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NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p block elements

NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p block elements

Edited By Shivani Poonia | Updated on Jun 25, 2025 05:04 PM IST | #CBSE Class 12th
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CBSE Class 12th  Exam Date : 15 Jul' 2025 - 15 Jul' 2025

Do you know what phosphorus means? Phosphorus glows on slow oxidation in the presence of air. It is present in the p-block of the modern periodic table. The elements present in the p-block have their last electron filled in the p-orbital. Because of the presence of different types of elements, like metals, non-metals and metalloids, the p-block is so versatile.

This Story also Contains
  1. NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: MCQ (Type 1)
  2. NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: MCQ (Type 2)
  3. NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Short Answer Type
  4. NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Matching Type
  5. NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Assertion and Reason Type
  6. NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Long Answer Type
  7. NCERT Exemplar Class 12 Chemistry Chapter 7: Higher Order Thinking Skills (HOTS) Questions
  8. Approach to Solve Questions of Class 12 Chemistry Chapter 7
  9. NCERT Exemplar Class 12 Chemistry Chapter 7: Important Trends
  10. Topics and Subtopics Covered in the NCERT Exemplar Class 12 Chemistry Chapter 7
  11. NCERT Exemplar Class 12 Chemistry Chapter 7: Learning Outcomes
  12. NCERT Exemplar Solutions Class 12 Chemistry Chapter-Wise
  13. NCERT Exemplar Solutions Class 12 Subject-Wise
  14. NCERT Solutions for Class 12 Chemistry Chapter-Wise
  15. NCERT Solutions for Class 12 Subject-Wise
  16. NCERT Notes Subject-Wise
  17. NCERT Books and NCERT Syllabus
NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p block elements
NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p block elements

This chapter has a discussion on the periodic trends, chemical and physical properties of the p-block elements, electronic configuration, and the chemistry of the important compounds of the p-block. It discusses the different allotropes of elements, the inert pair effect, electronegativity, and ionisation energy.

The NCERT Exemplar Solutions of p-block elements are designed by our subject matter experts, which ensures that the students understand key concepts by providing detailed explanations and easy-to-follow solutions. The NCERT Exemplar Solutions of Class 12 Chemistry strengthen their problem-solving skills and help them memorise and understand chemical reactions, reaction mechanisms, and chemical properties. The solutions will help the student prepare effectively for competitive exams like NEET, JEE Mains, CBSE Board exams. To enhance conceptual understanding and thinking, this article includes selected HOTS questions that strengthen students problem-solving abilities.

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NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: MCQ (Type 1)


All the MCQ (type 1) questions with solutions are given below:

Question 1. On addition of conc. H2SO4 to a chloride salt, colourless fumes are evolved but in case of an iodide salt, violet fumes come out. This is because
(i) H2SO4 reduces HI to I2
(ii) HI is of violet colour
(iii) HI gets oxidised to I2
(iv) HI changes to HIO3
Answer:

The answer is the option (iii). When concentrated H2SO4 is added to iodine salt, HI, which is a strong reducing agent, is formed. During the reaction, it is oxidized to I2 , which is violet in colour.
2NaCl + H2SO4Na2SO4+2HCl
In case of iodine the halogen acid HI obtained reduces H2SO4 to SO2 and itself is oxidised to free iodine.
2NaI + H2SO4Na2SO4+2HIH2SO42H2O + SO2+I2

Question 2. In qualitative analysis when H2S is passed through an aqueous solution of salt acidified with dil.HCl, a black precipitate is obtained. On boiling the precipitate with dil. HNO3, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives _________.
(i) a deep blue precipitate of Cu(OH)2
(ii) a deep blue solution of [Cu(NH3)4]2+
(iii) a deep blue solution of Cu(NO3)2
(iv) a deep blue solution of Cu(OH)2.Cu(NO3)2
Answer:

The answer is the option (ii). Black precipitate of copper sulphide is formed on passing H2S through acidified solution of salt. On boiling with dilute HNO3, the precipitate gives blue colour of copper nitrate. On addition of excess aqueous solution of ammonia, deep blue coloured compound is found. The reactions involved are:

Cu2++H2SCuS(black)+2H+


CuS+dil.HNO3Cu(NO3)2


Cu(NO3)2+4NH3[Cu(NH3)4]2+(Deepblue)

Question 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

Answer:

The answer is the option (i) 3 double bonds; 9 single bonds
In Cyclotrimetaphosphoric acid, there are 3 double bonds and 9 single bonds as present

Question 4. Which of the following elements can be involved in pπdπ bonding?
(i) Carbon
(ii) Nitrogen
(iii) Phosphorus
(iv) Boron
Answer:

The answer is the option (iii) Phosphorus
The involvement of phosphorous in pπ-dπ bonding is due to presence of vacant d orbitals. The elements C, N, and B do not have vacant d orbitals.

Question 5. Which of the following pairs of ions are isoelectronic and isostructural?
(i) CO32,NO3
(ii) ClO3,CO32
(iii) SO32,NO3
(iv) ClO3,SO32
Answer:

The answer is the option (i).

Isoelectronic means that two ions or molecules have the same number of electrons. Isostructural means they have the same shape or geometry. The number of electrons in both the isoelectronic species is 32 and they exhibit sp2 hybridisation and thus trigonal planar structure.

Question 6. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have the highest bond dissociation enthalpy?
(i)HF
(ii)HCl
(iii)HBr
(iv)HI
Answer:

The answer is the option (i).
On moving down the group, there is an increase in the atomic radii and a decrease in bond dissociation enthalpy. That is why; HF should have the highest bond dissociation enthalpy.

