Question is based on following passage:
p-block elements are placed in groups 13 to 18 of the periodic table. Group 15 is nitrogen family. Nitrogen makes various compounds like \(NH_3\), \(HNO_3\), \(N_2O\), \(NO_2\) etc. Phosphorus exists in the form of white and red phosphorus. White phosphorus glows in dark. \(PH_3\), gas is used as smoke screen. Group 16 is called oxygen family. Dry air contains 21% \(O_2\) gas by volume. Combined sulphur exists as sulphates such as gypsum, Epsom salt, baryte etc. The allotropic forms of sulphur are Rhombic and monoclinic sulphur. \(S_2\) is paramagnetic like \(O_2\) and it is stable at high temperature. Group 17 is Halogen family. \(F_2\), is highly reactive, \(Cl_2\), is greenish yellow gas. Both \(SO_2\), and \(Cl_2\), are used for bleaching. Group 18 is Noble gas family but some noble gases also form compounds. Complete hydrolysis of \(XeF_6\) gives \(HF\) and \(XeO_3\), as main products.
The product of complete hydrolysis of \(XeF_6\) in the following reaction is:
\(XeF_6 + H_2O \longrightarrow\) ____ \(+ HF\)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(XeO_3\)
Complete hydrolysis of \( \ce{XeF6} \) gives xenon trioxide and HF:
\( \displaystyle \ce{XeF6 + 3H2O -> XeO3 + 6HF} \)
Xenon stays in the +6 oxidation state; all six F atoms are replaced by three O atoms, so this is not a redox reaction.
Why the other options are incorrect
Option (1) XeOF4 — Incorrect. Formed in partial hydrolysis:
\( \displaystyle \ce{XeF6 + H2O -> XeOF4 + 2HF} \)
Option (2) XeO2F2 — Incorrect. Also a product of partial hydrolysis:
\( \displaystyle \ce{XeF6 + 2H2O -> XeO2F2 + 4HF} \)
Option (4) Xe — Incorrect. Free xenon forms only in a redox change (as in the hydrolysis of \( \ce{XeF4} \), where part of the Xe is reduced); in the hydrolysis of \( \ce{XeF6} \) the oxidation state stays +6.
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-oxygen Compounds – “Hydrolysis of XeF4 and XeF6 with water gives XeO3.”
\( \displaystyle \ce{6XeF4 + 12H2O -> 4Xe + 2XeO3 + 24HF + 3O2} \)
\( \displaystyle \ce{XeF6 + 3H2O -> XeO3 + 6HF} \)
“Partial hydrolysis of XeF6 gives oxyfluorides, XeOF4 and XeO2F2.”
\( \displaystyle \ce{XeF6 + H2O -> XeOF4 + 2HF} \)
\( \displaystyle \ce{XeF6 + 2H2O -> XeO2F2 + 4HF} \)