Animated Solution for Chemistry - s and p-Block Elements: In the following reactions, products $(A)$ and $(B)$, respectively, are $\text{NaOH}+{\text{Cl}}_2\to (A)+\text{side products}$ (hot and conc.) ${\text{Ca(OH)}}_2+{\text{Cl}}_2\to (B)+\text{side products}$ (dry)

Welcome, future engineers and doctors! Today, we are going to unravel a fascinating mystery from the world of inorganic chemistry. We are diving deep into the $s$ and $p$-block elements, specifically focusing on the halogens and their intriguing behavior when they encounter alkaline environments.

Have you ever wondered why the same reactants can produce entirely different products just by changing the temperature or concentration? This question perfectly captures that chemical drama. Let's break it down step by step and understand the beautiful logic behind these reactions.

Before we look at the specific reactions, we need to understand a core chemical concept: disproportionation. Imagine a seesaw where one person goes up while the other goes down. In a disproportionation reaction, a single element acts as both the oxidizer and the reducer. It simultaneously loses and gains electrons, splitting its identity into two different oxidation states.

Chlorine (${\text{Cl}}_2$), in its elemental form, has an oxidation state of $0$. When it reacts with alkalis, it loves to play this seesaw game. It will reduce itself to the chloride ion (${\text{Cl}}^{-}$) with a $-1$ state, and simultaneously oxidize itself to a positive oxidation state. But how high does it go on the positive side? That entirely depends on the conditions!

Let's look at the first reaction in our problem:

Here, we are using hot and concentrated sodium hydroxide. The high temperature and high concentration provide a lot of thermal energy. This energy pushes the chlorine to oxidize much further than it normally would at room temperature.

Instead of stopping at the $+1$ oxidation state (which forms hypochlorite, ${\text{OCl}}^{-}$), the chlorine is forced all the way up to the $+5$ oxidation state, forming the chlorate ion (${\text{ClO}}_3^{-}$).

So, the products are sodium chlorate (${\text{NaClO}}_3$), sodium chloride ($\text{NaCl}$), and water. In the context of our question, the main oxidized product $(A)$ is clearly ${\text{NaClO}}_3$.

A quick tip to remember: Hot conditions lead to higher oxidation states because the system has the energy to overcome the activation barriers required to form the more thermodynamically stable chlorate ion.

Bleaching Powder

Now, let's shift our focus to the second reaction:

This time, we are reacting chlorine gas with dry slaked lime, which is calcium hydroxide (${\text{Ca(OH)}}_2$). This isn't just a random textbook reaction; it is one of the most important industrial processes in history! This is exactly how commercial bleaching powder is manufactured.

When chlorine passes over dry slaked lime at room temperature, it again undergoes disproportionation. However, because we are not using extreme heat, the chlorine only oxidizes to the $+1$ state, forming the hypochlorite ion (${\text{OCl}}^{-}$), while the rest reduces to the $-1$ state (chloride, ${\text{Cl}}^{-}$).

The resulting mixture contains calcium hypochlorite (${\text{Ca(OCl)}}_2$) and calcium chloride (${\text{CaCl}}_2$). The active bleaching agent in this mixture—the chemical that actually does the work of removing stains and killing bacteria—is calcium hypochlorite. Therefore, our product $(B)$ is ${\text{Ca(OCl)}}_2$.

Let's summarize our findings: 1. Reaction with hot and concentrated $\text{NaOH}$ yields sodium chlorate, so $(A)={\text{NaClO}}_3$. 2. Reaction with dry slaked lime yields calcium hypochlorite, so $(B)={\text{Ca(OCl)}}_2$.

Matching these results with our given options, we can confidently select Option (a) as the correct answer.

Inorganic chemistry often feels like a massive list of reactions to memorize, but once you understand the underlying principles—like how temperature affects disproportionation—it becomes a logical and predictable science. Keep visualizing the electron flow, and you'll master these reactions in no time. Keep up the great work, and happy studying!