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NCERT Solutions for Class 12 Chemistry Chapter 7 - The P Block Elements

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NCERT Solutions for Class 12 Chemistry Chapter 7 - The p-Block Elements

The p-Block Elements consist of groups 13 to 18 of the periodic table. Chapter 7 Chemistry class 12 is a continuation of class 11 syllabus.  While the course of grade 11 deals with the first two groups of p-Block, 13 and 14; class 12 syllabus deals with groups 15 to 18. Group 15 is no longer a part of the syllabus as per the latest changes made by CBSE.


It is one of the book's longest chapters and has many subtopics. Thus, a solution book helps a student do well on their tests.


NCERT Solutions for Class 12


Class 12 Chemistry

Chapter Name:

Chapter 7 - The p-Block Elements 


Text, Videos, Images and PDF Format

Academic Year:



English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes

Important Topics Covered in Chapter 7 of Class 12 Chemistry NCERT Solutions

  1. Group 15 Elements

  • Occurrence

  • Electronic configuration

  • Atomic and ionic radii

  • Ionisation enthalpy

  • Electronegativity

  • Physical properties

  • Chemical properties

  1. Dinitrogen: Preparation, Properties and Uses

  2. Ammonia: Preparation, Properties and Uses

  3. Oxides of Nitrogen: Structure

  4. Nitric Acid: Preparation, Properties and Uses

  5. Phosphorus: Allotropic forms

  6. Phosphine: Preparation, Properties and Uses

  7. Phosphorus Halides

  • Phosphorus Trichloride: Preparation, Properties and Uses

  • Phosphorus Pentachloride: Preparation, Properties and Uses

  1. Oxoacids of Phosphorus 

  2. Group 16 Elements

  • Occurrence

  • Electronic configuration

  • Atomic and ionic radii

  • Ionisation enthalpy

  • Electronegativity

  • Physical properties

  • Chemical properties

  1. Dioxygen: Preparation, Properties and Uses

  2. Simple Oxides

  3. Ozone: Preparation, Properties and Uses

  4. Sulphur — Allotropic Forms

  5. Sulphur Dioxide: Preparation, Properties and Uses

  6. Oxoacids of Sulphur

  7. Sulphuric Acid

  8. Group 17 Elements

  • Occurrence

  • Electronic configuration

  • Atomic and ionic radii

  • Ionisation enthalpy

  • Electronegativity

  • Physical properties

  • Chemical properties

  1. Chlorine: Preparation, Properties and Uses

  2. Hydrogen Chloride: Preparation, Properties and Uses

  3.  Oxoacids of Halogens

  4. Interhalogen Compounds: Preparation, Properties and Uses

  5. Group 18 Elements

  • Occurrence

  • Electronic configuration

  • Atomic and ionic radii

  • Ionisation enthalpy

  • Electronegativity

  • Physical properties

  • Chemical properties

The p-Block Elements Chapter at a Glance - Class 12 NCERT Solutions

Group-15 Elements

  • General Electron Configuration: ns2np3 , extra stable electronic configuration due to half filled p-orbital.

  • Atomic and Ionic Radii: Increase down the group due to increase in the number of inner shells.

  • Ionisation Enthalpy(IE): Decreases down the group due to increase in atomic size.

  • Electronegativity : Decreases down the group. Melting & Boiling Point : Boiling points increase from top to bottom and Melting points increase upto Arsenic & then decrease upto Bismuth.

  • Oxidation State: The tendency to exhibit -3 oxidation state decreases down the group due to increase in size and metallic character (i.e electronegativity). +3 and +5 are generally the covalency and the tendency to show (+3) O.S. in comparison to (+5) O.S. increases down the group due to the inert pair effect.

Group-16 Elements

  • General Electron Configuration: ns2np4

  • Atomic and Ionic Radii : Increase down the group due to increase in the number of shells.

  • Ionisation Enthalpy(IE) : Decreases down the group due to increase in size.

  • Electron Gain Enthalpy (EGE) : Becomes less negative down the group (S < Se < Te < Po < O) due to increase in size. O is an exception due to interionic repulsion in its small size

  • Electronegativity : Decreases down the group. Melting & Boiling Point : Increase down the group as the strength of van der waals force increases with increase in molar mass & size.

  • Oxidation State : The stability of -2 oxidation state decreases down the group because of decrease in electronegativity. The tendency or stability of +4 oxidation state in comparison to +6 oxidation state increases down the group due to inert pair effect.

Group-17 Elements

  • General Electron Configuration : ns2np5

  • Atomic & Ionic Radii : F < Cl < Br < I due to increasing number of shells down the group.

  • Ionisation Enthalpy(IE) : F > Cl > Br > I due to increase in size down the group.

  • Electron Gain Enthalpy (EGE) : I > Br > F > Cl as size increases down the group. F is an exception because the incoming electron faces greater interelectronic repulsions in the small-

  • sized 2p orbital of fluorine(F).

  • Electronegativity: Fluorine is the most electronegative element.

  • Melting & Boiling Point : Their melting and boiling points increase down the group with atomic number. Because the strength of van der Waal force increases with increase in mass.

  • Order of Bond Dissociation Energy is Cl2> Br2> F2> I2

Group-18 Elements

  • General Electron Configuration : ns2np6 except helium which has 1s2.

  • Ionisation Enthalpy : Decreases down the group, Due to stable electronic configuration these gases exhibit very high ionisation enthalpy.

  • Atomic Radius : Increases down the group.

  • Electron Gain Enthalpy (EGE) : They have no tendency to accept the electron & therefore, have large +ve values of EGE.

  • Melting & Boiling Point : They have low melting point and boiling point. Boiling point increases down the group with increase in strength of dispersion forces. All noble gases are monatomic, colourless, they are sparingly soluble in water.

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Access NCERT Solutions for Class 12 Chemistry Chapter 7- p- Block elements

Intext Exercise 1 

1. Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionization and electronegativity. 

Ans: General trends in group 15 elements are as follows:

  • Electronic Configuration: All the elements in group 15 have 5 valence electrons. Their general electronic configuration is $n{{s}^{2}}n{{p}^{3}}$. 

  • 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. 

  • 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. 

  • 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. 

2. Why does the reactivity of nitrogen differ from phosphorus? 

Ans: Nitrogen is chemically less reactive. This is because of the high stability of its molecule, ${{N}_{2}}$. In ${{N}_{2}}$, the two nitrogen atoms form a triple bond. This triple bond has very high bond strength, which is very difficult to break. It is because of nitrogen’s small size that it is able to form $p\pi \text{ }-\text{ }p\pi $ bonds with itself. This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is more reactive than nitrogen. 

