09243500460  09243500460

Chapter 7: The p Block Elements

Q 1:Briefly describe the general characteristics of Group 15 elements with reference to their oxidation state, electronic configuration, atomic size, electronegativity andionisation enthalpy.

General trends in group 15 elements

(i) Electronic configuration: There are 5 valence electrons for all the elements in group 15.
\(ns^{2}np^{3}\) is their general electronic configuration.
(ii) Oxidation states: All these elements require three or more electrons to complete their octets and have 5 valence electrons. It is difficult in gaining electrons as the nucleus will have to attract three more electrons. This happens only with nitrogen as it is the smallest in size and the distance between the nucleus and the valence shell is relatively small. The remaining elements of this group show a formal oxidation state of −3 in their covalent compounds. In addition to the −3 state, N and P also show −1 and −2 oxidation states. All the elements present in this group show +3 and +5 oxidation states. However, the stability of +5 oxidation state decreases down a group, whereas the stability of +3 oxidation state increases. This happens because of the inert pair effect.
(iii) Ionization energy and electronegativity
Ionization decreases as we move down the group. This happens because of increase in atomic
sizes. Moving down the group, electronegativity decreases due to increase in size.
(iv) Atomic size: As we move down the group atomic size increases. This increase in
the atomic size is attributed to an increase in the number of shells.

Q 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,
\(N_{2}\). In \(N_{2}\), 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 isable to form \(p\pi−p\pi\) bonds with itself. This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is more reactive than nitrogen.


Q 3: Discuss the trends in chemical reactivity of group 15 elements.

General trends in chemical properties of group − 15
(i) Reactivity towards hydrogen:
The elements of group 15 react with hydrogen to form hydrides of type \(EH_{3}\), where E = N,P, As, Sb, or Bi. The stability of hydrides decreases on moving down from \(NH_{3}\;to \;BiH_{3}\).

(ii)Reactivity towards oxygen:
The elements of group 15 form two types of oxides: \(E_{2}O_{3}\) and \(E_{2}O_{5}\), where E = N, P, As, Sb, or Bi. The oxide with the element in the higher oxidation state is more acidic than the other. However, the acidic character decreases on moving down a group.
(iii) Reactivity towards halogens:
The group 15 elements react with halogens to form two series of salts: \(EX_{3}\;and\;EX_{5}\). However, nitrogen does not form \(NX_{5}\) as it lacks the d-orbital. All trihalides (except \(NX_{3}\)) are stable.
(iv) Reactivity towards metals:
The group 15 elements react with metals to form binary compounds in which metals exhibit
−3 oxidation states.
Q 4: Why does \(NH_{3}\) form hydrogen bond but \(PH_{3}\) does not?

When compared to phosphorus nitrogen is highly electronegative. This results in a greater
attraction of electrons towards nitrogen in \(NH_{3}\) than towards phosphorus in \(PH_{3}\). Hence, the extent of hydrogen bonding in \(PH_{3}\) is very less as compared to \(NH_{3}\).


Q 5: How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.

An aqueous solution of ammonium chloride is treated with sodium nitrite.

NH4Cl (aq ) + NaNO2 →N2(g) + 2H2O(l) + NaCl(aq)

NO and \(HNO_{3}\) are produced in small amounts. These are impurities that can be removed on passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate.

Q 6: How is ammonia manufactured industrially?
Ammonia is prepared on a large-scale by the Haber’s process.
N2(g) + 3H2(g) ⇌ 2NH3(g)

The optimum conditions for manufacturing ammonia are:
(i) Pressure (around \(200 × 10^{5}\) Pa)
(ii) Temperature (4700 K)
(iii) Catalyst such as iron oxide with small amounts of \(Al_{2}O_{3}\) and \(K_{2}O\)


Q 7: Illustrate how copper metal can give different products on reaction with \(HNO_{3}\).


Concentrated nitric acid is a strong oxidizing agent. It is used for oxidizing most metals.
The products of oxidation depend on the temperature, concentration of the acid, and also
on the material undergoing oxidation.

3Cu + 8HNO3( dil.) → 3Cu(NO3)2 + 2NO + 4H2O
CU + 4HNO3( conc. )  →Cu(No3)2 + 2NO2 +2H2O

Q 8: Give the resonating structures of \(NO_{2}\) and \(N_{2}O_{5}\).

(1) 8.1



 Q 9: The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be
explained on the basis of \(sp^{3}\) hybridisation in \(NH_{3}\) and only s−p bonding between hydrogen and other elements of the group].

