The p – Block Elements Exam Questions Class 12 Chemistry

Please see Chapter 7 The p – Block Elements Exam Questions Class 12 Chemistry below. These important questions with solutions have been prepared based on the latest examination guidelines and syllabus issued by CBSE, NCERT, and KVS. We have provided Class 12 Chemistry Questions and answers for all chapters in your NCERT Book for Class 12 Chemistry. These solved problems for The p – Block Elements in Class 12 Chemistry will help you to score more marks in upcoming examinations.

Exam Questions Chapter 7 The p – Block Elements Class 12 Chemistry

Very Short Answer Questions :

Question. Give reason :
Bond dissociation energy of F₂ is less than that of Cl₂. 
Answer : F₂ has lower bond dissociation enthalpy than Cl₂ because F atom is very small and hence the electron-electron repulsions between the lone pairs of electrons are very large.

Question. Account for the following :
The two O—O bond lengths in the ozone molecule are equal.
Answer : The two O — O bond lengths in the ozone molecule are equal as it is a resonance hybrid of two main forms :

Question. Write the formula and describe the structure of noble gas species which is isostructural with BrO^–₃
Answer : The central atom Br has seven electrons. Four of these electrons form two double bonds or coordinate bonds with two oxygen atoms while the fifth electron forms a single bond with O– ion. The remaining two electrons form one lone pair. Hence, in all there are three bond pairs and one lone pair around Br atom in BrO^–₃. therefore, according to VSEPR theory, BrO^–₃ should be pyramidal.
Here, BrO^–₃ has 26(7 + 3 × 6 + 1 = 26) valence electrons. A noble gas species having 26 valence electrons is XeO₃(8 + 3 × 6 = 26). Thus, like BrO^–₃, XeO₃ is also pyramidal.

Question. Write the balanced chemical equation for the reaction of Cl₂ with hot and conc.
NaOH solution. Justify that this reaction is a disproportionation reaction.
Answer : 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O
This reaction is a disproportionation reaction as chlorine from zero oxidation state is changed to – 1 and + 5 oxidation states.

Question. Account for the following :
Concentrated sulphuric acid has charring ction on carbohydrates. 
Answer : Concentrated H₂SO₄ removes water from organic compounds hence, it has charring action on carbohydrates.

Question. Answer the following :
Which neutral molecule would be isoelectronic with ClO^–? 
Answer : ClO^– has 17 + 8 + 1 = 26 electrons.
A neutral molecule with 26 electrons is OF₂ (8 + 2 × 9) = 18 + 8 = 26 electrons.

Question. Why is K_a2 << K_a1 for H₂SO₄ in water?
Answer : H₂SO_4(aq) + H₂O_(l) → H₃O⁺(_aq) + HSO^–_4(aq) K_a1 > 10, very large
HSO^–_4(aq) + H₂O_(l) → H₃O⁺_(aq) + SO^2–_4(aq) K_a2 = 1.2 × 10^–2
Ka₂ is smaller than K_a1 because dissociation of HSO₄^– is less probable due to presence of negative charge on the ion.

Question. Why are halogens coloured?
Answer : Halogens absorb radiations in visible region which results in excitation of outer electrons to higher level resulting in different colours.

Question. Complete the following equation :
SO₃ + H₂SO₄ →
Answer : SO₃ + H₂SO₄ → H₂S₂O₇

Question. Predict the shape and the asked angle (90° or more or less) in each of the following cases. ClF₃ and the angle F—Cl—F
Answer : Hybridisation – sp³d
Structure – Trigonal bipyramidal
Shape – Bent (T-shaped)
Angle F–Cl–F : less than 90°

Question. Complete the following reaction equation :
NaOH (cold & dilute) + Cl₂ →
Answer : 2NaOH _(dil.) + Cl_2(aq) → NaCl_(aq) + NaClO_(aq.) + H₂O_(l)

Question. With the help of chemical equation explain the principle of Contact process in brief for the manufacture of sulphuric acid by Contact process.
Answer : (a) Contact process : It involves three stages :
(i) Burning of sulphur or sulphide ore in air to generate SO₂.

