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Chemical Bonding and Molecular Structure are vital in the Class 11 Chemistry curriculum. Chemical Bonding explains how different atoms combine to form molecules. Apart from this, it explains the properties of chemical bonds, types of bonds, their nature, formation, and the molecular structure of different compounds. We can relate this chapter to our day-to-day life, like the water we drink is a molecule which is made up of atoms of Oxygen and Hydrogen. Only just basic understanding of Chemistry helps us to explain the phenomenon that occurs every day around us.
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NCERT solutions for Class 11 Chemistry Chapter 4 will offer a systematic and structured approach for the exercise problems in the NCERT textbook to prepare well for your exams by providing detailed solutions to all the exercise questions in the textbook. Our subject matter expert prepares these NCERT solutions of class 11, which also serve as a valuable resource for students to enhance their exam performance in board exams as well as in competitive exams like JEE, and NEET.
To get all the solved exercise questions, click below on the download PDF icon. In this PDF, you will get detailed solutions to all the questions that are given in the NCERT textbook.
Question 4.1 Explain the formation of a chemical bond.
Answer :
The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species is called a chemical bond . Different theories and concepts have been put forward from time to time to analyze the formation of the bond. These are Kössel-Lewis approach, Valence Shell Electron Pair Repulsion (VSEPR) Theory, Valence Bond (VB) Theory, and Molecular Orbital (MO) Theory.
And every system tends to be more stable and bonding is nature’s way of lowering the energy of the system to attain stability.
Atoms, therefore combine with each other and complete their respective octets or duplets to attain stable configuration of the nearest noble gases. As it was seen that the noble gases are very stable and were inert to react to others.
So, there is a sharing of electrons or transferring one or more electrons from one atom to another, as a result, a chemical bond is formed, known as a covalent bond or ionic bond.
Question 4.2(a) Write Lewis dot symbols for atoms of the following elements :
Answer :
The Lewis dot symbol of Mg atom is;
As there are two valence electrons in Mg atom.
Hence, the Lewis dot symbol for Mg is:
Question 4.2(b) Write Lewis dot symbols for atoms of the following elements :
Answer:
The Lewis dot symbol of Na atom is;
As there is only one valence electron in Na atom of Na.
Hence, the Lewis dot structure is
Question 4.2(c) Write Lewis dot symbols for atoms of the following elements :
Answer :
The Lewis dot symbol of
As there are three valence electrons in
Hence, the Lewis dot structure is
Question 4.2(d) Write Lewis dot symbols for atoms of the following elements :
Answer :
The Lewis dot symbol of
As there are six valence electrons in an atom of
Hence, the Lewis dot structure is
Question 4.2(e) Write Lewis dot symbols for atoms of the following elements :
Answer :
The Lewis dot symbol of
As there are five valence electrons in an atom of
Hence, the Lewis dot structure is
Question 4.2(f) Write Lewis dot symbols for atoms of the following elements :
Answer :
The Lewis dot symbol of
As there are seven valence electrons in an atom of
Hence, the Lewis dot structure is
Question 4.3(a) Write Lewis symbols for the following atoms and ions:
Answer :
As the number of valence electrons in sulphur is six.
Therefore its Lewis dot symbol of sulphur(S) is
And of
Question 4.3(b) Write Lewis symbols for the following atoms and ions:
Answer :
As the number of valence electrons in aluminium is three.
Therefore its Lewis dot symbol of aluminium(Al) is
And of
Hence. the Lewis symbol is
Question 4.3(c) Write Lewis symbols for the following atoms and ions:
Answer :
As the number of valence electrons in hydrogen is one.
