Question

Given an equation for the effect of heat on red lead and lead dioxide.

Answer 1

Hint: Lead exists as two oxides. One is Lead (II) oxide i.e., $PbO$ and the other one is Lead (IV) oxide i.e.,$Pb{O_2}$. Apart from this, there is Lead (II, IV) oxide which is known as the red lead or the minimum.

Complete Step by step answer: ead is an element of group 14 with an atomic number of 82. It is present in various minerals in minute quantities excluding sulphide and lead glance.
As we said the lead (IV) oxide which is thermally unstable and they tend to decompose into lead (II) oxide and oxygen when heated.
We can see the reaction:
$2Pb{O_2} \to 2PbO + {O_2}$
2 moles $Pb{O_2}$ when heated we get 2 moles of $PbO$ and oxygen gas.
Hence this is the reaction we get when we heat lead dioxide.
Now let’s look into a red lead. Also, we have lead(II) oxide we got from heating the lead dioxide. First, let us know what happens when we heat this lead (II) oxide.

$6PbO + {O_2} \to 2P{b_3}{O_4}$

We can see that when 6 moles of lead (II) oxide is heated to around 450-480 $^\circ C$ in presence of oxygen we get 2 moles of lead (II, IV) oxide known as the red lead.
Now let’s heat this red lead and see what we obtain
$2P{b_3}{O_4} \to 6PbO + {O_2}$
When we head red lead above 480$^\circ C$, the above reaction gets reversed and we obtain lead (II) oxide. Hence we can say that when we heat red heat we obtain lead (II) oxide.

Note: Red lead $\left( {P{b_3}{O_4}} \right)$ is also known as sindur. It is a mixed oxide and represented as $2PbO.Pb{O_2}$. It is better regarded as a plumbous ortho plumbate which contains Pb in +2 as well as +4 oxidation state.