Question. How would you account for the following?
Halogens are strong oxidizing agents
Answer : General electronic configuration of halogens is ns²np⁵. They easily accept one electron to complete their octet. This makes them a good oxidising agent.

Question. Describe the conditions and the steps involved in the manufacture of sulphuric acid by Contact process. Write the necessary reactions. (No diagram is required.) 
Answer :  (a) Contact process : It involves three stages :
(i) Burning of sulphur or sulphide ore in air to generate SO₂.

Question. Account for the following :
Chlorine water loses its yellow colour on standing.
Answer : Chlorine water on standing loses its yellow colour due to the formation of HCl and HClO.

Question. Explain the following situations :
SF₄ is easily hydrolysed whereas SF₆ is not easily hydrolysed. 
Answer : In SF₆, six F atoms protect the sulphur atom from attack by the reagent to such an extent that even thermodynamically most favourable reactions like hydrolysis do not occur. But in SF₄, S is not sterically protected as it is surrounded by only four F atoms hence, SF₄ is reactive.

Question. Write the structure of the following species :
H₂SO₅.
Answer : 

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

Question. Account for the following :
Acidic character increases from HF to HI.
Answer : The acidic strength of the hydrohalic acids in the order :
HF < HCl < HBr < HI
This order is a result of bond dissociation enthalpies of H — X bond decreases from H — F to H — I as the size of halogen atom increases

Question. Excess of SO₂ reacts with sodium hydroxide solution.
Answer : 2NaOH + SO₂ → Na₂SO3 + H₂O

Question. Describe the Contact process for the manufacture of sulphuric acid with special reference to the reaction conditions, catalysts used and yield in the process.
Answer : In Contact process, the rate determining step is 2SO₂ + O₂ →^V₂O₅ 2SO₃, Δ_fH° = –196.6 kJ mol^–1 This reaction is reversible and exothermic i.e., ΔH is negative. Thus, according to Le-Chatelier’s principle, the conditions to maximise the yield are as follows : 
(a) At lower temperature : As heat is evolved in the reaction so, at lower temperature the reaction proceeds more in forward direction.
(b) At higher pressure : As three moles of gaseous reactants give two moles of gaseous products thus, at higher pressure reaction moves in forward direction.
 (a) Contact process : It involves three stages :
(i) Burning of sulphur or sulphide ore in air to generate SO₂.

Question. Draw the structure of the following : HClO₃
Answer :

Question. Answer the following :
Why are halogens strong oxidising agents?
Answer : General electronic configuration of halogens is ns²np⁵. They easily accept one electron to complete their octet. This makes them a good oxidising agent.

Question. Ozone is thermodynamically unstable?
Answer : Ozone is thermodynamically unstable and decomposes into oxygen.
                            2O3 →^Δ 3O₂; ΔH = –ve
The above conversion is exothermic i.e., ΔH is negative. Also, entropy increases i.e., ΔS = + ve.
Thus, ΔG for the decomposition of ozone is negative. Hence, it is thermodynamically unstable.

Question. Account for the following :
Fluorine does not form oxoacids.
Answer : Fluorine forms only one oxoacid HOF. Because for the formation of other oxoacids d orbitals are required for the multiple pπ – dπ bonding between extra oxygen atoms and fluorine.
High electronegativity and small size of fluorine also favours only the formation of one oxoacid.

Question. With the help of chemical equations explain the principle of contact process in brief for the manufacture of sulphuric acid. (No diagram).
Answer :  (a) Contact process : It involves three stages :
(i) Burning of sulphur or sulphide ore in air to generate SO₂.

Question. What happens when :
SO₂ gas is passed through an aqueous solution Fe³⁺ salt?
Answer : 2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2– + 4H⁺

Question. How would you account for the following :
The oxidising power of oxoacids of chlorine follows the order :
                HClO₄ < HClO₃ < HClO₂ < HClO
Answer : As the stability of the oxoanion increases, its tendency to decompose to give O₂ decreases and hence its oxidising power decreases. Since the stability of the oxoanion decreases in the order :
ClO₄^– > ClO₃^– > ClO₂^– > ClO^– therefore oxidising power of their oxoacids increases in the reverse order :
             HClO₄ < HClO₃ < HClO₂ < HClO.

