Class 12 - Chemistry - P Block Elements

Question 7.1.

Discuss the general characteristics of Group 15 elements with reference to their electronic configuration,

oxidation state, atomic size, ionisation enthalpy and electronegativity.


The general characteristics of Group 15 elements are:

(i) Electronic configuration: All Group 15 elements have 5 electrons in their valence shell.

The general electronic configuration of these elements is ns2 np3.

(ii) Oxidation state: Group 15 elements have 5 valence electrons and they require 3 more electrons to complete their octet.

However, the gaining of 3 electrons is difficult

(iii) Atomic size: Atomic size increases as we move down the group due to increase in the number of shells.

(iv) Ionisation enthalpy: Ionisation enthalpy decreases as we move down the group because of increase in atomic sizes.

(v) Electronegativity: Electronegativity decreases on moving down the group due to increase in atomic radius.


Question 7.2.

Why does the reactivity of nitrogen differ from phosphorus?


Nitrogen is chemically less reactive. This is because of the high stability of its molecule, N2.

In N2, the two nitrogen atoms form a triple bond.

This triple bond has very high bond strength, which is very difficult to break.

It is because of nitrogen’s small size that it is able to form pπ−pπ bonds with itself.

This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is more reactive than nitrogen.


Question 7.3.

Discuss the trends in chemical reactivity of group 15 elements.


Chemical reactivity of group 15 elements towards hydrogen, oxygen, halogens, and metals are discussed below.

  • Reactivity towards hydrogen: Group 15 elements react with H to form hydrides of type EH3 where E=N, P, As, Sb or Bi.
  • On moving down from NH3 to BiH3, the stability of the hydrides decreases.
  • For example, the P-H bond in PH3 is less stable than the N-H bond in NH3.
  • The strength of the E-H bond gets weaker as the size of the central atom increases.

Stability order of E-H bond (where E is group 15 elements) can be represented as

N-H > P-H > As-H > Sb-H > Bi-H

  • Reactivity towards oxygen: Group 15 elements react with O to form oxides of the type E2O3 and E2O5 where E = N, P, As, Sb or Bi.
  • The oxide of the element with higher oxidation state is more acidic than those with lower oxidation states.
  • Down the group, the acidic character of the oxides decreases.
  • Reactivity towards halogens: Group 15 elements form halides of the type EX3 and EX5 where E = N, P, As, Sb or Bi.
  • However, N does not form pentahalides EX5 as it lacks d-orbital.
  • All group 15 elements form trihalides but trihalides of N are unstable.
  • Reactivity towards metals: Group 15 elements react with metals to form compounds of general formula M3E2 where E= N, P, As, Sb or Bi.
  • The binary compounds formed by the reaction of group 15 elements with metal show oxidation state -3.


Question 7.4.

Why NH3 form hydrogen bond but PH3 does not?


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.


Question 7.5.

How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved?


Nitrogen is prepared in the laboratory by heating aqueous ammonium chloride (NH4Cl) with sodium nitrite (NaNO2)

to form ammonium nitrite (NH4NO2), which is unstable.

Ammonium nitrite breaks down to form nitrogen and water.

NH4Cl (aq) + NaNO2 (aq)--> NH4NO2 + NaCl (aq)

NH4NO2  --> N2 (g) + H2O (l)

Small amounts of NO and HNO3 are also produced which can be removed by

passing nitrogen gas through aqueous sulphuric acid containing potassium dichromate.


Question 7.6.

How is ammonia manufactured industrially?


Ammonia is manufactured industrially by Haber’s process.

Nitrogen from gas is combined with hydrogen derived from natural gas (methane) in the ratio 1:3 giving rise to ammonia.

The reaction is reversible and exothermic.

N2 (g) + 3H2 (g) --> 2NH3 (g)   ∆ fHo= -461.1 kJ/mol


The optimum conditions for manufacturing ammonia are:

(i) Pressure (around 200 × 105 Pa)

(ii) Temperature (4700 K)

(iii) Catalyst such as iron oxide with small amounts of Al2O3 and K2O



Question 7.7.

