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

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Chapter7. The p Block Elements
TextbookChemistry I
CategoryNCERT Solutions for Class 12

NCERT Solutions for Class 12 Chemistry is an excellent tool to revise your syllabus quickly. These solutions are made by expert faculties and teachers of Chemistry. It provides you with complete information, making it easier for you to solve the problems. The NCERT textbook includes questions based on concepts and theories. They are important for testing students’ understanding of the topics. On this page, we have provided you with the NCERT Solutions for Class 12 Chemistry Chapter 7: The p Block Elements.

NCERT Solutions for Class 12 Chemistry Chapter 7

The p Block Elements Solutions

Q1) 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

  1. Electronic configuration: All the elements in group 15 have 5 valence electrons. Their general electronic configuration is ns2 np3.
  2. Oxidation states: All these elements have 5 valence electrons and require three more electrons to complete their octets. However, gaining electrons is very difficult as the nucleus will have to attract three more electrons. This can take place only with nitrogen as it is the smallest in size and the distance between the nucleus and the valence shell is relatively small. The remaining elements of this group show a formal oxidation state of −3 in their covalent compounds. In addition to the −3 state, N and P also show −1 and −2 oxidation states. 
    All the elements present in this group show +3 and +5 oxidation states. However, the stability of +5 oxidation state decreases down a group, whereas the stability of +3 oxidation state increases. This happens because of the inert pair effect.
  3. Ionization energy and electronegativity: First ionization decreases on moving down a group. This is because of increasing atomic sizes. As we move down a group, electronegativity decreases, owing to an increase in size.
  4. Atomic size: On moving down a group, the atomic size increases. This increase in the atomic size is attributed to an increase in the number of shells.

Q2) 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, N2. In N2, the two nitrogen atoms form a triple bond. This triple bond has very high bond strength, which is very difficult to break. It is because of nitrogen’s small size that it is able to form pπ−pπ bonds with itself. This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is more reactive than nitrogen.

Q3) Discuss the trends in chemical reactivity of group 15 elements.

Answer) General trends in chemical properties of group − 15

  1. Reactivity towards hydrogen: The elements of group 15 react with hydrogen to form hydrides of type EH3, where E = N, P, As, Sb, or Bi. The stability of hydrides decreases on moving down from NH3 to BiH3.
  2. Reactivity towards oxygen: The elements of group 15 form two types of oxides: E2O3 and E2O5, where E = N, P, As, Sb, or Bi. The oxide with the element in the higher oxidation state is more acidic than the other. However, the acidic character decreases on moving down a group.’
  3. Reactivity towards halogens: The group 15 elements react with halogens to form two series of salts: EX3 and EX5. However, nitrogen does not form NXas it lacks the d-orbital. All trihalides (except NX3) are stable.
  4. Reactivity towards metals: The group 15 elements react with metals to form binary compounds in which metals exhibit −3 oxidation states.

Q4) Why does NH3 form hydrogen bond but PH3 does not?

Answer) Nitrogen is highly electronegative as compared to phosphorus. This causes a greater attraction of electrons towards nitrogen in NH3 than towards phosphorus in PH3. Hence, the extent of hydrogen bonding in PH3 is very less as compared to NH3.

Q5) 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 ) + NaNO→N2(g) + 2H2O(l) + NaCl(aq)

NO and HNO3 are produced in small amounts. These are impurities that can be removed on passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate.

Q6) How is ammonia manufactured industrially?

Answer) Ammonia is prepared on a large-scale by the Haber’s process.

N2(g) + 3H2(g) ⇋ 2NH3(g)   ΔfHo = −46.1 kJ/mol

The optimum conditions for manufacturing ammonia are:

  1. Pressure (around 200 × 105 Pa)
  2. Temperature (700 K)
  3. Catalyst such as iron oxide with small amounts of Al2O3 and K2O

Q7) 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 concentration of the acid, temperature, and also on the material undergoing oxidation.

3Cu + 8HNO3( dil.) → 3Cu(NO3)+ 2NO + 4H2O
Cu + 4HNO3( conc. )  →Cu(NO3)2 + 2NO2 +2H2O

Q8) Give the resonating structures of NO2 and N2O5.


NCERT Solutions for Class 12 Chemistry Chapter 7

Q9) The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?

[Hint: Can be explained on the basis of sp3 hybridisation in NH3 and only s−p bonding between hydrogen and other elements of the group].


Hydride NH3 PH3 AsH3 SbH3
H−M−H angle 107° 92° 91° 90°

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

Q10) Why does R3P=O exist but R3N=O does not (R = alkyl group)?

Answer) N (unlike P) lacks the d-orbital. This restricts nitrogen to expand its coordination number beyond four. Hence, R3N=O does not exist.