Question 7. Bond dissociation enthalpy of EH(E=element) bonds are given below. Which of the compounds will act as the strongest reducing agent?

Compound
Δdiss(EH)/kJmol1

NH3
389

PH3
322

AsH3
297

SbH3

255

(i) NH3
(ii) PH3
(iii) AsH3
(iv) SbH3

Answer:

The answer is the option (iv). The bond dissociation enthalpy decreases down the group. SbH3 has the least bond enthalpy and thus acts as the strongest reducing agent.

Question 8. On heating with concentrated NaOH solution in an inert atmosphere of CO2, white phosphorus gives a gas. Which of the following statement is incorrect about the gas?
(i) It is highly poisonous and has smelled like rotten fish.
(ii) It’s a solution in water decomposes in the presence of light.
(iii) It is more basic than NH3
(iv) It is less basic than NH3
Answer:

The answer is the option (c)

P4+3NaOH+3H2OPH3(Phosphine)+3NaH2PO2
PH3 is less basic than NH3

Question 9. Which of the following acids forms three series of salts?
(i) H3PO2
(ii) H3BO3
(iii) H3PO4
(iv) H3PO3
Answer:

The answer is the option (iii) H3PO4
There are 3 ionisable H-atoms in H3PO4 as it has 3- OH groups.

Thus, it forms three series of salts are as follows: .
NaH2PO4,NaHPO4andNa3PO4.

Question 10. Strong reducing behaviour of H3PO2 is due to
(i) The low oxidation state of phosphorus
(ii) Presence of two OH groups and one PH bond
(iii) Presence of one OH group and two PH bonds
(iv) High electron gain enthalpy of phosphorus

Answer:

The answer is the option (iii). In the monobasic acid H3PO2, there are two PH bonds that impart reducing character to the acid.

Question 11. On heating lead, nitrate forms oxides of nitrogen and lead. The oxides formed are ______.
(i) N2O,PbO
(ii) NO2,PbO
(iii) NO,PbO
(iv) NO,PbO2
Answer:

The answer is the option (ii) 2Pb(NO3)22PbO+4NO2+O2

Question 12. Which of the following elements does not show allotropy?
(i) Nitrogen
(ii) Bismuth
(iii) Antimony
(iv) Arsenic
Answer:

The answer is the option (i). Nitrogen has weak N – N single bond because of high inter-electronic repulsion of non-bonding electrons. This is why, nitrogen does not exhibit allotropy.

Question 13. Maximum covalency of nitrogen is ______________.
(i) 3
(ii) 5
(iii) 4
(iv) 6

Answer:

The answer is the option (iii). Maximum covalency of nitrogen is four as it cannot extend its valency beyond four [NH4,R4N] due to absence of vacant d orbitals.

Question 14. Which of the following statements is wrong?
(i) Single NN bond is stronger than the single PP bond.
(ii) PH3 can act as a ligand in the formation of a coordination compound with transition elements.
(iii) NO2 is paramagnetic.
(iv) Covalency of nitrogen in N2O5 is four
Answer:

The answer is the option (i).Single PP bond is stronger than single NN bond, however, naturally occurring nitrogen forms p(pi)p(pi) multiple bonds and becomes much more stable than phosphorous.

Question 15. A brown ring is formed in the ring test for NO3 ion. It is due to the formation of
(i) [Fe(H2O)5(NO)]2+
(ii) FeSO4.NO2
(iii) [Fe(H2O)4(NO)2]2+
(iv) FeSO4.HNO3

Answer:

The answer is the option (i). On addition of freshly prepared solution of FeSO4 in a NO3 ion containing solution, a brown coloured complex is formed. This is also known as brown ring test of nitrate.


NO3+3Fe2++4H+NO+3Fe3++2H2O


[Fe(H2O)6]2++NO[Fe(H2O)5(NO)]2+(Brownring)+H2O

Hence, 2 moles of NO are produced from 2 moles of ammonia.

Question 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) Bi2O5
(ii) BiF5
(iii) BiCl5
(iv) Bi2S5
Answer:

The answer is the option (ii). Due to inert pair effect, bismuth generally exhibits +3 oxidation state. However, due to the small size and high electronegativity of the fluoride ion it forms the compound BiF3.

Question 17. On heating ammonium dichromate and barium azide separately we get
(i) N2 in both cases
(ii) N2 with ammonium dichromate and NO with barium azide
(iii) N2O with ammonium dichromate and N2 with barium azide
(iv) N2O with ammonium dichromate and NO2 with barium azide
Answer:

The answer is the option (i)
(NH4)2Cr2O7N2+Cr2O3+4H2O
Ba(N3)23N2+Ba

Question 18. In the preparation of HNO3, we get NO gas by catalytic oxidation of ammonia. The moles of NO produced by the oxidation of two moles of NH3 will be ______.
(i) 2
(ii) 3
(iii) 4
(iv) 6
Answer:

The answer is the option (i). In the process of preparation of nitric acid, catalytic oxidation of 2 moles of ammonia produces 2 moles of NO.
4NH3+5O2Δ4NO(g)+6H2O(l)
Hence, 2 moles of ammonia will produce 2 moles of NO.