Illustrate how copper metal can give different products on reaction with HNO3.


Concentrated nitric acid is a strong oxidizing agent. It is used for oxidizing most metals.

The products of oxidation depend on the concentration of the acid, temperature, and also on the material undergoing oxidation.

3Cu + 8HNO3 (dilute) --> 3 Cu (NO3)2 + 2NO2 + 4H2O

Cu + 4HNO3 (conc) --> Cu (NO3)2 + 2NO2 + 2H2O


Question 7.8.

Give the resonating structures of NO2 and N2O5.






Question 7.9.

The HNH angle value is higher than HPH, HAsH and HSbH angles.

Why? [Hint: Can be explained on the basis of sp3 hybridisation in NH3 and

only s–p bonding between hydrogen and other elements of the group].


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.


Question 7.10.

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


Nitrogen does not have d-orbital to expand its octet. So, it cannot have coordination number greater than 4.

But, phosphorous has d-orbital and can extend its octet and form R3P = O.

Therefore, R3P = O exist but R3N = O does not.


Question 7.11.

Explain why NH3 is basic while BiH3 is only feebly basic?


NH3 is distinctly basic while BiH3 is feebly basic.

Nitrogen has a small size due to which the lone pair of electrons is concentrated in a small region.

This means that the charge density per unit volume is high.

On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density.

Hence, the electron donating capacity of group 15 element hydrides decreases on moving down the group.

The basic strength in group 15 is in the following pattern:

NH3 > PH3 > AsH3 > SbH3 > BiH3


Question 7.12.

Nitrogen exists as diatomic molecule and phosphorus as P4. Why?


Nitrogen owing to its small size has a tendency to form pπ−pπ multiple bonds with itself.

Nitrogen thus forms a very stable diatomic molecule, N2.

On moving down a group, the tendency to form pπ−pπ bonds decreases (because of the large size of heavier elements).

Therefore, phosphorus (like other heavier metals) exists in the P4 state.


Question 7.13.

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


 White Phosphorous

Red Phosphorous

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.

 Class_12_Chemistry_p_Block_Elements_Tetrahedral_Shape  Class_12_Chemistry_p_Block_Elements_Tetrahedral_Shape_1



Question 7.14.

Why does nitrogen show catenation properties less than phosphorus?


Catenation is much more common in phosphorous compounds than in nitrogen compounds.

This is because of the relative weakness of the N−N single bond as compared to the P−P single bond.

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



Question 7.15.

Give the disproportionation reaction of H3PO3?


Disproportionation is a chemical reaction typically a redox reaction where a molecule is transformed into two or more dissimilar products.

On heating, orthophosphorus acid (H3PO3) disproportionate to give orthophosphoric acid (H3PO4) and phosphine (PH3).

The oxidation states of P in various species involved in the reaction are mentioned below.

4H3PO3 --> 3H3PO4 + PH3

[+3]            [+5]         [-3]     


Question 7.16.

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


PCl5 can only act as an oxidizing agent. The highest oxidation state that P can show is +5.

In PCl5, phosphorus is in its highest oxidation state (+5).

However, it can decrease its oxidation state and act as an oxidizing agent.


 Question 7.17.

Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration,

oxidation state and hydride formation?


The elements of group 16 are collectively called chalcogens.


(i) Elements of group 16 have six valence electrons each.

The general electronic configuration of these elements is ns2 np4, where n varies from 2 to 6.

(ii) Oxidation state: As these elements have six valence electrons [ns2np4], they should display an oxidation state of -2.

However, only oxygen predominantly shows the oxidation state of -2 owing to its high Electronegativity.

It also exhibits the oxidation state of -1 [H2O2], zero [O2], and +2 [OF2].

However, the stability of the -2 oxidation state decreases on moving down a group due to a decrease in the Electronegativity of the elements.

