Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer

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Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer

Class 12 Chemistry Chapter 7 Multiple Choice questions-

1. H2S is more acidic than H2O because
(a) oxygen is more electronegative than sulphur.
(b) atomic number of sulphur is higher than oxygen.
(c) H — S bond dissociation energy is less as compared to H — O bond.
(d) H — O bond dissociation energy is less also compared to H — S bond.

2. The boiling points of hydrides of group 16 are in the order
(a) H2O > H2Te > H2S > H2Se
(b) H2O > H2S > H2Se > H2Te
(c) H2O > H2Te > H2Se > H2S
(d) None of these

3. In the manufacture of sulphuric acid by contact process Tyndall box is used to
(a) convert SO2 and SO3
(b) test the presence of dust particles
(c) filter dust particles
(d) remove impurities

4. Fluorine differs from rest of the halogens in some of its properties. This is due to
(a) its smaller size and high electronegativity.
(b) lack of d-orbitals.
(c) low bond dissociation energy.
(d) All of the these.

5. The set with correct order of acidity is
(a) HClO < HClO2 < HClO3 < HClO4
(b) HClO4 < HClO3 < HClO2 < HClO
(c) HClO < HClO4 < HClO3 < HClO2
(d) HClO4 < HClO2 < HClO3 < HClO

6. When chlorine reacts with cold and dilute solution of sodium hydroxide, it forms
(a) Cl– and ClO–
(b) Cl– and ClO2–
(c) Cl– and ClO3–
(d) Cl– and ClO4–

7. The formation of O2+ [PtF6]– is the basis for the formation of first xenon compound. This is because
(a) O2 and Xe have different sizes.
(b) both O2 and Xe are gases.
(c) O2 and Xe have comparable electro-negativities.
(d) O2 and Xe have comparable ionisation enthalpies.

8. Partial hydrolysis of XeF4 gives
(a) XeO3
(b) XeOF2
(c) XeOF4
(d) XeF2

9. Helium is preferred to be used in balloons instead of hydrogen because it is
(a) incombustible
(b) lighter than hydrogen
(c) more abundant than hydrogen
(d) non polarizable

10. The increasing order of reducing power of the halogen acids is
(a) HF < HCl < HBr < HI
(b) HI < HBr < HCl < HF
(c) HBr < HCl < HF < HI
(d) HCl < HBr < HF < HI

Class 12 Chemistry Chapter 7 Very Short Questions-

1. Write the elements of group 15?
2.  Write chemical name & formulae of
a) Chile saltpetre
b) Indian saltpetre   
3.  What is special about the valence configuration of Group 15?
4.  The atomic radii increases considerably from N to P but very little increase is observed from As to Bi. why?
5.  Give reason for the following- the first ionization enthalpy of 15th group elements is higher than 16th group elements? 
6.  How does metallic character vary down the 15 group & why?
7.  What are the common oxidation states of this group?
8.  What is the maximum covalence shown by N?
9.  Bi (v) is a stronger oxidizing agent than Bi (III). Why?
10. Give an example showing disproportionation of oxidation state of nitrogen?
Short Questions-

1. Write the various steps for preparation of sulphuric acid by contact process?
2. Name different sulphates formed by sulphuric acid?
3. Why are pentahalides more covalent than trihalides?
4. Why is 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerthe strongest reducing agent amongst all the hydrides of Group 15 elements?
5. Why is 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerless reactive at room temperature?
6. How does ammonia react with a solution of 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer?
7. What is the covalence of nitrogen in
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer?
8. What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer?
9. Write a balanced equation for the hydrolytic reaction of 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerin heavy water.
10. What happens when 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answeris heated?
Long Questions-

1. Give reasons: –
(a) Oxygen molecule is diatomic whereas sulphur molecule is polyatomic.
(b) The most common oxidation state of oxygen is -2.
(c) H2O is liquid whereas 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answeris gas at room temperature.
(d) The increasing order of acidic character in 16th group hydrides is
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer.
(e) 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answeris exceptionally stable, 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerdoes not exist.
2. Discuss the different types of oxides.
3. Bond angle in 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answeris higher than that in 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer. Why?
4. Comment on the nature of two S-O bonds formed in 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answermolecule. Are the two S-O bonds in this molecule equal?
5. Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
6. Discuss the trends in chemical reactivity of group 15 elements.
7. Write main differences between the properties of white phosphorus and red phosphorus.
9. Describe the manufacture of 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerby contact process?
10. How is 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answeran air pollutant?