Question. How the supersonic jet aeroplanes are responsible for the depletion of ozone layers?
Answer : Nitrogen oxide emitted from the exhausts of supersonic jet aeroplanes readily combines with ozone to form nitrogen dioxide and diatomic oxygen. Since supersonic jets fly in the stratosphere near the ozone layer, they are responsible for the depletion of ozone layer.

Question. Account for the following
HClO₄ is stronger acid than HClO.
Answer : As the number of oxygen bonded to the central atom increases, the oxidation number of the oxidation atom increases causing a weakening of the O—H bond strength and an increase in the acidity. Hence, HClO₄ is stronger acid than HClO.

Question. HF is a weaker acid than HCl. Why?
Answer : HF is the weakest acid because of its high bond dissociation energy due to small size of fluorine atom.

Question. Account for the following :
HF is not stored in glass bottles but is kept in wax-coated bottles. 
Answer : HF acid attacks glass with the formation of fluoro silicate ions. Thus it is stored in wax-coated glass bottles to prevent the reaction.

Question. Explain the following observations.
Despite lower value of its electron gain enthalpy with negative sign, fluorine (F₂) is a stronger oxidising agent than Cl₂. 
Answer : Fluorine is the strongest oxidising agent as it accept electron easily. It oxidise other halide ions in solution or even in solid phase.
F₂ + 2X^– → 2F^– + X₂ (X = Cl, Br or I)
Cl₂ + 2X^– → 2Cl^– + X₂ (X = Br or I)
Br₂ + 2I^– → 2Br^– + I₂

Question. Account for the following :
O₃ acts as a powerful oxidising agent.
Answer : Ozone is a powerful oxidising agent because ozone has higher energy content than dioxygen hence, decomposes to give dioxygen and atomic oxygen.
                   O_3(g)   →   O_2(g)   +   O_(g)
                  Ozone      Dioxygen    Atomic oxygen 
The atomic oxygen thus liberated brings about the oxidation while molecular oxygen is set free.

Question. Fluorine exhibits only –1 oxidation state whereas other halogens exhibit +1, +3, +5 and +7 oxidation state also. Why is it so? 
Answer : This is due to non-availability of d-orbitals in valence shell of fluorine.

Question. F₂ has lower bond dissociation enthalpy than Cl₂. Why? 
Answer : F₂ has lower bond dissociation enthalpy than Cl₂ because F atom is very small and hence the electron-electron repulsions between the lone pairs of electrons are very large.

Question. Explain giving reason for the following situation.
In aqueous medium HCl is stronger acid than HF. 
Answer : In aqueous medium HCl is stronger acid than HF because bond dissociation enthalpy of H—Cl is lower than that of HF

Question. Assign reasons for the following :
The negative value of electron gain enthalpy of fluorine is less than that of chlorine.
Answer : The electron gain enthalpy of fluorine is less negative than that of chlorine due to the small size of fluorine atom.

Question. Give reasons for the following :
OF₆ compound is not known.
Answer : OF₆ compound is not known because oxygen cannot expand its octet due to unavailability of d-orbital.

Question. Arrange HClO₃, HClO₂, HOCl and HClO₄ in order of increasing acid strength. Give reason for your answer. 
Answer : Acid strength of oxoacids of the same halogen increases with increase in oxidation number of the halogen.
Thus the increasing order of acid strength is
              HOCl < HClO₂ < HClO₃ < HClO₄
                +1        +3         +5           +7

Question. Fluorine exhibits only – 1 oxidation state in its compounds whereas other halogens exhibit many other oxidation states. Why?
Answer : This is due to non-availability of d-orbitals in valence shell of fluorine.

Question. Compare the oxidizing action of F₂ and Cl₂ by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy. 
Answer : Oxidising power of a substance depends on the factors like bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy. Due to small ize of fluorine, its electron gain enthalpy is less than that of chlorine. However, its low bond dissociation enthalpy and high hydration enthalpy compensate the low electron gain enthalpy. Fluorine because of its small size has higher hydration enthalpy than chlorine. Also, due to repulsion between electrons it has lower bond dissociation energy. Thus, fluorine has better oxidising action than chlorine.

