Animated Solution for Chemistry - s and p-Block Elements: The number of halogen (s) forming halic (V) acid is

When we dive into the chemistry of halogens, one of the most fascinating areas is their ability to form oxoacids. Halogens, being highly electronegative, generally prefer negative oxidation states. However, when bonded to an even more electronegative element like oxygen, they are forced into positive oxidation states.

In this problem, we are asked to find the number of halogens that can form a halic (V) acid. The Roman numeral (V) is the key here—it tells us that the halogen must be in a $+5$ oxidation state. The general formula for these acids is ${\mathrm{HXO}}_3$, where $\mathrm{X}$ is the halogen.

Let's start at the top of Group 17 with Fluorine. Can Fluorine form a halic (V) acid? The answer is a resounding no.

Fluorine is the undisputed king of electronegativity in the periodic table. It never exhibits a positive oxidation state (except in the highly unstable $\mathrm{HOF}$, where it is technically still $-1$ while oxygen is $+1$, though some argue otherwise, fluorine's oxidation state is always $-1$ in its compounds). More importantly, Fluorine belongs to the second period, meaning its valence shell only has $2s$ and $2p$ orbitals. It strictly lacks vacant $d$-orbitals. Without $d$-orbitals, Fluorine cannot expand its octet to accommodate the multiple bonds required to reach a $+5$ oxidation state. Therefore, ${\mathrm{HFO}}_3$ does not exist.

As we move down the group to Chlorine, Bromine, and Iodine, the story changes. These elements belong to the third period and beyond, meaning they have accessible, vacant $d$-orbitals. They can easily promote electrons and expand their octets to form multiple bonds with oxygen.

1. Chlorine: Forms Chloric acid (${\mathrm{HClO}}_3$). This is a strong acid and a powerful oxidizing agent. 2. Bromine: Forms Bromic acid (${\mathrm{HBrO}}_3$). Similar to chloric acid, it exists only in aqueous solution and is a strong oxidant. 3. Iodine: Forms Iodic acid (${\mathrm{HIO}}_3$). Unlike the previous two, iodic acid can be isolated as a stable, white crystalline solid.

By analyzing the group, we see that out of the four common halogens, only Chlorine, Bromine, and Iodine possess the necessary atomic structure (vacant $d$-orbitals) and appropriate electronegativity to form halic (V) acids.

Thus, the total number of halogens forming halic (V) acid is exactly $3$.