Animated Solution for Chemistry - s and p-Block Elements: Identify the incorrect statement among the following

In this problem, we are presented with four distinct chemical reactions involving p-block elements. Our objective is to act as chemical detectives and identify the statement that contains an error. These types of questions are classic in JEE because they test your memory of specific reaction conditions and products. Let's break down each option systematically.

Let's evaluate the first statement: Ozone reacts with ${\text{SO}}_2$ to give ${\text{SO}}_3$.

Ozone (${\text{O}}_3$) is an incredibly powerful oxidizing agent, second only to fluorine among common chemicals. When it encounters a compound in a lower oxidation state, like sulfur in sulfur dioxide (${\text{SO}}_2$ where sulfur is $+4$), it readily donates an oxygen atom to oxidize it.

In this reaction, sulfur is oxidized to the $+6$ state in sulfur trioxide (${\text{SO}}_3$), and ozone is reduced to stable oxygen gas (${\text{O}}_2$). Therefore, this statement is perfectly correct.

Next, we look at the second statement: Silicon reacts with $\text{NaOH}(aq)$ in the presence of air to give ${\text{Na}}_2{\text{SiO}}_3$ and ${\text{H}}_2\text{O}$.

Silicon, being a metalloid, has a unique chemistry. It doesn't react with water under normal conditions due to a protective oxide layer. However, in a strong alkaline medium like aqueous sodium hydroxide, the oxide layer dissolves, and silicon reacts to form silicates. In the presence of air (oxygen), the overall reaction can be represented as forming sodium silicate and water.

This is a standard method for preparing soluble silicates, often called "water glass". Thus, this statement is also correct.

The third statement is a famous one: ${\text{Cl}}_2$ reacts with excess of ${\text{NH}}_3$ to give ${\text{N}}_2$ and $\text{HCl}$.

This reaction is highly dependent on which reactant is in excess. When ammonia is in excess, chlorine acts as an oxidizing agent, oxidizing the nitrogen in ammonia from $-3$ to $0$ (nitrogen gas). The initial step of this reaction produces nitrogen gas and hydrogen chloride.

While it is true that the $\text{HCl}$ formed will immediately react with the excess ammonia to form dense white fumes of ammonium chloride (${\text{NH}}_4\text{Cl}$), the primary products of the redox process are indeed ${\text{N}}_2$ and $\text{HCl}$. In the context of multiple-choice questions, this statement is considered acceptable because it correctly identifies the primary redox products before the subsequent acid-base neutralization.

Finally, let's examine the fourth statement: ${\text{Br}}_2$ reacts with hot and strong $\text{NaOH}$ solution to give $\text{NaBr}$, ${\text{NaBrO}}_4$ and ${\text{H}}_2\text{O}$.

This is where the trap lies! Halogens (except fluorine) undergo disproportionation reactions when treated with alkalis. The products depend heavily on the temperature.

With cold and dilute $\text{NaOH}$, bromine forms sodium bromide ($\text{NaBr}$) and sodium hypobromite ($\text{NaBrO}$). However, with hot and concentrated $\text{NaOH}$, the hypobromite ion is unstable and further disproportionates to form the halate ion.

The products are sodium bromide ($\text{NaBr}$) and sodium bromate (${\text{NaBrO}}_3$). The statement claims the formation of ${\text{NaBrO}}_4$, which is sodium perbromate. Perbromates are notoriously difficult to synthesize and are definitely not formed in this simple disproportionation reaction.

By carefully analyzing the reaction conditions and the specific formulas of the products, we have uncovered the error. The reaction of bromine with hot sodium hydroxide yields bromate (${\text{BrO}}_3^{-}$), not perbromate (${\text{BrO}}_4^{-}$). Therefore, statement (d) is the incorrect one, making it the right answer to our question.