State True or False.
(B) is $P b C l_{2}$ .

True

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

False

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A True
Lead nitrate (A) reacts with NaCl to form a white precipitate of lead chloride (B).
$P b (N O_{3})_{2} + 2 N a C l \to P b C l_{2} w h i t e p p t (B) + 2 N a N O_{3}$
Lead(II) ions react with hydrogen sulphide to from black ppt (C) of lead sulphide.
$P b^{2 +} + H_{2} S \to P b S b l a c k, C + 2 H^{+}$
Lead chloride reacts with KI to form yellow ppt D of lead iodide.
$P b C l_{2} + 2 K I \to P b I_{2} y e l l o w p p t, D + 2 K C l$
Lead nitrate on heating forms lead oxide and liberates a reddish brown gas which is nitrogen dioxide.
$2 P b (N O_{3})_{2} h e a t - - \to 2 P b O + 4 N O_{2} + O_{2}$

Suggest Corrections

Similar questions

S O_{2}

(C H_{3} O)_{3} B

_{2}

_{3}

H C l

P b C l_{2}

3 x + 2 y = 5, 2 x - 3 y = 7

Join BYJU'S Learning Program