HOTs The p – Block Elements Class 12 Chemistry

Please refer to The p – Block Elements HOTs Class 12 Chemistry provided below with The p – Block Elements. All HOTs for Class 12 Chemistry with answers provided below have been designed as per the latest syllabus and examination petter issued by CBSE, NCERT, KVS. Students of Standard 12 Chemistry should learn the solved HOTS for Class 12 Chemistry provided below to gain better marks in examinations.

The p – Block Elements Class 12 Chemistry HOTs

MULTIPLE CHOICE QUESTIONS

Question. The correct order of increasing electron affinity of halogens is:
(a) 1 < Br < Cl
(b) Br < 1 < u
(c) Cl < Br < 1
(d) 1 < Cl < Br

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

Question. Reduction potentials of same ions are given below. Arrange them in decreasing order of oxidising power.
Ion                                               ClO₄^–              IO₄^–               BrO₄^– 
Reduction potential E°/V     E^– = 1.19 V      E^–  = 1.65 V      E^– = 1.74 V
(a) ClO₄^– > IO₄^– > BrO₄^–
(b) 1O₄^– > BrO₄^– > ClO₄^–
(c) BrO₄^– > IO₄^– > ClO₄^–
(d) BrO₄^– > ClO₄^– > IO₄^–

Question. On addition of conc. H₂SO₄ to a chloride salt, colourless fumes are evolved but in case of iodise salt, violet fumes come out. This is because
(a) H₂SO₄ reduces Hl to l₂
(b) Hl is of violet colour
(c) Hl gets oxidised to l₂
(d) Hl changes to HlO₃

Question. Bond angle in H₂O (104.5°) is higher than the bond angle of H₂S (921.1°). The difference is due to

(a) O is diatomic and S is titra-atomic
(b) difference in electronegatively of S and O
(c) difference in oxidation states of S and O
(d) difference in shapes of hybrid orbitals of S and O

Question. What is the correct operation when Br₂ is treated with NaF, NaCl and Nal taken in three test tukes lavelled (X), (Y) and (Z)?

(a) F₂ is liberated in (X) and Cl₂ in (Y)
(b) Only l₂ is liberated in (Z).
(c) Only Cl₂ is liberated in (Y)
(d) Only F₂ is liberated in (X)

Question. The hybridisation of sulphur in sulphur tetrafluroide is
(a) sp³d
(b) sp³d²
(c) sp³d³
(d) sp³

Question. On heating KClO₃, we get
(a) KClO₂ + O₂
(b) KCl + O₂
(c) KCl + O₃
(d) KCl + O2+ O₃

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

Question. Which one is not a property of ozone?
(a) it acts an oxidising agent in dry state
(b) oxidation of K1 into KlO₂
(c) PbS is oxidised to PbSO₄
(d) Hg is oxidised to Hg₂O

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) both O₂ and Xe have same electron gain enthalpy
(c) both O₂ and Xe have almost same ionisation enthalpy
(d) both Xe and O₂ are gases.

Question. Complete the following reactions by filling the appropriate choice:

Question. Arrange the following hydrides of group 16 elements in order of increasing stalility.
(a) H₂S < H₂O < H₂Te > H₂Se
(b) H₂O < H₂Te < H₂Se < H₂S
(c) H₂O < H₂S < H₂Se < H₂Te
(d) H₂Te < H₂Se < H₂S < H₂O

Question. The oxidation state of sulphur in the anions SO₃^2–, S₂O₄^2– and S₂O₆ ^2– follows the order:
(a) S₂O₆^2– < S₂O₄^2– < SO₃^2–
(b) S₂O₄^2– < SO₃^2– < S₂O₆^2–
(c) SO₃^2– < S₂O₄^2– < S₂O₆^2–
(d) S₂O₄^2– < S₂O₆^2– < SO₃^2–

Question. Compound with the geometry square pyramidal and sp³d² hybridisation is
(a) XeOF₂
(b) XeOF₄
(c) XeO₄
(d) XeO₂F₂

