NCERT Solutions for Class 12 Chemistry Chapter 7 – Free PDF Download
*According to the latest CBSE Syllabus 2023-24, this chapter has been removed.
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 mastering the subject. Also, these solutions assist you in understanding the concept in-depth by giving P Block Elements Class 12 questions and answers in the textbook, question papers and sample papers.
The P Block Elements of Class 12 Chemistry NCERT Solutions also comprise exemplar questions, exercises, Chemistry HOTS (Higher Order Thinking Skills), and worksheets to help students excel in their board exams. All the solutions are curated by subject experts at BYJU’S following the latest CBSE Syllabus 2023-24 and its marking schemes. Grab the free PDF of this chapter’s NCERT Solutions for Class 12 Chemistry from the link provided below.
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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.
\(\begin{array}{l}ns^{2}np^{3}\end{array} \)
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 to gain 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 an increase in atomic sizes. Moving down the group, electronegativity decreases due to an increase in size.
(iv) Atomic size: As we move down the group, atomic size increases. This increase in atomic size is attributed to an increase in the number of shells.
Q 2: Why does the reactivity of nitrogen differ from phosphorus?
Answer
Nitrogen is chemically less reactive. This is because of the high stability of its molecule,
\(\begin{array}{l}N_{2}\end{array} \)
. In \(\begin{array}{l}N_{2}\end{array} \)
, 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 \(\begin{array}{l}p\pi−p\pi\end{array} \)
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.
Answer
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
\(\begin{array}{l}EH_{3}\end{array} \)
, where E = N,P, As, Sb, or Bi. The stability of hydrides decreases on moving down from \(\begin{array}{l}NH_{3}\;to \;BiH_{3}\end{array} \)
.
(ii)Reactivity towards oxygen:
The elements of group 15 form two types of oxides:
\(\begin{array}{l}E_{2}O_{3}\end{array} \)
and \(\begin{array}{l}E_{2}O_{5}\end{array} \)
, 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: \(\begin{array}{l}EX_{3}\;and\;EX_{5}\end{array} \)
. However, nitrogen does not form \(\begin{array}{l}NX_{5}\end{array} \)
as it lacks the d-orbital. All trihalides (except \(\begin{array}{l}NX_{3}\end{array} \)
) 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:Â Why does \(\begin{array}{l}NH_{3}\end{array} \)
form hydrogen bond but \(\begin{array}{l}PH_{3}\end{array} \)
does not?Answer
When compared to phosphorus, nitrogen is highly electronegative. This results in a greater attraction of electrons towards nitrogen in
\(\begin{array}{l}NH_{3}\end{array} \)
than towards phosphorus in \(\begin{array}{l}PH_{3}\end{array} \)
. Hence, the extent of hydrogen bonding in \(\begin{array}{l}PH_{3}\end{array} \)
is very less as compared to \(\begin{array}{l}NH_{3}\end{array} \)
.
Q 5: How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved
Answer
An aqueous solution of ammonium chloride is treated with sodium nitrite.
NH4Cl (aq ) + NaNO2 →N2(g) + 2H2O(l) + NaCl(aq)
NO and
\(\begin{array}{l}HNO_{3}\end{array} \)
are produced in small amounts. These are impurities that can be removed by passing nitrogen gas through aqueous sulphuric acid containing potassium dichromate.
Q 6: How is ammonia manufactured industrially?
Answer:
Ammonia is prepared on a large scale by Haber’s process.
N2(g) + 3H2(g) ⇌ 2NH3(g)
The optimum conditions for manufacturing ammonia are:
(i) Pressure (around
\(\begin{array}{l}200 × 10^{5}\end{array} \)
Pa)
(ii) Temperature (700 K)
(iii) Catalyst such as iron oxide with small amounts of \(\begin{array}{l}Al_{2}O_{3}\end{array} \)
and \(\begin{array}{l}K_{2}O\end{array} \)
Â
Q 7: Illustrate how copper metal can give different products on reaction with HNO3.
Answer
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.
3Cu + 8HNO3( dil.) → 3Cu(NO3)2 + 2NO + 4H2O
Cu + 4HNO3( conc. ) →Cu(NO3)2 + 2NO2 +2H2O
Â
Q 8: Give the resonating structures of \(\begin{array}{l}NO_{2}\end{array} \)
and \(\begin{array}{l}N_{2}O_{5}\end{array} \)
.
Answer
NO2
N2O5
Â
Â
 Q 9: The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of \(\begin{array}{l}sp^{3}\end{array} \)
hybridisation in \(\begin{array}{l}NH_{3}\end{array} \)
and only s−p bonding between hydrogen and other elements of the group].
Answer
Hydride
\(\begin{array}{l}NH_{3}\;PH_{3}\;AsH_{3}\;SbH_{3}\end{array} \)
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 a 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 \(\begin{array}{l}R_{3}P=O\end{array} \)
exist but \(\begin{array}{l}R_{3}N=O\end{array} \)
does not (R = alkylgroup)?