Hydride \(NH_{3}\;PH_{3}\;AsH_{3}\;SbH_{3}\)
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 electronegativitydecreases on moving down a group. Consequently, the repulsive interactions between the
electron pairs decrease, thereby decreasing the H−M−H bond angle.
Q 10: Why does \(R_{3}P=O\) exist but \(R_{3}N=O\) does not (R = alkylgroup)?

N (unlike P) lacks the d-orbital. This restricts nitrogen to expand its coordination number
beyond four. Hence, \(R_{3}N=O\) does not exist.
Q 11: Explain why \(NH_{3}\) is basic while \(BiH_{3}\) is only feebly basic.

\(NH_{3}\) is distinctly basic while \(BiH_{3}\) 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.
Q 12: Nitrogen exists as diatomic molecule and phosphorus as \(P_{4}\). Why?

Nitrogen owing to its small size has a tendency to form  multiple bonds with itself.
Nitrogen thus forms a very stable diatomic molecule, \(N_{2}\). On moving down a group, the tendency to form  bonds decreases (because of the large size of heavier elements). Therefore, phosphorus (like other heavier metals) exists in the \(P_{4}\) state.
Q 13: Write main differences between the properties of white phosphorus and red phosphorus.
Answer : 




Q 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.



Q 15:Give the disproportionation reaction of \(H_{3}PO_{3}\).

On heating, orthophosphorus acid (\(H_{3}PO_{3}\)) disproportionates to give orthophosphoric acid (\(H_{3}PO_{4}\)) and phosphine (\(PH_{3}\)). The oxidation states of P in various species involved in the reaction are mentioned below.
4H3P-3O3→ 3H3P+5O4 + P-3H3



Q 16: Can \(PCl_{5}\) act as an oxidising as well as a reducing agent? Justify.

\(PCl_{5}\)  can only act as an oxidizing agent. The highest oxidation state that P can show is +5. In \(PCl_{5}\), phosphorus is in its highest oxidation state (+5). However, it can decrease its oxidation state and act as an oxidizing agent.
Q 17: Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in
terms of electronic configuration, oxidation state and hydride formation.

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 \(ns^{2}np^{4}\), where nvaries from 2 to 6

(ii) Oxidation state:
As these elements have six valence electrons (\(ns^{2}np^{4}\)), 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 (\(H_{2}O_{2}\)), zero (\(O_{2}\)), and +2 (\(OF_{2}\)). 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 \(H_{2}E\), where E = O, S, Se, Te, PO. Oxygen and sulphur also form hydrides of type \(H_{2}E_{2}\). These hydrides are quite volatile in nature.


Q 18: Why is dioxygen a gas but sulphur a solid?

Oxygen is smaller in size when compared to sulphur. Since its size is small, it can form  bonds and form \(O_{2} (O==O)\) molecule. Also, the intermolecular forces in oxygen are weak van der Wall’s, which cause it to exist as gas. On the other hand, sulphurdoes not form \(M_{2}\) molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid.
Q 19: Knowing the electron gain enthalpy values for O → O -1 and O → O 2- as −141 and \(702\;kJ\;mol^{−1}\) respectively, how can you account for the formation of a large number of oxides having \(O^{2−}\; species\;and\; not\;O^{−}\)?
(Hint: Consider lattice energy factor in the formation of compounds).

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

Lattice energy is directly proportional to the charge carried by an ion. When a metal
combines with oxygen, the lattice energy of the oxide involving \(O^{2−}\) ion is much more thanthe oxide involving \(O^{−}\) ion. Hence, the oxide having \(O^{2−}\) ions are more stable than oxides having \(O^{−}\). Hence, we can say that formation of \(O^{2−}\) is energetically more favourable than
formation of \(O^{−}\).



Q 20. Which of the aerosol is responsible for the depletion of ozone?


The aerosol which is responsible for the depletion of ozone is: Freons or chlorofluorocarbons (CFCs)

The molecules of CFS breaks down when there is presence of ultraviolet radiations and forms chlorine free radicals which then combines with ozone to form oxygen.

\(2SO_{2\left(g\right)}+O_{2\left(g\right)}\overset{V_2 O_5}{\rightarrow}2SO_{3\left(g\right )}\)



Q 21. Describe the manufacture of \(H_2 SO_4\) by contact process?


The steps which are required in the production of Sulphuric Acid by the contact process

Step (1)

Sulphide ores or Sulphur are burnt in air to form \(SO_2\).

