How might you separation the following pairs of ions by addition of a single reagent :
$C r^{3 +}$ and $F e^{3 +}$ :

Excess of

N a O H

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Excess of

K O H

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M n O_{4}^{-}

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None of these

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Solution

The correct option is B Excess of $K O H$
$C r^{3 +}$ and $F e^{3 +}$ ions can be separated by using excess of $K O H$ . Excess of $K O H$ precipitates $F e^{3 +}$ as $F e (O H)_{3}$ while $C r^{3 +}$ remains soluble as $K C r (O H)_{4}$

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F e^{3 +}

N a O H

Question 14

The pair of ions having same electronic configuration is.............

(a) $C r^{3 +}, F e^{3 +}$ (b) $F e^{3 +}, M n^{2 +} +$ (c) $F e^{3 +}, C o^{3 +}$ (d) $S C^{3 +}, C r^{3 +}$

F e^{3 +}

F^{-}

C H_{3} C O O N a

F e^{3 +}

F e^{3 +}

C u^{2 +}

N H_{3} (a q)

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