Therefore its Lewis dot symbol of hydrogen (H) is
And of
Hence. the Lewis symbol is
Question 4.4(a) Draw the Lewis structures for the following molecules and ions :
Answer :
The Lewis structure of
Question 4.4(b) Draw the Lewis structures for the following molecules and ions :
Answer :
The Lewis structure of
Question 4.4(c) Draw the Lewis structures for the following molecules and ions :
Answer :
The Lewis structure of
Question 4.4(d) Draw the Lewis structures for the following molecules and ions :
Answer :
The Lewis structure of
Question 4.4(e) Draw the Lewis structures for the following molecules and ions :
Answer :
The Lewis structure of
Question 4.5 Define octet rule. Write its significance and limitations.
Answer:
Atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons in order to have an octet in their valence shells. This is known as the octet rule.
Significance: It is quite useful for understanding the structures of most of the organic compounds and it applies mainly to the second-period elements of the periodic table
Limitations: There are three types of exceptions to the octet rule.
Question 4.6 Write the favourable factors for the formation of ionic bond.
Answer :
The formation of ionic bond takes place by the transfer of one or more electrons from one atom to another.
So, ionic bond formation mainly depends upon the ease with which neutral atoms can lose or gain electrons.
The bond formation also depends upon the lattice energy of the compound formed.
Ionic bonds will be formed more easily between elements with comparatively low ionization enthalpies and elements with a comparatively high negative value of electron gain enthalpy.
Question 4.7(a) Discuss the shape of the following molecules using the VSEPR model:
Answer :
Using the VSEPR model we have,
THe central atom has no lone pair and there are two bond pairs.
hence, it has a linear shape .
Question 4.7(b) Discuss the shape of the following molecules using the VSEPR model:
(b)
Answer :
Using the VSEPR model we have,
The central atom has no lone pair and there are three bond pairs.
hence, it has trigonal planar shape. a
Question 4.7(c) Discuss the shape of the following molecules using the VSEPR model:
(c)
Answer :
Using the VSEPR model we have,
The central atom has no lone pair and there are four bond pairs.
hence, it has tetrahedral shape.
Question 4.7(d) Discuss the shape of the following molecules using the VSEPR model:
(d)
Answer:
Using the VSEPR model we have,
THe central atom has no lone pair and there are five bond pairs.
Hence, it has trigonal bipyramidal shape.
Question 4.7 (e) Discuss the shape of the following molecules using the VSEPR model:
Answer:
Using the VSEPR model we have,
The central atom has no lone pair and there are two bond pairs.
Hence, it has a bent shape.
Question 4.7(f) Discuss the shape of the following molecules using the VSEPR model:
(f)
Answer :
Using the VSEPR model we have,
THe central atom has no lone pair and there are three bond pairs.
Hence, it has trigonal bipyramidal shape.
Question 4.9 How do you express the bond strength in terms of bond order ?
Answer :
Bond Strength gives us that amount of energy needed to break a bond between atoms forming a molecule.
So, with an increase in bond order, bond enthalpy increases as a result bond strength increases.
Question 4.10 Define the bond length.
Answer :
Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.
The bond length in a covalent molecule AB.
Question 4.11 Explain the important aspects of resonance with reference to the
Answer :
The single Lewis structure based on the presence of two single bonds and one double bond between carbon and oxygen atoms is inadequate to represent the molecule accurately as it represents unequal bonds. According to the experimental findings, all carbon to oxygen bonds in
CO32- are equivalent. Therefore the carbonate ion is best described as a resonance hybrid of the canonical forms I, II, and III shown below.
Answer :
As per the rule, it is not having the same position as the atoms and is changed.
Hence the given structures cannot be taken as the canonical forms of the resonance hybrid.
Question 4.13 Write the resonance structures for
Answer :
The resonance structures
The resonance structures
The resonance structures
Question 4.14(a) Use Lewis symbols to show electron transfer between the following atoms to form cations and anions :
Answer :
K and S:
We have the electronic configurations of both:
So, there will be an electron transfer between them as follows:
Question 4.14(b) Use Lewis symbols to show electron transfer between the following atoms to form cations and anions :
Answer :
We have the electronic configurations of both:
So, there will be an electron transfer between them as follows:
Question 4.14(c) Use Lewis symbols to show electron transfer between the following atoms to form cations and anions :
Answer :
We have the electronic configurations of both:
So, there will be electron transfer between them as follows:
Answer :
H2O molecule, which has a bent structure, the two O–H bonds are oriented at an angle of 104.50. Net dipole moment of 6.17 × 10–30 C m is the resultant of the dipole moments of two O–H bonds.
