Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

ffImage

Chemical Bonding and Molecular Structure- Class 11 Chemistry Chapter 4 NCERT Solutions

Class 11 Chemistry Chapter 4 NCERT Solutions unravels the secrets behind the formation and stability of molecules. From understanding the forces that bind atoms together to molecular structures, this chapter offers a fascinating exploration of fundamental chemical principles. Chemical bonding NCERT solutions is a significant chapter that develops the conceptual foundation related to the chemical bonds between different atoms. Access your FREE PDF download now and embark on your journey through Ch 4 Chemistry Class 11, the dynamic world of chemical bonds with clear explanations and comprehensive solutions.

toc-symbol
Table of Content
1. Chemical Bonding and Molecular Structure- Class 11 Chemistry Chapter 4 NCERT Solutions
2. Quick Insights of Chemical Bonding and Molecular Structure Class 11 Chemistry
3. Access NCERT Solutions for Class 11 Chemistry Chemical Bonding and Molecular Structure
    3.1Limitations of the Octet Theory: 
4. Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure Quick Overview of Topics
5. Some Important Concepts for Chemical Bonding and Molecular Structure
6. Benefits of Referring to Vedantu’s NCERT Solutions for class 11 Chemistry Chapter 4
7. Related Study Materials for Class 11 Chemistry Chapter 4  
8. NCERT Solutions Class 11 Chemistry | Chapter-wise Links 
9. NCERT Solutions Class 11 Chemistry - Related Links
FAQs


Solutions for Chemical Bonding and Molecular Structure Class 11 NCERT in PDF is prepared by the Master teacher by referring to the updated NCERT textbook. Check out the Revised class 11 chemistry syllabus and get started with Vedantu to embark on a journey of academic excellence!


Quick Insights of Chemical Bonding and Molecular Structure Class 11 Chemistry

  • NCERT Solutions for Class 11 Chemistry Chapter 4 will give you insights about the General Introduction: Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character of covalent bond, covalent character of ionic bond, and valence bond theory.

  • Class 11 chemistry chapter 4 NCERT solutions section will give you crisp learnings on the resonance, geometry of covalent molecules, VSEPR theory, and concept of hybridisation involving s, p and d orbitals.

  • The understanding related to topics like the shapes of some simple molecules, the molecular orbital theory of homonuclear diatomic molecules, and Hydrogen bonds.

  • Using these chemical bonds, NCERT PDF can help students analyse their level of preparation and understanding of concepts.

  • Chemistry class 11, chapter 4 Chemical Bonding and Molecular Structure topics are included according to the revised academic year 2024-25 syllabus.

Competitive Exams after 12th Science
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow

Access NCERT Solutions for Class 11 Chemistry Chemical Bonding and Molecular Structure

1. Explain the formation of a chemical bond.

Ans:  A chemical bond is defined as an attractive force that holds the constituents (atoms, ions, etc.) together in a chemical species. 

Various theories have been suggested for the formation of chemical bonds such as the electronic theory, valence shell electron pair repulsion theory, valence bond theory, and molecular orbital theory. 

A chemical bond formation is attributed to the tendency of a system to attain stability. It was observed that the inertness of noble gases was because of their fully filled outermost orbitals. Hence, it was postulated that the elements having incomplete outermost shells are unstable (reactive). Atoms, therefore, combine with each other and complete their respective octets or duplets to attain the stable configuration of the nearest noble gases. This combination can occur either by sharing of electrons or by transferring one or more electrons from one atom to another. The chemical bond formed as a result of sharing of electrons between atoms is called a covalent bond. An ionic bond is formed as a result of the transfer of electrons from one atom to another.


2. Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.

Ans:  Mg: There are two valence electrons in an Mg atom. Hence, the Lewis dot symbol for Mg is: $\ddot{Mg}$

Na: There is only one valence electron in an atom of sodium. Hence, the Lewis dot structure is: $\dot{Na}$

B: There are 3 valence electrons in Boron atom. Hence, the Lewis structure is:

$\cdot\dot{B}\cdot$

O: There are six valence electrons in an atom of oxygen. Hence, the Lewis dot structure is:  $\colon\ddot{O}\colon$

N: There are five valence electrons in an atom of nitrogen. Hence, the Lewis dot structure is: $\colon\ddot{N}\cdot$

Br: There are seven valence electrons in bromine. Hence, the Lewis dot structure is:$\colon\ddot{\underset{\cdot}{Br}}\cdot$


3. Write Lewis symbols for the following atoms and ions 

  1.   S and $\mathbf{{{S}^{2-}}}$

Ans: The number of valence electrons in sulphur is 6. 

