Q. Noble gases are inert. Explain.
A: Due to stable ns2 np6 (Octet) configuration (He = 1s2), they have high ionisation
enthalpy and positive electron gain enthalpy they are inert. As they neither
lose, gain nor share the electrons, they are inert.
Q. Give 2 uses of Neon.
A: → Neon bulbs are used in botanical gardens.
→ Ne is used in fluorescent bulbs for advertisement display purposes.
Q. Write any 2 uses of Argon.
A: →Ar is used for filling electric bulbs (to increase life of the bulb).
→Ar is used in arc welding of metals or alloys to provide inert atmosphere.
Q. Helium is heavier than H2. Yet Helium is used instead of H2 in filling baloons
for meteorological observations. Why?
A: H2 is lighter but highly inflammable gas. Where as Helium is non-inflammable
and light gas (but heavier than H2). Hence 'He' is used in meteorological
baloons.
Q. Name a) Most abundant noble gas in atmosphere b) Radioactive noble gas not
found in atmosphere c) Noble gas with least boiling point d) Noble gas forming
large number of compounds.
A: a) Argon b) Radon c) Helium d) Xenon
Q. How are a) XeOF4 and b) XeO3 are prepared? (or) What happens when XeF6
is partially & completely hydrolysed?
A: Partial hydrolysis of XeF6 gives XeOF4
XeF6 + H2O → XeOF4 + 2 HF
Complete hydrolysis of XeF6 gives XeO3
Xe F6 + 3 H2 O → XeO3 + 6 HF
Q. Why do Noble gases form compounds with 'F' & 'O' only?
A: Due to high electronegativity of 'F' and 'O', they can form compounds with
noble gases like Xe (Xe has lowest ionization enthalpy among noble gases).
e.g.: XeO4 and XeF4.
Q. How does XeF2 & XeF4 reacts with water?
A: XeF2 on reaction with water gives Xe, O2 and HF.
2 Xe F2 + 2 H2 O → 2 Xe + O2 + 4 HF
XeF4 on hydrolysis gives XeO3
6 XeF4 + 12 H2O → 4 Xe + 3 O2 + 24 HF + 2 XeO3
Q. Noble gases have very low boiling points. Why?
A: Due to presence of weak dispersion forces between noble gas atoms, they can
be liquified at very low temperatures. Hence, they have low boiling points.
Q. Why has it been difficult to study the chemistry of Radon?
A: Radon is radioactive and having very short half-life period (3.82 days), which
makes the study of chemistry of Radon is difficult.
4 Marks
Q. Explain the structures of a) XeF4 and b) XeOF4
A: a) XeF4
Xe = 5s2 5px2 5py1 5pz1 5d1 5d1 (Second excited state).
Xe undergoes sp3d2 hybridization. Due to presence
of 2 lone pairs, 4 bond pairs. Xe the shape of
XeF4 is Square Planar.
b) XeOF4
Xe = 5s2 5px1 5py1 5pz1 5d1 5d1 5d1(3rd excited state)
Π bond
Xe undergoes sp3d2 hybridization. Due to presence
of one lone pair and 6 bond pairs (1 b.p. forms Π bond),
the shape of XeOF4 is Square Pyramidal.
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