While on the other hand. The dipole moment of carbon dioxide is zero. This may be because of linear shape of the molecule as it has two C-O bonds which has opposite dipole moments cancelling each other.
Question 4.16 Write the significance/applications of dipole moment.
Answer :
Some of the important significance of the dipole moment is as follows:
Question 4.17 Define electronegativity. How does it differ from electron gain enthalpy ?
Answer :
Electronegativity is the ability of an atom in a compound to attract a bond pair of electrons towards itself. It cannot be measured and it is a relative number.
The electron gain enthalpy,
An element has a constant value of the electron gain enthalpy that can be measured experimentally.
Question 4.18 Explain with the help of suitable example polar covalent bond.
Answer :
A polar covalent bond, when two different atoms are linked to each other by covalent bond, then the shared electron pair will not be in the centre just because the bonding atoms differ in electronegativities.
For examples, in
Here slightly positive charges are developed in hydrogen atoms and slightly negative charge developed in oxygen atom as oxygen is more electronegative than the hydrogen. Thus, opposite poles are developed in the molecule.
Hence the bond pair lies towards oxygen atom.
Question 4.19 Arrange the bonds in order of increasing ionic character in the molecules:
Answer :
The ionic character in a molecule depends on the electronegativity difference between the constituting atoms. More the difference more will be the ionic character of the molecule.
So, on this basis, we have the order of increasing ionic character in the given molecules.
Question 4.20 The skeletal structure of
Answer :
Here hydrogen atom is bonded to carbon with a double bond, which is not possible because hydrogen has only one electron to share with carbon.
Also, the second carbon does not have its valency satisfied.
Therefore, the correct skeletal structure of
Answer :
The electronic configuration of carbon atom is
Where it has s-orbital, p-orbital only and there is no d-orbital present.
Hence the carbon atom undergoes
And for a molecule to have a square planar structure it must have d orbital present.
But here the absence of d-orbital, as a result, it does not undergo
Also the reason that bond angle in square planar
Hence according to VSEPR theory
Question 4.22 Explain why
Answer :
Question 4.23 Which out of
Answer :
Here both have central atom Nitrogen and it has a lone pair of electrons with three bond pairs.
Hence both molecules have a pyramidal shape.
The electronegativity of fluorine is more as compared to the hydrogen. Hence it is expected that the net dipole moment of
However
This is because of the direction of the dipole moments of each individual bond in
The moments of the lone pair in
Question 4.24 What is meant by hybridisation of atomic orbitals ? Describe the shapes of
Answer :
Hybridisation which can be defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape.
The shapes of
Question 4.25 Describe the change in hybridisation (if any) of the
Answer :
Initially, the aluminium is in the ground state and the valence orbital can be shown as:
Then the electron gets excited so, the valence orbital can be shown as:
So, initially, aluminium
Then it reacts with chloride ion to form
Hence there is a shape change from trigonal planar to tetrahedral.
Question 4.26 Is there any change in the hybridisation of
Answer :
Initially boron atom
And nitrogen atom in
Then after the reaction has occured the product
Question 4.27 Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in
Answer :
We have the electronic configuration of C-atom in the excited state is:
Formation of an ethane molecule
The remaining two
Formation of
One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of the other carbon atom to form C–C sigma bond, while the other hybridised orbital of each carbon atom overlaps axially with the half-filled s orbital of hydrogen atoms forming σ bonds
Each of the two unhybridised p orbitals of both the carbon atoms overlaps sidewise to form two π bonds between the carbon atoms. So the triple bond between the two carbon atoms is made up of one sigma and two pi bonds as shown in Fig
Question 4.28(a) What is the total number of sigma and pi bonds in the following molecules?
