$Pb X^{2 +}$ is reported both in group I and in group II because

A $PbS$ is soluble in dilute $HCl$ , so it appears in group I
B $PbCl X_{2}$ is amphoteric and dissolves in excess $HCl$ only
C $Pb X^{2 +}$ exists in two oxidation states in solution, each belonging to a different group
D $PbCl X_{2}$ is appreciably soluble in cold water, so precipitation in group I is incomplete and the rest is precipitated as $PbS$ in group II

Answer: D

PbCl X_{2}

has a fairly large

K_{s p}

(of the order of

1 0^{- 5}

), so some

Pb X^{2 +}

stays in the filtrate after dilute

HCl

. That remainder gives black

PbS

with

H X_{2} S

in acid, because

PbS

is extremely insoluble.