A coloured solution known to contain two metal ions was treated with excess cold sodium hydroxide solution. When filtered a whitish solid, slowly changing to brown, was retained on the filter paper and a colourless solution collected as the filtrate. The dropwise addition of hydrochloric acid to the filtrate produced a white ppt. which dissolved in excess acid. Treatment of the residue on filter paper with a solution of a strong oxidiser produced a reddish-violet solution.

Indicate any pairs of ions:

Z n^{2 +}

and

M n^{2 +}

ions

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M g^{2 +}

and

Z n^{2 +}

ions

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M n^{2 +}

and

M g^{2 +}

ions

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F e^{2 +}

and

Z n^{2 +}

ions

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Solution

The correct option is A $Z n^{2 +}$ and $M n^{2 +}$ ions
Reactions:

With sodium hydroxide solution an initially white precipitate of manganese(II) hydroxide is obtained.

$M n^{2 +} + 2 O H^{-} ⟶ M n (O H)_{2} ↓ w h i t e p p t$

The precipitate is insoluble in the excess reagent. It rapidly oxidizes on exposure to air, becoming brown, when hydrated manganese dioxide, $M n O (O H)_{2} ↓$ is formed.

$2 M n (O H)_{2} ↓ + O_{2} ⟶ 2 M N (O H)_{2} B r o w n p p t ↓$
$Z n^{2 +} + ⟶ N z (O H)_{2} ↓ w h i t e g e l a t i o n p p t$

The precipitate is soluble in acids.

$Z n (O H)_{2} ↓ + 2 H^{+} ⟶ Z n^{2 +} + 2 H_{2} O$

So, the correct option is $A$

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Similar questions

Sodium hydroxide solution is added to the solution containing the Zn²⁺ ion.

Match the ions with their coloured precipitate.

Z n^{2 +}

P b^{2 +}

Sodium hydroxide solution is added to the solution containing the Fe³⁺ ion.

Match the ions with their coloured precipitate.

(F e^{2 +})

(F e^{3 +})

K_{2} C r_{2} O_{7}

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