Question 19. The oxidation state of a central atom in the anion of compound NaH2PO2 will be ______.
(i) +3
(ii) +5
(iii) +1
(iv) –3
Answer:

The answer is the option (c)

NaH2PO2:

(+1)+(+1×2)+x+(2×2)=0
x=+1

Question 20. Which of the following is not tetrahedral in shape ?
(i) NH4+
(ii) SiCl4
(iii) SF4
(iv) SO42
Answer:

The answer is the option (iii). SF4 is not tetrahedral in shape. It is a see saw structure.

Question 21. Which of the following are peroxoacids of sulphur?
(i) H2SO5andH2S2O8
(ii) H2SO5andH2S2O7
(iii) H2S2O7andH2S2O8
(iv) H2S2O6andH2S2O7
Answer:

The answer is the option (i).Peroxoacids of sulphur should contain one OO bond as shown below

Question 22. Hot conc. H2SO4 acts as a moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc. H2SO4 into two gaseous products?
(i) Cu
(ii) S
(iii) C
(iv) Zn
Answer:

The answer is the option (c)

[H2SO4H2O+SO2+O]×2
C+2OCO2
C+2H2SO4CO2+2SO2+2H2O
CO2 and SO2 are two gaseous products formed by oxidation of C by H2SO4

Question 23. A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When an excess of this gas reacts with NH3 an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from _________.
(i) – 3 to +3
(ii) – 3 to 0
(iii) – 3 to +5
(iv) 0 to – 3
Answer:

The answer is the option (i).

MnO2+4HClMnCl2+2H2O+Cl (greenish yellow gas)

on reaction of excess of Cl2 with NH3 we obtained NCl3 and HCl.


NH3+3Cl2NCl3+3HCl
O.S. (-3) O.S.(+3)

Question 24. In the preparation of compounds of Xe, Bartlett had taken O2+PtF6 as a base compound. This is because
(i) both O2 and Xe has the same size.
(ii) both O2 and Xe has the same electron gain enthalpy.
(iii) both O2 and Xe has almost same ionisation enthalpy.
(iv) both Xe and O2 are gases
Answer:

The answer is the option (iii). The ionization enthalpy of O2 is nearly equal to that of Xe, because of which Barlett chose O2 compounds as base.

Question 25. In solid state PCl5 is a _________.
(i) covalent solid
(ii) octahedral structure
(iii) ionic solid with [PCl6]+ octahedral and [PCl4] tetrahedra
(iv) ionic solid with [PCl4]+tetrahedral and [PCl6] octahedral
Answer:

The answer is the option (iv). Structure of PCl5 in solid state

Question 26.

Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.

Ion
Reduction potential E/V

ClO4
E=1.19V

IO4
E=1.65V

BrO4
E=1.74V

(i) ClO4>IO4>BrO4
(ii) IO4>BrO4>ClO4
(iii) BrO4>IO4>ClO4
(iv) BrO4>ClO4>IO4

Answer:

The answer is the option (iii).A higher reduction potential means that there is greater tendency to get reduced. Therefore, the order of oxidizing power is as follows
BrO4>IO4>ClO4

Question 27. Which of the following is isoelectronic pair?
(i) ICl2,ClO2
(ii) BrO2,BrF2+
(iii) ClO2,BrF
(iv) CN,O3
Answer:

The answer is the option (ii). Isoelectronic pair have same number of electrons
(i)ICl2=53+2×17

=87
ClO2=17+16

=33
(ii)BrO2=35+2×8+1

=52
BrF2+=35+9×21

=52
(iii)ClO2=17+16

=33
BrF=35+9

=44
(iv)CN=6+7+1

=14
O3=8×3

=24
Hence, option b is correct.

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: MCQ (Type 2)

All the MCQ (type 2) questions with solutions are given below:

Question 28. If chlorine gas is passed through a hot NaOH solution, two changes are observed in the oxidation number of chlorine during the reaction. These are ________ and _________.
(i) 0 to +5
(ii) 0 to +3
(iii) 0 to –1
(iv) 0 to +1
Answer:

The answer is the option (i) and (iii). The oxidation state changes from 0 to +5 and 0 to -1 respectively.

Question 29. Which of the following options are not in accordance with the property mentioned against them?
(i) F2>Cl2>Br2>I2 Oxidising power.
(ii) MI>MBr>MCl>MF Ionic character of metal halide.
(iii) F2>Cl2>Br2>I2 Bond dissociation enthalpy.
(iv) HI<HBr<HCl<HF Hydrogen-halogen bond strength.
Answer:

The answer is the option (ii, iii) The correct order of bond dissociation enthalpy is Cl2>Br2>F2>I2
There is an increase in the ionic character of metal halides with the electronegativity of halogen
MI>MBr>MCI>MF

Question 30. Which of the following is correct for P4 the molecule of white phosphorus?
(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.
Answer:

The answer is the option (ii, iv) Structure of P4 molecule can be represented as

There are 4 lone pairs of electrons over each P-atom. It has six P – P single bond.

Question 31. Which of the following statements are correct?
(i) Among halogens, radius ratio between iodine and fluorine is maximum.
(ii) Leaving FF bond, all halogens have weaker XX bond than XX bond in interhalogens.
(iii) Among interhalogen compounds, the maximum number of atoms are present in iodine fluoride.
(iv) Interhalogen compounds are more reactive than halogen compounds.
Answer:

The answer is the option (i, iii, iv)
Iodine is the largest halogen, while fluorine is the smallest. RIodineRfluorine is thus the maximum. A-B bond in interhalogen compound is weaker than A-A bond in halogen compounds; this also means that interhalogen compounds are more reactive.

