![]() ![]() Their salts have great practical significance – chromates and dichromates accordingly. To find the correct oxidation number for CrO2 (Chromium (IV) oxide), and each element in the compound, we use a few rules and some simple math. They can only exist in solutions and are practically not used. ![]() 2 acids are examined in this case as hydroxyls – chrome HCrO₄ and dechrome H₂Cr₂O₇. Chromium in the oxidation state of +6Ĭompounds of chromium in which it displays an oxidation state of +6 are strong oxidizers. Salts where chromium at an oxidation state of +3 acts as a cation display all typical properties of salts (most of them are soluble in water and hydrolyze – they decompose in water with the formation of chromium hydroxide Cr(OH)₃):Ĭhromium salts with an oxidation state of +3 can take part in oxidation-reduction reactions, for example in the following:ĢCrCl₃ + 3Zn + 4HCl = 2CrCl₂ + 3ZnCl₂ + 2H₂ (in the reaction between hydrochloric acid and zinc, atomic hydrogen is released, which reduces the chromium cation to the chromium cation). Chromium (II) oxideĬhromium (II) oxide is formed in the decomposition of chromium carbonyl (with heating):Ĭr(CO)₆ = CrO + 5CO + C. The reductive ability of Cr²⁺ salts is very high (in some cases these salts can even displace hydrogen from water). Hydrogen released in the course of reaction reduces Cr³⁺ to Cr²⁺. ![]() They are usually obtained by oxidation-reduction reactions from chromium (III). Salts of chromium (II) have a bluish color. The compounds are colored – chromium (II) oxide is black, and the hydroxide is yellow. Chromium in the oxidation state of +2Ĭhromium (II) oxide and hydroxide CrO and Cr(OH)₂ are compounds which display typical base properties. The same pattern is seen in all oxidationreduction reactions: the number of. electrons gained 6 O atoms × 2 e gained O atom 12 e gained. In compounds (which are brightly colored in the majority of cases), chromium displays several possible oxidation states - +2, +3, +4 (encountered quite rarely, chromium oxide CrO₂ is known), +6. In Equation 4.4.3, for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen: electrons lost 4 Al atoms × 3 e lost Al atom 12 e lost. ![]()
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