Answer :
Given molecule
So, there is three sigmas (2C-H bonds + 1 C-C bond) and two pi-bonds (2 C-C bonds) in
Question 4.28(b) What is the total number of sigma and pi bonds in the following molecules?
Answer :
Given molecule
So, there are five sigma (4C-H bonds + 1 C-C bond) and one pi-bonds (C-C bonds) in
Question 4.29(a) Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?
Answer :
Orbitals
Question 4.29(b) Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?
Answer :
Orbitals
Question 4.29(c) Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?
Answer :
Orbitals
Formation of pi bond takes place.
Question 4.29(d) Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?
Answer :
Orbitals
Question 4.30(a) Which hybrid orbitals are used by carbon atoms in the following molecules?
Answer :
There are four sigma bonds (single bond) each with the help of one s hybrid orbital and three p hybrid orbital, Hence C1 and C2 are
Question 4.30(b) Which hybrid orbitals are used by carbon atoms in the following molecules?
Answer :
While
Therefore they both are
Question 4.30(c) Which hybrid orbitals are used by carbon atoms in the following molecules?
Answer :
and
Therefore they both are
Question 4.30(d) Which hybrid orbitals are used by carbon atoms in the following molecules?
Answer :
and
Question 4.30(e) Which hybrid orbitals are used by carbon atoms in the following molecules?
Answer :
and
Question 4.31 What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one exmaple of each type.
Answer :
The shared pairs of electrons present between the bonded atoms are called bond pairs.
And all valence electrons may not participate in bonding that electron pairs that do not participate in bonding are called lone pairs of electrons.
For examples,
In
In
Question 4.32 Distinguish between a sigma and a pi bond.
Answer :
Difference between the sigma bond and the pi bond is shown in the table below:
Sigma
|
Pi
|
(a) Formed by end to end overlapping of orbitals.
|
Formed by the lateral overlapping of orbitals
|
(b) Sigma bonds are stronger than the pi bond.
|
Weak bond.
|
(c) The orbitals involved in the overlapping are s-s, s-p, p-p.
|
Bonds are formed only with overlapping of p-p orbitals.
|
(d) The electron cloud is symmetrical about an internuclear axis.
|
The electron cloud is not symmetrical.
|
(e) Free rotation is possible in case of a sigma bond.
|
Rotation is restricted in case of pi-bonds.
|
Question 4.33 Explain the formation of
Answer :
Formation of
Assume that two hydrogen atoms
When the two atoms are at a large distance, there is no interaction between them. As they approach each other, the attractive and repulsive forces start operating.
Attractive force arises between:
(a) The nucleus of one atom and its own electron i.e.,
(b) The nucleus of one atom and electron of another atom i.e.,
Repulsive force arises between:
(a) Electrons of two atoms i.e.,
(b) Nuclei of two atoms i.e.,
The force of attraction brings the two atoms together, whereas the force of repulsion tends to push them apart.
The attractive force overcomes the repulsive force. Hence, the two atoms approach each other. As a result, the potential energy decreases. Finally, a state is achieved when the attractive forces balance the repulsive forces and the system acquires minimum energy. This leads to the formation of a dihydrogen molecule.
Question 4.34 Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer :
The important conditions required for the linear combination of atomic orbitals to form molecular orbitals are as follows:
1. The combining atomic orbitals must have the same or nearly the same energy.
2..The combining atomic orbitals must have the same symmetry about the molecular axis.
3. The combining atomic orbitals must overlap to the maximum extent.
Question 4.35 Use molecular orbital theory to explain why the
Answer :
The electronic configuration of Be is
From the molecular orbital electronic configuration, we have for
We can calculate the bond order for
So, therefore we have,
Bond order of
that means that the molecule is unstable.