Question 32. Which of the following statements are correct for SO2 gas?
(i) It acts as a bleaching agent in moist conditions.
(ii) Its molecule has linear geometry.
(iii) Its dilute solution is used as a disinfectant.
(iv) It can be prepared by the reaction of dilute H2SO4 with metal sulphide.
Answer:

The answer is the option (i, iii). SO2 is used as a bleaching agent in presence of moisture because of the reducing nature of SO2
SO2+2H2OH2SO4+2H
It is also used as a preservative, disinfectant and anti-chlor.

Question 33. Which of the following statements are correct?
(i) All the three N—O bond lengths in HNO3 are equal.
(ii) All P—Cl bond lengths in PCl5 molecule in the gaseous state are equal.
(iii) P4 molecule in white phosphorus have angular strain therefore white phosphorus is very reactive.
(iv) PCl is ionic in the solid-state in which cation is tetrahedral and the anion is octahedral.
Answer:

The answer is the option (iii, iv).
(i) All the N – O bond in HNO3 are not of the same bond length.
(ii) In gaseous state, all P – Cl bond lengths in PCl5 molecule are not equal. Axial bond is longer than equatorial bond.
(iii) White phosphorous is more reactive than the other solid phases because of angular strain in the P4 molecule under normal conditions.
(iv) In Solid state PCl5 exists as ionic solid in which cation is tetrahedral and anion is octahedral.
Cation[PCl4]+Anion[PCl6]

Question 34. Which of the following orders are correct as per the properties mentioned against each?
(i) As2O3<SiO2<P2O3<SO2 Acid strength.
(ii) AsH3<PH3<NH3 Enthalpy of vaporization.
(iii) S<O<Cl<F More negative electron gain enthalpy.
(iv) H2O>H2S>H2Se>H2Te Thermal stability.

Answer:

The answer is the option (i, iv) There is an increase in the acidic strength of the group as we move up the group or left to right along a period. As we move down the group, thermal stability of hydrides decreases.

Question 35. Which of the following statements are correct?
(i) S–S bond is present in H2S2O6
(ii) In peroxosulphuric acid (H2SO5) sulphur is in +6 oxidation state.
(iii) Iron powder along with Al2O3 and K2O is used as a catalyst in the preparation of NH3 by Haber’s process.
(iv) Change in enthalpy is positive for the preparation of SO3 by catalytic oxidation of SO2
Answer:

The answer is the option (i, ii)
In H2SO5, there is a peroxo-linkage.

O in peroxide linkage has oxidation state - 1

Question 36. In which of the following reactions conc. H2SO4 is used as an oxidising reagent?
(i) CaF2+H2SO4CaSO4+2HF
(ii) 2HI+H2SO4I2+SO2+2H2O
(iii) Cu+H2SO4CuSO4+SO2+2H2O
(iv) NaCl+H2SO4NaHSO4+HCl
Answer:

The answer is the option (ii, iii). Amongst the given options, (ii). and (iii). represent oxidizing behaviour of H2SO4. The oxidizing agent reduces itself in the process. In option b, it oxidises HI and gets reduced to SO2 .Here,

Question 37. Which of the following statements are true?
(i) The only type of interactions between particles of noble gases is due to weak dispersion forces.
(ii) Ionisation enthalpy of molecular oxygen is very close to that of xenon.
(iii) Hydrolysis of XeF6 is a redox reaction.
(iv) Xenon fluorides are not reactive.
Answer:

The answer is the option (i, ii). Weak dispersion forces are responsible for attraction in noble gases. Ionization enthalpy of molecular oxygen and xenon are very similar. Also, Xenon fluorides are reactive in nature.

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Short Answer Type

All the short-answer type questions with solutions are given below:

Question 38. In the preparation of H2SO4 by Contact Process, why is SO3 not absorbed directly in water to form H2SO4 ?
Answer:

Formation of acid fog takes place which is difficult to condense.

Question 40. Write the structure of pyrophosphoric acid.

Answer:

H4P2O7 is pyrophosphoric acid. Its structure is

Question 41. PH3 forms bubbles when passed slowly in water but NH3 dissolves. Explain why?
Answer:

NH3 dissolves in water as it forms hydrogen bonds with water but PH3 cannot form hydrogen bonds with water so it escapes as gas.

Question 42. In PCl5, phosphorus is in sp3d hybridised state, but all its five bonds are not equivalent. Justify your answer with reason.
Answer:

PCl5 has a trigonal bipyramidal structure in liquid and gaseous phases.

The 3 equatorial P–Cl bonds are equivalent whereas 2 axial bonds are longer than equatorial bonds. This is because axial bond pairs suffer more repulsion compared to the equatorial bond pairs.

Question 43. Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic ?
Answer:

In the gaseous state, nitric oxide (NO) exists as a monomer having one unpaired electron as it is paramagnetic in nature.

However, in solid state it dimerises to N2O2 which is diamagnetic in nature.

Question 44. Give reason to explain why ClF3 exists but FCl3 does not exist.
Answer:

Though both ClandF are halogens, F has a configuration 2s22p5 and doesn’t have a d-orbital. Chlorine on the other hand can undergo sp3d-hybridisation, where three orbitals will have unpaired electrons as shown below. Three fluorine atoms with a single unpaired electron each can pair up with the sp3d-hybridised chlorine atom.

Question 45. Out of H2O and H2S, which one has a higher bond angle and why?
Answer:

Due to higher electronegativity of Oxygen than Sulphur, O-H bond will have a higher electron density closer to the Oxygen atom. The two O-H bonds will repel each other more strongly than S-H bond leading to a larger bond angle in H2O.