Intext Exercise 2 

3. Discuss the trends in chemical reactivity of group 15 elements. 

Ans: General trends in chemical properties of group 15:

  • Reactivity Towards Hydrogen: The elements of group 15 react with hydrogen to form hydrides of type$E{{H}_{3}}$, where E = N, P As, Sb, or Bi. The stability of hydrides decreases on moving down from $N{{H}_{3}}$ to$Bi{{H}_{3}}$ . 

  • Reactivity Towards Oxygen: The elements of group 15 form two types of oxides: ${{E}_{2}}{{O}_{3}}$ and${{E}_{2}}{{O}_{5}}$ , 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. 

  • Reactivity Towards Halogens: The group 15 elements react with halogens to form two series of salts: $E{{X}_{3}}$ and$E{{X}_{5}}$. However, nitrogen does not form $N{{X}_{3}}$ as it lacks the d-orbital. All trihalides (except $N{{X}_{3}}$) are stable.

  • Reactivity Towards Metals: The group 15 elements react with metals to form binary compounds in which metals exhibit -3 oxidation states. 

Intext Exercise 3 

4. Why does $\text{N}{{\text{H}}_{\text{3}}}$ form hydrogen bond but $\text{P}{{\text{H}}_{\text{3}}}$ does not? 

Ans: Hydrogen bond is always formed between highly electronegative atoms and H atoms. Nitrogen is highly electronegative compared to phosphorus as electronegativity decreases down the group. 

Hence, the extent of hydrogen bonding in $P{{H}_{3}}$ is very less as compared to$N{{H}_{3}}$. 

5. How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved. 

Ans: An aqueous solution of ammonium chloride is treated with sodium nitrite. 

\[N{{H}_{4}}C{{l}_{(aq)}}+NaN{{O}_{2(aq)}}\to {{N}_{2(g)}}+2{{H}_{2}}{{O}_{(l)}}+NaC{{l}_{(aq)}}\] 

NO and $HN{{O}_{3}}$ are produced in small amounts. These are impurities that can be removed on passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate. 

6. How is ammonia manufactured industrially? 

Ans: Ammonia is prepared on a large-scale by the Haber’s process. 

\[{{N}_{2(g)}}+3{{H}_{2(g)}}\to 2N{{H}_{3(g)}}\,\,\,{{\Delta }_{f}}{{H}^{o}}=-46.1kJ/mol\] 

The optimum conditions for manufacturing ammonia are: 

  • Pressure (around$200\text{ }\times 105\text{ }Pa$ ) 

  • Temperature (4700 K) 

  • Catalyst such as iron oxide with small amounts of $A{{1}_{2}}{{O}_{3}}$ and${{K}_{2}}O$.

7. Illustrate how copper metal can give different products on reaction with $\text{HN}{{\text{O}}_{\text{3}}}$ . 

Ans: Concentrated nitric acid is a strong oxidizing agent. It is used for oxidizing most metals. The products of oxidation depend on the concentration of the acid, temperature, and also on the material undergoing oxidation. 

\[3Cu+8HN{{O}_{3(dilute)}}\to 3Cu{{(N{{O}_{3}})}_{2}}+2NO+4{{H}_{2}}O\] 

\[Cu+4HN{{O}_{3(conc)}}\to Cu{{(N{{O}_{3}})}_{2}}+2N{{O}_{2}}+2{{H}_{2}}O\] 

8. Give the resonating structures of $\text{N}{{\text{O}}_{\text{2}}}$  and ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ 




9. The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?

  • Hint: Can be explained on the basis of $\text{s}{{\text{p}}^{\text{3}}}$ hybridisation in $\text{N}{{\text{H}}_{\text{3}}}$ and only s-p bonding between hydrogen and other elements of the group. 

Ans: In hydrides $N{{H}_{3}},P{{H}_{3}},As{{H}_{3}}\text{ ,}Sb{{H}_{3}}$ H-M-H angle is $107{}^\circ ,92{}^\circ ,91{}^\circ ,90{}^\circ $ respectively.

The above trend in the H-M-H bond angle can be explained on the basis of the electronegativity of the central atom. Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes greater repulsion between the electron pairs around nitrogen, resulting in maximum bond angle. We know that electronegativity decreases on moving down a group. 

Consequently, the repulsive interactions between the electron pairs decrease, thereby decreasing the H-M-H bond angle. 

10. Why does ${{\text{R}}_{\text{3}}}\text{P}$  = 0 exist but ${{\text{R}}_{\text{3}}}\text{N}$  = 0 does not (R = alkyl group)? 

Ans: N (unlike P) lacks the d-orbital. This restricts nitrogen to expand its coordination number beyond four. Hence, ${{R}_{3}}N$ = 0 does not exist. 

11. Explain why $\text{N}{{\text{H}}_{\text{3}}}$ is basic while $\text{Bi}{{\text{H}}_{\text{3}}}$ is only feebly basic. 

Ans: Nitrogen has a small size due to which the lone pair of electrons is concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density. Hence, the electron donating capacity of group 15 element hydrides decreases on moving down the group. This makes$N{{H}_{3}}$ more basic than$Bi{{H}_{3}}$.

12. Nitrogen exists as a diatomic molecule and phosphorus as${{\text{P}}_{\text{4}}}$ . Why? 

Ans: Nitrogen owing to its small size has a tendency to form$p\pi -p\pi $  multiple bonds with it. 

Nitrogen thus forms a very stable diatomic molecule, ${{N}_{2}}$ . On moving down a group, the tendency to form bonds decreases (because of the large size of heavier elements). Therefore, phosphorus (like other heavier metals) exists in the ${{P}_{4}}$  state. 

13. Write main differences between the properties of white phosphorus and red phosphorus. 


White Phosphorus

Red phosphorus

It is a soft and waxy solid. It possesses a garlic smell.
It is 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 
It is both solid and vapour states, it exists as a ${{P}_{4}}$ molecule.
It exists as a chain of tetrahedral${{P}_{4}}$ units.

14. Why does nitrogen show catenation properties less than phosphorus? 

Ans: Catenation is much more common in phosphorous compounds than in nitrogen compounds. This is because of the relative weakness of the N-N single bond as compared to the P-P single bond. 

Since the nitrogen atom is smaller, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the N-N single bond. 

15. Give the disproportionation reaction of ${{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{3}}}$ 

Ans: On heating, orthophosphorus acid $({{H}_{3}}P{{O}_{3}})$disproportionates to give orthophosphoric acid $({{H}_{3}}P{{O}_{4}})$ and phosphine $(P{{H}_{3}})$. 

The oxidation states of P in various species involved in the reaction are mentioned below. 

\[4{{H}_{3}}\overset{+3}{\mathop{P}}\,{{O}_{3}}\to 3{{H}_{3}}\overset{+5}{\mathop{P}}\,{{O}_{4}}+\overset{-3}{\mathop{P}}\,{{H}_{3}}\] 

16. Can $\text{PC}{{\text{l}}_{\text{5}}}$ act as an oxidizing as well as a reducing agent? Justify. 

Ans: $PC{{l}_{5}}$can only act as an oxidizing agent. The highest oxidation state that P can show is+5. 

