The correct option is B False
Lead nitrate (A) reacts with NaCl to form a white precipitate of lead chloride (B).
Pb(NO3)2+2NaCl→PbCl2whitepptB+2NaNO3
Lead(II) ions react with hydrogen sulphide to from black ppt (C) of lead sulphide.
Pb2++H2S→PbSblack,C+2H+
Lead chloride reacts with KI to form yellow ppt D of lead iodide.
PbCl2+2KI→PbI2yellowppt,D+2KCl
Lead nitrate on heating forms lead oxide and liberates a reddish brown gas which is nitrogen dioxide.
2Pb(NO3)2heat−−→2PbO+4NO2+O2