NCERT Solutions for Class 12 Chemistry Chapter 7 The P Block Elements
NCERT Solutions for Class 12 Chemistry Chapter 7 – Free PDF Download
NCERT Solutions Class 12 Chemistry Chapter 7 The p Block Elements is exclusively written for CBSE students of Class 12. These NCERT Solutions for Class 12 Chemistry provide an excellent approach to master the subject. Also, these solutions assist you in understanding the concept deeply by giving P Block Elements Class 12 questions and answers in the textbook, question papers and sample papers.
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Class 12 Chemistry NCERT Solutions Chapter 7 The p Block Elements – Important Questions
Q 1:Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity
Answer General trends in group 15 elements (i) Electronic configuration: There are 5 valence electrons for all the elements in group 15. ns2np3 is their general electronic configuration. (ii) Oxidation states: All these elements require three or more electrons to complete their octets and have 5 valence electrons. It is difficult in gaining electrons as the nucleus will have to attract three more electrons. This happens 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
Ionization decreases as we move down the group. This happens because of increase in atomic
sizes. Moving down the group, electronegativity decreases due to increase in size. (iv) Atomic size: As we move down the group atomic size increases. This increase in
the atomic size is attributed to an increase in the number of shells.
Q 2: Why does the reactivity of nitrogen differ from phosphorus?
Nitrogen is chemically less reactive. This is because of the high stability of its molecule, N2. In N2, 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 pπ−pπ bonds with itself. This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is more reactive than nitrogen.
Q 3: Discuss the trends in chemical reactivity of group 15 elements.
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 EH3, where E = N,P, As, Sb, or Bi. The stability of hydrides decreases on moving down from NH3toBiH3.
(ii)Reactivity towards oxygen:
The elements of group 15 form two types of oxides: E2O3 and E2O5, 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: EX3andEX5. However, nitrogen does not form NX5 as it lacks the d-orbital. All trihalides (except NX3) are stable. (iv) Reactivity towards metals:
The group 15 elements react with metals to form binary compounds in which metals exhibit
−3 oxidation states.
Q 4: Why does NH3 form hydrogen bond but PH3 does not?Answer
When compared to phosphorus nitrogen is highly electronegative. This results in a greater attraction of electrons towards nitrogen in NH3 than towards phosphorus in PH3. Hence, the extent of hydrogen bonding in PH3 is very less as compared to NH3.
Q 5: How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved
An aqueous solution of ammonium chloride is treated with sodium nitrite.
NH4Cl (aq ) + NaNO2 →N2(g) + 2H2O(l) + NaCl(aq)
NO and HNO3 are produced in small amounts. These are impurities that can be removed on passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate.
Q 6: How is ammonia manufactured industrially?
Ammonia is prepared on a large-scale by the Haber’s process.
N2(g) + 3H2(g) ⇌ 2NH3(g)
The optimum conditions for manufacturing ammonia are: (i) Pressure (around 200×105 Pa) (ii) Temperature (700 K) (iii) Catalyst such as iron oxide with small amounts of Al2O3 and K2O
Q 7: Illustrate how copper metal can give different products on reaction with HNO3.
Concentrated nitric acid is a strong oxidizing agent. It is used for oxidizing most metals.
The products of oxidation depend on the temperature, concentration of the acid, and also
on the material undergoing oxidation.
Q 8: Give the resonating structures of NO2 and N2O5.
Q 9: The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of sp3 hybridisation in NH3 and only s−p bonding between hydrogen and other elements of the group].
H−M−H angle 107° 92° 91° 90°
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.
Q 10: Why does R3P=O exist but R3N=O does not (R = alkylgroup)?
N (unlike P) lacks the d-orbital. This restricts nitrogen to expand its coordination number
beyond four. Hence, R3N=O does not exist.
Q 11: Explain why NH3 is basic while BiH3 is only feebly basic.
NH3 is distinctly basic while BiH3 is feebly basic.
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 decrease on moving down the group.
Q 12: Nitrogen exists as diatomic molecule and phosphorus as P4. Why?
Nitrogen owing to its small size has a tendency to form pπ−pπ multiple bonds with itself.
Nitrogen thus forms a very stable diatomic molecule, N2. On moving down a group, the tendency to form pπ−pπ bonds decreases (because of the large size of heavier elements). Therefore, phosphorus (like other heavier metals) exists in the P4 state.
Q 13: Write the main differences between the properties of white phosphorus and red phosphorus.
