Animated Solution for Chemistry - s and p-Block Elements: Choose the incorrect statement.

Halogens vs Interhalogens

When diving into the chemistry of Group 17 elements, one of the most fascinating topics is comparing the reactivity of pure halogens with their interhalogen counterparts. In this problem, we are tasked with finding the incorrect statement among four options, which primarily revolves around the reactivity order of $C{l}_2$, $ClF$, and $F_2$.

Let's break down the chemistry behind these molecules to see which statement doesn't hold up to scientific scrutiny.

The Case of $ClF$

First, let's compare a pure halogen, $C{l}_2$, with an interhalogen, $ClF$.

In a molecule of $C{l}_2$, both atoms are identical. They have the exact same electronegativity, meaning they share the bonding pair of electrons perfectly equally. This results in a non-polar covalent bond, which is relatively stable and strong.

However, when we look at $ClF$, the story changes. Fluorine is the most electronegative element on the periodic table, significantly more so than chlorine. This electronegativity difference causes the shared electron pair to be pulled closer to the fluorine atom, creating a polar covalent bond ($C{l}^{\delta +}-F^{\delta -}$).

Because the electron cloud is distorted, the bond in $ClF$ is inherently weaker than the perfectly balanced bond in $C{l}_2$. In chemistry, a weaker bond generally translates to higher reactivity because it requires less energy to break and participate in a chemical reaction. Therefore, we can confidently state that $ClF$ is more reactive than $C{l}_2$.

This immediately tells us that statement (a), which claims "$C{l}_2$ is more reactive than $ClF$," is factually incorrect!

The Fury of $F_2$

Now, you might be thinking: "If interhalogens are more reactive than pure halogens, then $ClF$ must be more reactive than $F_2$, right?"

There is a massive catch here. Fluorine is a notorious exception to many periodic trends.

While $F_2$ is indeed a pure halogen with a non-polar bond, it is exceptionally reactive—even more so than interhalogens. Why? The answer lies in its size. Fluorine atoms are incredibly small. When two fluorine atoms bond together, their non-bonding electrons (the lone pairs) are forced into very close proximity.

This extreme closeness results in massive lone pair-lone pair interelectronic repulsion. The atoms are essentially pushing each other apart even while bonded. This intense repulsion makes the $F-F$ bond exceptionally weak and incredibly easy to break. Consequently, $F_2$ is the most reactive of all halogens and interhalogens.

So, the true reactivity order is: $F_2>ClF>C{l}_2$.

Just to be thorough, let's quickly validate why the other statements are correct:

(b) $F_2$ is more reactive than $ClF$: As we just discussed, the intense lone pair repulsion in $F_2$ makes it the king of reactivity. This statement is true. (c) On hydrolysis $ClF$ forms $HOCl$ and $HF$: When an interhalogen hydrolyzes, the more electronegative halogen (Fluorine) forms the hydrohalic acid ($HF$), and the less electronegative halogen (Chlorine) forms the hypohalous acid ($HOCl$). The reaction is $ClF+H_{2}O\to HOCl+HF$. This statement is true. (d) $F_2$ is a stronger oxidising agent than $C{l}_2$ in aqueous solution:* Fluorine has the highest standard reduction potential of any element, driven by its low bond dissociation enthalpy and the extremely high hydration enthalpy of the small $F^{-}$ ion. This statement is also true.

Therefore, statement (a) stands alone as the incorrect statement.