Question 46. SF6 is known but SCl6 is not. Why?
Answer:

There are two reasons as to why SF6 exists in nature, but SCl6 doesn’t:

  • Fluorine is smaller in size than Chlorine and is easily accommodated around Sulphur atom.

  • Due to higher electronegativity possessed by fluorine, it can unpair the Sulphur atom to the highest degree (where it attains the oxidation state of +6).

Question 47. On reaction with Cl2, phosphorus forms two types of halides ‘A’ and ‘B’. Halide A is a yellowish-white powder, but halide ‘B’ is a colourless oily liquid. Identify A and B and write the formulas of their hydrolysis products.
Answer:

On passing dry chlorine over white phosphorus, first colourless oily phosphorus trichloride is formed. PCl3 again reacts with Cl2 and forms yellowish white powder like phosphorus pentachloride PCl5

(i)

(ii)

PCl5

PCl3

Yellowish

Colorless Oily

White Powder

Liquid

PCl3 reacts violently with water forming

phosphosphorous acid (H3PO3)
PCl3+3H2OH3PO3+3HCl

PCl5 undergoes a violent hydrolysis
PCl5+H2OPOCl3+2HCl
PCl5+4H2OH3PO4+5HCl

Question 49. Explain why the stability of oxoacids of chlorine increases in the order given below:
HClO<HClO2<HClO3<HClO4

Answer:

As number of oxygen atoms around chlorine increases, the electron density moves away from chlorine. It is so because, oxygen atoms are more electronegative than chlorine atoms.
The stability of ions will also increase with a higher number of Oxygen atoms:
ClO<ClO2<ClO3<ClO4
Thus due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the following order
HClO<HClO2<HClO3<HClO4

Question 50. Explain why ozone is thermodynamically less stable than oxygen.
Answer:

Ozone is thermodynamically unstable with respect to oxygen because its decomposition results in liberation of heat (ΔH=ve) and increase in entropy (ΔH=+ve)
2O33O2
These 2 factors reinforce each other resulting in large negative Gibbs free energy change (ΔG=ve) for its conversion into oxygen.
Therefore, the high concentration of ozone can result into dangerous explosion

Question 51. P4O6 reacts with water according to equation P4O6+6H2O4H3PO3.
Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of P4O6 in H2O.

Answer:

P4O6+6H2O4H3PO3


4H3PO3+8NaOH4Na2HPO3+8H2O

(H3PO3 can be neutralised with NaOH )


P4O6(1mol)+8NaOH(8mol)4Na2HPO3+2H2O
Moles of P4O6=1.1220

=0.005mol


Acid formed by 1 mol of P4O6 Require 8 mol NaOH


Acid formed by 0.005 mol of P4O6 required =8×0.005

=0.04mol


0.1 M NaOH means 0.1 mol of NaOH is present in 1000 mL solution


0.4 M of NaOH is present in solution =10000.1×0.04

=400mL

Question 52. White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of HCl obtained by the hydrolysis of the product formed by the reaction of 62 g of white phosphorus with chlorine in the presence of water.
Answer:

Phosphorus reacts with Cl2 as :

P4+6Cl24PCl3


PCl3+3H2OH3PO3+3HCl]×4(PCl3ishydrolysed)


P4+6Cl2+12H2O4H3PO3+12HCl


Moles of white P=62124

=0.5mol


1 mol of white P4 produce HCl=12mol


0.5 mol of white P4 will produce HCl=12×0.5

=6mol


Mass of HCl=6×36.5

=219.0g

Question 53. Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.

Answer:

The 3 oxyacids of nitrogen are:

  1. Nitric acid (HNO3)

  2. Nitrous acid (HNO2)

  3. Hyponitrous acid (H2N2O2)

The disproportionation reaction wherein the nitrogen is in +3 oxidation state is:
3HNO2HNO3+2NO+H2O

+3+5+2
Oxidation state changes from +3 to +5 and +2

Question 54. Nitric acid forms an oxide of nitrogen on reaction with P4O10. Write the reaction involved. Also, write the resonating structures of the oxide of nitrogen formed.

Answer:

P4O10+4NHO34HPO3+2N2O5

Reactivity: White phosphorous are more reactive compared to the Red phosphorous as it undergoes more angular strain since the bonds are angled at 60°.

Question 55. Phosphorus has three allotropic forms — (i) white phosphorus (ii) red phosphorus and (iii) black phosphorus. Write the difference between white and red phosphorus on the basis of their structure and reactivity.
Answer:

White phosphorous exists as a single molecule where the phosphorous atoms are sp3-hybridised and has a tetrahedral structure with 6 P-P bonds. It is highly reactive due to the strained bond angles (60o).

Red Phosphorous on the other hands also has a tetrahedral structure, but each tetrahedron is linked to two others as shown below. The bond angle is much more relaxed, leading to a lower reactivity than White phosphorus.

Question 56. Give an example to show the effect of concentration of nitric acid on the formation of oxidation product.

Answer:

We get different oxidation products on reaction of dilute and concentrated nitric acid with copper metals.

For example,

zinc with very dilute

HNO3 ( 6%) forms NH4NO3,

with dilute

HNO3 ( 20%) forms N2O

and with concentration.

HNO3 ( 70%) forms NO2.