In $PC{{l}_{5}}$, phosphorus is in its highest oxidation state (+5). Thus, it can only decrease its oxidation state and act as an oxidizing agent. 

17. 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: The elements of group 16 are collectively called chalcogens. 

  • Electronic Configuration:Elements of group 16 have six valence electrons each. The general electronic configuration of these elements is $n{{s}^{2}}n{{p}^{4}}$, where n varies from 2 to 6.

  • Oxidation State: As these elements have six valence electrons, 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 $({{H}_{2}}{{O}_{2}})$, zero (${{O}_{2}}$), and +2 ($O{{F}_{2}}$). 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. 

  • Formation of Hydrides: These elements form hydrides of formula ${{H}_{2}}E$ , where E = O, S, Se, Te, Po. Oxygen and sulphur also form hydrides of type ${{H}_{2}}{{E}_{2}}$. These hydrides are quite volatile in nature. 

18. Why is dioxygen a gas but sulphur a solid? 

Ans: Oxygen is smaller in size as compared to sulphur. Due to its smaller size, it can effectively form $p\pi -p\pi $ bonds and form ${{O}_{2}}(O=O)$molecule. Also, the intermolecular forces in oxygen are weak van der Waals, which cause it to exist as gas. On the other forces in oxygen are weak van der Waals, which cause it to exist as gas. On the other hand, sulphur does not form$p\pi -p\pi $ bonds but exists as a puckered structure held together by strong covalent bonds. Hence, it is solid. 

19. Knowing the electron gain enthalpy values for $\text{O}\to {{\text{O}}^{\text{-}}}$ and $\text{O}\to {{\text{O}}^{\text{-}}}$ and

$\text{O}\to {{\text{O}}^{\text{2-}}}$

as -141 and 702 $\text{kJ mo}{{\text{l}}^{\text{-1}}}$ respectively, how can you account for the formation of a large number of oxides having ${{\text{O}}^{\text{2-}}}$ species and not ${{\text{O}}^{\text{-}}}$? 

(Hint: Consider lattice energy factor in the formation of compounds). 

Ans: Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, the more stable it will be. Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving ${{O}^{2-}}$ion is much more than the oxide involving ${{O}^{-}}$ ion. Hence, the oxide having${{O}^{2-}}$ ions is more stable than oxides having${{O}^{-}}$. Hence, we can say that formation of ${{O}^{2-}}$  is energetically more favourable than formation of ${{O}^{-}}$.

20. Which aerosols deplete ozone? 

Ans: Freon’s or chlorofluorocarbons (CFCs) are aerosols that accelerate the depletion of ozone. In the presence of ultraviolet radiation, molecules of CFCs break down to form chlorine- free radicals that combine with ozone to form oxygen. 

21. Describe the manufacture of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$by contact process? 

Ans: 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 $S{{O}_{2}}$ . 

Step (ii): By a reaction with oxygen,$S{{O}_{2}}$  is converted into $S{{O}_{3}}$ in the presence of ${{V}_{2}}{{O}_{5}}$ as a catalyst. 


Step (iii): $S{{O}_{3}}$ produced is absorbed on ${{H}_{2}}S{{O}_{4}}$ to give${{H}_{2}}{{S}_{2}}{{O}_{7}}$  (oleum) 

\[S{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\to {{H}_{2}}{{S}_{2}}{{O}_{7}}\] 

This oleum is then diluted to obtain ${{H}_{2}}S{{O}_{4}}$ of 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.

22. How is $\text{S}{{\text{O}}_{\text{2}}}$ an air pollutant? 

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

  • 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. 

  • Even in very low concentrations,$S{{O}_{2}}$  causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness. 

  • 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. 

23. Why are halogens strong oxidising agents? 

Ans: The general outer electronic configuration of halogens is $n{{p}^{5}}$, where n = 2-6. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration. 

Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron. Hence, they act as strong oxidizing agents. 

24. Explain why fluorine forms only one oxoacid, HOF 

Ans: Fluorine forms only one oxoacid i.e., HOF because of its high electronegativity, small size and non availability of d- orbitals. 

25. Explain why in spite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not. 

Ans: Both chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume. 

26. Write two uses of $\text{Cl}{{\text{O}}_{\text{2}}}$.

Ans: Uses of 

  • It is used for purifying water

  •  It is used as a bleaching agent. 

27. Why are halogens coloured? 

Ans: Almost all halogens are coloured. This is because halogens absorb radiation in the visible region. This results in the excitation of valence electrons to a higher energy region. Since the amount of energy required for excitation differs for each halogen, each halogen displays a different colour. ${{F}_{2}}$-yellow, $C{{l}_{2}}$-greenish yellow, $B{{r}_{2}}$- red and ${{I}_{2}}$ - violet.

28. Write the reactions of${{\text{F}}_{\text{2}}}$  and $\text{C}{{\text{l}}_{\text{2}}}$with water. 


Reaction of ${{F}_{2}}$ with water is as follows:

\[2{{F}_{2(g)}}+2{{H}_{2}}O\to {{O}_{2(g)}}+4H{{F}_{(aq)}}\] 

Reaction of$C{{l}_{2}}$ with water is as follows:

\[C{{l}_{2}}+{{H}_{2}}O\to \underset{hydrochloric\,acid}{\mathop{HCl}}\,+\underset{hypochlorous\,acid}{\mathop{HOCl}}\,\] 

29. How can you prepare$\text{C}{{\text{l}}_{\text{2}}}$from HCl and HCl from $\text{C}{{\text{l}}_{\text{2}}}$? Write reasons only.


  • $C{{l}_{2}}$ can be prepared from HCl by Deacon’s process. 


  • HCl can be prepared from $C{{l}_{2}}$on treating it with water. 

\[C{{l}_{2}}+{{H}_{2}}O\to \underset{hydrochloric\,acid}{\mathop{HCl}}\,+\underset{hypochlorous\,acid}{\mathop{HOCl}}\,\] 

30. What inspired N. Bartlett to carry out a reaction between Xe and $\text{Pt}{{\text{F}}_{\text{6}}}$?

Ans: Neil Bartlett initially carried out a reaction between oxygen and $Pt{{F}_{6}}$ . This resulted in the formation of a red compound, ${{O}_{2}}^{+}[Pt{{F}_{6}}]$. 

Later, he realized that the first ionization energy of oxygen $(1175kJ/mol)$ and 

Xe $(1175kJ/mol)$ is almost the same. Thus, he tried to prepare a compound with Xe and $Pt{{F}_{6}}$. 

He was successful and a red-coloured compound,$X{{e}^{+}}{{[Pt{{F}_{6}}]}^{-}}$was formed.

