Class 12 Chemistry Sample Paper Set A

Please refer to Class 12 Chemistry Sample Paper Set A with solutions below. The following CBSE Sample Paper for Class 12 Chemistry has been prepared as per the latest pattern and examination guidelines issued by CBSE. By practicing the Chemistry Sample Paper for Class 12 students will be able to improve their understanding of the subject and get more marks.

Objective Type Questions

A. Multiple choice Questions:

Question. Hot conc. H₂SO₄ acts as moderately strong oxidising agent. It oxidises both metals and nonmetals.
Which of the following element is oxidised by conc. H₂SO₄ into two gaseous products?
(a) Cu
(b) S
(c) C
(d) Zn

Question. Which of the following are peroxoacids of sulphur?
(a) H₂SO₅ and H₂S₂O₈
(b) H₂SO₅ and H₂S₂O₇
(c) H₂S₂O₇ and H₂S₂O₈
(d) H₂S₂O₆ and H₂S₂O₇

B. Answer the following:

Question. Draw the structure of H₂S₂O₈.
Answer.

Question. Oxygen is a gas but sulphur a solid. Explain.
Answer. Oxygen being smallest in size is capable of forming pp – pp bond and exists as O=O(O₂ molecule) while sulphur being larger in size is not able to form pp – pp bond.

Question. Draw the structure of H₂S₂O₇.
Answer. Pyrosulphuric acid (oleum) :

Question. Account for the following:
Oxygen shows catenation behaviour less than sulphur. 
Answer. Bond energy of S—S bond (213 kJ mol^–1) is greater than O – O bond (138 kJ mol^–1). Due to small size of oxygen atom there is greater lp–bp repulsion in O – O, resulting in weakening of O – O bond more than in S–S bond. Therefore, the tendency of catenation in oxygen is lower than sulphur.

Question. Write the formula of the compound of sulphur which is obtained when conc. HNO₃ oxidises S₈.
Answer. H₂SO₄

Question. Predict the shape and the asked angle (90° or more or less) in the following case SO₃^2– and the angle O—S—O.
Answer. Pyramidal :

Angle : O—S—O more than 90° 

Question. Write the formulae of any two oxoacids of sulphur.
Answer. H₂SO₃ (Sulphurous acid) and H₂SO₄ (Sulphuric acid). 

Question. Arrange the following hydrides of Group-16 elements in the increasing order of their thermal stability : H₂O, H₂S, H₂Se, H₂Te.
Answer. H₂Te<H₂Se<H₂S<H₂O

Short Answer Type Questions 

Question. What happens when :
(i) Concentrated H₂SO₄ is added to calcium fluoride ?
(ii) SO₃ is passed through water ?
Answer. (i) H₂SO₄ + CaF₂ → 2HF + CaSO₄ 
           (conc.)
(ii) SO₃ + H₂O → H₂SO₄

Question. Write the conditions to maximise the yield of H₂SO₄ by contact process.
Answer. The conditions used to maximise the yield of H₂SO₄ by contact process :
(i) High pressure
(ii) Low temperature
(iii) V₂O₅ is used as catalyst

Question. What happens when :
(i) conc. H₂SO₄ is added to Cu ?
(ii) SO₃ is passed through water ?
Write the equations. 
Answer. (i) Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O 
(ii) SO₃ + H₂O → H₂SO₄

Question. How is O₃ estimated quantitatively?
Answer. When ozone reacts with potassium iodide (KI) solution buffered with a borate buffer (pH = 9.2) then iodine (I2) is liberated that can be titrated against a standard solution of sodium thiosulphate using starch as indicator. 

Long Answer Type Questions-I 

Question. How is SO₂ an air pollutant? 
Answer. Sulphur dioxide causes harm to the environment in many ways :
(i) When it combines with water vapour it forms sulphuric acid (H₂SO₄). This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble.
(ii) Even in very low concentrations, SO₂ causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness. 
(iii) It is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves. This condition is known as chlorosis.
This happens because the formation of chlorophyll is affected by the presence of sulphur dioxide.