Q 14: Why does nitrogen show catenation properties less than phosphorus?
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 nitrogen atom is smaller, there is greater repulsion of
electron density of two nitrogen atoms, thereby weakening the N−N single bond.
Q 15: Give the disproportionation reaction of H3PO3.
On heating, orthophosphorus acid (H3PO3) disproportionates to give orthophosphoric acid (H3PO4) and phosphine (PH3). The oxidation states of P in various species involved in the reaction are mentioned below.
4H3P+3O3→ 3H3P+5O4 + P-3H3
Q 16: Can PCl5 act as an oxidising as well as a reducing agent? Justify.
PCl5 can only act as an oxidizing agent. The highest oxidation state that P can show is +5. In PCl5, phosphorus is in its highest oxidation state (+5). However, it can decrease its oxidation state and act as an oxidizing agent.
Q 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.
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 ns2np4, where n varies from 2 to 6
(ii) Oxidation state:
As these elements have six valence electrons (ns2np4), 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 (H2O2), zero (O2), and +2 (OF2). 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 H2E, where E = O, S, Se, Te, PO. Oxygen and sulphur also form hydrides of type H2E2. These hydrides are quite volatile in nature.
Q 18: Why is dioxygen a gas but sulphur a solid?
Oxygen is smaller in size when compared to sulphur. Since its size is small, it can form pπ−pπ bonds and form O2(O=O) molecule. Also, the intermolecular forces in oxygen are weak van der Wall’s, which cause it to exist as gas. On the other hand, sulphurdoes not form M2 molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid.
Q 19: Knowing the electron gain enthalpy values for O → O -1 and O → O 2- as −141 and 702kJmol−1 respectively, how can you account for the formation of a large number of oxides having O2−speciesandnotO−?
(Hint: Consider lattice energy factor in the formation of compounds).
More the lattice energy of a compound, more stable it will be. Stability of an ionic compound depends on its lattice energy.
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 O2- ion is much more than the oxide involving O– ion. Hence, the oxide having O2- ions are more stable than oxides having O– ion. Hence, we can say that formation of O2- is energetically more favourable than formation of O–.
Q 20. Which aerosols deplete ozone?
The aerosol which is responsible for the depletion of ozone is: Freons or chlorofluorocarbons (CFCs).
The molecules of CFS break down when there is presence of ultraviolet radiations and forms chlorine free radicals which then combines with ozone to form oxygen.
Q 21. Describe the manufacture of H2SO4 by contact process.
The steps which are required in the production of Sulphuric Acid by the contact process
Sulphide ores or Sulphur are burnt in air to form SO2.
By a reaction with oxygen, SO2 is converted into SO3 in the presence of V2O5 as a catalyst.
SO3 produced is absorbed on H2SO4 to give H2S2O7 (oleum).
This oleum is then diluted to obtain H2SO4of 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.
Q 22: How is SO2 an air pollutant?
Soln: The environment is harmed by sulphur dioxide in many ways:
Sulphuric acid is formed, when it is combined with water vapour present in the atmosphere. This causes acid that damages plants, soil, buildings (those made of marble are more prone), etc.
SO2 causes irritation in respiratory tract, throat, eyes and can also affect the larynx to cause breathlessness.
The colour of the leaves of the plant gets faded when it is exposed to sulphur dioxide for a long time. This defect is known as chlorosis. The formation of chlorophyll is affected by the presence of sulphur dioxide.
Q 23: Why are halogens strong oxidising agents?
Soln: Halogens have an electronic configuration of np5, where n =2 to 6. Thus, halogens require only one more electron to complete their octet and to attain the stable noble gas configuration. Moreover, halogens have high negative electron gain enthalpies and are highly electronegative with low dissociation energies. As a result, they have a high tendency to gain an electron. Hence, they act as strong oxidising agents.
Q 24: Explain why fluorine forms only one oxoacid, HOF.
Soln: Fluorine has high electronegativity and small size, hence it forms only one oxoacid i.e HOF.
Q 25: Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.
Ans : Oxygen has a smaller size and due to which a higher electron density per unit volume. Hence, oxygen forms hydrogen bonds while chlorine does not despite having similar electronegative values.
Q 26. Write two uses of ClO2. Ans : Applications of ClO2
( a )Used for purification of water.
( b ) Used for bleaching.
Q 27. Why are halogens coloured? Ans : Halogens are coloured because they take in radiations from the visible spectrum. This excites the valence electrons to a higher energy level. The amount of energy required for excitation differs from halogen to halogen, thus they exhibit different colors.