The Lewis dot symbol of sulphur (S) is $\colon\ddot{{S}}\colon$

The dinegative charge infers that there will be two electrons more in addition to the six valence electrons. Hence, that Lewis dot symbol of ${{S}^{2-}}$ is $[\colon \ddot{S}\underset{..} \colon]^{-2}$

  1.  Al and $\mathbf{A{{l}^{+3}}}$

 Ans: The number of valence electrons in aluminium is 3. 

The Lewis dot symbol of aluminum (Al) is $\cdot\dot{AI}\cdot$

The tripositive charge on a species infers that it has donated its three electrons. Hence, the Lewis do symbol is ${{[Al]}^{+3}}$ 

  1.  H and $\mathbf{{{H}^{-}}}$

Ans: The number of valence electrons in hydrogen is 1. 

The Lewis do the symbol of hydrogen (H) is  $H\cdot$

This negative charge infers that there will be one electron more in addition to the one valence electron. Hence, the Lewis dot symbol is $[\ddot{H}]^-$


4. Draw the Lewis structure for the following molecules and ions:

  1.  $\mathbf{{{H}_{2}}S}$ 

Ans: 


seo images

  1.  $\mathbf{SiC{{l}_{4}}}$ 

Ans: 


seo images


  1.  $\mathbf{Be{{F}_{2}}}$

Ans: 


seo images

  1.  $\mathbf{C{{O}_{3}}^{2-}}$ 

Ans: 


seo images

5. Define the octet rule. Write its significance and limitations.

Ans: The octet rule or the electronic theory of chemical bonding was developed by Kossel and Lewis. According to this rule, atoms can combine either by transfer of valence electrons from one atom to another or by sharing their valence electrons in order to attain the nearest noble gas configuration.


seo images

The octet rule successfully explained the formation of chemical bonds depending upon the nature of the element.

Limitations of the Octet Theory: 

The Following are the Limitations of the Octet Rule:

  1. The rule failed to predict the shape and relative stability of molecules. 

  2. It is based upon the inert nature of noble gases. However, some noble gases like xenon and krypton form compounds such as $Xe{{F}_{2}}$ , $Kr{{F}_{2}}$  etc.

  3. The octet rule cannot be applied to the elements in and beyond the third period of the periodic table. The elements present in these periods have more than eight valence electrons around the central atom. For example: $P{{F}_{5}}$, $S{{F}_{6}}$ etc.


seo images

  1. The octet rule is not satisfied for all atoms in a molecule having an odd number of electrons. For example, NO and $N{{O}_{2}}$ do not satisfy the octet rule. 


seo images

  1. This rule cannot be applied to those compounds in which the number of electrons surrounding the central atom is less than eight. For example, $LiCl,Be{{H}_{2}},AlC{{l}_{3}}$  etc. do not obey that octet rule. 


seo images

6. Write the favourable factors for the formation of ionic bonds.

Ans: An ionic bond is formed by the transfer of one or more electrons from one atom to another. Hence, the formation of ionic bonds 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. 

Hence, favourable factors for ionic bond formation are as follows:

  1. Low ionization enthalpy of metal atoms. 

  2. High electrons gain enthalpy of a nonmetal atom. 

  3. The high lattice energy of the compound formed


7. Discuss the shape of the following molecules using the VSEPR model:

  1. $\mathbf{BeC{{l}_{2}}}$

Ans: $Cl \;\;\; \colon \;\;Be \;\;\;Cl$


seo images


The central atom has no lone pair and there are two bond pairs. i.e., $BeC{{l}_{2}}$ is of the type$A{{B}_{2}}$ . Hence, it has a linear shape.

  1. $\mathbf{BC{{l}_{3}}}$ 

Ans: $\;\underset{\cdot\cdot}Cl\\Cl\;\;\colon \;Be\;\; \colon\;\;Cl$


seo images


The central atom has no lone pair and there are three bond pairs. Hence, it is of the type $A{{B}_{3}}$ . Hence, it is trigonal planar.


seo images


  1. $\mathbf{SiC{{l}_{4}}}$ 


seo images

Ans: 


seo images


The central atom has no lone pair and there are four bond pairs. Hence, the shape of $SiC{{l}_{4}}$ is tetrahedral being the $A{{B}_{4}}$ type molecule.