31. What are the oxidation states of phosphorus in the following:

  • ${{\text{H}}_{\text{2}}}\text{P}{{\text{O}}_{\text{3}}}$ 

Ans: Let the oxidation state of P be x. 

3 + x +3(-2) = 0 

3 + x -6 = 0 

x = +3 

  • $\text{PC}{{\text{l}}_{\text{3}}}$

Ans: Let the oxidation state of P be x. 

 x+ 3(-1) = 0 

x – 3 = 0 

x = + 3 

  • $\text{C}{{\text{a}}_{\text{3}}}{{\text{P}}_{\text{2}}}$ 

Ans: Let the oxidation state of P be x. 

3 (+2) + 2(x) = 0 

6 + 2x = 0 

2x = 6 

x = -3

  • $\text{N}{{\text{a}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}$ 

Ans: Let the oxidation state of P be x. 

3(+1)+ x +4(-2) = 0 

3+ x – 8 = 0 

x – 5 = 0 

x = + 5 

  • $\text{PO}{{\text{F}}_{\text{3}}}$

Ans: Let the oxidation state of P be x. 

 x +(-2) +3(-1) = 0 

x – 5 = 0 

x = + 5 

32. Write balanced equations for the following: 

  • $\text{NaCl}$ is heated with sulphuric acid in the presence of$\text{Mn}{{\text{O}}_{\text{2}}}$.

Ans: $4NaCl+Mn{{O}_{2}}+4{{H}_{2}}S{{O}_{4}}\to MnC{{l}_{2}}+4NaHS{{O}_{4}}+2{{H}_{2}}O+C{{l}_{2}}$  

  • Chlorine gas is passed into a solution of NaI in water. 

Ans: $C{{l}_{2}}+2NaI\to 2NaCl+{{I}_{2}}$ 

33.How are xenon fluorides$\text{Xe}{{\text{F}}_{\text{2}}}\text{,Xe}{{\text{F}}_{\text{4}}}$and$\text{Xe}{{\text{F}}_{\text{6}}}$obtained? 

Ans: $Xe{{F}_{2}},Xe{{F}_{4}}$ and $Xe{{F}_{6}}$ are obtained by a direct reaction between Xe and ${{F}_{2}}$. The condition under which the reaction is carried out determines the product.

\[\underset{(excess)}{\mathop{X{{e}_{(g)}}}}\,+{{F}_{2(g)}}\to Xe{{F}_{2(g)}}\] 

\[\underset{(1:5\,ratio)}{\mathop{X{{e}_{(g)}}}}\,+2{{F}_{2(g)}}\to Xe{{F}_{4(g)}}\] 

\[\underset{(1:20\,ratio)}{\mathop{X{{e}_{(g)}}}}\,+3{{F}_{2(g)}}\to Xe{{F}_{6(g)}}\] 

34. With what neutral molecule is$\text{Cl}{{\text{O}}^{\text{-}}}$  isoelectronic? Is that molecule a Lewis base? 

Ans: $Cl{{O}^{-}}$ is isoelectronic to ClF. Also, both species contain 26 electrons in all as shown. 

Total electrons 

$Cl{{O}^{-}}$ =17+ 8+ 1= 26 

In HF =17 +9 =26 

ClF acts like a Lewis base as it accepts electrons from F to form $Cl{{F}_{3}}$.

35. How are$\text{Xe}{{\text{O}}_{\text{3}}}$and $\text{XeO}{{\text{F}}_{\text{4}}}$prepared? 


$Xe{{O}_{3}}$ can be prepared in two ways as follows:

\[6Xe{{F}_{4}}+12{{H}_{2}}O\to 4Xe{{O}_{3}}+24HF++4Xe+3{{O}_{2}}\]

\[Xe{{F}_{6}}+3{{H}_{2}}O\to Xe{{O}_{3}}+6HF\] 

$XeO{{F}_{4}}$ can be prepared using $Xe{{F}_{6}}$ as follows:

\[Xe{{F}_{6}}+{{H}_{2}}O\to XeO{{F}_{4}}+2HF\]  

36. Arrange the following in the order of property indicated for each set: 

  • ${{\text{F}}_{\text{2}}}\text{,}{{\text{C}}_{\text{2}}}\text{,B}{{\text{r}}_{\text{2}}}{{\text{I}}_{\text{2}}}\text{-}$increasing bond dissociation enthalpy. 

Ans: Bond dissociation energy usually decreases on moving down a group as the atomic size increases. However, the bond dissociation energy of ${{F}_{2}}$is lower than that of$C{{l}_{2}}$ and$B{{r}_{2}}$. 

This is due to the small atomic size of fluorine. Thus, the increasing order for bond dissociation energy among halogens is as follows:

\[{{I}_{2}} < {{F}_{2}} < B{{r}_{2}} < C{{l}_{2}}\] 

  • HF, HCl, HBr, HI - increasing acid strength. 

Ans: Order of increasing acidic strength is:

\[HF < HCl < HBr < HI\] 

The bond dissociation energy of H-X molecules where X = F, Cl, Br, I, decreases with an increase in the atomic size. Since the H-I bond is the weakest, HI is the strongest acid. 

  • $\text{N}{{\text{H}}_{\text{3}}}\text{,P}{{\text{H}}_{\text{3}}}\text{,As}{{\text{H}}_{\text{3}}}\text{,Sb}{{\text{H}}_{\text{3}}}\text{,Bi}{{\text{H}}_{\text{3}}}$- increasing base strength

Ans: The order is:

\[Bi{{H}_{3}}\le Sb{{H}_{3}} < As{{H}_{3}} < P{{H}_{3}} < N{{H}_{3}}\] 

On moving from nitrogen to bismuth, the size of the atom increases while the electron density on the atom decreases. Thus, the basic strength decreases. 

37. Which one of the following does not exist? 





Ans: $Ne{{F}_{2}}$does not exist. 

38. Give the formula and describe the structure of a noble gas species which is isostructural With: 

  • $\text{IC}{{\text{l}}_{\text{4}}}^{\text{-}}$ 

Ans:   $Xe{{F}_{4}}$ is isostructural with$IC{{l}_{4}}^{-}$ and has square planar geometry. 

  • $\text{IB}{{\text{r}}_{\text{2}}}^{\text{-}}$ 

Ans: $Xe{{F}_{2}}$is isostructural to$IB{{r}_{2}}^{-}$ and has a linear structure.

  • $\text{Br}{{\text{O}}_{\text{3}}}^{\text{-}}$ 

Ans: $Xe{{F}_{3}}$ is isostructural with$Br{{O}_{3}}^{-}$and has a pyramidal geometry.  

