NCERT Exemplar Class 11 Chemistry Solutions Chapter 4 Chemical Bonding and Molecular Structure focuses on the chemical bonding and molecular structure. It helps understanding the several fundamental concepts in the field of chemistry. This chapter of NCERT Class 11 chemistry solutions deals with different topics that are related to chemical bonding and molecular structure. Class 11 Chemistry Exemplar solutions Chapter 4 also aids students in solving different questions and finding the appropriate answers accordingly.
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In ion the formal charge on the oxygen atom of bond is
(i) + 1
(ii) – 1
(iii) – 0.75
(iv) + 0.75
The answer is the option (ii) -1
Explanation: Formal charge in an atom of a polyatomic molecule or ion can also be defined as the difference between the number of valence electrons of that atom in an isolated or free state and the number of electrons assigned to that atom in the Lewis structure.
The formal charge of the atom in the molecule or ion = (Number of the valence electron in the free atom) – (Number of lone pair electrons + ½ number of bonding electrons)
Therefore, the formal charge on each O-atom
= 6 – (6+ 1/2 × 2) = 6 – 7 = -1
Number of bonds and σ bonds in the following structure is–
(i) 6, 19
(ii) 4, 20
(iii) 5, 19
(iv) 5, 20
The answer is the option (iii) 5, 19
Explanation: Here, each C atom is sp2 hybridized and surrounded by 3 sigma bonds and 1 pi bond between two C atoms.
8 C – H + 11 C – C bonds = 19 bonds.
There are - bonds and 19 bonds.
Which of the following statements are correct about ?
(i) The hybridization of central atom is sp3.
(ii) Its resonance structure has one C–O single bond and two C=O double bonds.
(iii) The average formal charge on each oxygen atom is 0.67 units.
(iv) All C–O bond lengths are equal.
The answer is the option (iii) The average formal charge on each oxygen atom is 0.67 units and (iv) All C-O bond lengths are equal.
Explanation: In resonating, structures bonds are not fixed, and as a result, all bond lengths are equal.
Which of the following statements are not correct?
(i) NaCl being an ionic compound is a good conductor of electricity in the solid state.
(ii) In canonical structures there is a difference in the arrangement of atoms.
(iii) Hybrid orbitals form stronger bonds than pure orbitals.
(iv) VSEPR Theory can explain the square planar geometry of
The answer is the option (i) NaCl being an ionic compound is a good conductor of electricity in solid-state and (ii) In canonical structures, there is a difference in the arrangement of atoms
Explanation: In (i) ions are not free in solid-state but are strongly bonded through electrostatic forces, and in (ii) it does not show canonical structures being ionic compounds.
Explain the shape of
The central atom in is bromine. Its hybridization is .
In total, there are 7 valence electrons in a Br atom out of which 5 valence electrons are used for making a pair with the F atoms, and two valence electrons are used for making a lone pair of electrons. We know that the bond pair and lone pair repel each other. Hence, the shape is square pyramidal.
Write Lewis structure of the following compounds and show formal charge on each atom.
The formula for calculation of the formal charge is as follows: -
Formal charge = ½ [total no: of bonding or shared electrons]
Oxygen with single bond will have formal charge of = 6-6-2/2 = -1
Oxygen with double bond will have formal charge of = 6-4-4/2 = 0
Therefore, nitrogen will have a formal charge of = 5-2-6/2 = 0
Oxygen 1 and 4 will have a formal charge = 6-4-4/2= 0
Oxygen 2 and 3 will have a formal charge = 6-4-4/2=0
Hydrogen 1 and 2 will have a formal charge = 1-0-2/2=0
Sulfur will have a formal charge =6-0-12/2 = 0
Give reasons for the following :
(i) Covalent bonds are directional bonds while ionic bonds are nondirectional.
(ii) The water molecule has a bent structure whereas carbon dioxide molecule is linear.
(iii) Ethyne molecule is linear.
