NCERT Exemplar for Class 12 Chemistry - The p-Block Elements - Free PDF Download
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Chemistry requires a lot of revision to make sure that students do not forget important formulas. The p-Block Elements is an important Chapter from Inorganic Chemistry that has a massive weightage in most Examinations. What makes this Chapter unique is that it is fairly easy to comprehend and understand and therefore makes it a perfect way for students to score more in their upcoming Exams.






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Chapter 7 – p- Block Elements
(Examples, Easy Methods and Step by Step Solutions)
I. Multiple Choice Questions (Type-I)
1. On addition of conc.
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (iii)
Hydrogen iodide is a more powerful reducing agent than sulphuric acid, it reduces the amount of iodine in the solution from
When chloride salts are treated with sulfuric acid,
Violet fumes are produced during the reaction due to the creation of iodine $(I2)$ gas.
2. In qualitative analysis when
(i) deep blue precipitate of
(ii) deep blue solution of
(iii) deep blue solution of
(iv) deep blue solution of
Ans. Correct option (ii)
When
3. In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present?
(i) 3 double bonds; 9 single bonds
(ii) 6 double bonds; 6 single bonds
(iii) 3 double bonds; 12 single bonds
(iv) Zero double bonds; 12 single bonds
Ans. Correct option is (iii)

4. Which of the following elements can be involved in
(i) Carbon
(ii) Nitrogen
(iii) Phosphorus
(iv) Boron
Ans. Correct option (iii)
If an electron pair is donated to a vacant orbital on one atom and a lone pair of electrons on the other, the bonding is designated
Because their valence shells lack d-orbitals, nitrogen, carbon, and boron cannot form a pi bond. Phosphorus, on the other hand, can produce pi bonds.
5. Which of the following pairs of ions are isoelectronic and isostructural?
(i)
(ii)
(iii)
(iv)
Ans. Correct option (i)
The compound has the following number of electrons:
Hence,
6. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy?
(i)
(ii)
(iii)
(iv)
Ans. Correct option (i)
Fluorine is the most reactive element, forming hydrogen fluoride when it combines with hydrogen gas.
Due to the shorter bond length between the two atoms, it takes a lot of energy to break the link between hydrogen and fluorine.
As a result, as we move down the group, the bond dissociation energy drops.
7. Bond dissociation enthalpy of E—H (E = element) bonds is given below. Which of the compounds will act as the strongest reducing agent?
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (iv)
The E-H bond dissociation energy of
8. On heating with concentrated
(i) It is highly poisonous and smells like rotten fish.
(ii) It’s solution in water decomposes in the presence of light.
(iii) It is more basic than
(iv) It is less basic than
Ans. Correct option is (iii)
When heated with concentrated
As we progress from the top to the bottom in a group, the basic character of hydrides reduces
9. Which of the following acids forms three series of salts?
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (iii)
When it reacts with a metal atom, it can produce three different forms of salts, referred known as the three series of salts.
10. Strong reducing behaviour of
(i) Low oxidation state of phosphorus
(ii) Presence of two –OH groups and one P–H bond
(iii) Presence of one –OH group and two P–H bonds
(iv) High electron gain enthalpy of phosphorus
Ans. Correct option is (iii)
The existence of a P-H link gives phosphorus oxyacids their reducing characteristics. It has a strong proclivity for releasing protons. As a result, it demonstrates a decreasing nature.
11. On heating lead nitrate forms oxides of nitrogen and lead. The oxides formed are ______.
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (ii)
On heating, lead nitrate decomposes into lead monoxide, nitrogen dioxide, and oxygen.
12. Which of the following elements does not show allotropy?
(i) Nitrogen
(ii) Bismuth
(iii) Antimony
(iv) Arsenic
Ans. Correct option is (i)
The single N-N bond is weak because of high interelectronic repulsion of the non-bonding electrons, owing to the small bond length. As a result the catenation tendency is weaker in nitrogen that is why it does not show allotropy.
13. Maximum covalency of nitrogen is ______________.
(i) 3
(ii) 5
(iii) 4
(iv) 6
Ans. Correct option is (ii)
The valence electrons of nitrogen are two 2s and three 2p. Nitrogen can make three covalent bonds by sharing its three 2p electrons. Consider the
As a result, nitrogen can create four different bonds.
14. Which of the following statements is wrong?
(i) Single N–N bond is stronger than the single P–P bond.
(ii)
(iii)
(iv) Covalency of nitrogen in
Ans. Correct option is (i)
The N-N sigma bond (single bond) is weaker than the P-P sigma bond (single bond) due to the smaller size of nitrogen which increases the interelectronic repulsions and thus making the bond weaker.
15. A brown ring is formed in the ring test for NO3- ion. It is due to the formation of
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (i)
The 'Brown Ring Test' is an identification and conformation test of the nitrate ion
16. Elements of group-15 form compounds in +5 oxidation state. However, bismuth forms only one well characterised compound in +5 oxidation state. The compound is
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (ii)
Bismuth is the sole element that has an inert pair effect. Only generates trihalides and has a +3 oxidation state. Florine, on the other hand, is small and has a high electronegativity.
17. On heating ammonium dichromate and barium azide separately we get
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (i)
We get
18. In the preparation of
will be ______.
(i) 2
(ii) 3
(iii) 4
(iv) 6
Ans. Correct option is (i)
The catalytic oxidation of ammonia produces
19. The oxidation state of central atom in the anion of compound
(i) +3
(ii) +5
(iii) +1
(iv) –3
Ans. Correct option is (iii)
20. Which of the following is not tetrahedral in shape?
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (iii)
The C-atom in the