38. Why do noble gases have comparatively large atomic sizes? 

Ans: Noble gases do not form molecules. In case of noble gases, the atomic radii correspond to van der Waals radii. On the other hand, the atomic radii of other elements correspond to their covalent radii. By definition, van der Waals radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period. 

39. List the uses of Neon and argon gases. 

Ans: Uses of neon gas: 

  • It is mixed with helium to protect electrical equipment from high voltage 

  • It is filled in discharge tubes with characteristic colours. 

  • It is used in beacon lights. 

Uses of Argon gas: 

  • Argon along with nitrogen is used in gas-filled electric lamps. This is because Ar is more inert than N. 

  • It is usually used to provide an inert temperature in a high metallurgical process. 

  • It is also used in laboratories to handle air-sensitive substances.

Intext Questions 

1. Why are pentahalides more covalent than trihalides? 

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

2. Why is $\text{Bi}{{\text{H}}_{\text{3}}}$the strongest reducing agent amongst all the hydrides of Group 15 elements? 

Ans: 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 $N{{H}_{3}}$ to$Bi{{H}_{3}}$, the reducing character of the hydrides increases on moving from $N{{H}_{3}}$to$Bi{{H}_{3}}$. 

3. Why is ${{\text{N}}_{\text{2}}}$less reactive at room temperature? 

Ans: The two N atoms in ${{N}_{2}}$are bonded to each other by very strong triple covalent bonds. The bond dissociation energy of this bond is very high. As a result, ${{N}_{2}}$is less reactive at room temperature. 

4. Mention the conditions required to maximize the yield of ammonia. 

Ans: Ammonia is prepared using the Haber’s process. The yield of ammonia can be maximized under the following conditions: 

  • High pressure $\left( \sim 200atm \right)$ 

  • A temperature of $\left( \sim 4700K \right)$ 

  • Use of a catalyst such as iron oxide mixed with small amounts of ${{K}_{2}}O$ and $A{{l}_{2}}{{O}_{3}}$.

5. How does ammonia react with a solution of $\text{C}{{\text{u}}^{\text{2+}}}$? 

Ans: $N{{H}_{3}}$acts as a Lewis base. It donates its electron pair and forms a linkage with metal ion. The reaction is as follows:

\[\underset{blue}{\mathop{C{{u}^{2+}}_{(aq)}}}\,+4N{{H}_{3(g)}}\leftrightarrow \underset{deep\,blue}{\mathop{{{\left[ Cu{{(N{{H}_{3}})}_{4}} \right]}^{2+}}_{(aq)}}}\,\] 

6. What is the covalence of nitrogen in ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ ?


From the structure of${{N}_{2}}{{O}_{5}}$, it is evident that the covalency of nitrogen is 4. 

7. Bond angle in $\text{P}{{\text{H}}_{\text{4}}}^{\text{+}}$ is higher than that in$\text{P}{{\text{H}}_{\text{3}}}$.Why? 

Ans: In$P{{H}_{3}}$,P is $s{{p}^{3}}$hybridized. 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$s{{p}^{3}}$ bonding is changed to pyramidal. $P{{H}_{3}}$ combines with a proton to form $P{{H}_{4}}^{+}$ in which the lone pair is absent. Due to the absence of lone pairs in $P{{H}_{4}}^{+}$ there is no lone pair-bond pair repulsion. Hence, the bond angle in $P{{H}_{3}}$ is lesser than in$P{{H}_{4}}^{+}$.

8. What happens when white phosphorus is heated with concentrated solution in an inert atmosphere of $\text{C}{{\text{O}}_{\text{2}}}$ ? 

Ans: White phosphorus dissolves in boiling NaOH solution (in a $C{{O}_{2}}$atmosphere) to give phosphine,$P{{H}_{3}}$. The reaction is as follows:

\[{{P}_{4}}+3NaOH+3{{H}_{2}}O\to \underset{phosphine}{\mathop{P{{H}_{3}}}}\,+\underset{sodium\,hypophosphine}{\mathop{3Na{{H}_{2}}P{{O}_{2}}}}\,\] 

9. What happens when$\text{PC}{{\text{l}}_{\text{5}}}$ is heated? 

Ans: All the bonds that are present in$PC{{l}_{5}}$, are not similar. It has three equatorial and two axial bonds. The equatorial bonds are stronger than the axial ones. Therefore, when$PC{{l}_{5}}$, is heated strongly, it decomposes to form $PC{{l}_{3}}$and$C{{l}_{2}}$ as follows:

\[PC{{l}_{5}}\xrightarrow{\Delta }PC{{l}_{3}}+C{{l}_{2}}\] 

10. Write a balanced equation for the hydrolytic reaction of $\text{PC}{{\text{l}}_{\text{5}}}$, in heavy water. 

Ans: Hydrolytic reaction of$PC{{l}_{5}}$ is as follows:

\[PC{{l}_{5}}+{{D}_{2}}O\to POC{{l}_{3}}+2DC{{l}_{2}}\] 

\[POC{{l}_{3}}+3{{D}_{2}}O\to {{D}_{3}}P{{O}_{4}}+3DCl\] 

Therefore, the net reaction can be written as:

\[POC{{l}_{5}}+4{{D}_{2}}O\to {{D}_{3}}P{{O}_{4}}+5DCl\] 


11. What is the basicity of ${{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}$? 


Since there are three OH groups present in,${{H}_{3}}P{{O}_{4}}$its basicity is three i.e., it is a tribasic acid. 

12. What happens when ${{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{3}}}$is heated? 

Ans: ${{H}_{3}}P{{O}_{3}}$, on heating, undergoes disproportionation reaction to form $P{{H}_{3}}$and ${{H}_{3}}P{{O}_{4}}$. The oxidation numbers of P in${{H}_{3}}P{{O}_{3}},P{{H}_{3}}$ and${{H}_{3}}P{{O}_{4}}$ are +3, -3, and +5 respectively. As the oxidation number of the same element is decreasing and increasing during a particular reaction, the reaction is a disproportionation reaction. 

 \[\underset{\text{ortho phosphorous}\text{acid (+3)}}{\mathop{4{{H}_{3}}P{{O}_{3}}}}\,\xrightarrow{\Delta }\underset{\text{orthophosphoric}\text{acid (+5)}}{\mathop{3{{H}_{3}}P{{O}_{4}}}}\,+\underset{\text{phosphine (-3)}}{\mathop{P{{H}_{3}}}}\,\]


13. List the important sources of sulphur. 

Ans: Sulphur mainly exists in combined form in the earth’s Crust primarily as sulphates, gypsum $\left( CaS{{O}_{4}}.2{{H}_{2}}O \right)$, Epsom salt$\left( MgS{{O}_{4}}.7{{H}_{2}}O \right)$ , baryte$(BaS{{O}_{4}})$ and sulphides galena (PbS) , zinc blends (ZnS) , copper pyrites$(CuFe{{S}_{2}})$.