Q11) Explain why NH3 is basic while BiH3 is only feebly basic.

Answer) NH3 is distinctly basic while BiH3 is feebly basic.

Nitrogen has a small size due to which the lone pair of electrons is concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density. Hence, the electron donating capacity of group 15 element hydrides decreases on moving down the group.

Q12) Nitrogen exists as diatomic molecule and phosphorus as P4. 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, N2. On moving down a group, the tendency to form pπ−pπ bonds decreases (because of the large size of heavier elements). Therefore, phosphorus (like other heavier metals) exists in the P4 state.

Q13) Write main differences between the properties of white phosphorus and red phosphorus.


White phosphorusRed Phosphorus
It is a soft and waxy solid. It possesses a garlic smell.It is a hard and crystalline solid, without any smell.
It is poisonous.It is non-poisonous.
It is insoluble in water but soluble in carbon disulphide.It is insoluble in both water and carbon disulphide.
It undergoes spontaneous combustion in air.It is relatively less reactive.
In both solid and vapour states, it exists as a P4 molecule.It exists as a chain of tetrahedral P4 units.

Q14) 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 nitrogen atom is smaller, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the N−N single bond.

Q15) Give the disproportionation reaction of H3PO3.

Answer) On heating, orthophosphorus acid (H3PO3) disproportionates to give orthophosphoric acid (H3PO4) and phosphine (PH3). The oxidation states of P in various species involved in the reaction are mentioned below.

4H3P+3O3→ 3H3P+5O4 + P-3H3

Q16) Can PCl5 act as an oxidising as well as a reducing agent? Justify.

Answer) PCl5 can only act as an oxidizing agent. The highest oxidation state that P can show is +5. In PCl5, phosphorus is in its highest oxidation state (+5). However, it can decrease its oxidation state and act as an oxidizing agent.

Q17) 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.

  1. Elements of group 16 have six valence electrons each. The general electronic configuration of these elements is ns2 np4, where n varies from 2 to 6.
  2. Oxidation state: As these elements have six valence electrons (ns2 np4), they should display an oxidation state of −2. However, only oxygen predominantly shows the oxidation state of −2 owing to its high electronegativity. It also exhibits the oxidation state of −1 (H2O2), zero (O2), and +2 (OF2). However, the stability of the −2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements of the group show an oxidation state of +2, +4, and +6 due to the availability of d-orbitals.
  3. Formation of hydrides: These elements form hydrides of formula H2E, where E = O, S, Se, Te, PO. Oxygen and sulphur also form hydrides of type H2E2. These hydrides are quite volatile in nature.

Q18) Why is dioxygen a gas but sulphur a solid?

Answer) Oxygen is smaller in size as compared to sulphur. Due to its smaller size, it can effectively form pπ−pπ bonds and form O2 (O==O) molecule. Also, the intermolecular forces in oxygen are weak van der Wall’s, which cause it to exist as gas. On the other hand, sulphur does not form M2 molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid.

Q19) Knowing the electron gain enthalpy values for O → O and O → O2− as −141 and 702 kJ mol−1 respectively, how can you account for the formation of a large number of oxides having O2− species and not O?

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

Answer) Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be. 

Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving O2− ion is much more than the oxide involving O ion. Hence, the oxide having O2− ions are more stable than oxides having O. Hence, we can say that formation of O2− is energetically more favourable than formation of O.

Q20) Which aerosols deplete ozone?

Answer) Freons or chlorofluorocarbons (CFCs) are aerosols that accelerate the depletion of ozone. In the presence of ultraviolet radiations, molecules of CFCs break down to form chlorine-free radicals that combine with ozone to form oxygen.

Q21) Describe the manufacture of H2SO4 by contact process?

Answer) Sulphuric acid is manufactured by the contact process. It involves the following steps:

  1. Sulphur or sulphide ores are burnt in air to form SO2
  2. By a reaction with oxygen, SO2 is converted into SO3 in the presence of V2O5 as a catalyst.
  3. SO3 produced is absorbed on H2SO4 to give H2S2O7 (oleum).

This oleum is then diluted to obtain H2SO4 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.

Q22) How is SO2 an air pollutant?

Answer) Sulphur dioxide causes harm to the environment in many ways:

  • It combines with water vapour present in the atmosphere to form sulphuric acid. This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble.
  • Even in very low concentrations, SO2 causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.
  • It is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves. This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of sulphur dioxide.

Q23) Why are halogens strong oxidising agents?

Answer) The general electronic configuration of halogens is np5, where n = 2-6. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron. Hence, they act as strong oxidizing agents.

Q24) Explain why fluorine forms only one oxoacid, HOF.

Answer) Fluorine forms only one oxoacid i.e., HOF because of its high electronegativity and small size.

