haq+→CrOH3s+3H2Ol

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