Question. Give reasons for the following :
(i) (CH₃)₃ P = O exists but (CH₃)₃ N = O does not.
(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.
(iii) H₃PO₂ is a stronger reducing agent than H₃PO₃.
Answer. (i) Because as N can’t form 5 covalent bonds as its maximum covalency is three. The octet cannot be extended as it doesn’t have d orbital, while P can extend its octet as it has empty d orbital. 
(ii) This is due to very small size of oxygen atom,repulsion between electrons is large in relatively small 2p sub-shell. 
(iii) In H₃PO₂ there are 2P – H bonds, whereas in H₃PO₃ there is 1 P-H bond. 

Topic-2
Group-17 Elements, Properties and Some Important Compounds

A. Multiple choice Questions:

Question. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy?
(a) HF
(b) HCl
(c) HBr
(d) HI

Question. Reduction potentials of some ions are given below.Arrange them in decreasing order of oxidising power.

(a) ClO₄^– > IO₄^– > BrO₄^–
(b) IO₄^–> BrO₄^– > ClO₄^–
(c) BrO₄^–> IO₄^– > ClO₄^––
(d) BrO₄^– > ClO₄^– > IO₄^–

Question. Which of the following is iso-electronic pair?
(a) ICl₂, ClO₂
(b) BrO₂^–, BrF₂⁺
(c) ClO₂, BrF
(d) CN^–, O₃

Question. A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When excess of this gas reacts with NH3 an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from :
(a) – 3 to +3.
(b) – 3 to 0.
(c) – 3 to +5.
(d) 0 to – 3.

B. Answer the following:

Question. HF is a weaker acid than HCl why ?
Answer. Since H—F bond is strongest with higher bond dissociation energy than HCl, hence it is weakest acid among all the halogen acids. 

Question. Give reasons: Fluoride ion has higher hydration enthalpy than chloride ion.
Answer. Due to small size of fluoride ion/high charge density of fluoride ion/high charge size ratio of fluoride ion.

Question. Bond enthalpy of fluorine is lower than that of chlorine why ? 
OR
Bond enthalpy of F₂ is less than of Cl₂.
Answer. Bond enthalpy of F—F is smaller due to greater repulsive interactions between the lone pair of one F atom with those of other. The repulsive interaction arise due to greater concentration of electron density on each F atom because of its extremely small size. 

Question. Draw the structure of BrF₃ molecule.
Answer.

Question. Write the formula of the compound of iodine which is obtained when conc. HNO₃ oxidises I₂.
Answer. HIO₃ 

Question. Give reason for the following :
F₂ is more reactive than ClF₃ but ClF₃ is more reactive than Cl₂. 
Answer. Interhalogen compounds are more reactive than halogen compounds. But in case of fluorine due to the small size of fluorine, it has high electronegativity and low bond energy so it is more reactive than ClF₃.Therefore ClF₃ is more reactive than Cl₂. 

Question. Account for the following: Iron on reaction with HCl forms FeCl₂ and not FeCl₃.
Answer. Because HCl is a mild oxidising agent and formation of hydrogen gas prevents the formation of FeCl₃. 

Question. Give reason for the following :
PbCl₄ is more covalent than PbCl₂.
Answer. Due to high oxidising power, halogens combine directly with most metals to form their corresponding halides. But if the metal exhibit more than one oxidation state, the halide in higher oxidation state will be more covalent than the one in lower oxidation state. Therefore PbCl₄ is more covalent than PbCl₂, as Pb exhibits more than one oxidation states.

Question. Name two poisonous gases which can be prepared from chlorine gas.
Answer. Phosgene (COCl₂), tear gas (CCl₃NO₂), mustard gas [C₄H₈Cl₂S].

Question. Arrange the following in the order of property indicated against each set:
(a) F₂, Cl₂, Br₂, l₂ (increasing bond dissociation enthalpy)
(b) H₂O, H₂S, H₂Se, H₂Te (increasing acidic character) 
OR
Answer the following questions :
(i) Write the formula of the neutral molecule which is isoelectronic with ClO.
(ii) Draw the shape of H₂S₂O₇.
Answer. (a) I₂ < F₂< Br₂ < Cl₂
(b) H₂O < H₂S < H₂Se < H₂Te
OR
(i) CIF

Question. Fluorine exhibits only –1 oxidation state whereas other halogens exhibit +1, +3, +5 and +7 oxidation states also. Why is it so ?
Answer. Fluorine has no d-orbital for excitation of electrons and it is most electronegative element. Hence, it shows oxidation state of –1 only.