(Very dilute) HNO3:4Zn+10HNO34Zn(NO3)2+NH4NO3+3H2O


(Dilute) HNO3:4Zn+10HNO34Zn(NO3)2+N2O+5H2O


(Concentrated) HNO3:Zn+HNO3Zn(NO3)2+2NO2+2H2O

Question 57. PCl5 reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH3 solution. Write the reactions involved to explain what happens.

Answer:

PCl5 reacts with finely divided silver on heating to form white silver salt (AgCl). This dissolves on addition of excess aqueous ammonia to form soluble complex.
PCl5+2Ag2AgCl+PCl3
AgCl+2NH3(aq)[Ag(NH3)2]+Cl

Question 58. Phosphorus forms a number of oxoacids. Out of these oxoacidsphosphinic acid has strong reducing property. Write its structure and also write a reaction showing its reducing behaviour.
Answer:

One of the oxoacids of phosphorous that has strong reducing property is phosphinic acid or hypophosphorus acid (H3PO2). This is due to possession of P – H bonds.

Its reducing properties are :


4AgNO3+H3PO2+2H2O4Ag+H3PO4+4HNO3


2HgCl2+H3PO2+2H2O2Hg+H3PO4+4HCl


H3PO2+2H2O+2Cl2H3PO4+4HCl

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Matching Type

All the matching type questions with solutions are given below:

Question 59. Match the compounds given in Column I with the hybridization and shape given in Column II and mark the correct option.

Column I

Column II

(A)

XeF6

(1)

sp3d2 - distorted octahedral

(B)

XeO3

(2)

sp3d2-square planar

(C)

XeOF4

(3)

sp3 - pyramidal

(D)

XeF4

(4)

sp3d2-square pyramidal

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(a) A (1) B(3) C(4) D(2)
(b) A (1) B(2) C(4) D(3)
(c) A (4) B(3) C(1) D(2)
(d) A (4) B(1) C(2) D(3)

Answer:

(a) (A1),(B3),(C4),(D2)

Compound

Hybridisation

(i).

(1)

(sp3d2)- distorted octahedral

(ii).

(2)

(sp3) - pyramidal

(iii).

(3)

(sp3d2)- square pyramidal

(iv).

(4)

(sp3d2) - square planar

Question 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)

Pb3O4

(1)

Neutral Oxide

(B)

N2O

(2)

Acidic Oxide

(C)

Mn2O7

(3)

Basic Oxide

(D)

Bi2O3

(4)

Mixed Oxide

(a) A (1) B(2) C(3) D(4)
(b) A (4) B(1) C(2) D(3)
(c) A (3) B(2) C(4) D(1)
(d) A (4) B(3) C(1) D(2)

Answer:

(b). (A 4), (B 1), (C 2), (D 3)

Formulas

Type of Oxide

(i)

Pb3O4 (PbO.Pb2O3)

(4)

Mixed Oxide

(ii)

N2O

(1)

Neutral Oxide

(iii)

Mn2O7

(2)

Acidic Oxide

(iv)

Bi2O3

(3)

Basic Oxide

Question 61. Match the items of Column I and Column II and mark the correct option.

Column I

Column II

(A)

H2SO4

(1)

Highest electron gain enthalpy

(B)

CCl3NO2

(2)

Chalcogen

(C)

Cl2

(3)

Tear gas

(D)

Sulphur

(4)

storage batteries

(a) A (4) B(3) C(1) D(2)
(b) A (3) B(4) C(1) D(2)
(c) A (4) B(1) C(2) D(3)
(d) A (2) B(1) C(3) D(4)

Answer:

(a). (A 4), (B 3), (C 1), (D 2)

Question 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)

SF4

(1)

Tetrahedral

(B)

BrF3

(2)

Pyramidal

(C)

BrO3

(3)

Sea-Saw shaped

(D)

NH4+

(4)

Bent-T-shaped

(a) A (3) B(2) C(1) D(4)
(b) A (3) B(4) C(2) D(1)
(c) A (1) B(2) C(3) D(4)
(d) A (1) B(4) C(3) D(2)

Answer:

(b) (A 3), (B 4), (C 2), (D 1)

Species

Shape

Structure

SF4

Sea-Saw shaped

BrF3

Bent-T-shaped

BrO3

Pyramidal

NH4+

Tetrahedral

Question 63. Match the items of Column I and Column II and mark the correct option.

Column I

Column II

(A)

Its partial hydrolysis does not change oxidation state of central atom

(1)

He

(B)

It is used in modern diving apparatus

(2)

XeF6

(C)

It is used to provide inert atmosphere for filling electrical bulbs

(3)

XeF4

(D)

Its central atom is in sp3d2 hybridisation

(4)

Ar

(a) A (1) B(4) C(2) D(3)
(b) A (1) B(2) C(3) D(4)
(c) A (2) B(1) C(4) D(3)
(d) A (1) B(3) C(2) D(4)

Answer:

(c). (A 2), (B 1), (C 4), (D 3)

  1. Oxidation state of Xe is unaffected by partial hydrolysis of XeF6

  2. He is utilized in the modern driving apparatus.

  3. Air is used for filling electrical bulbs to provide inert atmosphere.

  4. Xe has 4 bond pair and 2 lone pair and a hybridisation state of sp3d2 in XeF4

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Assertion and Reason Type

All the assertion-reason type questions with solutions are given below:

Question 64. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (I): N2 is less reactive than P4
Reason (R): Nitrogen has more electron gain enthalpy than phosphorus.

(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but the reason is wrong.
(iv) Assertion is wrong but the Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is the option (iii). Nitrogen has a higher electron gain enthalpy than Phosphorous. N-N atoms in Nitrogen are bonded by triple bond, whereas Phosphorous atoms are bonded by single P-P bond, due to which Nitrogen has a high bond dissociation energy and is less reactive.

