Questions - Answers
Group - 16 Elements
2 Marks
1. Explain the structures of SF4 and SF6.
2. Why is H2O a liquid while H2S is a gas at room temperature?
A: Due to small atomic size and high electronegativity of O, H2O molecules associate together to form a liquid. This is due to intermolecular Hydrogen bonding.
O − H bonds are polar, but H − S bonds are unable to form Hydrogen bonds, so H2S is a gas.
3. Give one example each for a) Neutral oxide b) Peroxide c) Super oxide and
d) Amphoteric oxide.
A: a) NO b) H2O2 c) KO2 d) H2O
4. Give the hybridization of 'S' in the following.
a) SO2 b) SO3 c) SF4 d) SF6
A: a) sp2 b) sp2 c) sp3d d) sp3d2
5. What is tailing of mercury? How is it removed?
A: The loss of lustre and meniscus of Hg when it reacts with O3. It sticks to the walls of the glass (due to Hg2O).
2 Hg + O3 → Hg2O + O2
Hg2O is removed by dissolving in water by shaking.
6. Give the reactions of O3 with a) C2H2 and b) C2H4.
A: a) O3 gives acetylene ozonide with acetylene.
C2H2 + O3 → C2H2 . O3
b) O3 gives ethylene ozonide with ethylene.
C2H4 + O3 → C2H4 . O3
7. Write the names, formulae, structures of any two oxyacids of sulphur.
A: H2SO4 − Sulphuric acid H2S2O7 − Pyrosulphuric acid
8. Write any two uses of O3.
A: * Used in sterilization of water.
* Used as germicide.
9. SO2 can be used as antichlor. Explain.
A: SO2 removes Cl2 to form sulphuryl chloride. It can be used as antichlor.
SO2 + Cl2 SO2Cl2.
10. Out of O2 and O3, which is paramagnetic?
A: According to MOT, due to presence of 2 unpaired electrons O2 is paramagnetic, O3 is diamagnetic.
11. Explain the structures of SO4−2 and SO3.
12. Mention 2 threats to ozone layer.
A: * Use of freons (present in refrigirator).
* Releasing of NO from supersonic jet aeroplanes.
NO + O3 → NO2 + O2
4 Marks
1. Absorption of SO3 by H2SO4 to give oleum.
SO3 + H2SO4 → H2S2O7 (oleum)
Oleum on dilution with water gives H2SO4.
H2S2O7 + H2O → 2 H2SO4.
2. How is ozone prepared from oxygen?
Explain its reaction with a) PbS b) moist KI c) Hg and d) Ag. Give equations.
A: By passing silent electric discharge through pure, dry, cold oxygen 10% ozone is formed. Pure O3 can be condensed in a vessel surrounded by liquid oxygen.
3 O2 → 2 O3; Δ H = +142 K.J./ mole
a) O3 oxidises black PbS to white PbSO4.
PbS + 4 O3 → PbSO4 + 4 O2
b) O3 oxidises moist KI to I2.
2 KI + H2O + O3 → I2 + O2 + 2 KOH
c) Hg loses lustre and meniscus and sticks to glass surface when reacts with O3.
2 Hg + O3 → Hg2O + O2
d) O3 oxidises Ag to Ag2O.
2 Ag + O3 → Ag2O + O22
Group - 17 Elements
2 Marks
1. Electron gain enthalpy of F2 is less than that of Cl2. Explain.
A: Due to small atomic size of fluorine, repulsions between electron pairs already present in it and the added electron, the tendency of addition of electron becomes less in F. So the electron gain enthalpy of F2 is less (−333 K.J./ mole) than that of Cl2 (−349 K.J./ mole).
2. HF is a liquid while HCl is a gas. Explain.
A: Due to high electronegativity of F, H - F bonds are more polar than H - Cl bonds. Due to inter molecular hydrogen bonding, HF molecules associate together to get liquid state, HCl does not form H-bonding, hence it is a gas.
...... H - F ...... H - F ...... H - F ...... H - F ......
3. Write the balance equations for the following.
a) NaCl is heated with conc. H2SO4 in the presence of MnO2.
b) Cl2 is passed into the solution of NaI in water.
A: a) By heating NaCl with conc. H2SO4 and MnO2, chlorine gas is liberated.
b) Cl2 oxidises NaI to I2.
4. Explain the structure of ClF3.
A: Cl = [Ne] 3s2 3px2 3py1 3pz1 3d1 (1st excited state)
Cl undergoes sp3d hybridization.
shape of ClF3 is T.
5. Explain the structures of a) BrF5 and b) IF7.
A: a) BrF5
‘Br’ undergoes sp3d2 hybridization.
It has square pyramidal shape.
b) IF7
‘I’ undergoes sp3d3 hybridization.
It has pentagonal bipyramidal shape.
6. Describe the structure of I3−.
A: Central atom of I in I3− undergoes sp3d hybridization. Shape of I3− is linear (due to presence of 3 lone pairs in equitorial positions).