Q25) Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.

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

Q26) Write two uses of ClO2

Answer) Two main Uses of ClO2:

  1. It is used for purifying water.
  2. It is used as a bleaching agent.

Q27) Why are halogens coloured?

Answer) Almost all halogens are coloured. This is because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region. Since the amount of energy required for excitation differs for each halogen, each halogen displays a different colour.

Q28) Write the reactions of F2 and Cl2 with water.


  1. Cl2 + H2O  → HCL + HOCL
  2. 2 F2 + 2H2O → 4H+ + 4F­- + O2

Q29) How can you prepare Cl2 from HCl and HCl from Cl2? Write reactions only.


  1. HCl is prepared from Cl2 by reacting it with water.
    Cl2 + H2O → HCL + HOCL
  2. Cl2 is prepared by Deacon’s process from HCL
    4HCL + O2 → 2Cl2 + 2H2O

Q30) What inspired N. Bartlett for carrying out reaction between Xe and PtF6?

Answer) N. Barlett observed that PtF6 and O2 react to produce a compound O2+[ PtF6].

Later, he realized that the first ionization energy of oxygen (1175 kJ/mol) and Xe (1170 kJ/mol) is almost the same. Thus, he tried to prepare a compound with Xe and PtF6. He was successful and a red-coloured compound, Xe+[ PtF6] was formed.

Q31) What are the oxidation states of phosphorus in the following:

(a) H3PO3
(b) PCl3
(c) Ca3P2
(d) Na3PO4
(e) POF3?

Answer) 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

x + ( -2 ) + 3( -1) = 0
x -5 = 0
x = 5

Q32) 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.


(i) 4NaCl + MnO+ 4H2SO4→ MnCl2 + 4NaHSO4 + 2H2O +Cl2
(ii) Cl2 + NaI → 2NaCl + I2

Q33) How are xenon fluorides XeF2, XeF4 and XeF6 obtained?

Answer) XeF2, XeF4, and XeF6 are obtained by a direct reaction between Xe and F2. The condition under which the reaction is carried out determines the product.

Xe + F2→   XeF2

When Xe reacts with F2 under the condition of 673K and 1 bar XeFis 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 XeFis produced.
Xe + 3F→   XeF6
(1 : 20 ratio)

When Xe reacts with F2 in the ratio of 1:20 under the condition of 573K and 60-70 bar XeFis produced.

Q34) With what neutral molecule is ClO isoelectronic? Is that molecule a Lewis base?

ClO− is isoelectronic to ClF. Also, both species contain 26 electrons in all as shown.

Total electrons ClO = 17 + 8 + 1 = 26
In ClF = 17 + 9 = 26

ClF acts like a Lewis base as it accepts electrons from F to form ClF3.

Q35) How are XeO3 and XeOF4 prepared?

Answer) XeOcan be obtained using two methods :

  • 6XeF4 + 12H2O  → 4Xe + 2XeO3 + 24HF + 3O2
  • 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.

Answer i) Bond dissociation energy usually decreases on moving down a group as the atomic size increases. However, the bond dissociation energy of F2 is lower than that of Cl2 and Br2. This is due to the small atomic size of fluorine. Thus, the increasing order for bond dissociation energy among halogens is as follows:
I2 < F2 < Br2 < Cl2

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

Answer iii) BiH3 ≤ SbH3 < AsH3 < PH3 < NH3
On moving from nitrogen to bismuth, the size of the atom increases while the electron density on the atom decreases. Thus, the basic strength decreases.

Q37) Which one of the following does not exist?

(i) XeOF4 (ii) NeF2 (iii) XeF2 (iv) XeF6

Answer) 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

Answer a) XeF4 is isoelectronic to ICl4 . And it is square planar in geometry:

NCERT Solutions for Class 12 Chemistry Chapter 7

Answer b) XeF2 is isoelectronic with IBr2 . It has a linear structure

NCERT Solutions for Class 12 Chemistry Chapter 7

Answer c) XeO3 is isoelectric and isostructural to BrO3. It has a pyramidal structure.

NCERT Solutions for Class 12 Chemistry Chapter 7

Q39) Why do noble gases have comparatively large atomic sizes?

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

Q40) List the uses of Neon and argon gases.


Uses of neon gasUses of Argon gas
It is mixed with helium to protect electrical equipments from high voltageArgon along with nitrogen is used in gas-filled electric lamps.
It is filled in discharge tubes with characteristic colours.It is usually used to provide an inert temperature in a high metallurgical process.
It is used in beacon lights.It is also used in laboratories to handle air-sensitive substances.

That’s it. These were the solutions of NCERT Class 12 Chemistry Chapter 7 – The p Block Elements. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.

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