Animated Solution for Chemistry - s and p-Block Elements: Chlorine on reaction with hot and oncentrated sodium hydroxide gives

Chlorine's Dance with Sodium Hydroxide

When halogens encounter an alkaline environment, they don't just undergo a simple displacement or neutralization. Instead, they engage in a fascinating chemical phenomenon known as disproportionation.

Disproportionation is a specific type of redox reaction where a single element is simultaneously oxidized and reduced. Imagine a group of identical atoms where some decide to give away their electrons, while others eagerly snatch them up. This is exactly what happens when chlorine gas ($C{l}_2$) is bubbled through a solution of sodium hydroxide ($NaOH$). However, the final destination of these electrons—and consequently, the products formed—depends entirely on the thermal energy available in the system.

Hot and Concentrated

Let's first look at the conditions specified in our problem: hot and concentrated sodium hydroxide.

When the temperature is high, the system possesses abundant kinetic energy. This extra energy allows the chlorine atoms to be pushed to a much higher oxidation state. The general reaction for halogens (let's call them $X$) under these intense conditions is:

Substituting our specific halogen, chlorine ($Cl$), into this framework, we get:

Notice what happened to the oxidation states. The elemental chlorine starts at an oxidation state of $0$. In the sodium chloride ($NaCl$) product, it has been reduced to $-1$. But in the sodium chlorate ($NaCl{O}_3$) product, it has been heavily oxidized to a $+5$ state!

If we break these resulting salts down into their constituent ions, we find that the solution is swimming with chloride ions ($C{l}^{-}$) and chlorate ions ($Cl{O}_3^{-}$). This perfectly matches option (b) of our question.

Cold and Dilute

Now, what if we dial down the heat? This is a classic trap set by examiners. If the sodium hydroxide is cold and dilute, the system lacks the thermal energy required to push chlorine all the way up to the $+5$ oxidation state.

Instead, the reaction stops at a much lower oxidation state:

Here, the chlorine is still reduced to $-1$ in sodium chloride ($NaCl$), but it is only oxidized to $+1$ in sodium hypochlorite ($NaClO$). The resulting ions in this cooler scenario are chloride ($C{l}^{-}$) and hypochlorite ($Cl{O}^{-}$).

To master p-block chemistry, you must always pay microscopic attention to the reaction conditions. A simple change from "hot" to "cold" completely alters the chemical landscape.

Remember the mnemonic: Hot = High (Higher oxidation state, $+5$, Chlorate) Cold = Low (Lower oxidation state, $+1$, Hypochlorite)

By keeping this principle in mind, you will effortlessly navigate through these classic disproportionation reactions.