21. Which of the following are peroxoacids of sulphur?
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (i)
Peroxy linkage refers to the presence of a bond between oxygen and oxygen (O-O) in a molecule.
Only

22. Hot conc.
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (iii)
Conc.
23. A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When excess of this gas reacts with
(i) – 3 to +3
(ii) – 3 to 0
(iii) – 3 to +5
(iv) 0 to – 3
Ans. Correct option is (iv)
Greenish yellow color
24. In the preparation of compounds of Xe, Bartlett had taken
(i) both
(ii) both
(iii) both
(iv) both
Ans. Correct option is (iii)
Bartlett noticed that Xenon's first ionisation potential is nearly identical to that of oxygen and used Born-Haber calculations to predict the presence of a stable compound:
25. In solid state
(i) covalent solid
(ii) octahedral structure
(iii) ionic solid with
(iv) ionic solid with
Ans. Correct option is (iv)
The ionic bonding promotes the crystalline structure of

26. Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.
Ion | |||
Reduction potential |
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (iii)
The ability of a substance to be decreased is known as its reduction potential. As the substance's reduction potential rises, so does its power to reduce. It signifies the chemical's oxidising power (the ability of the material to cause other substances to lose electrons, which is known as oxidising power) grows.
As a result, the oxidising power is listed in decreasing order
27. Which of the following is an isoelectronic pair?
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (ii)
52 electrons are present in both $BrO_{2}^{-\text{ }}BrF_{2}$
II. Multiple Choice Questions (Type-II)
Note : In the following questions two or more options may be correct.
28. If chlorine gas is passed through hot
(i) 0 to +5
(ii) 0 to +3
(iii) 0 to –1
(iv) 0 to +1
Ans. Correct option is (i) and (iii)
Chlorine gas oxidation number ranges from 0 to –1 to 0 to +5.
29. Which of the following options are not in accordance with the property mentioned against them?
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (ii), (iii)
30. Which of the following is correct for
(i) It has 6 lone pairs of electrons.
(ii) It has six P–P single bonds.
(iii) It has three P–P single bonds.
(iv) It has four lone pairs of electrons.
Ans. Correct option is (ii) and (iv)
Tetrahedral geometry characterises the structure of a white phosphorus molecule.
Phosphorus has a valency of 5, and there are four phosphorus atoms in total. As a result, there will be a total of
As a result, each P atom will have one lone pair of electrons, and each covalent bond (two electrons) will be established.
In a molecule of white phosphorus, there will be four lone pairs of electrons and six P-P single bonds in total.