14. Write the order of thermal stability of the hydrides of Group 16 elements. 

Ans: The thermal stability of hydrides decreases on moving down the group. This is due to a decrease in the bond dissociation enthalpy (H – E) of hydrides on moving down the group. 

15. Why is ${{\text{H}}_{\text{2}}}\text{O}$ a liquid and ${{\text{H}}_{\text{2}}}\text{S}$ a gas? 

Ans: ${{H}_{2}}O$ has oxygen as the central atom. Oxygen has smaller size and electronegativity as compared to sulphur. Therefore, there is extensive hydrogen bonding in ${{H}_{2}}O$, which is absent in ${{H}_{2}}S$.

Molecules of${{H}_{2}}S$  are held together only by weak van der Waal’s forces of attraction. Hence, ${{H}_{2}}O$exists as a liquid while ${{H}_{2}}S$ as a gas.

16. Which of the following does not react with oxygen directly? 

  • Zn, Ti, Pt, Fe 

Ans: Platinum (Pt) is a noble metal and does not react very easily. All other elements, Zn, Ti, Fe, are quite reactive. Hence, oxygen does not react with platinum (Pt) directly. 

17. Complete the following reactions 

  • ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{+}{{\text{O}}_{\text{2}}}\to $ 

Ans: $\underset{ethene}{\mathop{{{C}_{2}}{{H}_{4}}}}\,+\underset{oxygen}{\mathop{3{{O}_{2}}}}\,\to \underset{carbon\,dioxide}{\mathop{2C{{O}_{2}}}}\,+\underset{water}{\mathop{2{{H}_{2}}O}}\,$ 

  • $\text{Al+}{{\text{O}}_{\text{2}}}\to $ 

Ans: $\underset{alu\min ium}{\mathop{4Al}}\,+\underset{oxygen}{\mathop{3{{O}_{2}}}}\,\to \underset{alu\min a}{\mathop{2A{{l}_{2}}{{O}_{3}}}}\,$ 

18. Why does ${{\text{O}}_{\text{3}}}$act as a powerful oxidizing agent? 

Ans: Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free radical, is very reactive.

\[\underset{ozone}{\mathop{{{O}_{3}}}}\,\xrightarrow{\Delta }\underset{oxygen}{\mathop{{{O}_{2}}}}\,+\underset{nascent\,oxygen}{\mathop{[O]}}\,\] 

Therefore, ozone acts as a powerful oxidizing agent 

19. How is ${{\text{O}}_{\text{3}}}$ estimated quantitatively? 

Ans: Quantitatively, ozone can be estimated with the help of potassium iodide. When ozone is made to react with potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated. This liberated iodine can be titrated against a standard solution of sodium thiosulphate using starch as an indicator. The reactions involved in the process are given below.

\[\underset{iodide}{\mathop{2{{I}^{-}}}}\,+\underset{ozone}{\mathop{{{O}_{3}}}}\,\to 2O{{H}^{-}}+\underset{iodine}{\mathop{{{I}_{2}}}}\,+{{O}_{2}}\] 

\[{{I}_{2}}+\underset{\text{Sodium thiosulphate}}{\mathop{2N{{a}_{2}}{{S}_{2}}{{O}_{2}}}}\,\to \underset{\text{Sodium tetrathionate}}{\mathop{N{{a}_{2}}{{S}_{2}}{{O}_{6}}}}\,+2NaI\] 

20. What happens when sulphur dioxide is passed through an aqueous solution of Fe(III)?

Ans: Sulphur dioxide $\left( S{{O}_{2}} \right)$ acts as a reducing agent when passed through an aqueous solution containing Fe(III) salt. It reduces Fe(III) to Fe(II) i.e., ferric ions to ferrous ions. 

The reaction is as follows:

\[2F{{e}^{3+}}+S{{O}_{2}}+2{{H}_{2}}O\to 2F{{e}^{2+}}+S{{O}_{4}}^{2-}+4{{H}^{+}}\] 

21. Comment on the nature of two S-O bonds formed in$\text{S}{{\text{O}}_{\text{2}}}$molecule. Are the two S-O bonds in this molecule equal? 

Ans: The electronic configuration of S is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{4}}$ 

During the formation of $S{{O}_{2}}$, one electron from the 3p orbital goes to the 3d orbital and S undergoes $s{{p}^{2}}$ hybridization. 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 a $p\pi -p\pi $ bond with one oxygen atom and the other forms a $p\pi -d\pi $bond with the other molecule. This is the reason $S{{O}_{2}}$ has 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. 

22. How is the presence of $\text{S}{{\text{O}}_{\text{2}}}$detected? 

Ans: Sulphur dioxide $\left( S{{O}_{2}} \right)$is a colourless and pungent smelling gas. 

It can be detected with the help of potassium permanganate solution. When 

$S{{O}_{2}}$ is passed through an acidified potassium permanganate solution, it decolorizes the solution by reducing $Mn{{O}_{4}}^{-}$ ions to $M{{n}^{2+}}$ions. The reaction taking place is as follows:

\[5S{{O}_{2}}+2Mn{{O}_{4}}^{-}+2{{H}_{2}}O\to 5S{{O}_{4}}^{2-}+4{{H}^{+}}+2M{{n}^{2+}}\] 

23. Mention three areas in which ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ plays an important role? 

Ans: Sulphuric acid is an important industrial chemical and is used for a lot of purposes. Some important uses of sulphuric acid are given below. 

  • It is used in the fertilizer industry. It is used to make various fertilizers such as ammonium sulphate and calcium super phosphate. 

  • It is used in the manufacture of pigments, paints, and detergents. 

  • It is used in the manufacture of storage batteries. 

24. Write the conditions to maximize the yield of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$by Contact process. 

Ans: Manufacture of sulphuric acid by Contact process involves three steps. 

  • Burning of ores to form $S{{O}_{2}}$.

  • Conversion of $S{{O}_{2}}$ to$S{{O}_{3}}$ by the reaction of the former:

(${{V}_{2}}{{O}_{5}}$ is used in this process as a catalyst.) 

  • Absorption of $S{{O}_{3}}$ in ${{H}_{2}}S{{O}_{4}}$ to give oleum$\left( {{H}_{2}}{{S}_{2}}{{O}_{7}} \right)$ 

The key step in this process is the second step. In this step, two moles of gaseous reactants combine to give one mole of gaseous product. Also, this reaction is exothermic. Thus, in accordance with Le Chatelier’s principle, that means, to obtain the maximum amount of $S{{O}_{3}}$gas, temperature should be low and pressure should be high. 

25. Why is${{\text{K}}_{\text{a2}}}\text{}{{\text{K}}_{\text{a1}}}$for ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$in water? 

