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In the fourth group, $Mn(OH)_2$ on heating with $PbO_2$ and conc. $HNO_3$ gives pink colour due to the formation of

(a) $KMnO_4$
(b) $K_2MnO_4$
(c) $Pb(MnO_4)_2$
(d) $PbMnO_4$
Can you answer this question?
 
 

1 Answer

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Answer: $KMnO_4$
 
$Mn(OH)_2 + 2HNO_3 \longrightarrow Mn(NO_3)_2 + 2H_2O$
On heating,
$2Mn(NO_3)_2 + 5PbO_2 + 6HNO_3 \overset{\Delta}{\longrightarrow} $
$2KMnO_4\text{{Potassium permanganate(pink)}} + 5Pb(NO_3)_2 \text{{Lead nitrate(colourless)}} + 2H_2O$
answered Jan 28, 2014 by mosymeow_1
edited Jan 28, 2014
 

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