31. Which of the following statements are correct?
(i) Among halogens, the radius ratio between iodine and fluorine is maximum.
(ii) Leaving the F—F bond, all halogens have weaker X—X bonds than X—X'bonds in interhalogens.
(iii) Among interhalogen compounds maximum number of atoms are present in iodine fluoride.
(iv) Interhalogen compounds are more reactive than halogen compounds.
Ans. Correct option is (i)
Due to its high electronegativity and tiny size, F is the only interhalogen with weaker X-X’ bonds and no X-X bonds.
32. Which of the following statements are correct for
(i) It acts as a bleaching agent in moist conditions.
(ii) It’s molecule has linear geometry.
(iii) It’s dilute solution is used as disinfectant.
(iv) It can be prepared by the reaction of dilute
Ans. Correct option is (i) and (iii)
33. Which of the following statements are correct?
(i) All the three N—O bond lengths in
(ii) All P—Cl bond lengths in
(iii)
(iv)
Ans. Correct option is (iii) and (iv)
The
34. Which of the following orders are correct as per the properties mentioned against each?
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (i) and (iv)
(i)
(ii)
(iii)
(iv)
35. Which of the following statements are correct?
(i) S–S bond is present in
(ii) In peroxosulphuric acid
(iii) Iron powder along with
(iv) Change in enthalpy is positive for the preparation of
Ans. Correct option is (i)
Oxidation state of
(iii) Iron oxide with
(iv) Change in enthalpy is negative for the preparation of
36. In which of the following reactions conc.
(i)
(ii)
(iii)
(iv)
Ans. Correct option is (ii) and (iii)

Here fig (b) contains one S-S bond.
The oxidising behaviour of
The oxidation state of sulphur's core atom drops from +6 to +4. It oxidises copper in (iii) and is reduced to
37. Which of the following statements are true?
(i) Only type of interactions between particles of noble gases are due to weak dispersion forces.
(ii) Ionisation enthalpy of molecular oxygen is very close to that of xenon.
(iii) Hydrolysis of
(iv) Xenon fluorides are not reactive.
Ans. Correct option is (i) and (ii)
(i) The modest dispersion force causes attraction in noble gases.
(ii) In nature, xenon fluorides are reactive.
III. Short Answer Type
38. In the preparation of
Ans.
39. Write a balanced chemical equation for the reaction showing catalytic oxidation of
Ans.
40. Write the structure of pyrophosphoric acid.
Ans.

41.
Ans.
42. In
Ans. Three P—Cl bonds are arranged in one plane at a

43. Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic?
Ans.
44. Give reason to explain why
Ans. Chlorine has vacant d-orbitals, which become excited upon bonding when electrons from the 3p-orbital are promoted to the 3d-orbital, giving it a covalency of three.
Due to the lack of unoccupied d-orbitals in the second energy shell, fluorine cannot expand its octet. As a result, it can only have one covalency.
45. Out of
Ans. Oxygen is more electronegative than sulphur, the bond angle of
46.
Ans. Six F atoms can be accommodated around sulphur due to fluorine's tiny size, however the chloride ion is comparably bigger, resulting in interatomic repulsion.
47. In reaction with
Ans. A is
B is
Hydrolysis of the following product
48. In the ring test of
Ans.
49. Explain why the stability of oxoacids of chlorine increases in the order given below:
Ans. Oxygen is more electronegative than chlorine, the negative charge on chlorine disperses from C
As the stability of the conjugate base rises, the acidic strength of the corresponding acid increases in the following order:
50. Explain why ozone is thermodynamically less stable than oxygen
Ans. Due its disintegration into oxygen results in the release of heat
51.
Ans.
Next is neutralization
On addition of 2 equations
The product formed is neutralized 8 moles of
Molarity of
52. White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of
Ans.
On adding eq. (i) and (ii)
Mass of
53. Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.
Ans. Three oxoacids are
54. Nitric acid forms an oxide of nitrogen on reaction with
Ans. On interaction with