Question. Which is the correct arrangement of the compounds based on their bond strength?
(a) HF > HCl > HBr > Hl
(b) Hl > HBr > HCl > HF
(c) HCl > HF > HBr > H1
(d) HF > HBr > HCl > H1 

Question. Which of the following increasing order is not correct as mentioned in the property with it?
(a) HClO < HClO₂ < HClO₃ < HClO₄ (thermal stalrlity)
(b) HClO₄ < HClO₃ < HClO₂ < HClO (oxidising power)
(c) F^– < Cl^– < Br^– < 1^– (reducing nature)
(d) HlO₄ < lCl < l₂ < Hl (oxidation number of iodine)

Question. The oxyacid of sulphur that contains a lone pair of electrons on sulphur is:
(a) sulphurous acid
(b) sulphuric acid
(c) peroxodisulphuric acid
(d) pyrosulphuric acid

Question. Among the following molecules (i) XeO₃, (ii) XeOF₄, (iii) XeF₆ those having same number of lone pairs on Xe are
(a) (i) and (ii) only
(b) (i) and (iii) only
(c) (ii) and (iii) only
(d) (i), (ii) and (iii) 

Question. The correct order of acidic strength is:
(a) K₂O > CaO > MgO
(b) CO₂ > N₂O₅ > SO₃
(c) Na₂O > Mgo > Al2O₃
(d) Cl₂O₇ > SO₂ > P₄O₁₀ 

More than one correct Response

Question. Which of the following statements are correct for SO₂ gas?
(a) It act as bleaching agent in moist conditions
(b) It’s molecule has linear geometry
(c) It’s dilute solution is used as disinfectant.
(d) It can be prepared by the reaction of dilute H₂SO₄ with metal sulphide.

Question. Which of the following statements are true?
(a) Only type of interactions between particles of noble gases are due to weak dispersion forces.
(b) Ionisation enthalpy of molecular oxygen is very close to that of numon.
(c) Hydrolysis of XeF₆ is a redox reaction.
(d) Xenon fluorides are not reactive.

Question. Which of the following statements are correct?
(a) All the three N—O bond lengths in HNO₃ are equal.
(b) All P–Cl bond lengths in PCl₅ molecule in gaseous state are equal.
(c) P₄ molecule in white phosphorous have angular strain therefore white phosphours is very reactive.
(d) PCl₅ is ionic in solid state in which cation is tetrahedral and anion is octahedral.

Question. Which of the following statements are correct?
(a) Among halogens, radius ratio between iodine and fluorine is maximum
(b) Leaving F–F bond, all halogens have weaker X—X bond than X—X’ bond in interhalogens
(c) Among interhalogen compounds maximum number of atoms are present in iodine fluoride.
(d) Interhalogen compounds are more reactive than halogen compounds.

Question. Which of the following orders are correct as per the properties mentioned against each?
(a) As₂O₃ < SiO₂ < P₂O₃ < SO₂          Acid strength
(b) AsH₃ < PH₃ < NH₃                        Enthalpy of vapourisation
(c) S < O < Cl < F                             More negative electron gain enthalpy
(d) H₂O > H₂S > H₂Se > H₂Te            Thermal stability

Question. Match the items of column 1 and column 2 and mark the correct option.

(a) A–1, B–4, C–2, D–3
(b) A–1, B–2, C–3, D–4
(c) A–2, B–1, C–4, D–3
(d) A–1, B–3, C–2, D–4

Question. Match the items of column 1 and column 2 and mark the correct option
Column 1                     Column 2
(A) H₂SO₄                 (1) Highest electron gain enthalpy
(B) CCl₃NO₂              (2) Chalcogen
(C) Cl₂                      (3) Tear gas
(D) Sulphur              (4) Storage batteries
(a) A–4, B–3, C–1, D–2
(b) A–3, B–4, C–1, D–2
(c) A–4, B–1, C–2, D–3
(d) A–2, B–1, C–3, D–4

Assertion and Reason Type
Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choice.
(a) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
(b) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
(c) Assertion is correct, but reason is wrong statement.
(d) Assertion is wrong but reason is correct statement.
(e) Both assertion and reason are wrong statements.

