Question

The chemical formula of lead sulphate is

• A. ${\mathrm{Pb}}_2{\mathrm{SO}}_4$
• B. $\mathrm{Pb}{\left({\mathrm{SO}}_4\right)}_2$
• C. ${\mathrm{PbSO}}_4$
• D. ${\mathrm{Pb}}_2{\left({\mathrm{SO}}_4\right)}_3$

Answer 1

The correct option is C

${\mathrm{PbSO}}_4$

Explanation for correct answer

(C) ${\mathbf{PbSO}}_{\mathbf{4}}$

1. A lead sulfate is a white solid that is crystalline in nature, not dissolve in water and when introduced to water density is heavier than water.
2. Lead has two positive ions and two negative ions present in the bond.
3. Lead exhibits positive two ions due to inert pair effect which states that two electrons in the outermost s-orbital to remain unshared in compounds of post-transition metals, as when the s electrons remain paired the oxidation state is lower than than characteristic oxidation state of that group.
4. Hence, option (C) is correct.

Explanation for incorrect option

(A) ${\mathbf{Pb}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$

1. This is lead sulfate where lead is in +4 oxidation state and four negative and positive ions are present, which due to inert pair effect does not holds true.
2. Hence option (A) is incorrect.

(B) $\mathbf{Pb}{\mathbf{\left(}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}$

1. This compound is Lead(II) Sulfate where lead is in +2 oxidation state and Four negative ions are present as therefore to remain neutral sulfur has to exhibit -2 oxidation state as lead is in +2.
2. Hence Option (B) is incorrect.

(D) ${\mathbf{Pb}}_{\mathbf{2}}{\mathbf{\left(}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{3}}$

1. This compound is lead(III) sulfate where lead is in +3 oxidation state and six negative ions are present as due to inert pair effect the maximum oxidation state of lead is +2.
2. Hence, option (D) is incorrect.

Therefore, The formula for lead sulfate is ${\mathrm{PbSO}}_4$, Hence option (C) is correct.