  1. $\mathbf{As{{F}_{5}}}$ 

Ans: 

seo images

The central atom has no lone pair and there are five bond pairs. Hence, AsF5 is of the type $A{{B}_{5}}$ . Therefore, the shape is trigonal bipyramidal.

  1. $\mathbf{{{H}_{2}}S}$ 

Ans: 


seo images

The central atom has one lone pair and there are two bond pairs. Hence, ${{H}_{2}}S$ is of the type $A{{B}_{2}}E$ . The shape is Bent

  1. $\mathbf{P{{H}_{3}}}$ 


seo images


Ans: 

seo images


The central atom has one lone pair and there are three bond pairs. Hence, $P{{H}_{3}}$ is of the $A{{B}_{3}}E$ type. Therefore, the shape is trigonal bipyramidal.


8. Although geometries of $\mathbf{N{{H}_{3}}}$  and $\mathbf{{{H}_{2}}O}$  molecules are distorted tetrahedral, the bond angle in water is less than that of ammonia. Discuss.

Ans: The molecular geometry of $N{{H}_{3}}$ and ${{H}_{2}}O$ can be shown as:


seo images

The central atom (N) in $N{{H}_{3}}$ has one lone pair and three bond pairs. In ${{H}_{2}}O$, there are two lone pairs and two bond pairs. 

The two lone pairs present on the oxygen atom of ${{H}_{2}}O$ molecule repel the two bond pairs. This repulsion is stronger than the repulsion between the lone pair and the three bond pairs on the nitrogen atom. 

Since the repulsions on the bond pairs in ${{H}_{2}}O$ molecule are greater than that in $N{{H}_{3}}$, the bond angle in water is less than that of ammonia.


9. How do you express the bond strength in terms of bond order?

Ans: Bond strength represents the extent of bonding between two atoms forming a molecule. The larger the bond energy, the stronger is the bond and the greater is the bond order.


10. Define Bond Length.

Ans: Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.

Bond length are expressed in terms of Angstrom (${{10}^{-10}}m$) or picometer (${{10}^{-12}}pm$) and are measured by spectroscopic X-ray diffractions and electron-diffraction techniques.

In an ionic compound, the bond length is the sum of the ionic radii of the constituting atoms ($d={{r}_{+}}+{{r}_{-}}$).

In a covalent compound, it is the sum of their covalent radii $(d={{r}_{A}}+{{r}_{B}})$ 


seo images

11. Explain the important aspects of resonance with reference to the $\mathbf{C{{O}_{3}}^{2-}}$ ion.

Ans: According to experimental findings, all carbon to oxygen bonds in $C{{O}_{3}}^{2-}$ are equivalent.

Hence, it is inadequate to represent $C{{O}_{3}}^{2-}$ ion by a single Lewis structure having two single bonds and one double bond.

Therefore, carbonate ion is described as a resonance hybrid of the following structures


12. $\mathbf{{{H}_{3}}P{{O}_{3}}}$ can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing$\mathbf{{{H}_{3}}P{{O}_{3}}}$? If not, give a reason for the same.


seo images

Ans: The given structures cannot be taken as the canonical forms of the resonance hybrid of ${{H}_{3}}P{{O}_{3}}$ because the positions of the atoms have changed.


13. Write the resonance structures for the following molecules. 

  1. $\mathbf{S{{O}_{3}}}$

Ans: 

seo images

  1. $\mathbf{N{{O}_{2}}}$ 

Ans:

seo images

  1. $\mathbf{N{{O}_{3}}^{-}}$ 

Ans: 


14. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions:

  1. K and S

Ans: the electronic configuration of K and S are as follows:

K: 2, 8, 8, 1

S: 2, 8, 6


seo images


   $K\cdot  \;\;\;\; \colon \ddot{S} \colon $  

Sulphur (S) requires 2 more electrons to complete its octet. Potassium (K) requires one electron more than the nearest noble gas i.e., Argon. Hence, the electron transfer can be shown as:


seo images

  1. Ca and O

Ans: The electronic configurations of Ca and O are as follows: 

Ca: 2, 8, 8, 2 

O: 2, 6 

Oxygen requires two electrons more to complete its octet, whereas calcium has two electrons more than the nearest noble gas i.e., Argon. Hence, the electron transfer takes place as:


seo images

  1. Al and N

Ans: The electronic configurations of Al and N are as follows: 

Al: 2, 8, 3 

N: 2, 5 

Nitrogen is three electrons short of the nearest noble gas (Neon), whereas aluminum has three electrons more than Neon. Hence, the electron transference can be shown as:


seo images

15. Although both $\mathbf{C{{O}_{2}}}$  and $\mathbf{{{H}_{2}}O}$ w are triatomic molecules, the shape of  $\mathbf{{{H}_{2}}O}$molecule is bent while that of $\mathbf{C{{O}_{2}}}$ is linear. Explain this on the basis of the dipole moment.

