Another Chemistry Doubt

My friends at school were debating about the oxidizing power of the oxy-acids of chlorine : $HClO_4 , HClO_3 , HClO_2$ and $HClO$ . They claimed the following:

$HClO_4$ : Cl is in +7 oxidation state and hence will oxidize easily.

$HClO_2$ : No particular reason. Some exam paper had this as answer.

$HClO$ : Cl is in +1 oxidation state and hence can reach octet configuration easily.

Guys , Please help.

#Chemistry

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Comments

The accepted order is HOCl>HClO2>HClO4.

The argument that HClO4 is the most oxidising due to chlorine, a highly electronegative element being in such a high oxidation state as +7; would be correct if oxidation state were the only factor involved.

The other very important factor is $stability$ . A species acts as a good oxidising agent, or equivalently, readily undergoes reduction precisely when it is in an unstable state. Here, as the number of oxygen atoms surrounding Cl increase, so does the stability of the oxide.

In other words, for HClO4 to undergo reduction, the oxygen bonds must be broken(which are in fact strongest in HClO4 due to the higher polarising power of the higher oxidation state and the resultant stronger covalent bond) and this just doesn't happen.

@Vishwak Srinivasan

Shashwat Shukla - 6 years ago

Thanks a lot bro.

Vishwak Srinivasan - 6 years ago

You're welcome :)

Shashwat Shukla - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$