Hence,
Question 4.36 Compare the relative stability of the following species and indicate their magnetic properties;
Answer :
The electronic configuration of
Here the number of bonding electrons is
Therefore,
The electronic configuration of
Here the number of bonding electrons is
Therefore,
The electronic configuration of
Here the number of bonding electrons is
Therefore,
The electronic configuration of
Here the number of bonding electrons is
Therefore,
Therefore, the bond dissociation energy is directly proportional to the bond order.
Thus, the higher the bond order, the greater will be the stability.
We get this order of stability:
Question 4.37 Write the significance of a plus and a minus sign shown in representing the orbitals.
Answer :
Wave functions can be used to represent molecular orbitals. The plus and minus represent the positive wave function while negative wave function respectively.
Question 4.38 Describe the hybridisation in case of
Answer :
The initial ground state and final excited state electronic configuration of phosphorus (P) are:
So, the phosphorus atom is
The resultant shape is trigonal bipyramidal and the five
The five P-Cl sigma bonds, three lies in one plane and make
So, just because of more repulsion from the equatorial bond pairs, the axial bonds are slightly longer than equatorial bonds.
Question 4.39 Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Answer :
Hydrogen bond can be defined as the attractive force acting between the hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule.
Because of the difference between electro-negativities, the bond pair between hydrogen and the electronegative atom gets drifted towards a more electronegative atom. As a result, the hydrogen atom becomes slightly positively charged.
Hydrogen bonds are stronger than the van der Waals forces because H-bonds are considered as an extreme form of dipole-dipole interaction.
Question 4.40 What is meant by the term bond order? Calculate the bond order of :
Answer :
Bond order (B.O.) is defined as one half the difference between the number of electrons present in the bonding and the antibonding orbitals of a molecule.
Where
So, bond order for different molecules are:
Where, the number of bonding electrons
So, Bond order of nitrogen molecule
Where, the number of bonding electrons
So, Bond order of nitrogen molecule
Where, the number of bonding electrons
So, Bond order of
Where, the number of bonding electrons
So, Bond order of
Question: Which of the following molecules(s) show/s paramagnetic behavior ?
(A)
(B)
(C)
(D)
(E)
Choose the correct answer from the options given below :
1) B only
2) A & C only
3) A & E only
4) A & D only
Answer:
No. of umpaired | ||
(A) | 2 | |
(B) | 0 | |
(C) | 0 | |
(D) | 2 | |
(E) | 0 |
If species contain unpaired electron than it is paramagnetic.
So A & D are paramagnetic.
Hence, the correct answer is option (4).
Question: Which of the following is not a postulate of VSEPR theory?
(1) The shape of a molecule depends upon the number of valence shell electron pairs (bonded or non-bonded) around the central atom.
(2) Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged.
(3) The positions of the electron pairs in space around the central atom are such that they maximise repulsion and thus minimise the distance between them.
(4) The magnitudes of the different types of electronic repulsions follow the order given below: Lone pair - Lone pair > Lone pair - Bonding pair > Bonding pair - Bonding pair
Answer:
The main postulates of VSEPR theory are:
The actual shape of a molecule depends upon the number of electron pairs (bonded or non–bonded) around the central atom.
The electron pairs tend to repel each other due to their negative charge.
Electron pairs arrange themselves in such a way that there exists a minimum repulsion between them.
The valence shell is considered as a sphere with the electron pairs placed at a distance.
A multiple bond is treated as if it is a single electron pair & the electron pairs that constitute the bond as a single pair.
The repulsive interaction of electron pairs decreases in the order as mentioned below:
Lone pair (lp) – Lone pair (lp) > Lone pair (lp) –
Bond pair (bp) > Bond pair (bp) – Bond pair (bp).
Double bonds cause more repulsion than single bonds, and triple bonds cause more repulsion than double bonds. This repulsion decreases sharply with increasing bond angle between the electron pairs.