Assertion and Reason Questions-

1. In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.

Assertion: Caro’s acid has S atom in +6 oxidation state.

Reason: Caro’s acid contains one peroxo O22 − group.

2. In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.

Assertion: HOF bond angle in HFO is higher than HOCl bond angle in HClO.

Reason: Oxygen is more electronegative than all halogens except fluorine.

Case Study Questions-

1. Read the passage given below and answer the following questions:

Chlorine is a greenish yellow gas with pungent and suffocating odour. With dry slaked lime, it gives bleaching powder. Bleaching powder is a mixture of calcium hypochlorite and basic calcium chloride:

[Ca(OCl)2· CaCl2· Ca(OH)2· 2H2O].

The amount of chlorine obtained from a sample of bleaching powder by the treatment with excess of dilute acids or CO2 is called available chlorine. Chlorine is a powerful bleaching agent. Bleaching effect of chlorine is permanent.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Chlorine gas reacts with _____ to form bleaching powder.
  1. Ca(OH)2
  2. CaCl2
  3. CaSO4
  4. dry CaO
  5. Chlorine reacts with cold and dilute alkali to form:
  1. Chloride
  2. Hypochlorite
  3. Chlorate
  4. Both (a) and (b)
  5. Which of the following is produced on the reaction of bleaching powder with a few drops of cone. HCl?
  1. Hypochlorous acid
  2. Oxygen
  3. Chlorine
  4. Calcium oxide
  5. Chlorine is used as a bleaching agent. The bleaching action is due to.
  1. Oxidation
  2. Chlorination
  3. Hydrogenation
  4. Reduction
  5. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is:
  1. Cl2O
  2. Cl2O7
  3. ClO2
  4. Cl2O6

2. Read the passage given below and answer the following questions:

Noble gases are inert gases with general electronic configuration of ns2np6. These are monoatomic, colourless, odourless and tasteless gases. The first compound of noble gases was obtained by the reaction of Xe with PtF6. A large number of compounds of Xe and fluorine have been prepared till now. The structure of these compounds can be explained on the basis of VSEPR theory as well as concept of hybridisation. The compounds of krypton are fewer. Only the diftuoride of krypton (KrF2) has been studied in detail. Compounds of radon have not isolated but only identified by radio tracer technique. However, no true compounds of helium, neon or argon are yet known.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. The formula of the compound when Xe and PtF6 are mixed, is:
  1. XeF6
  2. XeF4
  3. Xe2PtF6
  4. Xe+[PtF6]–
  5. The shape and hybridisation of some xenon oxy-fluoride and fluoride compounds are given below. Find the incorrect one.
  1. XeOF2 – T-shape – sy3d
  2. XeOF4 – square pyramidal – sp3d2
  3. XeF2 – linear – sp3d
  4. XeF6 – square planar – dsp2
  5. Which of the following is not formed by Xe?
  1. XeF5
  2. XeF
  3. XeF3
  4. All of these.
  5. The number oflone pairs and bond pairs of electrons around Xe in XeOF4 respectively are,
  1. 0 and 5
  2. 1 and 5
  3. 1 and 4
  4. 2 and 3
  5. Which of the following compounds has more than one lone pair of electrons around central atom?
  1. XeO3
  2. XeF2
  3. XeOF4
  4. XeO2F2

Class 12 Chemistry Chapter 7 MCQ Answers-

  1. Answer: b
  2. Answer: b
  3. Answer: b
  4. Answer: b
  5. Answer: b
  6. Answer: a
  7. Answer: d
  8. Answer: b
  9. Answer: a
  10. Answer: a

Class 12 Chemistry Chapter 7 Very Short Answers-

  1. Ans. The elements of group 15 are Nitrogen (N), Phosphorous (P) , Arsenic (As) , 

Antimony (Sb) and Bismuth (Bi).