Question 65. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (I): HNO3 makes iron passive.
Reason (R): HNO3 forms a protective layer of ferric nitrate on the surface of iron.

(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is the option (iii). Iron attains passivity due to formation of a thin layer of oxide.

Question 66. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (I): HI cannot be prepared by the reaction of KI with concentrated H2SO4.
Reason (R): HI has lowest H – X bond strength among halogen acids.

(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is the option (ii). Though correct, both the statements are independent of each other.

Question 67. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (i): Both rhombic and monoclinic sulphur exist as S8 but oxygen exists as O2.
Reason (R): Oxygen forms pπpπ multiple bond due to small size and small bond length but pπpπ bonding is not possible in sulphur.

(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is the option (i). Sulphur cannot form pπpπ multiple bonds like oxygen, so it exists as S8.

Question 68. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (I): NaCl reacts with concentrated H2SO4 to give colourless fumes with pungent smell. But on adding MnO2 the fumes become greenish yellow.
Reason (R): MnO2 oxidises HCl to chlorine gas which is greenish yellow.

(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is the option (i). MnO2 is a strong oxidising agent and it oxidises colourless fumes of HCl to greenish yellow Cl2

Question 69. In the following question, a statement of Assertion (i). followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (i): SF6 cannot be hydrolysed but SF4 can be.
Reason (R): Six F atoms in SF6 prevent the attack of H2O on sulphur atom of SF6 .

(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is the option (i). H2O can’t react with SF6 because the fluorine atoms don’t allow H2O to attack SF6, and thus, is sterically protected.

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Long Answer Type

All the long-answer type questions with solutions are given below:

Question 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 KMnO4 solution and reduces Fe3+ to Fe+2. Identify the solid “A” and the gas “B” and write the reactions involved.

Answer:

A is sulphur (S8) while B is sulphur dioxide (SO2).
S8+8O28SO2


2ZnS+3O22ZnO+2SO2


2KMnO4+3H2SO4K2SO4+2MnSO4+3H2O+5[O]


[SO2+[O]+H2OH2SO4]×5


2KMnO4+3H2SO4K2SO4+2MnSO4+2H2SO4

SO2+2H2OH2SO4+2[H]


Fe2(SO4)3+2[H]2FeSO4+H2SO4


Fe2(SO4)3+SO2+2H2O2FeSO4+2H2SO4

NCERT Exemplar Class 12 Chemistry Chapter 7: Higher Order Thinking Skills (HOTS) Questions


Some higher-order thinking skills questions with solutions are given below:

Question 1. When a salt is treated with a sodium hydroxide solution, it gives gas X. On passing gas X through reagent Y, a brown-colored precipitate is formed. X and Y, respectively, are

  1. X=NH3 and Y=HgO
  2. X=NH3 and Y=K2HgI4+KOH
  3. X=NH4Cl and Y=KOH
  4. X=HCl and Y=NH4Cl

Answer:

NH4++NaOHH2O+NH3

NH3 is identify by K2[HgI4]+KOH

Hence, the correct answer is option (2).

Question 2. Evaluate the following statements related to group 14 elements for their correctness.
(A)Covalent radius decreases down the group from C to Pb in a regular manner.
(B) Electronegativity decreases from C to Pb down the group gradually.
(C) Maximum covalence of C is 4 whereas other elements can expand their covalence due to the presence of d orbitals.
(D) Heavier elements do not form pπpπ bonds.
(E) Carbon can exhibit negative oxidation states.

Choose the correct answer from the options given below:

  1. (C), (D) and (E) Only
  2. (A) and (B) Only
  3. (A), (B) and (C) Only
  4. (C) and (D) Only

Answer:

(A) Down the group; the radius increases
(B) Electronegativity does not decrease gradually from C to Pb.
(C) Correct.
(D) Correct.
(E) Range of oxidation state of carbon: -4 to +4

Hence, the correct answer is option (1).

Question 3. The species that do not contain peroixde ions is:

(1) PbO2

(2) H2O2

(3) SrO2

(4) BaO2

Answer:

All members, when heated with oxygen, form oxides. There are mainly two types of oxides, i.e., monoxide and dioxide of formula MO and MO2, respectively. SiO only exists at high temperatures. Oxides in higher oxidation states of elements are generally more acidic than those in lower oxidation states. The dioxides — CO2, SiO2, and GeO2 are acidic, whereas SnO2 and PbO2 are amphoteric in nature. Among monoxides, CO is neutral, GeO is distinctly acidic, whereas SnO and PbO are amphoteric.

In PbO2 , Pb is in +4 oxidation state and oxygen is in a -2 oxidation State. In all other cases, the peroxide ion (O22) is present.

Hence, the correct answer is option (1).

Approach to Solve Questions of Class 12 Chemistry Chapter 7

To effectively solve questions of the chapter "P-Block Element", a systematic and structured approach that helps understand the concepts and the theories is given below:

1. Focus on Key Concepts

Each group has similar themes. Focus on:

  • Electronic configuration

  • Oxidation states and variable valency

  • Trends in physical and chemical properties (electronegativity, metallic character, ionization enthalpy)

  • Important compounds (e.g., NH3, HNO3, SO2, H2SO4, PCl5, etc.)

  • Anomalous behaviour of the first element (like N or O)

  • Trends down the group (acidic/basic nature, thermal stability, etc.)