Question. Assertion: Hl cannot be prepared by the reaction of Kl with concentrated H₂SO₄. 
Reason : Hl has lowest H—X bond strength among halogen acids.

Question. Assertion: Both rhombic and monollinic sulphur exist as S8 but oxygen exist as O₂. 
Reason: Oxygen forms Pπ–Pπ multiple bond due to small size and small bond length but Pπ–Pπ bonding is not possible is sulphur.

Question. The difference in the oxidisation numbers of the two types of sulphur atoms in Na₂S₄O₆ is

GROUP 16 ELEMENTS

VERY SHORT ANSWER TYPE QUESTIONS

Question. Thermal stability of hydrides of group 16 elements decreases down the group. Why ?
Answer. Because down the group E – H bond dissociation enthalpy decreases. 

Question. Draw the structure of H₂S₂O₈ and find the number of S – S bond, if any.
Answer.

Number of S – S bond ⇒ 0.

Question. In the preparation of H₂SO₄ by contact process, why is SO₃ not absorbed directly in water to form H₂SO₄ ?
Answer. Because it forms a dense fog of sulphuric acid which does not condense easily.

Question. H₂O is a liquid while inspite of a higher molecular mass, H₂S is a gas.Explain.
Answer. H₂O molecules are stabilized by intermolecular hydrogen bonding,while H₂S by weak van der Waal’s forces.

Question. All the bonds in SF₄ are not equivalent. Why ?
Answer. It is having see-saw shape. (4BP + 1LP) 

Question. Explain why SF₄ is easily hydrolysed, whereas _SF6 is resistant to hydrolysis ?
Answer. Water molecule cannot attack ‘S’ atom due to steric hinderance and ‘S’ atom is also coordinately saturated in SF₆ molecule.

Question. Which one of the following is not oxidized by O₃ ? State the reason :
KI, FeSO₄, K₂MnO₄, KMnO₄
Answer. KMnO₄, since Mn is showing maximum oxidation state of + 7.

Question. In group 16, the stability of + 6 oxidation state decreases and that of + 4 oxidation state increases down the group. Why ?
Answer. Due to inert pair effect.

Question. Dioxygen O₂ is a gas while sulphur (S₈) is a solid. Why ?
Answer. Because oxygen is smaller in size hence have capacity to form pπ–pπ multiple bond, exists as dioxygen (O₂), whereas due to bigger size sulphur do not form multiple bond and exist as S₈.

Question. Why does oxygen not show an oxidation state of + 4 and + 6 ?
Answer. Due to absence of vacant d-orbitals in the octet of oxygen.

Question. O₃ acts as a powerful oxidizing agent. Why ?
Answer. Due to the ease with which it liberates atoms of nascent oxygen.
O₃ → O₂ + [O]

Question. Oxygen and sulphur in vapour phases are paramagnetic in nature. Explain why ?
Answer. Due to presence of unpaired electrons in anti-bonding molecular orbitals in them. 

Question. Why are the two S – O bonds in SO₂ molecule have equal strength ?
Answer. Due to resonance, two S – O bonds have partial double bond character, hence have equal strength.

Question. What happens when sulphur dioxide gas is passed through an aqueous solution of a Fe(III) salt ?
Answer. It converts Fe³⁺ ions to Fe²⁺ ions.
2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2– + 4H⁺

Question. The electron gain enthalpy with negative sign for oxygen (− 141 KJ mol⁻¹) is numerically less than that for sulphur (− 200 KJ mol⁻¹). Give reason.
Answer. Due to smaller size of oxygen than sulphur electron-electron repulsion is more in oxygen than sulphur. 

Question. Ka₂ << Ka₁ for H₂SO₄ in water, why ?
Answer. H₂SO₄ (aq) + H₂O (l) → H₃O+ (aq) + HSO₄⁻ (aq); Ka₁ > 10
HSO₄ (aq) + H₂O (l) → H₃O+ (aq) + SO₄⁻² (aq); Ka₂ = 10⁻²
Ka₂ is less than Ka₁ because HSO₄⁻ ion has much less tendency to donate a proton. 

