Answer the question on basis of passage given below:
The noble gases \(He\) to \(Rn\) have closed shell configuration and are monoatomic gases under normal conditions. Though mostly unreactive, 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 guest atoms.
In what molar ratio must \(Xe\) and \(F_2\) be taken to give \(XeF_4\)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1:5
Xenon reacts directly with fluorine, and the product depends on the ratio of the gases taken:
\( \ce{XeF2} \): xenon in excess
XeF4: Xe and F2 in the ratio 1:5 (873 K, 7 bar)
\( \ce{XeF6} \): Xe and \( \ce{F2} \) in the ratio 1:20 (573 K, 60-70 bar)
\( \ce{Xe(g) + 2F2(g)} \xrightarrow{\text{873 K, 7 bar}} \ce{XeF4(s)} \)
Although the equation needs only 1:2, fluorine is taken in excess (1:5) so that the reaction goes beyond \( \ce{XeF2} \).
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Xenon-Fluorine Compounds – “Xenon forms three binary fluorides, XeF2, XeF4 and XeF6 by the direct reaction of elements under appropriate experimental conditions.
Xe (g) + 2F2 (g) → XeF4(s) (873 K, 7 bar) (1:5 ratio)”