Step (2)

By a reaction with oxygen, \(SO_2\) is converted into \(SO_3\) in the presence of \(V_2 O_5\) as a catalyst.

\(2 SO_{2\left ( g \right )} + O_{2\left ( g \right )} \overset{V_2 O_5}{\rightarrow} 2SO_{3 \left ( g \right )}\)

Step (3)

\(SO_3\) produced is absorbed on \(H_2 SO_4\) to give \(H_2 S_2 O_7\) (oleum).

\(SO_3 + H_2 S_4 \rightarrow H_2S_2O_7\)

This oleum is then diluted to obtain \(H_2 SO_4\)of the desired concentration.

In practice, the plant is operated at 2 bar (pressure) and 720 K (temperature). The sulphuric acid thus obtained is 96-98% pure.



Q 22: How does SO2 cause air pollution?

Soln: The environment is harmed by sulphur dioxide in many ways:

  1. Sulphuric acid is formed, when it is combined with water vapour present in the atmosphere. This causes acid that damages plants, soil, buildings (those made of marble are more prone) etc.
  2. SO2 causes irritation in respiratory tract, throat, eyes and can also affect the larynx to cause breathlessness.
  3. The color of the leaves of the plant gets faded when it is exposed to sulphur dioxide for a long time. This defect is known as chlorosis. The formation of chlorophyll is affected by the presence of sulphur dioxide.



Q 23: Halogens are strong oxidizing agents. Explain.

Soln: Halogens have an electronic configuration of np5, where n =2-6. Thus, halogens require only one more electron to complete their octet and to attain the stable noble gas configuration. Moreover, halogens have high negative electron gain enthalpies and are highly electronegative with low dissociation energies. As a result, they have a high tendency to gain an electron. Hence, they act as strong oxidisong agents.



Q 24: Fluorine forms only one oxoacid, HOF. Explain.

Soln:Flourine has high electronegativity and a small size, hence it forms only one oxoacid i.e HOF.



Q 25: Oxygen and Chlorine has nearly the same electronegativity, still oxygen forms hydrogen bonds while chlorine does not. Why?

Soln : Oxygen has a smaller size and due to which a higher electron density per unit volume. Hence, oxygen forms hydrogen bonds while chlorine does not despite having the similar electronegative values.



Q 26. State two applications of CIO2
Ans : Applications of ClO2
( a )Used for purification of water.
( b ) Used for bleaching.


Q 27. What is the reason for  halogensbeing colored?
Ans : Halogens  arecolored because they take in radiations from the visible spectrum. This  excites the valence electrons to a higher energy level. The amount of energy required for excitation differs from halogento  halogen, thus they exhibit different colors.


Q28. Give the reactions of Cl2 and F2 with water.
Ans:  ( i ) Cl2 + H2O  → HCL + HOCL
( ii ) 2 F2 + 2H2O → 4H+ + 4F­- + O2 + 4HF


Q29.Write the reactions involved in the  preparations of  HCl from Cl2 and Cl2 from HCl ?

( i ) HCl is prepared from Cl2by reacting it with water.
Cl2 + H2O → HCL + HOCL

( ii ) Cl2 is prepared by Deacon’s process from HCL
4HCL + O2 → 2Cl2 + 2H2O



Q30. What was the inspiration for N. Bartlett to carry out the reaction between PtF6 and Xe?
Ans:  N.Barlett observed that PtF6 and O2 reacts to produce a compound O2+[ PtF6].
As the first ionization enthalpy of Xe( 1170 kJ/mol  ) is very close to that of O2 , he figured that PtF6 could also oxidize Xe to Xe+. Thus, he reacted PtF6 and Xe to form a red colored compound   Xe+[ PtF6] .



Q31. In the compounds given below, find the oxidation states of phosphorus:
( a ) H3PO3
( b ) PCl3
( c ) Ca3P2
( d ) Na3PO4
( e ) POF3?