Answer
N (unlike P) lacks the d-orbital. This restricts nitrogen from expanding its coordination number beyond four. Hence,
\(\begin{array}{l}R_{3}N=O\end{array} \)
does not exist.
Q 11: Explain why \(\begin{array}{l}NH_{3}\end{array} \)
is basic while \(\begin{array}{l}BiH_{3}\end{array} \)
is only feebly basic.
Answer
\(\begin{array}{l}NH_{3}\end{array} \)
is distinctly basic while \(\begin{array}{l}BiH_{3}\end{array} \)
is feebly basic.
Nitrogen has a small size, due to which the lone pair of electrons are 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.
Q 12: Nitrogen exists as a diatomic molecule and phosphorus as \(\begin{array}{l}P_{4}\end{array} \)
. Why?
Answer
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,
\(\begin{array}{l}N_{2}\end{array} \)
. 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 \(\begin{array}{l}P_{4}\end{array} \)
state.
Q 13: Write the main differences between the properties of white phosphorus and red phosphorus.
Answer :Â
Q 14: Why does nitrogen show catenation properties less than phosphorus?
Answer
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.
Q 15:Â Give the disproportionation reaction of \(\begin{array}{l}H_{3}PO_{3}\end{array} \)
.
Answer
On heating, orthophosphorus acid (
\(\begin{array}{l}H_{3}PO_{3}\end{array} \)
) disproportionates to give orthophosphoric acid (\(\begin{array}{l}H_{3}PO_{4}\end{array} \)
) and phosphine (\(\begin{array}{l}PH_{3}\end{array} \)
). The oxidation states of P in various species involved in the reaction are mentioned below.
4H3P+3O3→ 3H3P+5O4 + P-3H3
Q 16: Can \(\begin{array}{l}PCl_{5}\end{array} \)
act as an oxidising as well as a reducing agent? Justify.
Answer
\(\begin{array}{l}PCl_{5}\end{array} \)
 can only act as an oxidizing agent. The highest oxidation state that P can show is +5. In \(\begin{array}{l}PCl_{5}\end{array} \)
, 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.
Answer
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
\(\begin{array}{l}ns^{2}np^{4}\end{array} \)
, where n varies from 2 to 6
(ii) Oxidation state:
As these elements have six valence electrons (
\(\begin{array}{l}ns^{2}np^{4}\end{array} \)
), 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 (\(\begin{array}{l}H_{2}O_{2}\end{array} \)
), zero (\(\begin{array}{l}O_{2}\end{array} \)
), and +2 (\(\begin{array}{l}OF_{2}\end{array} \)
). 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 \(\begin{array}{l}H_{2}E\end{array} \)
, where E = O, S, Se, Te, PO. Oxygen and sulphur also form hydrides of type \(\begin{array}{l}H_{2}E_{2}\end{array} \)
. These hydrides are quite volatile in nature.
Q 18: Why is dioxygen a gas but sulphur a solid?
Answer
Oxygen is smaller in size when compared to sulphur. Since its size is small, it can form pπ−pπ bonds and form
\(\begin{array}{l}O_{2} (O=O)\end{array} \)
molecule. Also, the intermolecular forces in oxygen are weak van der Wall’s, which cause it to exist as a gas. On the other hand, sulphur does not form \(\begin{array}{l}M_{2}\end{array} \)
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 \(\begin{array}{l}702\;kJ\;mol^{−1}\end{array} \)
respectively, how can you account for the formation of a large number of oxides having \(\begin{array}{l}O^{2−}\; species\;and\; not\;O^{−}\end{array} \)
?
(Hint: Consider lattice energy factor in the formation of compounds).
Answer
The more the lattice energy of a compound, the more stable it will be. The 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 the formation of O–.
Q 20. Which aerosols deplete ozone?
Ans:
The aerosols which are responsible for the depletion of ozone are Freons or chlorofluorocarbons (CFCs).
The molecules of CFCs break down when there is the presence of ultraviolet radiations and form chlorine free radicals, which then combine with ozone to form oxygen.
Q 21. Describe the manufacture of \(\begin{array}{l}H_2 SO_4\end{array} \)
by contact process.
Ans:
The steps which are required in the production of Sulphuric Acid by the contact process
Step (1)
Sulphide ores or Sulphur are burnt in the air to form
\(\begin{array}{l}SO_2\end{array} \)
.
Step (2)
By a reaction with oxygen,
\(\begin{array}{l}SO_2\end{array} \)
is converted into \(\begin{array}{l}SO_3\end{array} \)
in the presence of \(\begin{array}{l}V_2 O_5\end{array} \)
as a catalyst.