SHORT ANWER-I TYPE QUESTIONS 

Question. Mention the favourable conditions for the manufacture of sulphuric acid by contact process.
Answer. (i) Low temperature (operating temperature is 720 K).
(ii) High pressure (2 bar).
(iii) Presence of catalyst (V₂O₅).
2SO₂ (g) + O₂ (g) → 2SO₃ (g) ∆Hθ = − 196.6 KJ mol⁻¹

Question. Draw the structure of :
(i) H₂SO₅
(ii) SO₃²⁻
Answer.

Question. Explain why :
(a) H₂S is more acidic than H₂O.
(b) Two S – O bonds in SO₂ are identical.
(c) SF₆ is inert and stable but SF₄ is reactive.
(d) Sulphur has greater tendency for catenation than oxygen.
Answer. (iii) Because six F atoms protect the sulphur atom from attack by any reagent due to steric hindrance but four F atoms in SF₄ cannot offer much steric hindrance, hence reactive.

Question. Write the chemical equations of the following reactions :
(a) Sucrose is heated with conc. H₂SO₄.
(b) Sodium nitrate is heated with conc. H₂SO₄.
Answer.

(b) NaNO₃ + H₂SO₄ → NaHSO₄ + HNO₃

Question. How is O₃ estimated quantitatively ?
Answer. O₃ reacts with an excess of KI solution buffered with a borate buffer, I₂ is liberated which is titrated against standard solution of sodium thiosulphate.
2I− (aq) + H₂O (l) + O₂ (g) → I₂ (s) + O₂ (g) + 2OH⁻ (aq)

Question. Explain why O₃ is thermodynamically less stable than O₂ ?
Answer. Because O₃ is endothermic compound/decomposition of O₃ is exothermic and ∆G is negative/decomposition of O₃ is spontaneous. 

SHORT ANSWER-II TYPE QUESTIONS 

Question. An amorphous solid ‘A’ burns in air to form a gas ‘B’ which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aq. KMnO₄ solution. Identify the solid ‘A’ and the gas ‘B’ and write the reaction involved.
[Answer. A = S₈; B = SO₂ (g)]
(i) How is SO₂ prepared in laboratory ?
(ii) What happens when SO₂ is passed through water and reacts with NaOH ? Write balanced equation.
(iii) Write its any two uses.
Answer.(i) Na₂SO₃ (s) + H₂SO₄ (aq) → SO₂ (g) + Na₂SO₄ (aq) + H₂O (l)
(ii) 2NaOH + SO₂ (g) → Na₂SO₃ (aq) + H₂O
Na₂SO₃ (aq) + SO₂ + H₂O → 2NaHSO₃ (aq)
(iii) It is used as bleaching agent and disinfectant.

Question. (i) How does O₃ react with lead sulphide ? Write chemical equation.
(ii) What happens when SO₂ is passed in acidified KMnO₄ solution ?
(iii) SO₂ behaves with lime water similar to CO₂. Explain why ?
Answer. (i) PbS (s) + 4O₃ (g) → PbSO₄ (s) + 4O₂ (g)
(ii) It decolourises acidified KMnO₄ solution.
SO₂ + 2MnO₄⁻ + 2H₂O → 5SO₄²⁻ + 4H⁺ + 2Mn²⁺
(iii) It turns lime water milky due to the formation of insoluble CaSO₃.
Ca(OH)₂ + SO₂ → CaSO₃ + H₂O
                               (milkiness)

Question. Write contact process for the manufacture of king of chemicals.
Answer. (i) 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
(ii) 2SO₂ (g) + O₂ (g) → 2SO₃ (g)
(iii) SO₃ + H₂SO₄ (98%) → H₂S₂O₇ (oleum)
(iv) H₂S₂O₇ + H₂O → 2H₂SO₄

Question. Assign reason for the following :
(i) Sulphur in vapour state exhibits paramagnetism.
(ii) H₂O is less acidic than H₂Te.
(iii) In spite of having same electronegativity, oxygen forms hydrogen bond while chlorine does not.
Answer. (iii) Due to bigger size of Cl.