Question. Arrange the following in order property indicated for each set.
(i) F₂, Cl₂, Br₂, I₂ increasing bond dissociation enthalpy.
(iii) HF, HCl, HBr, HI increasing acid strength
Answer : (i) Increasing bond dissociation enthalpy order is
I₂< F₂ < Br₂ < Cl₂
Bond dissociation enthalpy of F₂ is less than that of Br₂ and Cl₂ due to the lone pair – lone pair repulsions.
(ii) The acidic strength of the hydrohalic acids in the order :
HF < HCl < HBr < HI
This order is a result of bond dissociation enthalpies of H — X bond decreases from H — F to H — I as the size of halogen atom increases

Question. Complete the following chemical equation :
F₂(Excess) + Cl₂ →3^00°C 
Answer : 3F₂ (excess) + Cl₂ →3^00°C  2ClF₃

Question. Complete the following reaction equation :
I₂ + H₂O + Cl₂ →
Answer : I₂ + 6H₂O + 5Cl₂ → 2HIO₃ + 10HCl
                                            Iodic acid

Question. Describe the following about halogens (Group 17 elements):
(i) Relative oxidising power of halogens.
(ii) Relative acidic strength of the hydrogen halides. 
Answer : (i) From top to bottom in group – 17 oxidising power of halogens decreases
               F₂ > Cl₂ > Br₂ > I₂.
(ii) F₂ has lower bond dissociation enthalpy than Cl₂ because F atom is very small and hence the electron-electron repulsions between the lone pairs of electrons are very large.

Question. Write balanced equation for the following reactions :
Chlorine reacts with dry slaked lime.
Answer : 2Cl₂ + 2Ca(OH)_2(dil.) →^Cold CaCl₂ + Ca(OCl)₂ + 2H₂O
                                                             Calcium hypochlorite

Question. Name two poisonous gases which can be prepared from chlorine gas.
Answer : (i) Phosgene (ii) Mustard gas

Question. Give reasons : ICl is more reactive than I₂.
Answer : Interhalogen compounds are more reactive than halogens (except flourine) because X — X′ bond (I—Cl bond) in interhalogens is weaker than X—X bond (I—I bond) in halogens except F — F bond. In other words I —Cl bond is weaker than I — I bond. That’s why ICl is more reactive than I₂.

Question. Draw the structure of O₃ molecule.
Answer :

Question. Arrange the following in the order of property indicated against each set :
HF, HCl, HBr, HI – increasing bond dissociation enthalpy 
Answer : HI < HBr < HCl < HF

Question. Draw the structure of the following : ClF₃
Answer : Hybridisation – sp³d
Structure – Trigonal bipyramidal
Shape – Bent (T-shaped)
Angle F–Cl–F : less than 90°

Question. Account for the following :
Fluorine forms only one oxoacid HOF.
Answer : Fluorine forms only one oxoacid HOF. Because for the formation of other oxoacids d orbitals are required for the multiple pπ – dπ bonding between extra oxygen atoms and fluorine.
High electronegativity and small size of fluorine also favours only the formation of one oxoacid.

Question. How would you account for the following?
The value. of electron gain enthalpy with negative sign for sulphur is higher than that for oxygen.
Answer : The electron gain enthalpy of oxygen is less negative than sulphur. This is due to its small size. As a result of which the electron-electron repulsion in the relatively small 2p-subshell are comparatively larger and hence the incoming electrons are not accepted with same ease as in case of other (sulphur) elements of this group.

Question. Account for the following :
BrCl₃ is more stable than BrCl₅.
Answer : O.S. of Br in BrCl₅ is +5, whereas in case of BrCl₃ is +3. As Br is more stable in +3 oxidation state than +5, due to inert pair effect. Therefore, it is unstable and readily reduces from +5 to +3 oxidation state.

