Answer the question on basis of passage given below:

The noble gases \(He\) to \(Rn\) have close shell configuration and are monoatomic gases under normal conditions. Though mostly unreactive, but direct reaction of Xe with \(F_2\) leads to a series of compounds with Xe oxidation numbers +2, +4 and +6. When certain compounds like phenol and quinol are crystallised in the presence of noble gas such as Ar, Kr and Xe a category of compound called clathrates are formed. In these the noble gas atoms are trapped in the cavities of the crystal lattices. The compounds which crystallise are known as hosts (H) while the noble gas atoms are called guests (G). The general formula of a clathrate compound is nHmG, where n is the number of host molecules and m is the number of atoms a molecule of guest.

\(XeF_4\) and \(XeF_6\) are

oxidizing agents

The correct answer is option 2. oxidizing agents.

Xenon Fluorides as Oxidizing Agents

Xenon Tetrafluoride (\(XeF_4\))

Structure and Bonding: \(XeF_4\) has a square planar geometry with the xenon atom in the center bonded to four fluorine atoms. The xenon atom in \(XeF_4\) has a formal oxidation state of +4.

Oxidizing Properties: As a fluorinating agent, \(XeF_4\) can oxidize other substances. The xenon atom in \(XeF_4\) can accept electrons, which makes it an oxidizing agent. This is due to the high electronegativity of fluorine, which pulls electron density away from the xenon atom, making it capable of oxidizing other elements or compounds.

Xenon Hexafluoride (\(XeF_6\))

Structure and Bonding: \(XeF_6\) has a distorted octahedral structure with xenon in the +6 oxidation state. The xenon atom is surrounded by six fluorine atoms, which are highly electronegative.

Oxidizing Properties: \(XeF_6\) is even more reactive than \(XeF_4\) and exhibits strong oxidizing behavior. The high oxidation state of xenon (+6) in \(XeF_6\) makes it very electron-deficient. As a result, \(XeF_6\) can accept electrons readily and oxidize other substances.

Why They Are Oxidizing Agents

High Oxidation State: Both \(XeF_4\) and \(XeF_6\) involve xenon in high oxidation states (+4 and +6, respectively). In such high oxidation states, xenon is electron-deficient and can accept electrons from other substances, which is a key characteristic of oxidizing agents.

Fluorine's High Electronegativity: Fluorine is the most electronegative element, which means it strongly attracts electrons. In compounds with fluorine, xenon becomes even more electron-deficient, enhancing its ability to act as an oxidizing agent

Reaction Examples:

Both \(XeF_4\) and \(XeF_6\) can oxidize substances. For example:

\(XeF_4\) can oxidize water to form \(XeO_3\) and \(HF\).

\(XeF_6\) can react with water to form \(XeO_4\) (xenon tetroxide), \(HF\), and \(O_2\), showing its strong oxidizing nature

Conclusion

Both \(XeF_4\) and \(XeF_6\) are indeed oxidizing agents due to their high oxidation states and the strong electron-withdrawing effect of fluorine. They can readily accept electrons and oxidize other substances.