Ans: The reaction of sulphuric acid and water is as follows:

\[{{H}_{2}}S{{O}_{4(aq)}}+{{H}_{2}}{{O}_{(l)}}\to {{H}_{3}}{{O}^{+}}_{(aq)}+HS{{O}_{4}}^{-};{{K}_{a1}}>10\] 

\[HS{{O}_{4}}^{-}+{{H}_{2}}{{O}_{(l)}}\to {{H}_{3}}{{O}^{+}}_{(aq)}+S{{O}_{4}}{{^{-}}_{(aq)}};{{K}_{a2}}=1.2\times {{10}^{-2}}\] 

It can be noticed that ${{K}_{a1}} < < {{K}_{a2}}$. This is because a neutral${{H}_{2}}S{{O}_{4}}$ has a much higher tendency to lose a proton than the negatively charged $HS{{O}_{4}}^{-}$.

Thus, the former is a much stronger acid than the latter. 

25. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidizing power of ${{\text{F}}_{\text{2}}}$and $\text{C}{{\text{l}}_{\text{2}}}$.

Ans: Fluorine is a much stronger oxidizing agent than chlorine. The oxidizing power depends on three factors:

  • Bond dissociation energy 

  • Electron gain enthalpy 

  • Hydration enthalpy 

The electron gain enthalpy of chlorine is more negative than that of fluorine. However, the bond dissociation energy of fluorine is much lesser than that of chlorine Also, because of its small size, the hydration energy of fluorine is much higher than that of chlorine Therefore, the latter two factors more than compensate for the less negative electron gain enthalpy of fluorine. Thus, fluorine is a much stronger oxidizing agent than chlorine. 

27. Give two examples to show the anomalous behaviour of fluorine. 

Ans: Examples that show anomalous behaviour of fluorine are:

  • It forms only one oxoacid as compared to other halogens that form a number of oxoacids. 

  • Ionization enthalpy, electronegativity and electrode potential of fluorine are much higher than expected. 

28. Sea is the greatest source of some halogens. Comment. 

Ans: Sea water contains chlorides, bromides, and iodides of Na, K, Mg, and Ca. However, it primarily contains NaCl. The deposits of dried up sea beds contain sodium chloride and carnallite$\left( KCl.MgC{{l}_{2}}.6{{H}_{2}}O \right)$  Marine life also contains iodine in their systems. For example, sea weeds contain up-to 0.5% iodine as sodium iodide. Thus, the sea is the greatest source of halogens. 

29. Give the reason for the bleaching action of$\text{C}{{\text{l}}_{\text{2}}}$.

Ans: When chlorine reacts with water, it produces nascent oxygen. This nascent oxygen then combines with the coloured substances present in the organic matter to oxidise them into colourless substances. The reaction that happens is as follows:

\[C{{l}_{2}}+{{H}_{2}}\to 2HCl+\left[ O \right]\]

Coloured substances + $\left[ O \right]\to $ oxidised colourless substance.

29. Name two poisonous gases which can be prepared from chlorine gas. 

Ans: Two poisonous gases that can be prepared from chlorine gas are: 

  • Phosgene $\left( COC{{l}_{2}} \right)$ 

  • Mustard gas $\left( ClC{{H}_{2}}C{{H}_{2}}SC{{H}_{2}}C{{H}_{2}}Cl \right)$ 

30. Why is ICl more reactive than${{\text{I}}_{\text{2}}}$? 

Ans: ICl is more reactive than ${{I}_{2}}$because I – CI bond in ICl is weaker than I – I bond in${{I}_{2}}$. 

32. Why is helium used in diving apparatus? 

Ans: Air contains a large amount of nitrogen and the solubility of gases in liquids increases with increase in pressure. When sea divers dive deep into the sea, a large amount of nitrogen dissolves in their blood. When they come back to the surface, solubility of nitrogen decreases and it separates from the blood and forms small air bubbles. This leads to a dangerous medical condition called bends. Therefore, air in oxygen cylinders used for diving is diluted with helium gas. This is done as ‘He’ is sparingly less soluble in blood. 

32. Balance the following equation: $\text{Xe}{{\text{F}}_{\text{6}}}\text{+}{{\text{H}}_{\text{2}}}\text{O}\to \text{Xe}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}\text{+HF}$ 

Ans: The given equation when balanced is as follows:

\[Xe{{F}_{6}}+{{H}_{2}}O\to Xe{{O}_{2}}{{F}_{2}}+4HF\] 

33. Why has it been difficult to study the chemistry of radon? 

Ans: It is difficult to study the chemistry of radon because it is a radioactive substance that has a half life of only 3.82 days, also compounds of radon such as $Rn{{F}_{2}}$have not been isolated. They have only been identified.

Important Points Learned from the Chapter The p-Block Elements

  • The p-block elements are elements that belong to groups 13 through 18 of the periodic table.

  • Electronic configuration of p-block members in general: The ns2np1-6 valence shell electrical configuration defines the p-block components.

  • Representative elements, also known as main group elements, are elements that belong to the s and p-blocks of the periodic table.

  • The tendency of the ns2 electron pair to engage in bond formation reduces as the atomic size increases. As the atomic number grows, the higher oxidation state becomes less stable in comparison to the lower oxidation state within a group. The 'inert pair effect' is the name given to this pattern. In other words, the energy required to unpair the electrons is more than the energy released when two new bonds are formed.

The p-Block Elements Class 12 NCERT PDF

The p-Block Element was previously one of the most scoring topics of the syllabus, but now after removal of Group 15, its scope has narrowed. Moreover, students have to prepare in-depth to be able to answer the questions. They can take the help of these PDF solutions which are readily available in online sites.

Students can check the website of Vedantu for p block elements class 12 NCERT solutions PDF download. The answers are accurate, specially curated by teachers with years of knowledge and expertise. Moreover, these PDFs are free and easily accessible and students can download them easily as per their convenience.

Class 12 Chemistry Chapter 7 NCERT Solutions

The p-Block Element is a crucial chapter for boards and other competitive exams as well. Students should aim at achieving a comprehensive understanding of all the subtopics. Moreover, they can refer p-Block Elements Class 12 NCERT Solution for acquiring in-depth knowledge and solving the following types of questions:

  • Intext Questions - There are approximately thirty-four intext questions. These questions will examine a student’s understanding of a particular subsection.

  • Exercise Questions - These questions are incorporated at the end of the chapter. It helps to assess the overall understanding of a student regarding the overall topic.

The class 12 Chemistry p-Block Elements NCERT solutions PDF provides answers to these questions in a detailed manner. The answers are written in a concise and easy-to-understand language, and an elaborate explanation is given for every answer. Moreover, students referring to the solutions book will get to know about the following things:

  • Physical, chemical and atomic properties of the elements of group 16,17 and 18.

  • Allotropic forms of sulphur.

  • Structure of sulphur’s oxoacids.