Q28. Write the reactions of F2 and Cl2 with water.
Ans: ( i ) Cl2 + H2O → HCL + HOCL
( ii ) 2 F2 + 2H2O → 4H+ + 4F- + O2
Q29. How can you prepare Cl2 from HCl and HCl from Cl2? Write reactions only Ans:
( i ) HCl is prepared from Cl2by reacting it with water.
Cl2 + H2O → HCL + HOCL
( ii ) Cl2 is prepared by Deacon’s process from HCL
4HCL + O2 → 2Cl2 + 2H2O
Q30. What inspired N. Bartlett for carrying out reaction between Xe and PtF6?
Ans: N.Barlett observed that PtF6 and O2 react to produce a compound O2+[ PtF6]–.
As the first ionization enthalpy of Xe( 1170 kJ/mol ) is very close to that of O2 , he figured that PtF6 could also oxidize Xe to Xe+. Thus, he reacted PtF6 and Xe to form a red coloured compound Xe+[ PtF6]– .
Q31. What are the oxidation states of phosphorus in the following:
( a ) H3PO3
( b ) PCl3
( c ) Ca3P2
( d ) Na3PO4
( e ) POF3?
Ans: Let the oxidation state of phosphorous be x
(a) H3PO3 3 + x + 3( -2) = 0
x -3 = 0
x + 3( -1) = 0
x = 3
(c) Ca3P2 3( 2 ) + 2 (x) = 0
2x = -6
x = -3
3( 1 ) + x + 4( -2 ) = 0
x -5 =0
x + ( -2 ) + 3( -1) = 0
x -5 = 0
x = 5
Q 32. Write balanced equations for the following:
(i) NaCl is heated with sulphuric acid in the presence of MnO2.
(ii) Chlorine gas is passed into a solution of NaI in water.
Q33. How are xenon fluorides XeF2, XeF4 and XeF6 obtained?
Ans: XeF2, XeF4 and XeF6are obtained through direction reactions between Xe and F2. The product depends upon the conditions of the reaction :
Xe + F2→ XeF2 (excess)
When Xe reacts with F2 under the condition of 673K and 1 bar XeF2 is produced.
Xe + 2F2→ XeF4
( 1:5 ratio )
When Xe reacts with F2 in the ratio of 1:5 under the condition of 873K and 7 bar XeF4 is produced.
Xe + 3F2 → XeF6
(1 : 20 ratio)
When Xe reacts with F2 in the ratio of 1:20 under the condition of 573K and 60-70 bar XeF6 is produced.
Q34. With what neutral molecule is ClO– isoelectronic? Is that molecule a Lewis base?
Ans: ClO– is isoelectronic with ClF.
Total electrons in ClO– = 17 + 8 + 1 =26
Total electrons in ClF = 17 + 9 = 26
As ClF accepts electrons from F to form ClF3 , ClF behaves like a Lewis base.
Q35. How are XeO3 and XeOF4 prepared? Ans: XeO3 can be obtained using two methods :
( 1 ) 6XeF4 + 12H2O → 4Xe + 2XeO3 + 24HF + 3O2
( 2 ) XeF6 + 3H2O → XeO3 + 6HF
XeOF4 is obtained using XeF6
XeF6 + H2O → XeOF4 + 2HF
Q36. Arrange the following in the order of property indicated for each set:
(i) F2, Cl2, Br2, I2 – increasing bond dissociation enthalpy.
(ii) HF, HCl, HBr, HI – increasing acid strength.
(iii) NH3, PH3, AsH3, SbH3, BiH3 – increasing base strength.
(1) Bond dissociation energy normally lowers on moving down a group because of increase in the atomic size. However, F2has a lower bond dissociation energy than Cl2 and Br2. This is because the atomic size of fluorine is very small.
Therefore, the increasing order for bond dissociation enthalpy is:
I2< F2< Br2< Cl2
(2) Bond dissociation energy of a H-X molecule ( where X = F, Cl, Br, I ) lowers with an increase in the size of an atom. As, H-I bond is the weakest it will be the strongest acid.
Therefore, the increasing order acidic strength is :
HF <HCl<HBr< HI
(3) BiH3≤ SbH3<AsH3< PH3< NH3
On moving from nitrogen to bismuth, the atomic size increases but the electron density of the atom decreases. Hence, the basic strength lowers.
Q37. Which one of the following does not exist?
(i) XeOF4 (ii) NeF2 (iii) XeF2 (iv) XeF6 Ans: The one that does not exist is NeF2.
