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.

In what molar ratio must \(Xe\) and \(F_2\) be taken to give \(XeF_4\)?

1:5

The correct answer is option 4. 1:5.

XeF4 is obtained in good amounts by heating a mixture of xenon and fluorine in the molar ratio 1:5 at 873 K and 7 bar pressure in an enclosed nickel vessel for a few hours.

\(Xe(g) + 2F_2 \overset{873 K, 7 bar}{\longrightarrow} XeF_4\)