Question. Give reasons: ICl is more reactive than I₂.
Answer. Inter halogen compound ICl are more reactive than halogens Cl₂ because X–X (I–Cl) bond in inter halogens is weaker than Cl–Cl bond in halogens except F–F bond.

Short Answer Type Questions

Question. Arrange the following in order of property indicated for each set :
(i) F₂, Cl₂, Br₂, I₂ – increasing bond dissociation enthalpy.
(ii) PH₃, AsH₃, BiH₃, SbH₃, NH₃ – increasing base strength.
Answer. (i) I₂ < F₂ < Br₂ < Cl₂ 
(ii) BiH₃ <SbH₃ < AsH₃ < PH₃ < NH₃

Question. Complete the following chemical equations :
(i) Ca₃P₂ + H₂O →
(ii) 3Cu + H₂SO₄ (conc.) → 
OR
Arrange the following in the order of property indicated against each set :
(i) HF, HCl, HBr, HI – increasing bond dissociation enthalpy.
(ii) H₂O, H₂S, H₂Se, H₂Te – increasing acidic character. 
Answer. (i) Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃
(ii) Cu + 2H₂SO₄ (conc.) → CuSO₄ + 2H₂O + SO₂
OR
(i) H—I < H—Br < H—Cl < H—F 
(ii) H₂O < H₂S < H₂Se < H₂Te 

Question. Draw the structures of the following:
(i) H₂SO₃
(ii) HClO₃ 
Answer.

Question. What happens when
(i) HCl is added to MnO₂?
(ii) PCl₅ is heated ?
Write the equation involved.
Answer. (i) MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O

Question. Draw the structures of the following:
(i) H₂S₂O₈
(ii) ClF₃ 
Answer. 

Question. Account for the following:
(i) Two S-O bond lengths in SO₂ are equal.
(ii) Fluorine shows only -1 oxidation state in its compounds.
Answer. (i) Due to resonance the two S-O bond lengths are identical. 
(ii) Absence of d-orbitals and most electronegative element. 

Question. Draw the structures of the following :
(i) XeF₄
(ii) HClO₃ 
Answer.

Question. Account for the following:
(i) The two oxygen-oxygen bond lengths in ozone molecule are identical.
(ii) Most of the reactions of fluorine are exothermic.
Answer. (i) Due to resonance the two O-O bond lengths are identical. 
(ii) Due to strong bond formed by it with other elements.

Question. Give reasons :
(i) When Cl₂ reacts with excess of F₂, ClF₃ is formed and not FCl₃.
(ii) Dioxygen is a gas while Sulphur is a solid at room temperature.
Answer. (ii) F cannot show positive oxidation state as it has highest electronegativity/Because Fluorine cannot expand its covalency / As Fluorine is a small sized atom, it cannot pack three large sized Cl atoms around it. 
(iii) Oxygen has multiple bonding whereas sulphur shows catenation / Due to pπ-pπ bonding in oxygen whereas sulphur does not / Oxygen is diatomic therefore held by weak intermolecular force while sulphur is polyatomic held by strong intermolecular forces. 

Question. Account for the following:
(i) Bond angle in NH₄⁺ is higher than that in NH₃.
(ii) ICl is more reactive than I₂.
Answer. (i) In NH₄⁺, all are bond pairs whereas in ammonia the lone pair of electron on nitrogen repels the bond pairs and reduces the bond angle.
(ii) I-Cl bond is weaker than I-I bond / low bond dissociation enthalpy in I-Cl. 

Long Answer Type Questions-I 

Question. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidizing power of F₂ and Cl₂. 
Answer. F₂ is a stronger agent than Cl₂ because
(i) Bond dissociation enthalpy of fluorine is less than that of chlorine. 
(ii) Electron gain enthalpy of fluorine is less than that of chlorine. 
(iii) F– has high hydration enthalpy. 