Q38. Give the formula and describe the structure of a noble gas species which is isostructural with:
( a ) ICl4–
( b ) IBr2– ( c ) BrO3–
(a) XeF4 is isoelectronic to ICl4– . And it is square planar in geometry :
(b) XeF2 is isoelectronic with IBr2– . It has a linear structure.
(c)XeO3 is isoelectric and isostructural to BrO3–. It has a pyramidal structure.
Q39. Why do noble gases have comparatively large atomic sizes?
Noble gases have atomic radii that correspond to van der Waal’s radii. Whereas, other elements have a covalent radius. Now, by definition, van der Waal’s radii are bigger than covalent radii. This is the reason why noble gases have relatively bigger atomic sizes.
Q40. List the uses of neon and argon gases.
Ans: Uses of Argon gas:
(a) Argon is used to keep an inert atmosphere in high temperature metallurgical operations like arc welding.
(b) It is used in fluorescent and incandescent lamps where it is required to check the sublimation of the filament. Thereby, increasing the life of the lamp.
(c) Argon is used in laboratories to handle substances that are air-sensitive.
Uses of neon gas:
(a) Neon is filled in discharge tubes for advertising or decoration.
(b) Neon is used for making beacon lights.
(c) It is used alongside helium to protect electrical equipment against high voltage.
The NCERT Solutions for Class 12 Chemistry Chapter 7 deals with the chemistry of inorganic ring systems of the p-block elements has a long and venerable history that dates back to the early 19th century. The elements of Group 13, 14, 15, 16, 17 and 18 are called P block elements. They are represented by the general outer electronic configuration ns2np1-6. P block elements exist in all three physical states and maybe metals, non-metals or metalloids. Access the NCERT Solutions of this chapter to learn more in detail.
Class 12 Chemistry NCERT Solutions for Chapter 7 The p Block Elements
Chapter 7 p-Block Elements of Class 12 Chemistry is categorized under the term – I CBSE Syllabus for the session 2021-22. The NCERT Solutions for Class 12 are provided here for better understanding and clarification of the concepts. It is the best reference material when it comes to learning and understanding complex concepts/problems.
Subtopics of Class 12 Chemistry Chapter 7 – The P Block Elements
Group 15 Elements
Oxides of Nitrogen
Phosphorus — Allotropic Forms
Oxoacids of Phosphorus
Group 16 Elements
Sulphur — Allotropic Forms
Oxoacids of Sulphur
Group 17 Elements
Oxoacids of Halogens
Group 18 Elements
Nitrogen comprises 78% by volume of the atmosphere. Oxygen which is the most abundant element on earth forms 46.6% of the mass of the earth’s crust. Fluorine is present in insoluble fluorides. Seawater contains bromides, chlorides and iodides of potassium, calcium, magnesium and sodium. It is mainly sodium chloride solution. Learn more by accessing the NCERT Solutions for Class 12 Chemistry for free.
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By referring to the NCERT Solutions for Class 12 Chemistry Chapter 7, you can answer both intext and exercise questions. The intext questions will improve the understanding of students about a particular subsection whereas the exercise questions cover the topics present in the chapter. Students can download the NCERT Solutions available in the form of PDF and refer to them while solving the textbook questions to obtain a grip on the concepts. The solutions are present in both chapter-wise and exercise formats to help students in learning new concepts without any time constraints.
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The benefits of using the NCERT Solutions for Class 12 Chemistry Chapter 7 from BYJU’S are –
1. The solutions created are completely based on the latest term – I CBSE Syllabus and its guidelines.
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3. Students obtain a strong foundation of fundamental concepts which are important from the first term exam point of view.
4. By regular practice, students will be able to answer complex questions without any difficulty.
What are the important topics covered in Chapter 7 of NCERT Solutions for Class 12 Chemistry?
The important topics covered in Chapter 7 of NCERT Solutions for Class 12 Chemistry are –
1. Group 15 Elements
4. Oxides of Nitrogen
5. Nitric Acid
6. Phosphorus — Allotropic Forms
8. Phosphorus Halides
9. Oxoacids of Phosphorus
10. Group 16 Elements
12. Simple Oxides
14. Sulphur — Allotropic Forms
15. Sulphur Dioxide
16. Oxoacids of Sulphur
17. Sulphuric Acid
18. Group 17 Elements
20. Hydrogen Chloride
21. Oxoacids of Halogens
22. Interhalogen Compounds
23. Group 18 Elements