  • Preparation and properties of chlorine and hydrochloric acids.

  • Properties of interhalogens.

  • Uses of noble gases.

The p-Block elements have a share of approximately 8 marks in the class 12 exam. The table below shows the marks distribution:

Weightage of Topics

Short Answers (II) - 3 Marks

Long Answers – 5 Marks

Total Marks




As it can be observed that this topic has a considerable share in the total marks, students should study the chapter thoroughly. Moreover, they can refer to the p-block elements class 12 solutions for quickly understanding this topic.

Benefits of NCERT Solutions for p-Block Elements

NCERT books are integral to the CBSE curriculum. However, despite its multiple benefits, one cannot deny the fact that students must read references to score better. When it comes to studying p-Block elements NCERT solutions of chapter p Block elements Class 12 is crucial.

The Reasons Behind it are as follows:

  • Easy to read 

  • Covers the basics of a topic

  • Adheres to the pattern of CBSE

Students while preparing for class 12 boards often get confused as to how to prepare their p-Block Elements. In such cases, students should consider the following p-Block Elements class 12 NCERT solutions. All the topics are extensively covered in the form of questions and answers, which helps a student to prepare well.


In conclusion, the NCERT Solutions for Class 12 Chemistry Chapter 7 on p-Block Elements provide comprehensive guidance and solutions to the questions and exercises mentioned in the chapter. This chapter focuses on the elements belonging to the p-block of the periodic table, including Group 15, Group 16, Group 17, and Group 18 elements.

By studying and practising the NCERT solutions for Class 12 Chemistry Chapter 7, students can enhance their understanding of p-block elements, their properties, and their significance in different contexts. These solutions serve as valuable resources for self-study, homework, and exam preparation, enabling students to develop a strong foundation in chemistry and excel in their academic pursuits.

NCERT Solutions for Class 12 Chemistry PDF Download

FAQs on NCERT Solutions for Class 12 Chemistry Chapter 7 - The P Block Elements

1. How to prepare p-Block Elements for CBSE Class 12?

The p-Block Elements is one of the lengthiest chapters of class 12 NCERT, but at the same time, the topic is scoring. It includes Group 15 to Group 18 elements. To prepare this chapter for board examinations, students are advised to follow the NCERT Chemistry books strictly. They can also refer to any one reference book for a depth study. Also, they can refer to Vedantu Notes for preparing the p-Block Elements. Not only notes, Vedantu provides them with detailed NCERT solutions to help them to write answers appropriately.

These solutions are easily accessible and are designed to make learning easy. The answers are strictly according to the pattern followed by CBSE, which enables a student to fetch more marks in their exams.

2. Why should I refer to p-Block Elements NCERT Solutions Class 12?

NCERT Solutions for p-Block Elements makes learning easier by offering a straightforward approach. Undoubtedly, students should start their preparation with NCERT books. However, to score well, they must refer to solutions. These guide books are curated by experienced teachers who have adequate expertise regarding this subject.

The books are written in a straightforward language which students can grasp quickly.  And the answers to every question is as per the pattern followed by the CBSE. Moreover, NCERT solution of chapter 7 chemistry class 12 is easily accessible as students can download these PDFs in a hassle-free way. These downloads are free of cost and highly reliable.

3. What are the essential topics for p-Block Elements?

The most important topics for p-Block Elements include physical and chemical properties of each group, preparation of Dinitrogen, Ammonia, Phosphine, Dioxygen. Besides, they should also focus on Inter-halogen compounds and xenon-fluorine compounds and their structure as well. While these are the most important topics, students should not leave out any other part of the chapter if they want to score well.

Moreover, while preparing, they should thoroughly go through the previous years papers to understand which topics are most relevant for their exams. They can then refer to the answers of NCERT solutions for class 12 Chemistry chapter p Block Elements accordingly.

4. What will students learn from Chapter 7 The p-Block Elements of CBSE Class 12 Chemistry?

Chapter 7 of CBSE Class 12 Chemistry discusses the properties of the p-Block elements in a detailed manner. The elements from Group 13 to 18 in the periodic table are characterized as the p-Block elements. The study of the elements and their electronic configuration is an important part of Physical Chemistry. The chapter also has a high weightage in the competitive exams. In this chapter, electronic configuration, atomic and ionic radii, ionization enthalpy, electronegativity, physical and chemical properties of Group 15 to Group 18 elements are explained in detail. Students can refer to online NCERT Solutions designed by experts on Vedantu for better understanding of the chapter.

5. What are the questions that can be answered by referring to the NCERT Solutions for Chapter 7 The p-Block Elements of Class 12 Chemistry?

With several subtopics, Chapter 7 in the Class 12 Chemistry is a long chapter. You can find a variety of questions and topics when you refer to the NCERT Solutions in an easy to understand manner. These answers are correct and tailormade by experts with many years of experience and knowledge. So, when you refer to the NCERT Solutions thoroughly, you will be able to answer all the questions from that particular chapter correctly.

6. What are the benefits of using the NCERT Solutions for Chapter 7 The p-Block Elements of Class 12 Chemistry from Vedantu?

NCERT Solutions for Class 12 Chemistry from Vedantu has various significant advantages. They are accurate, written according to the marking scheme of CBSE, and curated carefully by subject matter experts as per the NCERT syllabus. These solutions are also very easy to comprehend and are written in simple language. From using them to make your revision notes to going through the solutions by Vedantu as your last-minute preparation, they are absolutely beneficial.

7. What is the reason for the difference in reactivity of Nitrogen to that of phosphorus?

Nitrogen’s molecules are of high stability so it is less reactive. The two atoms of nitrogen form a triple bond which is very strong. This bond is very difficult to break because of the small size of nitrogen and therefore it can form pπ bonds by itself. Phosphorus does not have this property hence it is more reactive than Nitrogen. This chapter is an important chapter and has to be perfectly understood to answer the questions relevantly.

These solutions are available on Vedantu's official website( and mobile app free of cost.

8. Is Chapter 7 The p-Block Elements of Class 12 Chemistry easy?

Class 12 Chemistry Chapter 7 is not very difficult but it surely includes important topics that have to be studied well. This chapter also carries 8 marks weightage in the exam. This implies that good solutions will come in handy for the students to learn and understand the concepts well. Here's where Vedantu's NCERT solutions for Class 12 Chemistry can become really helpful. Going through these solutions will make the concept easy for you.

9. What is the reason for NH3  to form hydrogen bonds but not PH3?

Hydrogen bond will be always found between the atom of electronegative and H atom. Nitrogen is highly electronegative and phosphorus is not electronegative in comparison with Nitrogen. Therefore the Hydrogen bonding in PH3 is less when compared to NH3. These topics should be understood with the right concepts to get the proper grasp of the chapter. The concepts of the elements in the p- and f-block to be understood and practised with the proper characteristics.