(i) A covalent bond is formed by the overlapping of the atomic orbitals, therefore, it’s a directional bond. Ionic bond is formed by the transference of electrons and electrostatic field of an ion is there which is non-directional, hence it’s a non-directional bond. Positive ion is surrounded by number of anions in any direction depending upon the size and similarly, negative ion is surrounded by no. of cations in any directional depending upon the size.
(ii) In , oxygen atom is sp3 hybridized and we already know that it is surrounded by two lone pairs and two bonded pairs. There are four sp3 hybrid orbitals which give distorted tetrahedral geometry in which the two corners are occupied by hydrogen atoms and other two by lone pairs. The bond angle is reduced to 104.5° and 109.5°due to greater repulsive forces between 1p-1p and molecules acquire a bent or V-shaped structure. Carbon atom is sp hybridized in and the two sp hybrid orbitals forms linear shape, i.e., an angle of 180° and hence are oriented in opposite direction.
(iii) Both the carbon atoms are sp hybridized in ethyne molecule, and there are two unhybridized orbitals. Hybridized orbitals of both the carbon atoms are directed in opposite directions forming an angle of 180°, and hence, a linear structure is formed.
Predict the shapes of the following molecules based on hybridization.
(i) In , Boron is sp2 hybridized and thus, the shape of is trigonal planar.
(ii) The shape of is tetrahedral, carbon is sp3 hybridized.
(iii) The shape of molecule is pyramidal, nitrogen atom is sp3 hybridized nut lone pair on N gives pyramidal shape.
(iv) is linear because carbon is sp hybridized.
NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Matching Type
Match the species in Column I with the type of hybrid orbitals in Column II.
(i)→ (c); (ii) →(a); (iii) →(e); (iv) →(d)
S is surrounded by 4 bond pairs and 1 lone pair (sp3d hybridization)
I is surrounded by 5 bond pairs and 1 lone pair (sp3d2 hybridization)
N has 2 bond pairs and no lone pair (sp hybridization)
N has 4 bond pairs and no lone pair (sp3 hybridization)
Match the species in Column I with the geometry/shape in Column II.
(i) →(e); (ii) →(a); (iii) →(b); (iv) →(c)
Oxygen has 3 bond pairs and 1 lone pair (pyramidal shape)
Linear sp hybridization
Cl has 2 bond pairs and two lone pairs (Angular shape)
N has 4 bond pairs and no lone pair (tetrahedral)
NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Assertion and Reason Type
In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound.
Reason (R): This is because sodium and chloride ions acquire octet in sodium chloride formation.
(i) A and R both are correct, and R is the correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false.
The answer is the option (i) A and R both are correct, and R is the correct explanation of A
2,8,1 2,8,7 (2,8 X 2,8,8)
In NaCl, Sodium and Chlorine ions have complete octet. Therefore, NaCl is a stable compound.
NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Long Answer Type
i) Discuss the significance/ applications of dipole moment.
(ii) Represent diagrammatically the bond moments and the resultant dipole moment in and .
(i) The formula of calculating dipole moment is,
Dipole moment = charge(Q) × distance of separation(r)
Dipole moment is expressed in Debye units (D).
Significance of Dipole moment is:
Helps predicting polar and non-polar nature of compounds
Percentage of ionic character can be calculated by the formula:
% of ionic character = observed × 100/ , ionic
It helps to know the symmetry of the molecule.
Symmetrical molecules have zero dipole moment, although they have two or more polar bonds. For example,
The dipole moment in case of is zero. This is because the two equal bond dipoles point in opposite directions and cancel the effect of each other.
Briefly describe the valence bond theory of covalent bond formation by taking an example of hydrogen. How can you interpret energy changes taking place in the formation of dihydrogen?
Valence bond theory was introduced by Heilter and London in 1927. It was developed by Pauling. It is based on knowledge of atomic orbitals, electronic configuration of elements, overlapping of atomic orbitals and hybridization of orbitals.
Let us consider two hydrogen atoms A and B. Let’s assume that they are approaching each other and let their nuclei be NA and NB and electrons present in them be eA and eB.
