Q4. Predict the shape of XeO3 and the oxidation state of xenon in it.
Xenon is +6 (x + 3(−2) = 0). Of its 8 valence electrons, 6 go into the three Xe=O bonds, leaving one lone pair: three bonding directions and a lone pair give a trigonal pyramidal molecule, like NH3 (reasoned from VSEPR theory, NCERT Class 11 Chemistry, Unit 4).
Q5. What volume of fluorine, at 273.15 K and 1 bar, reacts completely with 0.040 mol of xenon to give XeF4? (Molar volume 22.7 L mol−1.)
Xe + 2F2 → XeF4, so 0.040 mol Xe needs 0.080 mol F2: 0.080 × 22.7 ≈ 1.82 L (reasoned; equation from Wikipedia, “Noble gas”).
Q6. Assertion: helium is placed in group 18 although its configuration is 1s. Reason: its valence shell is completely filled. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: a full 1s shell makes helium behave like the other noble gases, though strictly it belongs to the s-block (NCERT Class 11 Chemistry, page 82).
Q7. Why did Bartlett try xenon, rather than another noble gas, with platinum hexafluoride?
Platinum hexafluoride was known to oxidise the oxygen molecule, and xenon’s ionisation enthalpy is close to that of O2; so PtF6 should be able to take an electron from xenon too. In 1962 it gave a xenon–platinum fluoride product initially formulated as XePtF, establishing noble-gas compound formation (Wikipedia, “Noble gas”).
Q8. Arrange He, Ne, Ar and Kr in order of increasing boiling point, and say why helium cannot be solidified by cooling alone.
He < Ne < Ar < Kr (4.2, 27, 87 and 120 K; data pages): London forces grow with atomic size. At atmospheric pressure helium does not solidify at any temperature; about 25 atm is needed (Wikipedia, “Noble gas”).