55. Phosphorus has three allotropic forms — (i) white phosphorus (ii) red phosphorus and (iii) black phosphorus. Write the difference between white and red phosphorus since their structure and reactivity.
Ans. The crystal structure of red phosphorus features a sophisticated network of bonding, whereas white phosphorus is made up of

56. Give an example to show the effect of concentration of nitric acid on the formation of oxidation products.
Ans. When nitric acid is mixed with copper metal, it produces distinct oxidation products.
57.
Ans.
58. Phosphorus forms several oxoacids. Out of these oxoacids phosphinic acid has strong reducing properties. Write its structure and write a reaction showing its reducing behaviour
Ans. The following reaction with silver nitrate demonstrates phosphinic acid reducing behaviour:
IV. Matching Type
Note : Match the items of Column I and Column II in the following questions.
59. Match the compounds given in Column I with the hybridisation and shape given in Column II and mark the correct option

Column I | Column II |
(A) | (1) |
(B) | (2) |
(C) | (3) |
(D) | (4) |
Code :
(i) A (1) | B (3) | C (4) | D (2) |
(ii) A (1) | B (2) | C (4) | D (3) |
(iii) A (4) | B (3) | C (1) | D (2) |
(iv) A (4) | B (1) | C (2) | D (3) |
Ans. Correct option is (i)
In
In
In
In
60. Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.
Column I | Column II |
(A) | (1) Neutral oxide |
(B) | (2) Acidic oxide |
(C) | (3) Basic oxide |
(D) | (4) Mixed oxide |
Code :
(i) A (1) | B (2) | C (3)0 | D (4) |
(ii) A (4) | B (1) | C (2) | D (3) |
(iii) A (3) | B (2) | C (4) | D (1) |
(iv) A (4) | B (3) | C (1) | D (2) |
Ans. Correct option is (ii)
A.
B.
C.
D.
61. Match the items of Columns I and II and mark the correct option.
Column I | Column II |
(A) | (1) Highest electron gain enthalpy |
(B) | (2) Chalcogen |
(C) | (3) Tear gas |
(D) Sulphur | (4) Storage batteries |
Code:
(i) A (4) | B (3) | C (1) | D (2) |
(ii) A (3) | B (4) | C (1) | D (2) |
(iii) A (4) | B (1) | C (2) | D (3) |
(iv) A (2) | B (1) | C (3) | D (4) |
(i) A (4) | B (3) | C (1) | D (2) |
(ii) A (3) | B (4) | C (1) | D (2) |
(iii) A (4) | B (1) | C (2) | D (3) |
(iv) A (2) | B (1) | C (3) | D (4) |
Ans. Correct option is (i)
62. Match the species given in Column I with the shape given in Column II and mark the correct option.
Column I | Column II |
(A) | (1) Tetrahedral |
(B) | (2) Pyramidal |
(C) | (3) Seesaw shaped |
(D) | (4) Bent T-shaped |
Code:
(i) A (3) | B (2) | C (1) | D (4) |
(ii) A (3) | B (4) | C (2) | D (1) |
(iii) A (1) | B (2) | C (3) | D (4) |
(iv) A (1) | B (4) | C (3) | D (2) |
Ans. Correct option is (ii)
63. Match the items of Columns I and II and mark the correct option.
Column I | Column II |
(A) Its partial hydrolysis does not change oxidation state of central atom | (1) |
(B) It is used in modern diving apparatus | (2) |
(C) It is used to provide inert atmosphere for filling electrical bulbs | (3) |
(D) Its central atom is in | (4) |
Code:
(i) A (1) | B (4) | C (2) | D (3) |
(ii) A (1) | B (2) | C (3) | D (4) |
(iii) A (2) | B (1) | C (4) | D (3) |
(iv) A (1) | B (3) | C (2) | D (4) |
Ans. Correct option is (iii)
V. Assertion and Reason Type
Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(i) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
(iii) Assertion is correct, but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Both assertion and reason are wrong statements.
64. Assertion :
Reason : Nitrogen has more electron gain enthalpy than phosphorus.
Ans. Correct option is (iii)
Nitrogen gas has a complete octet structure for both atoms and is unreactive due to the presence of a strong triple bond. On the other hand, phosphor has bonds with unstable angles strains compared to nitrogen, therefore it burns quickly, thus readily reacts. Nitrogen has a lower electron gain enthalpy than phosphorus.
65. Assertion :
Reason :
Ans. Correct option is (iii)
66. Assertion :
Reason :
Ans. Correct option is (ii)
Because
67. Assertion : Both rhombic and monoclinic sulphur exist as
Reason : Oxygen forms pπ – pπ multiple bonds due to small size and small bond length but
Ans. Correct option is (i)
The rhombic and the monoclinic forms of sulphur exist as
68. Assertion :
Reason :
Ans. Correct option is (i)
When
69. Assertion :
Reason : Six F atoms in
Ans. Correct option is (i)
Here, the
VI. Long Answer Type
70. An amorphous solid “A” burns in air to form a gas “B” which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous
Ans. Gas 'B' is produced as a by-product during the roasting of sulphide, so gas 'B' must be
Since gas 'B' is formed when amorphous solid 'A' burns in air, amorphous solid 'A' must be sulphur,
71. On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with
Ans. Here, ‘A’ is
‘B’ is
‘C’ is
A brown gas is produced when lead nitrate(II) is heated