7. Bond dissociation enthalpy of F2 is less than that of Cl2. Explain.
A: Atomic size of F is small, hence the F − F bond distance small (1.48 A°).
Due to lone pair, lone pair electron repulsions, the bond becomes weak. Hence bond dissociation enthalpy of F2 is less than that of Cl2.
8. What happens when Cl2 reacts with dry slaked lime?
A: Bleaching powder CaOCl2 (calcium chloro hypo chlorite) is formed when Cl2 reacts with Ca(OH)2.
9. Name the halogen that produces O2 & O3 on passing through water.
A: F2 liberates O2 & O3 on reaction with water.
10. Though electron gain enthalpy of F2 is less negative than Cl2, F2 is stronger oxidising agent than Cl2. Why?
A: * Due to high hydration enthalpy of F− (515 K.J./ mole).
* Low enthalpy of dissociation of F − F bond (158.8 K.J./ mole).
11. Name 2 poisonous gases which can be prepared from Cl2 gas.
A: Tear gas (CCl3NO2)
Phosgene (COCl2)
4 Marks
1. How is Cl2 prepared in the laboratory? How does it react with a) Iron b) acidified
FeSO4 c) H2S and d) Na2S2O3.
A: Cl2 can be prepared by heating MnO2 with conc. HCl.
Reactions:
a) Cl2 reacts with Fe to give FeCl3.
b) Cl2 oxidises acidified ferrous sulphate to ferric sulphate.
c) Cl2 reacts with H2S to produce HCl & S.
d) Cl2 reacts with Na2S2O3 to give Na2SO4.
2. How is chlorine prepared by electrolytic method?
Expalin its reactions with a) NaOH b) NH3 under different conditions.
A: Cl2 is liberated at anode by the electrolysis of brine solution.
Reactioins:
a) with cold, dil. NaOH: Cl2 gives NaCl & NaOCl with Cold, dil. NaOH.
with hot, conc. NaOH: Cl2 gives NaCl and NaClO3 with hot, conc. NaOH.
b) with excess of Cl2: NH3 reacts with excess of Cl2 to give NCl3.
with excess of NH3: Cl2 reacts with excess of NH3 to give NH4Cl.
3. What are inter halogen compounds? How are they classified? Illustrate with some examples.
A: The compounds which are formed between 2 different halogens are called inter halogen compounds.
* The general formula of these compounds is AXn where n = 1, 3, 5 or 7.
* A is larger halogen, X is smaller halogen.
* Oxidation state of X is always -1, but A could be +1, +3, +5 or +7.
* AX type interhalogens : ClF, BrF, BrCl, ICl
AX3 type interhalogens : ClF3, BrF3, ICl3
AX5 type interhalogens : BrF5, IF5
AX7 type interhalogens : IF7.
4. Write the names, formulae of oxyacids of Cl. Explain their structures and relative acidic nature.
A:
Acid Strength: HClO4 > HClO3 > HClO2 > HClO1
Group - 18 Elements
2 Marks
1. 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.
2. A mixture of He & O2 is used in modern diving apparatus. Why?
A: Due to very low solubility of Helium, it is used as diluent for oxygen in modern diving apparatus.
3. Give 2 uses of Neon.
A: * Neon bulbs are used in botanical gardens.
* Ne is used in fluorescent bulbs for advertisement display purposes.
4. 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.
5. 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.
6. 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
7. 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.
8. 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.
9. How are XeF2 and XeF4 are prepared?
A: Excess of Xe reacts with F2 at 673 K and 1 bar gives XeF2.
1 part of Xe, 5 parts of F2 reacts at 873 K, 7 bar gives XeF4.
10. How does XeF2 & XeF4 reacts with water?
A: XeF2 on reaction with water gives Xe, O2 and HF.
XeF4 on hydrolysis gives XeO3.
6 XeF4 + 12 H2O → 4 Xe + 3 O2 + 24 HF + 2 XeO3
11. 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.
12. 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
1. 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 in Xe. The shape of XeF4 is Square Planar.
b) XeOF4
Xe = 5s2 5px1 5py1 5pz1 5d1 5d1 
(3rd excited state)
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.
2. Explain the structures of a) XeF2 and b) XeF6.
A: a) XeF2
Xe = 5s2 5px2 5py2 5pz1 5d1 (First excited state).
Xe undergoes sp3d hybridization. The shape of XeF2 is linear.
b) XeF6:
Xe = 5s2 5px1 5py1 5pz1 5d1 5d1 5d1
(Third excited state).
Xe undergoes sp3d3 hybridization.
The shape of Xe F6 is distorted Octahedral.
3. How is XeO3 prepared? Explain the structure of XeO3.
A: Complete hydrolysis of XeF6 gives XeO3.
(Third excited state)
Xe has one lone pair and six bond pairs. 3 bond pairs form 3 Π (Pi) bonds. Xe undergoes sp3 hybridization.
The shape of XeO3 is Pyramidal.
Bond angle is 103°.