Question. Explain why the stability of oxoacids of chlorine increases in the order given below : HClO < HClO₂ < HClO₃ < HClO₄
Answer. Oxygen is more electronegative than chlorine;therefore, dispersal of negative charge present on chlorine increases from ClO to ClO₄ ion because numbers of oxygen atoms are attached to chlorine increases. Therefore, stability of ions will increase in the order given below :
ClO < ClO₂ < ClO₃ < ClO₅ 
Thus due to increase in stability of conjugate base,acidic strength of corresponding acid increases in the following order :
HClO < HClO₂ < HClO₅ < HClO₄

Topic-3
Group-18 Elements, Properties and Some Important Compounds

Very Short Answer-Objective Type Questions 

A. Multiple choice Questions:

Question. Which one of the following does not exist?
(a) XeOF₄
(b) NeF₂
(c) XeF₂
(d) XeF₆

Question. In the preparation of compounds of Xe, Bartlett had taken O₂⁺PtF₆^– as a base compound. This is because
(a) both O₂ and Xe have same size.
(b) bothO₂ and Xe have same electron gain enthalpy.
(c) both O₂ and Xe have same ionisation enthalpy.
(d) both Xe and O₂ are gases.

B. Match the following :

Question. Match the species given in Column I with those mentioned in Column II.

C. Answer the following:

Question. XeF₆ undergoes complete hydrolysis?
Answer. When XeF₆ undergoes complete hydrolysis, it forms
XeF₆ + 3H₂O → 6HF + XeO₃

Question. Complete the following chemical equation :
XeF₄ + SbF₅ → 
Answer. XeF₄ + SbF₅ → [XeF₃]+[SbF₆]^– 

Question. Helium is used in diving equipment.
Answer. Helium is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood. 

Question. What inspired N. Bartlett for carrying out reaction between Xe and PtF₆ ?
Answer. First ionisation enthalpy of molecular oxygen was almost similar with that of Xenon. Thus, after preparing red coloured compound O₂⁺ [PtF₆ ] ⁻,
Bartlett got inspired for carrying out reaction between Xe and PtF6 and made efforts to prepare Xe⁺[PtF₆]^– by mixing Xe and PtF₆. 

Question. Draw the structure of XeF₂ molecule.
Answer. XeF₂ :

Question. Draw the structures of the following :
XeF₄ 
Answer.

Question. Draw the structures of the following :
(i) XeF₄ (ii) BrF₅
Answer.

Question. Draw the structures of the following :
(i) XeO₃ (ii) H₂SO₄ 
Answer.

Question. Write the structures of the following :
(i) BrF₃ (ii) XeF₄ 
OR
What happens when :
(i) SO₂ gas is passed through an aqueous solution Fe³⁺ salt ?
(ii) XeF₄ reacts with SbF₅ ?
Answer.

OR
(i) When SO₂ gas is passed through an aqueous solution of Fe³⁺ salt, SO₂ acts as a reducing agent and reduces Fe³⁺ to Fe²⁺. The brown colour of iron
(III) solution turns into green. 
2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺+ SO₄^2– + 4H⁺
(ii) When XeF₄ reacts with strong Lewis acids like SbF₅ it forms complexes (addition compounds).
XeF₄ + SbF₅ → [XeF₃]⁺[SbF₆]^–

Question. Draw the structures of the following:
(i) XeF₂ (ii) BrF₅ 
Answer.

Question. Complete the following reactions:
(i) NH₃ + 3Cl₂ (excess) →
(ii) XeF₆ + 2H₂O →
Answer. (i) NH₃ + 3Cl₂ (excess) → NCl₃ + 3HCl 
(ii) XeF₆ + 2H₂O → XeO₂F₂

Question. Draw the molecular structure of XeF₆.
Answer. XeF₆ :

Question. Explain the following giving an appropriate reason in each case :
(i) O₂ and F₂ both stabilize higher oxidation states of metals but O₂ exceeds F₂ in doing so.
(ii) Structures of Xenon fluoride cannot be explained by Valence Bond approach.
Answer. (i) This is due to the ability of oxygen to form multiple bonds with metals. Oxygen has -2 charge for each atom while F₂ bears -1 for each atoms. Thus the force of attraction between the metal atom and O^2– ion is greater than the force between same metal atom and F–. Hence O₂ gets higher oxidation state of metals.
(ii) This is because the energy required for the promotion of electrons in Xenon is very high.Energy factor does not favour VB approach.

