Animated Solution for Chemistry - s and p-Block Elements: Chlorine reacts with hot and concentrated NaOH and produces compounds $(X)$ and $(Y)$. Compound $(X)$ gives white precipitate with silver nitrate solution. The average bond order between Cl and O atoms in $(Y)$ is ....... .

Let's decode a very classic and high-yield reaction in inorganic chemistry. We are reacting chlorine gas ($C{l}_2$) with hot and concentrated sodium hydroxide ($NaOH$). This is a textbook example of a disproportionation reaction, meaning the same element (chlorine) is simultaneously oxidized and reduced.

When chlorine reacts with hot and concentrated $NaOH$, it forms sodium chloride ($NaCl$) and sodium chlorate ($NaCl{O}_3$), along with water. The balanced chemical equation is:

The problem states that compound $(X)$ gives a white precipitate when treated with a silver nitrate ($AgN{O}_3$) solution. This is a classic qualitative analysis test! We know that chloride ions react with silver ions to form a highly insoluble white precipitate of silver chloride ($AgCl$).

This confirms that compound $(X)$ is definitely sodium chloride ($NaCl$). Consequently, the other major product, compound $(Y)$, must be sodium chlorate ($NaCl{O}_3$).

Now, our main objective is to find the average bond order between the chlorine and oxygen atoms in compound $(Y)$, specifically within the chlorate ion ($Cl{O}_3^{-}$). Let's visualize its Lewis structure. Chlorine acts as the central atom, bonded to three surrounding oxygen atoms.

To complete the octets and minimize the formal charge across the molecule, chlorine expands its octet (thanks to its empty d-orbitals) and forms two double bonds and one single bond with the oxygen atoms. The single-bonded oxygen carries a formal charge of $-1$.

Because all three oxygen atoms are chemically equivalent, there is no reason for the single bond to be fixed at one specific oxygen atom. Instead, the single bond and the negative charge resonate across all three positions. This creates three equivalent resonance structures.

In the actual molecule—the resonance hybrid—the double bond character is equally distributed among all three chlorine-oxygen bonds. None of the bonds are pure single or pure double bonds; they are all identical partial double bonds.

To calculate the average bond order in a resonating symmetrical molecule like this, we don't need complex quantum mechanics. We can use a very simple and elegant formula:

If we look at any single resonance structure of the $Cl{O}_3^{-}$ ion, we count two double bonds (which is 4 bonds) and one single bond (which is 1 bond), giving us a total of 5 bonds connected to the central chlorine atom. There are 3 surrounding oxygen atoms.

So, the average bond order between the Cl and O atoms is 1.67.

Always be careful with halogen reactions! If the question had specified cold and dilute $NaOH$ instead of hot and concentrated, the products would have been sodium chloride ($NaCl$) and sodium hypochlorite ($NaClO$). In the hypochlorite ion ($Cl{O}^{-}$), there is only one single bond, making the bond order exactly 1. Temperature changes everything in thermodynamics!