72. On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen (
Ans.
NCERT Exemplar is a great addition to a student’s preparation. Many times, questions appear directly from the textbook not only in school-level Exams but also in competitive Exams. This pattern has occurred several times over and over again and has led to students putting a lot of faith in the NCERT Exemplar. By using the solutions from Vedantu that cover all the important solutions to every single question from the Chapter The p-Block Elements, students can study the Chapter in a more effective way.
FAQs on NCERT Exemplar for Class 12 Chemistry Chapter-7 (Book Solutions)
1. Do the p-Block Elements have a high weightage in the Class 12 CBSE Examinations?
The p-Block Element is the Chapter that carries the most marks in the Class 12 CBSE board Examinations. Fortunately, it is not a very difficult topic to study and understand. By reading through the basic theory and understanding everything that the Chapter wants to convey, students can score full marks in The p-Block Elements in their Class 12 Board Examinations. It is recommended to not skip this Chapter at any cost as it has a very high weightage.
2. Should I use the NCERT Exemplar solutions (Class 12) for Chapter 7 The p-Block Elements if I’m preparing for JEE?
JEE Mains and Advance test their students on the basis of their subject understanding and clarity of concepts. The Exam has some really complicated questions from several Chapters but a huge chunk of the questions are considered to be easy. The NCERT Exemplar is a great tool as many times, questions appear directly from the book in the Exams, even in JEE. So yes, by using this free solution PDF, you can get good marks in JEE.
3. How do I use the free solution PDF of NCERT Exemplar for Chapter 7 The p-Block Element?
After downloading the free Solution PDF from Vedantu, students are expected to use it as a way to revise the main content of the Chapter. These solutions can be referred to at any moment before the Exam and be used as an extra asset in the overall preparation. Use these solutions to check the answers that you write in your mock tests so that you can change your original answer to an answer that can guarantee you better marks on the test.
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Doubts make us better learners. Every doubt is important and must be solved immediately. Vedantu offers easy doubt clearing sessions that allow students to continue their preparation without overthinking on previous doubts. In order to fill the gaps in the basic concepts of the Chapter The p-Block Element for students of Class 12, watch this video from Vedantu to go through the entire core ideas from the Chapter The p-Block Elements. All the basic concepts of the Chapter are explained properly and easily in the video.
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