\(\begin{array}{l}2 SO_{2\left ( g \right )} + O_{2\left ( g \right )} \overset{V_2 O_5}{\rightarrow} 2SO_{3 \left ( g \right )}\end{array} \)
Step (3)
\(\begin{array}{l}SO_3\end{array} \)
produced is absorbed on \(\begin{array}{l}H_2 SO_4\end{array} \)
to give \(\begin{array}{l}H_2 S_2 O_7\end{array} \)
(oleum).
\(\begin{array}{l}SO_3 + H_2 S_4 \rightarrow H_2S_2O_7\end{array} \)
This oleum is then diluted to obtain
\(\begin{array}{l}H_2 SO_4\end{array} \)
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.
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 the respiratory tract, throat, and 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:Â Fluorine has high electronegativity and a small size. Hence it forms only one oxoacid, i.e., HOF.
Q 25: Explain why in spite 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 colours.
Q28. Write the reactions of F2 and Cl2 with water.
Ans: ( i ) Cl2 + H2O → HCL + HOCL
( ii ) 2 F2 + 2H2O → 4H+ + 4FÂ- + O2
Q29. How can you prepare Cl2 from HCl and HCl from Cl2? Write reactions only
Ans:
( i ) HCl is prepared from Cl2 by 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 to carry 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
(b) PCl3
x + 3( -1) = 0
x = 3
(c) Ca3P2
3( 2 ) + 2 (x) = 0
2x = -6
x = -3
(d) Na3PO4
3( 1 ) + x + 4( -2 ) = 0
x -5 =0
x =5
(e)POF3
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.
Ans:
(a) 4NaCl + MnO2 + 4H2SO4→ MnCl2 + 4NaHSO4 + 2H2O +Cl2
(b) Cl2 + NaI → 2NaCl + I2
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.
Ans:
(1) Bond dissociation energy normally lowers on moving down a group because of increase in the atomic size. However, F2 has 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–
Ans:
(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?
Ans:
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.
Â
The NCERT Solutions for Class 12 Chemistry Chapter 7 deals with the chemistry of inorganic ring systems of the p-block elements and has a long and venerable history that dates back to the early 19th century. The elements of Groups 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 curated as per the CBSE Syllabus for 2023-24. 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
- Dinitrogen
- Ammonia
- Oxides of Nitrogen
- Nitric Acid
- Phosphorus — Allotropic Forms
- Phosphine
- Phosphorus Halides
- Oxoacids of Phosphorus
- Group 16 Elements
- Dioxygen
- Simple Oxides
- Ozone
- Sulphur — Allotropic Forms
- Sulphur Dioxide
- Oxoacids of Sulphur
- Sulphuric Acid
- Group 17 Elements
- Chlorine
- Hydrogen Chloride
- Oxoacids of Halogens
- Interhalogen Compounds
- Group 18 Elements
Nitrogen comprises 78% by the 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 a sodium chloride solution. Learn more by accessing the NCERT Solutions for Class 12 Chemistry for free.
Why Opt for BYJU’S?
NCERT Solutions is well known among students for the quality of learning it provides. Also, BYJU’S has some of the most reputed teachers in the country on its board, who teach the subjects in an interesting manner by using videos, animations and study materials. Students can ask doubts anytime, and our team of experts responds to your queries immediately.
BYJU’S also provides feedback on your preparation by giving assignments and exercises. This will help you stay ahead of the competition.
Frequently Asked Questions on NCERT Solutions for Class 12 Chemistry Chapter 7
Q1
What type of questions can I answer by referring to the NCERT Solutions for Class 12 Chemistry Chapter 7?
By referring to the NCERT Solutions for Class 12 Chemistry Chapter 7, you can answer both intext and exercise questions. The in-text 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.
Q2
What are the benefits of using the NCERT Solutions for Class 12 Chemistry Chapter 7 from BYJU’S?
The benefits of using the NCERT Solutions for Class 12 Chemistry Chapter 7 from BYJU’S are as follows:
1. The solutions created are completely based on the latest CBSE Syllabus and its guidelines.
2. The subject matter experts use simple and understandable language to help students grasp the concepts effectively.
3. Students obtain a strong foundation of fundamental concepts which are important from the exam point of view.
4. By regular practice, students will be able to answer complex questions without any difficulty.
Q3
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 the following:
1. Group 15 Elements
2. Dinitrogen
3. Ammonia
4. Oxides of Nitrogen
5. Nitric Acid
6. Phosphorus — Allotropic Forms
7. Phosphine
8. Phosphorus Halides
9. Oxoacids of Phosphorus
10. Group 16 Elements
11. Dioxygen
12. Simple Oxides
13. Ozone
14. Sulphur — Allotropic Forms
15. Sulphur Dioxide
16. Oxoacids of Sulphur
17. Sulphuric Acid
18. Group 17 Elements
19. Chlorine
20. Hydrogen Chloride
21. Oxoacids of Halogens
22. Interhalogen Compounds
23. Group 18 Elements
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