Question. Complete the following chemical equation :
(i) F₂+2Cl- →
(ii) 2XeF₂ + 2H₂O →
Answer. (i) F₂ + 2Cl- → 2F- + Cl₂ 
(ii) 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂

Question. Account for the following :
(a) Cl₂ acts as a bleaching agent.
(b) Noble gases have very low boiling points. 
OR
(a) What happens when F₂ reacts with water ?
(b) Write the formula of a noble gas species which is isostructural with IBr₂.
Answer. (a) Because Cl₂ in presence of moisture liberates nascent oxygen. 
(b) Interatomic interactions are weak.
OR
(a) 2F₂+2H₂O → 4HF + O₂
HF and O₂ are produced. 
(b) XeF₂

Question. Complete the following equations :
(i) C + conc. H₂SO₄ →
(ii) XeF₂ + H₂O →
Answer. (i) C + 2H₂SO₄(conc.) → CO₂ + 2SO₂ + 2H₂O 
(ii) 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂

Question. Draw the structures of the following :
(i) H₂SO₄ (ii) XeF₂ 
Answer.

Question. Complete the following equations :
(i) P₄ + H₂O →
(ii) XeF₄ + O2F₂ →
Answer. (i) P₄ + H₂O → no reaction or if attempted in any form, award one mark. 
(ii) XeF₄ + O₂F₂ → XeF₆ + O₂

Question. Draw the structures of the following molecules :
(a) H₂S₂O₇
(b) XeOF₄ 
Answer.

Question. Account for the following :
(i) Acidic character increases from HF to HI.
(ii) There is a large difference between the melting and boiling points of oxygen and sulphur.
Answer. (a) Acidic character increases from HF to HI due to increase in size as a result attraction force decreases and acidity increase.

(ii) There is large difference between the melting point and boiling point of oxygen and sulphur due to small size and high electronegativity of oxygen.

Question. Complete the following reactions:
(i) Cl₂ + H₂O →
(ii) XeF₆ + 3H₂O → 
Answer. (i) Cl₂ + H₂O → 2HCl + [O] / HCl +HOCl 
(ii) XeF₆ + 3H₂O → XeO₃ + 6HF

Question. (i) Complete the following chemical equations :
(a) NaOH + Cl₂ →
(hot and conc.)
(b) XeF₄ + O₂F₂ → 
Answer. (i) (a) 6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O

Long Answer Type Questions-I 

Question. (i) F₂ has lower bond dissociation enthalpy than Cl₂. Why ?
(ii) Which noble gas is used in filling balloons for meteorological observations ?
(ii) Complete the following equation :
XeF₂ + PF₅ →
Answer. (i) F₂ has lower bond dissociation energy than Cl₂ because the size of F₂ is much smaller than Cl₂ as a result interelectronic repulsion works and make F₂ weak. 
(ii) Helium is used in filling balloons for meteorological observations. 
(iii) XeF₂ + PF₅ → [XeF]+ [PF₆]^–

Question. (i) What happens when :
(a) chlorine gas reacts with cold and dilute solution of NaOH ?
(b) XeF₂ undergoes hydrolysis ?
(ii) Assign suitable reasons for the following :
(a) Out of noble gases only Xenon is known to form established chemical compounds.
Answer. (i) (a) 2NaOH + Cl₂ → NaCl + NaOCl + H₂O 
(b) 2XeF₂(s) + 2H₂O(l) → 2Xe(g) + 4HF(aq) + O₂(g)
(ii) Xe has least ionization energy among the noble gases and hence it forms chemical compounds particularly with O₂ and F₂.

Question. (i) Draw the structure of a noble gas species which is isostructural with BrO3–.
(ii) Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and Cl₂.
(iii) Why is K_a2<< K_a1 for H₂SO₄ in water ?
OR
Explain the following :
(i) Hydrogen fluoride is a weaker acid than hydrogen chloride in aqueous solution.
(ii) SF₆ is inert towards hydrolysis.
(iii) Out of noble gases only Xenon is known to form established chemical compounds.
Answer. (i) XeO₃ is isostructural with BrO₃^–. (pyramidal structure)

(ii) The bond dissociation enthalpy of F—F bond is lower than that of Cl—Cl bond and hydration enthalpy of F– ion is much higher than that of Cl– ion.These two factors more than compensate the less negative electron gain enthalpy of F₂. Thus, F₂ is a stronger oxidizing agent than Cl₂. 
(iii) H₂SO₄ ionises in two stages and hence has two dissociation constants. K_a2 << K_a1.
This is because the negatively charged HSO₄^– ion resonance stabilized which has much less tendency to donate a proton to H2O as compared to neutral H₂SO₄. 
OR
(i) Due to stronger H-F bond than HCl bond, HF ionises less readily than HCl in aqueous solution to give H⁺ ions. Therefore, HF is a weaker acid than HCl. 
(ii) In SF₆, S is sterically protected by six F atoms and hence does not allow H₂O molecules to attack the S molecule. Also, F does not have d-orbitals to accept the electrons donated by H₂O molecules.