The heavier elements of the group show an oxidation state of +2, +4, and +6 due to the availability of d-orbitals.

(iii) Formation of hydrides: These elements form hydrides of formula H2E, where E = O, S, Se, Te, PO.

Oxygen and sulphur also form hydrides of type H2E2.

These hydrides are quite volatile in nature.


Question 7.18.

Why is dioxygen a gas but sulphur a solid?


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.


Question 7.19.

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)?


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.


Question 7.20.

Which aerosols deplete ozone?


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.


Question 7.21.

Describe the manufacture of H2SO4 by contact process?


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

Step (i):

Sulphur or sulphide ores are burnt in air to form SO2.

Step (ii):

By a reaction with oxygen, SO2 is converted into SO3 in the presence of V2O5 as a catalyst.

2SO2 (g) + O2 (g) à 2SO3 (g) (In the presence of catalysts (V2O5)

Step (iii):

SO3 produced is absorbed on H2SO4 to give H2S2O7 (oleum).

SO3 + H2SO4 à H2S2O7

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.


 Question 7.22.

How is SO2 an air pollutant??


SO2 is a highly irritating gas and causes serious respiratory problems, and may cause a fit of coughing.

  1. It reacts with water vapour present in the atmosphere to form sulphuric acid.
  2. This causes acid rain. Acid rain damages soil [soil become more acidic], plants, and buildings get corroded, especially those made of marble.

In the air, SO2 is oxidized to SO3 which is also an irritant.

2SO2 + O2 → 2SO3

SO3 + H2O → H2SO4

  1. Even in very low concentrations, SO2 causes irritation in the respiratory tract.
  2. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.
  3. It is extremely harmful to plants. Plants exposed to sulphur dioxide for a prolong time lose colour from their leaves.
  4. This condition is known as chlorosis. This is because the formation of chlorophyll is retarded in presence of sulphur dioxide.



Question 7.23.

Why are halogens strong oxidising agents?


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.


Question 7.24.

Explain why fluorine forms only one oxoacid, HOF?


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


Question 7.25.

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


Both chlorine and nitrogen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding.

This is because in comparison to chlorine, nitrogen has a smaller size and as a result, a higher electron density per unit volume.

Hence nitrogen forms hydrogen bonding more readily.


Chlorine has larger atomic size as compared to nitrogen and so has lower electron density per unit volume.

Hence chlorine does not readily form hydrogen bonding.



Question 7.26.

Write two uses of ClO2.


ClO2 which is called as Chlorine Dioxide has following uses:

1) It is used as a bleaching agent in paper pulp and textile industries.

2) It is used as disinfectant in sewage and for purification of drinking water.

3) Chlorine dioxide is used to control tastes and odours associated with algae and decaying vegetation.


Question 7.27.

Why are halogens coloured?


 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.


Question 7.28.

Write the reactions of F2 and Cl2 with water?


Chlorine gas reacts with water to give Hydrochloric acid and Hypochlorous acid

Cl2 + H2O → HCl (Hydrochloric Acid) + HOCl (Hypochlorous acid)

Fluorine gas reacts with water to give Hydrogen ions, Fluorine ions, Oxygen gas and Hydrofluoric acid.

2F2 + 2H2O → 4H+ + 4F- + O2 + 4HF


Question 7.29.

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


  • Cl2 can be prepared from HCl by Deacon’s process.

4HCl + O2 --> 2Cl2 + 2H2O (In presence of CuCl2)

  • HCl can be prepared from Cl2 on treating it with water.

Cl2 + H2O -->      HCl           + HOCl

                    Hydrochloric acid Hypochlorous acid


Question 7.30.

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


Neil Bartlett first performed an experiment in which reaction between oxygen and PtF6 was carried out

which lead to the formation of a red coloured compound O2+[PtF6]-.

He observed that the first ionization energy of Oxygen and Xenon is almost same (~1170 kJ/mol).

So, when he tried to react Xe and PtF6 in which he was successful to obtain a red coloured compound Xe+ [PtF6]-.