  1. Ans. (a) Chile saltpetre – Sodium nitrate – NaNO3
    (b) Indian saltpetre – Potassium nitrate – KNO3
  2. Ans. The valence configuration of 15 group is ns2np3 the s-orbital is completely filled, and p-orbital is half filled. This half-filled orbital gives extra stability to elements of this group.
  3. Ans. There is a considerable increase in size from N to P as expected but due to the presence of completely filled d- orbitals which have very poor shielding effects, the increase in size is very little from As to Bi.
  4. Ans. Due to extra stability of half-filled configuration, the first Ionisation enthalpy of 15th group elements is higher than 16th group configuration 
  5. Ans. The metallic character increases down the group due to decrease in ionization enthalpy and increase in size of atom.
  6. Ans. The common oxidation states of the group are -3, +3 & +5.
  7. Ans. Nitrogen shows a maximum covalence of +4 because only four orbitals, one S and three P- orbitals are available for bonding in Nitrogen.
  8. Ans. Bi is more stable in +3 oxidation state in comparison to +5 due to inert pair effect therefore Bi (v) has a strong tendency to act as oxidizing agent.
  9. Ans.
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer

Here Nitrogen is getting oxidized to a higher oxidation state as well as reduced to a lower oxidation state.

Short Answers-

Ans 1. Contact process for sulphuric acid: –

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer

Ans 2. The two type of sulphates are –

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer

Ans 3. In pentahalides, the oxidation state is +5 and in trihalides, the oxidation state is +3. Since the metal ion with a high charge has more polarizing power, pentahalides are more covalent than trihalides.

Ans 4. As we move down a group, the atomic size increases and the stability of the hydrides of group 15 elements decreases. Since the stability of hydrides decreases on moving from 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerto , the reducing character of the hydrides increases on moving from 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerto .

Ans 5. The two N atoms in 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerare bonded to each other by very strong triple covalent bonds. The bond dissociation energy of this bond is very high. As a result, 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answeris less reactive at room temperature.

Ans 6. NH3 acts as a Lewis base. It donates its electron pair and forms a linkage with metal ion.

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer

Ans 7.

From the structure of , it is evident that the covalence of nitrogen is 4.

Ans 8. White phosphorous dissolves in boiling NaOH solution (in a atmosphere) to give phosphine, .

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer

Ans 9. All the bonds that are present in 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerare not similar. It has three equatorial and two axial bonds. The equatorial bonds are stronger are stronger than the axial ones. Therefore, when 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answeris heated strongly, it decomposes to form PCl3.

Ans 10. 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer

Long Answers-

Ans 1. (a) Oxygen being small in size forms effective and strong P

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer-P
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer bonds with other oxygen atom. Therefore, oxygen molecule is diatomic and discrete whereas Sulphur due to its larger size, its orbitals cannot overlap effectively to form P
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer-P
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer bonds & completes valency by forming 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerbonds with many sulphur, atom. Therefore, sulphur molecule is polyatomic solid.

(b) Since oxygen is highly electronegative, it has little tendency to give electrons. Therefore, its most common oxidation state is -2.

(c) 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answeris liquid at room temperature due to presence of intermolecular Hydrogen bonding which is absent in
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer.

(d) As we move down the group, the size of atom increases this make the bond of the element with hydrogen weak. Due to weaker bonds, the bond dissociation enthalpy decreases making the molecule more acidic. Therefore, the order of acidic strength is.

