Animated Solution for Chemistry - s and p-Block Elements: Choose the correct statement(s) among the following.

In this problem, we are tasked with evaluating four distinct chemical statements related to the compounds of tin (Sn) and lead (Pb), both of which belong to Group 14 of the periodic table. Let's break down the chemistry behind each option to uncover the truth.

Let's start with option (A), which claims that ${\text{SnCl}}_2\cdot 2{\text{H}}_2\text{O}$ is a reducing agent. To understand this, we must look at the stability of oxidation states in Group 14. As we move down the group, the $+2$ oxidation state becomes increasingly stable due to the inert pair effect. However, for tin, which is higher up than lead, the $+4$ oxidation state is thermodynamically more stable than the $+2$ state.

Because of this preference, ${\text{Sn}}^{2+}$ has a strong tendency to lose two electrons and oxidize to ${\text{Sn}}^{4+}$:

Since it readily undergoes oxidation, it forces another substance to be reduced. Therefore, ${\text{SnCl}}_2$ acts as a powerful reducing agent. Statement (A) is absolutely correct.

Moving on to option (B), we examine ${\text{SnO}}_2$. Tin dioxide is a classic example of an amphoteric oxide, meaning it can react with both acids and bases. When it is treated with a strong alkali like potassium hydroxide ($\text{KOH}$), it dissolves to form a soluble complex salt known as potassium hexahydroxostannate(IV):

This reaction perfectly demonstrates its acidic character when reacting with a base. Thus, statement (B) is also correct.

Option (C) states that a solution of ${\text{PbCl}}_2$ in $\text{HCl}$ contains ${\text{Pb}}^{2+}$ and ${\text{Cl}}^{-}$ ions. While ${\text{PbCl}}_2$ is sparingly soluble in cold water, its behavior in hydrochloric acid is quite interesting. In the presence of excess $\text{HCl}$, the high concentration of chloride ions drives the formation of a soluble complex ion, tetrachloroplumbate(II):

Because the lead is tied up in this complex, the solution predominantly contains $[{\text{PbCl}}_4{]}^{2-}$ ions rather than free ${\text{Pb}}^{2+}$ ions. Therefore, statement (C) is considered incorrect in the context of standard qualitative analysis.

Finally, let's look at option (D) regarding ${\text{Pb}}_3{\text{O}}_4$, commonly known as red lead. ${\text{Pb}}_3{\text{O}}_4$ is not a simple oxide; it is a mixed oxide composed of two moles of $\text{PbO}$ and one mole of ${\text{PbO}}_2$ (i.e., $2\text{PbO}\cdot {\text{PbO}}_2$).

When treated with hot dilute nitric acid, only the basic $\text{PbO}$ component reacts to form soluble lead(II) nitrate, while the ${\text{PbO}}_2$ component remains unreacted as an insoluble brown-black precipitate:

If we analyze the oxidation states, lead is present in $+2$ and $+4$ states in the reactant, and it remains in $+2$ (in ${\text{Pb(NO}}_3{)}_2$) and $+4$ (in ${\text{PbO}}_2$) in the products. Since there is no change in oxidation states, this is purely an acid-base reaction, not a redox reaction. Statement (D) is incorrect.

After a thorough chemical analysis, we can confidently conclude that only statements (A) and (B) are correct.