(iii) Except radon which is radioactive, Xenon has least ionisation energy among noble gases and hence it readily forms chemical compounds particularly with O₂ and F₂. 

Question. (i) Arrange the hybrids of group 16 in increasing order of their acidic character. Justify your answer.
(ii) Draw structure of XeOF₄.
Answer. (i) H₂O < H₂S < H₂Se < H₂Te, because of decrease in bond dissociation enthalpy.

Question. (i) Draw the structure of the following
(a) BrF₃ (b) XeO₃
(c) XeOF₄ 
Answer.

Question. (i) Compare the oxidizing action of F₂ and Cl₂ by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
(ii) Write the conditions to maximize the yield of H₂SO₄ by contact process.
Answer. (i) F₂ is the stronger oxidising agent than chlorine :
(a) Low enthalpy of dissociation of F-F bond.
(b) Less negative electron gain enthalpy of F.
(c) High hydration enthalpy of F– ion. 
(ii) The main reaction in contact process is 2SO₂ + O₂ ⇌ 2SO₃ ↑
This reaction is reversible and conditions to increase the yield of SO₃ are :
(a) High pressure
(b) Low temperature (optimum temperature 720 K) should be maintained and
(c) V₂O₅ as catalyst. 

Long Answer Type Questions-II

Question. (i) Write balanced equations for the following reactions :
(a) Chlorine reacts with dry slaked lime.
(b) Carbon reacts with concentrated H₂SO₄.
(ii) Describe the contact process for the manufacture of sulphuric acid with special reference to the reaction conditions, catalysts used and the yield in the process.
Answer. (i) (a) 2Ca(OH)₂ + 2Cl₂ → Ca(OCl)₂ + CaCl₂ + 2H₂O
(b) C + 2H₂SO₄(conc.) → CO₂ + 2SO₂ + 2H₂O 
(ii) It is manufactured by Contact Process which involves following steps :
(a) burning of sulphur or sulphide ores in air to generate SO₂.
(b) conversion of SO₂ to SO₃ by the reaction with oxygen in the presence of a catalyst (V₂O₅).
(c) absorption of SO₃ in H₂SO₄ to give Oleum (H₂S₂O₇). The oleum obtained is diluted to give sulphuric acid.

Reaction condition – pressure of 2 bar and temperature of 720 K.
Catalyst used is V₂O₅.
Yield – 96 – 98% pure.

Question. (i) Account for the following:
(a) Reducing character decreases from SO₂ to TeO₂.
(b) HClO₃ is a stronger acid than HClO.
(c) Xenon forms compounds with fluorine and oxygen only.
(ii) Complete the following equations:
(a) 4NaCl + MnO₂ + 4H₂SO₄ →
(b) 6XeF₄ + 12H₂O →
Answer. (i) (a) Stability of higher oxidation state decreases down the group from S to Te/Stability of lower oxidation state increases down the group from S to Te. 
(b) ClO₃ is more stable than ClO^–/ClO₃^– is a weak conjugate base than ClO-/Due to higher oxidation state of chlorine in HClO₃ 
(c) Fluorine and oxygen are most electronegative and very reactive. 
(ii)
(a) 4NaCl + MnO₂ + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂
(b) 6XeF₄+ 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂

Question. (i) Complete the following chemical reaction equations :
(a) P₄ + SO₂Cl₂ →
(b) XeF₆ + H₂O →
(ii) Predict the shape and the asked angle (90° or more or less) in each of the following cases :
(a) SO₃^2– and the angle O—S—O
(b) ClF₃ and the angle F—Cl—F
(c) XeF₂ and the angle F—Xe—F
Answer. (i) (a) P₄ + 10SO₂Cl₂ → 4PCl₅ + 10SO₂. 
(b) XeF₆ + H₂O → XeOF₄ + 2HF
or
XeF₆ + 2H₂O → XeO₂F₂ + 4HF
or
XeF₆ + 3H₂O → XeO₃ + 6HF 
(ii) (a) Pyramidal or