Ans: According to experimental results, the dipole moment of carbon dioxide is zero. This is possible only if the molecule is linear so that the dipole moments of C-O bonds are equal and opposite to nullify each other.


seo images

Resultant $\mu =0D$ 


seo images

${{H}_{2}}O$, on the other hand, has a dipole moment value of 1.84 D (though it is a triatomic molecule as $C{{O}_{2}}$ ). The value of the dipole moment suggests that the structure of ${{H}_{2}}O$ molecule is bent where the dipole moment of O-H bonds are unequal.


16. Write the significance/applications of dipole moment.

Ans: In heteronuclear molecules, polarization arises due to a difference in the electronegativities of the constituents of atoms. As a result, one end of the molecule acquires a positive charge while the other end becomes negative. Hence, a molecule is said to possess a dipole. 

The product of the magnitude of the charge and the distance between the centers of positive-negative charges is called the dipole moment ($\mu $) of the molecule. It is a vector quantity and is represented by an arrow with its tail at the positive center and head pointing towards the negative center.

Dipole moment ($\mu $) = charge (Q) X distance separation (r)

The S I unit of dipole moment is ‘esu’

\[1esu=3.335\times {{10}^{-30}}cm\] 

Dipole moment is the measure of the polarity of a bond. It is used to differentiate between polar and nonpolar bonds since all non-polar molecules (e.g, ${{H}_{2}},{{O}_{2}}$) have zero dipole moments. It is also helpful in calculating the percentage ionic character of a molecule.


seo images

17. Define electronegativity? How does it differ from electron gain enthalpy?

Ans: Electronegativity is the ability of an atom in a chemical compound to attract a bond pair of electrons towards itself. The electronegativity of any given element is not constant. It varies according to the element to which it is bonded. It is not a measurable quantity. It is only a relative number. 

On the other hand, electron gain enthalpy is the enthalpy change that takes place when an electron is added to a neutral gaseous atom to form an anion. It can be negative or positive depending upon whether the electron is added or removed. An element has a constant value of the electron gain enthalpy that can be measured experimentally.


18. Explain with the help of suitable example covalent bonds.

Ans: When two dissimilar atoms having different electronegativities combine to form a covalent bond, the bond pair of electrons are not shared equally. The bond pair shifts towards the nucleus of the atom having greater electronegativity. As a result, electron distribution gets distorted and the electron cloud is displaced towards the electronegative atom. 

As a result, the electronegative atom becomes slightly negatively charged while the other atom becomes slightly positively charged. Thus, opposite poles are developed in the molecule and this type of bond is called a polar covalent bond. 

HCl, for example, contains a polar covalent bond. Chlorine atoms are more electronegative than hydrogen atoms. Hence, the bond pair lies towards chlorine and therefore, it acquires a partial negative charge.


seo images

19. Arrange the bonds in order of increasing ionic character in the molecules: \[\mathbf{LiF,{{K}_{2}}O,{{N}_{2}},S{{O}_{2}} \And Cl{{F}_{3}}}\] 

Ans: The ionic character in a molecule is dependent upon the electronegativity difference between the constituting atoms. The greater the difference, the greater will be the ionic character of the molecule. 

On this basis, the order of increasing ionic character in the given molecules is

\[{{N}_{2}} < S{{O}_{2}} < Cl{{F}_{3}} < {{K}_{2}}O < LiF\] 


20. The skeletal structure of $C{{H}_{3}}COOH$ as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.


seo images


Ans: The correct Lewis structure for acetic acid is as follows:


seo images

21. Apart from tetrahedral geometry, another possible geometry for $\mathbf{C{{H}_{4}}}$  is square planar with the four H atoms at the corners of the square and the C atom at its center. Explain why $\mathbf{C{{H}_{4}}}$ is not square planar?