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer


(e)

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer is exceptionally stable due to steric reasons. Hydrogen being electropositive or less electronegative than fluorine cannot make the s- electrons of sulphur to participate in bonding. Therefore 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerdoes not exist.

Ans 2. A binary compound of oxygen with another element is called oxide. Oxides can be simple or mixed. Simple oxides can be classified as acidic, basic Amphoteric or neutral.
An oxide that combines with water to give an acid is termed acidic oxide e.g.

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer, 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answeretc.
An oxide that combines with water to give a base is called basic oxide e.g.
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer, CaO, BaO etc.
An oxide that shows characteristics of both acids and bases is Amphoteric oxide e.g. 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer.
 An oxide that shows characteristic of neither acid nor base is called neutral oxide e.g. CO, NO  and
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer.

Ans 3. In , P is 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerhybridized. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the tetrahedral shape associated with sp3bonding is changed to pyramidal. combines with a proton to form in which the lone pair is absent. Due to the absence of lone pair in, there is no lone pair-bond pair repulsion. Hence, the bond angle in is higher than the bond angle in 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer.

Concept insite: the long pair-bond pair repulsion is more than bond pair-bond pair repulsion.

Ans 4. The electronic configuration of S is 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer.
During the formation of

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer, one electron from 3p orbital goes to the 3d orbital and S undergoes 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerhybridization. Two of these orbitals form sigma bonds with two oxygen atoms and the third contains a lone pair. p-orbital and d-orbital contain an unpaired electron each. One of these electrons forms
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerbond with one oxygen atom and the other forms 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerbond with the other oxygen. This is the reason 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerhas a bent structure. Also, it is a resonance hybrid of structures I and II.
Both S-O bonds are equal in length (143 pm) and have a multiple bond character.

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer

Ans 5. General trends in group15 elements

(i) Electronic configuration: All the elements in group 15 have 5 valence electrons. Their general electronic configuration is 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer.

(ii) Oxidation states: All these elements have 5 valence electrons and require three more electrons to complete their octets. However, gaining electrons is very difficult as the nucleus will have to attract three more electrons. This can take place only with nitrogen as it is the smallest in size and the distance between the nucleus and the valence shell is relatively small. The remaining elements of this group show a formal oxidation state of -3 in their covalent compounds. In addition to the -3 state, N and P also show -1 and -2 oxidation states.
All the elements present in this group show +3 and +5 oxidation states. However, the stability of +5 oxidation state decreases down a group, whereas the stability of +3 oxidation state increases. This happens because of the inert pair effect.

(iii) Ionization energy and electronegativity
First ionization decreases on moving down a group. This is because of increasing atomic sizes. As we move down a group, electronegativity decreases, owing to an increase in size.

(iv) Atomic size: On moving down a group, the atomic size increases. This increase in the atomic size is attributed to an increase in the number of shells.

Ans 6. General trends in chemical properties of group – 15

(i) Reactivity towards hydrogen:
The elements of group 15 react with hydrogen to form hydrides of type 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer, where E = N, P, As, Sb, or Bi. The stability of hydrides decreases on moving down from 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerto 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer.

(ii) Reactivity towards oxygen:
The elements of group 15 form two types of oxides: 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerand 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer, where E = N, P, As, Sb, or Bi. The oxide with the element in the higher oxidation state is more acidic than the other. However, the acidic character decreases on moving down a group.

(iii) Reactivity towards halogens: The group 15 elements react with halogens to form two series of salts: 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerand 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer. However, nitrogen does not form 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer as it lacks the d-orbital. All trihalides (except 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer) are stable.

(iv) Reactivity towards metals: The group 15 elements react with metals to form binary compounds in which metals exhibit -3 oxidation states.

Ans 7.

