Colour of M + 2 ion & M + 3 ion respectively when obtain by dissolving MO & M 2 | TestHub

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Metal (M) is a white, crystalline metal and is not appreciably ductile or malleable. It has a very high melting point. The metal is soluble in dilute or concentrated hydrochloric acid. If air is excluded, Metal(II) ions are formed:

$M + 2 H^{+} \to M^{+ 2} + H_{2} ↑$ .

$M + 2 HCl \to M^{+ 2} + 2 Cl^{-} + H_{2} ↑$

Metal(II) ions are rather unstable as they are strong reducing agents; they decompose even in acidic solution with the formation of hydrogen and $M^{+ 3}$ .

In the presence of atmospheric oxygen, metal (M) gets partly or wholly oxidized to the trivalent state:

$4 M^{+ 2} + O_{2} + 4 H^{+} \to 4 M^{+ 3} + 2 H_{2} O$ .

Dilute sulfuric acid attacks metal (M) slowly, with the formation of hydrogen. In hot, concentrated sulfuric acid, metal (M) dissolves readily to form metal(III) ions and sulfur dioxide: $2 M + 6 H_{2} SO_{4} \to 2 M^{+ 3} + 3 SO_{4}^{- 2} + 3 SO_{2} ↑ + 6 H_{2} O$ .

Highly concentrated nitric acid renders metal (M) passive.

Colour of $M^{+ 2}$ ion & $M^{+ 3}$ ion respectively when obtain by dissolving MO & $M_{2} O_{3}$ in dil. HCl .

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Answer:

A

Solution:

$Cr + 2 H^{+} ⟶ Cr^{+ 2} + H_{2} ↑$

$Cr + 2 HCl \to Cr^{+ 2} + 2 Cl^{-} + H_{2} 4 Cr^{+ 2} + O_{2} + 4 H^{+} \to 4 Cr^{+ 3} + 2 H_{2} O 2 Cr + 6 H_{2} SO_{4} \to 2 Cr^{+ 3} + 3 SO_{4}^{- 2} + 3 SO_{2} ↑ + 6 H_{2} O$

Colour of $CrCl_{2}$ (aq.) and $CrCl_{3}$ (aq.) are blue and green, respectively.

Stream:JEESubject:ChemistryTopic:Salt AnalysisSubtopic:TEST OF CATIONS Dry Test, Group Analysis, Group 1,2

⏱ 2mℹ️ Source: QB

Chemistry - Salt Analysis Question with Solution | TestHub

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