2. Use a Comparative Approach

When solving theory questions:

  • Compare elements down the group (e.g., N vs P vs As…)

  • Compare group-wise behaviours (Group 15 vs Group 16)

3. Focus on Structures and Hybridization

  • Practice how to identify molecular structures (e.g., NH3, SO2, PCl5, ClF3)

  • Understand types of hybridization (e.g., sp3d for PCl5)

4. Practice Conceptual and Memory-Based Questions

Examples:

  • Why is N2 less reactive than P4?

  • Write one use of H2SO4.

  • Compare the thermal stability of H2O, H2S, H2Se.

  • What is the oxidation state of Cl in ClO3?

5. Revise with Tables and Charts

  • Make group-wise comparison tables for similar trends

  • Prepare short flashcards for compounds, oxidation states, and trends.

6. Practice previous year questions and solve mock tests.

7. Make suitable short notes and revise them timely.

NCERT Exemplar Class 12 Chemistry Chapter 7: Important Trends

Some Important general trends in properties across the p-block are given below:

  • Electronegativity: Increases from left to right across a period and generally decreases down a group.
  • Ionization Enthalpy: Generally increases from left to right across a period and decreases down a group, with some exceptions.
  • Metallic Character: Decreases from left to right and increases down a group.
  • Atomic Radius: Generally increases down a group and decreases from left to right across a period.
  • Oxidation States: Elements in the p-block exhibit a variety of oxidation states, generally related to their group number such as, group 15 elements can show +3 and +5 oxidation states.

Topics and Subtopics Covered in the NCERT Exemplar Class 12 Chemistry Chapter 7

All the topics and subtopics covered in the NCERT exemplar are listed below:

  • Occurrence
  • Electronic Configuration
  • Atomic and Ionic Radii
  • Ionisation Enthalpy
  • Electronegativity
  • Physical Properties
  • Chemical Properties
  • Group 15 Elements (Nitrogen Family)
  • Dinitrogen
  • Ammonia
  • Oxides of Nitrogen
  • Nitric Acid
  • Phosphorus — Allotropic Forms
  • Phosphine
  • Phosphorus Halides
  • Oxoacids of Phosphorus
  • Group 16 Elements (oxygen Family)
  • Dioxygen
  • Simple Oxides
  • Ozone
  • Sulphur And Its Allotropic Forms
  • Sulphur Dioxide
  • Oxoacids of Sulphur
  • Sulphuric Acid
  • Group 17 Elements (halogen Family)
  • Chlorine
  • Hydrogen Chloride
  • Oxoacids of Halogens
  • Interhalogen Compounds
  • Group 18 Elements

NCERT Exemplar Class 12 Chemistry Chapter 7: Learning Outcomes

In Class 12 NCERT exemplar solutions chapter 7, the students learn about the different topics and subtopics covered in this chapter, which they need to know and understand. The p-block elements are found on the right side of the periodic table. The p-block is the region of the periodic table that includes columns 3A to 8A and does not include helium. P-block elements are the elements in which the last electron enters any of the three p-orbitals of their respective shells. P block elements are shiny and usually good conductors of electricity.

NCERT Exemplar Solutions Class 12 Chemistry Chapter-Wise

Class 12 NCERT exemplar chapter-wise solutions are given below:

NCERT Exemplar Solutions Class 12 Subject-Wise

Class 12 NCERT exemplar subject-wise solutions are given below:

NCERT Solutions for Class 12 Chemistry Chapter-Wise

Class 12 NCERT chemistry chapter-wise solutions are given below:

NCERT Solutions for Class 12 Subject-Wise

Class 12 NCERT subject-wise solutions are given below:

NCERT Notes Subject-Wise

Class 12 NCERT subject-wise notes are given below:

NCERT Books and NCERT Syllabus

The NCERT books and syllabus links for class 12 are given below:

Frequently Asked Questions (FAQs)

1. What types of elements are found in the p-block?

Types of elements are found in the p-block 

  • Metals: Located, lower left of the block (Aluminum, Gallium, Indium, Tin, Lead, and Bismuth).
  • Nonmetals: Located, upper right of the block (Carbon, Nitrogen, Oxygen, Fluorine, Sulfur, Chlorine, Bromine, and Iodine).
  • Metalloids: Elements with properties intermediate between metals and nonmetals (Boron, Silicon, Germanium, Arsenic, Antimony, and Tellurium).
  • Noble Gases: Group 18 ( He, Ne, Ar, Kr, and Xe).
2. How does the reactivity of p-block elements change across a period?
  • Reactivity of nonmetals increases from left to right across a period due to increasing electronegativity and tendency to gain electrons. 
  • Reactivity of metals decreases from left to right. Noble gases are largely unreactive.
3. How do the oxidation states of elements in Group 15 (Nitrogen group) vary?

Group 15 elements (Nitrogen, Phosphorus, etc.) can exhibit a range of oxidation states from -3 to +5.

4. What are the general trends of electronegativity and ionization enthalpy across the p-block?

The general trends of electronegativity and ionization enthalpy across the p-block are:

  • Electronegativity: Increases from left to right across a period and generally decreases down a group.
  • Ionization Enthalpy: Generally increases from left to right across a period and decreases down a group (with some exceptions).
5. What is the significance of the "staircase line" in the p-block?

The staircase line or diagonal line separating metals and nonmetals in the p-block. It is important because the elements immediately bordering this line are metalloids (Elements with properties intermediate between metals and nonmetals).

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Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
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    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.
  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
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Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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