Browse Questions

# 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) $PbMnO_4$

Answer: $KMnO_4$

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