Question

The chemical formula of lead sulphate is?
A.\[P{b_2}S{O_4}\]
B.\[Pb{(S{O_4})_2}\]
C.\[PbS{O_4}\]
D.\[P{b_2}{(S{O_4})_3}\]

Answer 1

Hint: Lead sulphate is an ionic salt and will therefore be an overall neutral molecule. Find the charge on the cation as well as anion, balance the charges (if not already balanced) by the crisscross method and determine the formula.

Complete answer:
Lead sulphate is an ionic compound, which contains lead cations and sulphate anions. Lead belongs to the fourteenth group of the periodic table and has a total of four valence electrons in its outermost shell. It is a member of the P-block.
P-block elements experience an inert pair effect due to which the group shows two types of oxidation states and the lower lying members of the groups tend to be more stable in the lower oxidation state.
The group fourteen has an electronic configuration of \[n{s^2}n{p^2}\] and its members show \[ + 4\] as well as \[ + 2\] oxidation states, but the heavier members preferably show \[ + 2\] oxidation state.
Lead is one of the heavier members of the fourteenth group and prefers \[ + 2\] oxidation state. It is easier to form \[P{b^{2 + }}\] ion as the energy required in removing two electrons is more than compensated by the energy released during the ionic bond formation of lead salt. But losing more than two electrons i.e. forming \[P{b^{3 + }}\] or \[P{b^{4 + }}\] ion is an energetically unfavorable process.
The sulphate ions carry \[ - 2\] charge. Thus the charges on the cation as well as anion are balanced and the formula of lead sulphate comes out to be \[PbS{O_4}\].
Hence, the correct option is \[(C)\] \[PbS{O_4}\].

Note:
The charge present on a cation is not always equal to the number of valence electrons present in its outermost shell. The inert pair effect that causes this discrepancy is a consequence of the poor shielding effect of the electrons present in d and f orbitals. The pair of electrons present in the s-orbital becomes inaccessible due to the strong nuclear force.