White phosphorusRed Phosphorus
It is a soft and waxy solid. It possesses a garlic smell.It is a hard and crystalline solid, without any smell.
It is poisonous.It is non-poisonous.
It is insoluble in water but soluble in carbon disulphide.It is insoluble in both water and carbon disulphide.
It undergoes spontaneous combustion in air.It is relatively less reactive.
In both solid and vapour states, it exists as a P4 molecule. 
A picture containing antennaDescription automatically generated		It exists as a chain of tetrahedral P4 units. 
A picture containing triangleDescription automatically generated

8.  Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.

Ans 8. The elements of group 16 are collectively called chalcogens.

(i) Elements of group 16 have six valence electrons each. The general electronic configuration of these elements is 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer, where n varies from 2 to 6.

(ii) Oxidation state:
As these elements have six valence electrons 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer, they should display an oxidation state of -2. However, only oxygen predominantly shows the oxidation state of -2 owing to its high electronegativity. It also exhibits the oxidation state of -1 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer, zero 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer, and +2
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer. However, the stability of the -2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements of the group show an oxidation state of +2, +4, and +6 due to the availability of d-orbitals.

(iii) Formation of hydrides:
These elements form hydrides of formula 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer, where E = O, S, Se, Te, PO. Oxygen and sulphur also form hydrides of type 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer. These hydrides are quite volatile in nature.

Ans 9. Sulphuric acid is manufactured by the contact process. It involves the following steps:
Step (i): Sulphur or sulphide ores are burnt in air to form 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer.

Step (ii): By a reaction with oxygen, 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answeris converted into 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerin the presence of 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answeras a catalyst.

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer

Step (iii): 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerproduced is absorbed on 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerto give 
Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer(oleum).

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer



This oleum is then diluted to obtain 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answerof the desired concentration.
In practice, the plant is operated at 2 bar (pressure) and 720 K (temperature). The sulphuric acid thus obtained is 96-98% pure.

Ans 10. Sulphur dioxide causes harm to the environment in many ways:

1. It combines with water vapour present in the atmosphere to form sulphuric acid. This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble.
2. Even in very low concentrations, 

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answercauses irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.
3. 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.
Assertion and Reason Answers-

1. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:

This can be explained through structure of caro’s acid (peroxomonosulphuric acid).

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer

Oxidation no. of S = x, oxidation no. of H = +1,

Oxidation no. of O in peroxo linkage = -1 (each),

Oxidation no. of other oxygen atoms = -2 (each).

2 + x – 6 – 2 = 0 or x = +6.

2. (d) Assertion is wrong statement but reason is correct statement.

Explanation:

HOF bond angle in HFO is lesser than that of HOCl bond angle in HClO. Oxygen is more electronegative than all halogens except fluorine.

Case Study Answers-

1. Answer :

  1. (a) Ca(OH)2
  2. (d) Both (a) and (b)

Explanation:

In cold, chlorine reacts with dilute alkalies to form chlorides and hypochlorites.

  1. (c) Chlorine
  2. (a) Oxidation

Explanation:

Cl2 + H2O → 2HCl + [O]

  1. (a) Cl2O

Explanation:

Bleaching powder contains OCl– ion, hence the oxoacid is HOCl. Anhydride of HOCl is Cl2O.

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer

2. Answer :

  1. (d) Xe+[PtF6]–
  2. (d) XeF6 – square planar – dsp2

Explanation:

XeF6 has sp3d3 hybridisation and distorted octahedral shape.

  1. (d) All of these.

Explanation:

Xe has completely filled 5p-orbital. As a result, when it undergoes bonding with an odd number (1, 3 or 5) of fluorine atoms, it leaves behind one unpaired electron. This causes the molecule to become unstable. As a result, XeF, XeF3 and XeF5 do not exist.

  1. (b) 1 and 5

Explanation:

Class 12 Chemistry Chapter 7 The p-Block Elements Question Answer
  1. (b) XeF2

Explanation: XeF2 has 3 lone pairs on Xe atom.

Class 12 Chemistry All Chapter Notes & Question Answer

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