Question 7.31.

What are the oxidation states of phosphorus in the following?

  • H3PO3
  • PCl3
  • Ca3P2
  • Na3PO4
  • POF3


 Let the oxidation state of p be x.

(i) H3PO3

[3 + x + 3(-2)] = 0

(3 + x – 6) = 0

(x – 3) = 0

x = 3

(ii) PCl3

[x + 3(-1)] = 0

(x – 3) = 0

x = 3

(iii) Ca3P2

3(+2) + 2(x)   = 0

(6 + 2x) = 0

2x = -6

x = (-6 / 2)

x = -3

(iv) Na3PO4

3(+1) + x + 4(-2) = 0

(3 + x – 8) = 0

(x – 5) = 0

x = 5

(v) POF3:

[x + (-2) + 3(-1)] = 0

(x - 2 – 3)  = 0

(x  - 5)  = 0

x = 5


Question 7.32.

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.


1) 4NaCl + MnO2 + 4H2SO4→ MnCl2 + 4NaHSO4 + 2H2O + Cl2

Manganese (IV) oxide reacts with sodium chloride and sulfuric acid to produce manganese (II) chloride,

chlorine, sodium bisulfate and water.

This reaction takes place at a temperature near 100°C.

2) Cl2 + NaI → 2NaCl + I2

Chlorine reacts with sodium iodide to produce sodium chloride and iodine.

Chlorine - diluted solution.

Sodium iodide - cold solution


Question 7.33.

How are xenon fluorides XeF2, XeF4 and XeF6 obtained?


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 (g)   + F2 (g) à XeF2(s) (Temperature (673K), Pressure (1 bar))


Xe (g)      + 2F2 (g) à XeF4(s) (Temperature (873K), Pressure (7 bar))

(1:5 ratio)

Xe (g)      + 3F2 (g) à XeF6(s) (Temperature (573K), Pressure (60-70 bar))

(1:5 ratio)


Question 7.34.

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.


Question 7.35.

How are XeO3 and XeOF4 prepared??


1) XeO3 can be produced by hydrolysis of XeF4 and XeF6 under controlled pH of the medium in which reaction is taking place as shown below:

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

XeF6 + 3H2O → XeO3 + 6HF

2) XeOF4 can be obtained on partial hydrolysis of XeF6 as shown below:

XeF6 + H2O → XeOF4 + 2HF


Question 7.36.

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.


(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

(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.

(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.


Question 7.37.

Which one of the following does not exist?

  • XeOF4
  • NeF2
  • XeF2
  • XeF6


 NeF2 does not exists as it would require d orbital for bonding which Ne does not have as

it belongs to period 2 and moreover a lot of energy would be used to excite the electrons to a higher energy level as

the valence electrons are very close to the nucleus thus they experience strong electrostatic attraction.

Question 7.38.

Give the formula and describe the structure of a noble gas species which is isostructural with:

  • ICl4
  • IBr2
  • BrO3-


  • XeF4 is isoelectronic with ICl4-.

By isoelectronic, we mean that the two have the same number of valence electrons.

It has square planar geometry as shown below:


  • XeF2 is isoelectronic with IBr2-. And the geometry is linear structure.


  • XeF3 is isoelectronic with BrO3-. And the geometry is pyramidal structure.



Question 7.39.

Why do noble gases have comparatively large atomic sizes?


Noble gases do not form molecules. In case of noble gases, the atomic radii correspond 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.


Question 7.40.

List the uses of neon and argon gases.


Uses of neon gas:

(i) It is mixed with helium to protect electrical equipment’s from high voltage.

 (ii) It is filled in discharge tubes with characteristic colours.

(iii) It is used in beacon lights.

Uses of Argon gas:

(i) Argon along with nitrogen is used in gas-filled electric lamps. This is because Ar is more inert than N.

(ii) It is usually used to provide an inert temperature in a high metallurgical process.

(iii) It is also used in laboratories to handle air-sensitive substances.

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