Short Answer Questions :

Question. Draw the structures of the following : XeF₂
Answer : XeF₂ :
Total valence electron pair
= 8 + 2/2 = 5
Bond pairs = 2
Lone pairs = 5 – 2 = 3
Hybridisation = sp³d
Geometry = Trigonal bipyramidal
Shape = Linear

Question. Write the formulae and the structure of noble gas species which are isostructural with
(i) ICl^–₄
(ii) BrO^–₃
Answer :

Question. Draw the structure of the following : XeO₃
Answer : XeO₃ :
Hybridisation = sp³
Geometry = Tetrahedral
Shape = Pyramidal

Question. Noble gases have low boiling points. Why?
Answer : Noble gases being monoatomic gases are held together by weak London dispersion forces, therefore they have low boiling points.

Question. Complete the following reactions :
(i) XeF₄ + O₂F₂
(ii) XeF₄ + SbF₅ 
Answer : (i) XeF₄ + O₂F₂ →^{143 K} XeF₆ + O₂
(ii) XeF₄ + SbF₅ → [XeF₃]+ [SbF₆]^–

Question. Complete the following equation : XeF₂ + H₂O 
Answer : 2XeF_2(s) + 2H₂O_(l) → 2Xe_(g) + 4HF_(aq) + O_2(g)

Question. Give reasons for the following :
Helium is used in diving apparatus as a diluent for oxygen. 
Answer : Helium is used in diving apparatus as diluent for oxygen because of its low solubility (as compared to N₂) in blood, a mixture of oxygen and helium is used in diving apparatus used by deep sea divers.

Question. Account for the following :
Helium is used in diving apparatus.
Answer : Helium is used in diving apparatus as diluent for oxygen because of its low solubility (as compared to N₂) in blood, a mixture of oxygen and helium is used in diving apparatus used by deep sea divers.

Question. Write the chemical equation for the following process : PtF₆ and xenon are mixed together
Answer : PtF₆ + Xe → Xe⁺[PtF₆]^–

Question. Complete the following chemical reaction equation : XeF₄ + H₂O
Answer : 6XeF₄ + 12H₂O → 2XeO₃ + 24HF +3O₂ + 4Xe

Question. Write the balanced chemical equations for obtaining XeO₃ and XeOF₄ from XeF₆.
Answer : XeF₆ + 3H₂O → XeO₃ + 6HF
XeF₆ + H2O → XeOF₄ + 2HF

Question. Complete the following chemical reactions equations :
XeF₆ + H₂O 
Answer : XeF₆ + H₂O → XeOF₄ + 2HF

Question. Predict the shape and asked angle (90° or more or less) in each of the following case.
XeF₂ and the angle F – Xe – F
Answer : XeF₂ :
Total valence electron pair
= 8 + 2/2 = 5
Bond pairs = 2
Lone pairs = 5 – 2 = 3
Hybridisation = sp³d
Geometry = Trigonal bipyramidal
Shape = Linear

Question. (i) How does xenon atom form compounds even though the xenon atom has a closed shell electronic configuration?
(ii) Draw the structure of XeOF₄.
(iii) Complete and balance the following equation : XeF₄ + H₂O
Answer : (i) Except radon which is radioactive, Xe has least ionisation energy among noble gases and hence it readily forms chemical compounds particularly with oxygen and fluorine.
(ii) XeOF₄ is square pyramidal.

(iii) 6XeF₄ + 12H₂O → 2XeO₃ + 24HF +3O₂ + 4Xe

Question. Complete the following reaction equation :
XeF₂ + PF₅
Answer : XeF₂ + PF₅ → [XeF]⁺[PF₆]^–

Question. Draw structure of
(i) XeOF₄
(ii) XeO₃.
Answer : (i) XeOF₄ is square pyramidal.

(ii) XeO₃ :
Hybridisation = sp³
Geometry = Tetrahedral
Shape = Pyramidal

Question. Account for the following :
Unlike xenon, no distinct chemical compound of helium is known. 
Answer : Extremely small size and fully filled outer orbital makes helium very stable and reistant to chemical reaction and hence, it does not form compounds unlike bigger atoms of other elements of noble gas family.

