The black coloured compound may be:

N i S

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C d S

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P b S

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C o S

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Solution

The correct options are
B $P b S$
C $N i S$
D $C o S$

$A + d i l . H_{2} {S O}_{4} ⟶ B$
(black)
$C + B ⟶ D$
(acid) (white turbidity)
${B a C l}_{2} (q) + E ⟶ w h i t e p p t d i l . {H N O}_{3} ⟶ X$
insoluble
$∙$ Clearly white ppt is ${B a S O}_{4}$
$∙$ So $E =$ contains ${S O}_{4}^{- 2}$ ion
$∙$ Clearly chocolate brown ppt indicates ${C u}^{+ 2}$ in $(G)$
$∙$ Since $G$ came from $F$ , $F$ contains ${C u}^{+ 2}$
$∙$ $F$ cam from gas $B$ and acidified solution $(E)$ So $E$ has ${C u}^{+ 2}$
$∙$ Hence $E = {C u S O}_{4}$
$∙$ White ppt clearly indicates sulphur. Gas $B$ is $H_{2} S$

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