The positions of the electron pairs in space around the central atom are such that they minimize repulsion and thus minimize the distance between them.
Hence, the correct answer is option (3).
Question: Which of the following compounds has the lowest melting point?
(1) NaCl
(2) MgCl2
(3) AlCl3
(4) BeCl2
Answer:
AlCl3 has the lowest melting point due to its covalent character, which is a result of the high charge density of Al3+. BeCl2 also has a lower melting point but higher than AlCl3.
Hence, the correct answer is option (3).
To effectively solve questions of the chapter on Chemical Bonding and Molecular Structure, follow a systematic and structured approach that helps understand the concepts and the theories:
All the topics and subtopics covered in the NCERT textbook are listed below:
4.1 Kössel-Lewis Approach to Chemical Bonding
4.1.1 Octet Rule
4.1.2 Covalent Bond
4.1.3 Lewis Representation of Simple Molecules( the Lewis Structures)
4.1.4 Formal Charge
4.1.5 Limitations of the Octet Rule
4.2 Ionic or Electrovalent Bond
4.2.1 Lattice Enthalpy
4.3 Bond Parameters
4.3.1 Bond Length
4.3.2 Bond Angle
4.3.3 Bond Enthalpy
4.3.4 Bond Order
4.3.5 Resonance Structures
4.3.6 Polarity of Bonds
4.4 The Valence Shell Electron Pair Repulsion (VSEPR) Theory
4.5 Valence Bond Theory
4.5.1 Orbital Overlap Theory
4.5.2 Directional Properties of Bonds
4.5.3 Overlapping of Atomic Orbitals
4.5.4 Types of Overlapping and Nature of Covalent Bonds
4.5.5 Strength of Sigma and Pi Bonds
4.6 Hybridisation
4.6.1 Types of Hybridisation
4.6.2 Other Examples of sp3, sp2 and sp Hybridisation
4.6.3 Hybridisation of Elements Involving d Orbitals
4.7 Molecular Orbital Theory
4.7.1 Formation of Molecular Orbitals Linear Combination of Atomic Orbitals (LCAO)
4.7.2 Conditions for the Combination of Atomic Orbitals
4.7.3 Types of Molecular Orbitals
4.7.4 Energy Level Diagram for Molecular Orbitals
4.7.5 Electronic Configuration and Molecular Behaviour
4.8 Bonding in Some Homonuclear Diatomic Molecules
4.9 Hydrogen Bonding
4.9.1 Cause of Formation of Hydrogen Bond
4.9.2 Types of H-Bonds
Class 11 NCERT chapter-wise solutions are given below:
Class 11 NCERT Subject-wise solutions are given below:
NCERT solutions for Class 11 Biology |
NCERT solutions for Class 11 Chemistry |
NCERT solutions for Class 11 Maths |
NCERT solutions for Class 11 Physics |
The NCERT books and syllabus links for class 11 are given below:
NCERT Books Class 11 Chemistry |
NCERT Books Class 11 |
NCERT Syllabus Class 11 Chemistry |
NCERT Syllabus Class 11 |
Also Read
To achieve a more stable electronic configuration, typically resembling that of a noble gas, atoms form chemical bonds.
The octet rule states that atoms achieve a full outer shell of eight valence electrons; for that, they tend to gain, lose, or share electrons to resemble the electron configuration of a noble gas (Hydrogen is an exception, aiming for two electrons).
Properties of ionic compounds:
It is a measure of the polarity of a bond or molecule. It is the product of the magnitude of the charge and the distance between the charges. Nonpolar molecules have a zero or very small net dipole moment, while polar molecules have a net dipole moment.
There are five types of chemical bonds: ionic, covalent, hydrogen, coordinate/dative, and metallic. We have also discussed the electronic theory of chemical bonding, in which a chemical bond is formed between atoms to get the nearest inert gas configuration.
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