Question. Complete the following equation : XeF₄ + O₂F₂ 
Answer : XeF₄ + O₂F₂ →^{143 K} XeF₆ + O₂

Question. Explain the following observations:
Helium forms no real chemical compound.
Answer : Helium has completely filled ns² electronic configurations in its valence shell. Due to its small size and high IE, helium is chemically unreactive. That’s why it forms no real chemical compound.

Question. Explain the following situations :
XeF₂ has a straight linear structure and not a bent angular structure. 
Answer : Since there are two Xe—F covalent bonds and three one pairs in XeF₂. According to VSEPR theory, the shape of XeF₂ is linear.

Question. Draw the structures of the following molecules: XeF₆ 
Answer : XeF₆
Total valence electron pairs = 8 + 6/2 = 7
Bond pair = 6
Lone pair = 7 – 6 = 1
Hybridisation = sp³d³
Shape = Distorted octahedral
Geometry = Pentagonal bipyramidal

Question. What happens when XeF₄ reacts with SbF₅?
Answer : XeF₄ + SbF₅ → [XeF₃]+ [SbF₆]^–

Question. Assign reasons for the following :
Of the noble gases only xenon is known to form well-established chemical compounds.
Answer : Except radon which is radioactive, Xe has least ionisation energy among noble gases and hence it readily forms chemical compounds particularly with oxygen and fluorine.

Question. Why do some noble gases form compounds with fluorine and oxygen only? 
Answer : Fluorine and oxygen are the most electronegative elements and hence are very reactive. Therefore, they form compounds with noble gases particularly with xenon.

Question. List the uses of neon and argon gases.
Answer : Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes.
Argon is used to provide an inert atmosphere in high temperature metallurgical processes and for filling electric bulbs.

Question. Explain the following observations :
The majority of known noble gas compounds are those of Xenon.
Answer : Since, xenon (Xe) has least ionization energy among noble gases hence it readily forms chemical compounds particularly with O₂ and F₂.

Question. Write balanced chemical equations for the following reaction : XeF₆ is hydrolysed
Answer : XeF₆ + H₂O → XeOF₄ + 2HF

Question. What inspired N. Bartlett for carrying out reaction between Xe and PtF₆? 
Answer : Neil Bartlett first prepared a red compound which is formulated as O₂⁺PtF₆^–. He then realised that the first ionisation enthalpy of molecular oxygen (1175 kJ/mol) is almost identical with Xe (1170 kJ/mol). He made efforts to prepare same type of compound with Xe and was successful in preparing another red compound Xe⁺PtF₆^–.

Question. (i) Which noble gas is used in filling balloons for meteorological observations?
(ii) Complete the equation :
XeF₂ + PF₅ 
Answer : (i) Helium is used for filling balloons for meteorological observations because it is non-inflammable.
(ii) XeF₂ + PF₅ → [XeF]⁺[PF₆]^–

Question. Complete the following reaction :
XeF₆ + KF 
Answer : XeF₆ + KF → K⁺[XeF₇]^–

Question. Draw the structures of XeF₄ and predict their shapes. 
Answer : 

Question. Explain the following :
(i) Xenon does not form such fluorides as XeF₃ and XeF₅.
(ii) Out of noble gases, only Xenon is known to form real chemical compounds
Answer : (i) 

All the orbital of Xe have paired electrons. The promotion of one, two or three electrons from 5p-filled orbitals to the 5d-vacant orbitals will give rise to two, four and six-half filled orbitals. Therefore, xenon can combine with even number of fluorine atoms, not add. Thus, it cannot form XeF₃ and XeF₅.
(ii) Except radon which is radioactive, Xe has least ionisation energy among noble gases and hence it readily forms chemical compounds particularly with oxygen and fluorine.

Question. Write the structures of the following molecule : XeOF₄. 
Answer: XeOF₄ is square pyramidal.

Question. Explain the following :
XeF₂ is linear molecule without a bend.
Answer : Since there are two Xe—F covalent bonds and three one pairs in XeF₂. According to VSEPR theory, the shape of XeF₂ is linear.

Question. How is XeO₃ obtained? Write the related chemical equations. Draw the structure of XeO₃.
Answer : XeO₃ can be obtained by hydrolysis of XeF₄ and XeF₆.