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.
Match List I with List II
|
List I (Xenon Compounds) |
List II (Shape) |
| A. \(XeF_2\) | I. Square planar |
| B. \(XeF_4\) | II. Square pyramidal |
| C. \(XeOF_4\) | III. Linear |
| D. \(XeO_3\) | IV. Pyramidal |
Choose the correct answer from the options given below
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-III, B-I, C-II, D-IV
Xenon has 8 valence electrons. In each compound the bonded atoms and the remaining lone pairs set the shape (VSEPR); every O is counted as a double bond (two electrons of Xe) and every F as a single bond (one electron of Xe).
A. \( \ce{XeF2} \) — This matches III. Linear. Lone pairs = (8 − 2)/2 = 3; 2 bond pairs + 3 lone pairs = 5 domains (trigonal bipyramidal); the three lone pairs take the equatorial positions, leaving F–Xe–F in a straight line.
B. \( \ce{XeF4} \) — This matches I. Square planar. Lone pairs = (8 − 4)/2 = 2; 4 bond pairs + 2 lone pairs = 6 domains (octahedral); the lone pairs sit opposite each other, so the four F atoms lie in a square.
C. \( \ce{XeOF4} \) — This matches II. Square pyramidal. Lone pairs = (8 − 2 − 4)/2 = 1; 5 bonded atoms + 1 lone pair = 6 domains; one corner of the octahedron is the lone pair.
D. \( \ce{XeO3} \) — This matches IV. Pyramidal. Lone pairs = (8 − 6)/2 = 1; 3 bonded atoms + 1 lone pair = 4 domains (tetrahedral), so the molecule is pyramidal, like \( \ce{NH3} \).
Note 1: 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 – “… The structures of the three xenon fluorides can be deduced from VSEPR and these are shown in Fig. 7.9. XeF2 and XeF4 have linear and square planar structures respectively. XeF6 has seven electron pairs (6 bonding pairs and one lone pair) and would, thus, have a distorted octahedral structure as found experimentally in the gas phase.”
Xenon-oxygen Compounds – “… XeO3 is a colourless explosive solid and has a pyramidal molecular structure (Fig. 7.9). XeOF4 is a colourless volatile liquid and has a square pyramidal molecular structure (Fig. 7.9).”
Note 2: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 14-06 S3 05-06 S1 | – |