Ans: Let the oxidation state of phosphorous be x
(a) H3PO3
3 + x + 3( -2) = 0
x -3 = 0
x =3

(b) PCl3
x + 3( -1) = 0
x = 3

(c) Ca3P2
3( 2 ) + 2 (x) = 0
2x = -6
x = -3

(d) Na3PO4
3( 1 ) + x + 4( -2 ) = 0
x -5 =0
x =5

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


Q 32. Give balanced equations for the reactions below:
( a ) NaCl being heated with H2SO4 in the presence of MnO2.
( b ) Chlorine gas passed through aNaIand water solution.­


(a) 4NaCl + MnO2 + 4H2SO4→ MnCl2 + 4NaHSO4 + 2H2O +Cl2
(b) Cl2 + NaI → 2NaCl + I2



Q33. State the reaction that gives xenon fluorides XeF2, XeF4 and XeF6.
Ans: XeF2, XeF4 and XeF6are obtained through direction reactions between Xe and F2. The product depends upon the conditions of the reaction :
Xe       +      F2→   XeF2

Xe       +     2F2→   XeF4
( 1:5 ratio )
Xe       +      F2XeOF4.
(1 : 20 ratio)


Give the neutral molecule with whichClO isoelectronic. Is this molecule a Lewis base?
Ans: ClO is isoelectronic with ClF.
Total electrons in ClO  = 17 + 8 + 1 =26
Total electrons in ClF = 17 + 9 = 26
As ClF accepts electrons from F to form ClF3 , ClF behaves like a lewis base.


Q35.  State the preparation reaction of XeO3 and XeOF4.
Ans: XeO3 can be obtained using two methods :
( 1 ) 6XeF4 + 12H2O  → 4Xe + 2XeO3 + 24HF + 3O2
( 2 ) XeF6 + 3H2O  → XeO3 + 6HF
XeOF4 is obtained using XeF6
XeF4 + H2O  → XeOF4 + 2HF



Q36. Rearrange the given sets in the order as mentioned :
( 1 ) Cl2, F2, I2, Br2 – increasing bond dissociation enthalpy.
( 2 ) HCl, HI, HBr, HF- increasing acidic strength.
( 3 ) PH3, NH3, AsH3, BiH3, SbH3 − increasing base strength.


(1) Bond dissociation energy normally lowers on moving down a group becauseof increase in the atomic size. However, F2has a lower bond dissociation energy than Cl2 and Br2. This is because the atomic size of fluorine is very small.
Therefore, the increasing order for bond dissociation enthalpy is:
I2< F2< Br2< Cl2
(2)  Bond dissociation energy of  a H-X molecule ( where X = F, Cl, Br, I ) lowers with an increase in the size of an atom. As, H-I bond is the weakest it will be the strongest acid.
Therefore, the increasing order acidic strength is :

(3) BiH3≤ SbH3<AsH3< PH3< NH3
On moving from nitrogen to bismuth, the atomic size increases but the electron density of the atom decreases. Hence, the basic strength lowers.



Q37. Identify the one that does not exist, from among the following.
(a) XeOF4
(b) NeF2
(c) XeF2
(d) XeF6
Ans: The one that does not exist is NeF2.


Q38. Present the structure and write the formula of a noble gas speciethat is isostructural with:
( a ) ICl4
( b )  IBr2
( c ) BrO3


(a) XeF4 is isoelectronic toICl4 . And it square planar in geometry :38.1

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


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

Q39. What is the reason for noble gases having relatively bigger atomic sizes ?


Noble gases haveatomic radii that corresponds to van der Waal’s radii. Whereas,  other elements have a covalent radii. Now, by definition van der Waal’s radii are bigger than covalent radii. This is the reason why noble gases have relatively bigger atomic sizes.



Q40. Give some uses of argon and neon gases.
Ans: Uses of Argon gas:
(a)Argon is used to keep an inert atmosphere in high temperature metallurgical operations like arc welding.
(b)It is used in fluorescent and incandescent lamps where it is required to  check the sublimation of the filament. Thereby, increasing the life of the lamp.
(c) Argon is used in laboratories to handle substances that are air-sensitive.

Uses of neon gas:
(a) Neon is filled in discharge tubes for advertising or decoration.
(b) Neon is used for making beacon lights.
(c) It is used alongside helium to protect electrical equipments against high voltage.

Practise This Question

More NCERT Solutions
NCERT Solutions Class 12 Maths Chapter 11 3 Dimensional GeometryNCERT Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions
NCERT Solutions Class 12 Maths Chapter 6 Applications Of DerivativeNCERT Solutions Class 12 Physics Chapter 10 Wave Optics
NCERT Solutions Class 12 Physics Chapter 15 Communication SystemNCERT Solutions Class 12 Physics Chapter 4 Moving Charges And Magnetism
NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic WavesNCERT Solutions Class 6 Maths Chapter 12 Ratio Proportion

No Comments Yet

Leave a Reply

Your email address will not be published. Required fields are marked *

Practise This Question

Join BYJU’S NCERT Learning Program

NCERT Solutions

NCERT Solutions