Now let us imagine that the two atoms are at a large distance from each other, hence, there is no interaction between them. Now think that they are coming close to each other, and new attractive and repulsive forces begin to operate.
Now we know that attractive forces arises between:
Nucleus of one atom and its own electron that is and .
Nucleus of one atom and electron of other atom, i.e., , NB-eA. Similarly, repulsive forces arise between
(a) Electrons of two atoms like
(b) Nuclei of two atoms .
It has been found that the magnitude of the new attractive force is more than the new repulsive forces. As a result, two atoms approach each other and potential energy decreases. Eventually the net force of attraction balances the force of repulsion and system acquires minimum energy. At this point two hydrogen atoms are said to be bonded together to form a stable molecule having bond length 74pm.
Since the energy is released when the bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than that of the isolated hydrogen atoms. The energy so released is called the bond enthalpy, which is corresponding to minimum in the curve depicted in figure. Conversely, 435.8kJ of energy is required to dissociate one mole of H2 molecule.
The potential energy curve for the formation of H2 molecule as a function of internuclear distance of H atoms. The minimum in the curve corresponds to the most stable state of H2.
The students can use NCERT Exemplar Class 11 Chemistry Solutions Chapter 4 pdf download to save the copy of solutions in their device. class 11 Chemistry NCERT exemplar solutions chapter 4 will help the students for understanding the topics and concepts in a better manner. Thus they can solve any questions with confidence.
Topics and Sub-topics of NCERT Exemplar Class 11 Chemistry Solutions Chapter 4
Here is the list of topics included in NCERT Exemplar Solutions for Class 11 Chemistry Chapter 4:
Valence bond theory
The polar character of covalent bonds
The concept of hybridization
The molecular orbital theory of homonuclear diatomic molecules
Main Subtopics in NCERT Exemplar Class 11 Chemistry Solutions Chapter 4 Chemical Bonding and Molecular Structure
Ionic or Electrovalent Bond
Polarity of Bonds
The Valence Shell Electron Pair Repulsion (VSEPR) Theory
Valence Bond Theory
Orbital Overlap Concept
Directional Properties of Bonds
Overlapping of Atomic Orbitals
Types of Overlapping and Nature of Covalent Bonds
The Strength of Sigma and Pi Bonds
Molecular Orbital Theory
Formation of Molecular Orbitals Linear Combination of Atomic Orbitals (Lcao)
Conditions for The Combination of Atomic Orbitals
Types of Molecular Orbitals
Energy Level Diagram for Molecular Orbitals
Electronic Configuration and Molecular Behaviour
Bonding in Some Homonuclear Diatomic Molecules
What will students learn in NCERT Exemplar Class 11 Chemistry Solutions Chapter 4?
The NCERT Exemplar Class 11 Chemistry Chapter 4 solutions will help the students in getting a better understanding of the topics and the concepts that are included in the book.
NCERT Class 11 Chemistry Book is equipped with all the subject details with the probable questions from the exam.
In NCERT Exemplar Class 11 Chemistry Solutions Chapter 4, The students will learn about how to deal with a chemical bond and its long-lasting attractive force.
It’s working on the atoms, ions, compounds, etc. as a group in a chemical reaction. It is also explained in several theories which include valence shell electron pair repulsion theory, etc.
NCERT Exemplar Class 11 Chemistry Solutions Chapter Wise
Important topics to cover for exams from NCERT Exemplar Class 11 Chemistry Solutions Chapter 4 Chemical Bonding and Molecular Structure
Given below are some of the important topics that the students will learn from this chapter:
The lessons and topics will help you to learn about the chemical bond and the reaction. A chemical bond has been explained in the several theories and also these bonds which will gain stability that may result in the formation of electrostatic force.
Class 11 NCERT Exemplar Chemistry Solutions Chapter 4 explains the chemical bond and its importance and its usefulness. The solutions incorporate simple methods that are explained for easy understanding of concepts.
The students will also learn about the theories that are explained by different theorists in the book. It involves different concepts that will develop more understanding among the students.
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