Ans: The electronic configuration of carbon atom is:

\[_{6}C=1{{s}^{2}}2{{s}^{2}}2{{p}^{2}}\] 

In the excited state, the orbital picture of carbon can be represented as:


seo images

Hence, the carbon atom undergoes $S{{p}^{3}}$ hybridization in $C{{H}_{4}}$ molecule and takes a tetrahedral shape.


seo images

For a square planar shape, the hybridization of the central atom has to be $ds{{p}^{2}}$. 

However, an atom of carbon does not have d-orbitals to undergo $ds{{p}^{2}}$ hybridization. 

Hence, the structure of $C{{H}_{4}}$ cannot be square planar. 

Moreover, with a bond angle of ${{90}^{0}}$  in square planar, the stability of $C{{H}_{4}}$ will be very less because of the repulsion existing between the bond pairs. Hence, VSEPR theory also supports a tetrahedral structure for$C{{H}_{4}}$.


22. Explain why the $\mathbf{Be{{H}_{2}}}$  molecule has a zero dipole moment although the Be-H bonds are polar.

Ans: the Lewis structure for $Be{{H}_{2}}$is as follows:


seo images

 There is no lone pair at the central atom (Be) and there are two bond pairs. Hence, $Be{{H}_{2}}$ is of the type $A{{B}_{2}}$ . It has a linear structure. 


seo images

Dipole moments of each H-Be bond are equal and are in opposite directions. Therefore, they nullify each other. Hence, $Be{{H}_{2}}$ molecule has zero dipole moment.


23. Which out of $\mathbf{N{{H}_{3}}}$  and $\mathbf{N{{F}_{3}}}$  has a higher dipole moment and why?

Ans: In both molecules, i.e $N{{H}_{3}}$ and $N{{F}_{3}}$, the central atom (N) has a lone pair electron and there are three bond pairs. Hence, both molecules have a pyramidal shape. Since fluorine is more electronegative than hydrogen, it is expected that the net dipole moment of $N{{F}_{3}}$ is greater than $N{{H}_{3}}$. However, the net dipole moment of $N{{H}_{3}}$ (1.46 D) is greater than that of $N{{F}_{3}}$ (0.24 D). 

This can be explained on the basis of directions of the dipole moments of each individual bond is $N{{H}_{3}}$ and $N{{F}_{3}}$. These directions can be shown as:


seo images

Thus, the resultant moment of the N-H bonds adds up to the bond moment of the lone pair (the two being in the same direction), whereas that of the three N – F bonds partly cancels the moment of the lone pair. 

Hence, the net dipole moment of $N{{F}_{3}}$ is less than that of $N{{H}_{3}}$.


24. What is meant by hybridization of atomic orbitals? Describe the shapes of $\mathbf{sp,s{{p}^{2}},s{{p}^{3}}}$  hybrid orbitals.

Ans: Hybridization is defined as an intermixing of a set of atomic orbitals of slightly different energies, thereby forming a new set of orbitals having equivalent energies and shapes. 

For example, one 2s-orbital hybridizes with two 2p-orbitals of carbon to form three new $s{{p}^{2}}$ hybrid orbitals. 

These hybrid orbitals have minimum repulsion between their electron pairs and thus are more stable. Hybridization helps indicate the geometry of the molecule. 

The shape of sp hybrid orbitals: sp hybrid orbitals have a linear shape. They are formed by the intermixing of s and p orbitals as:


seo images

Shape of $s{{p}^{2}}$ hybrid orbitals: 

$s{{p}^{2}}$ hybrid orbitals are formed as a result of the intermixing of one s-orbital and two 2p-orbitals. The hybrid orbitals are oriented in a trigonal planar arrangement as: 


seo images

Shape of $s{{p}^{3}}$  hybrid orbitals: 

Four $s{{p}^{3}}$ hybrid orbitals are formed by intermixing one s-orbital with three p-orbitals. 

The four $s{{p}^{3}}$ hybrid orbitals are arranged in the form of a tetrahedron as:


seo images

25. Describe the change in hybridization (if any) of the Al atom in the following reaction.\[\mathbf{AlC{{l}_{3}}+C{{l}^{-}}\to AlC{{l}_{4}}^{-}}\] 

Ans: The valence orbital picture of aluminum in the ground state can be represented as:


seo images

The orbital picture of aluminum in the excited state can be represented as:


seo images

Hence, it undergoes $s{{p}^{2}}$  hybridization to give a trigonal planar arrangement (in $AlC{{l}_{3}}$ ). To form, the empty $3{{p}_{z}}$  orbital also gets involved and the hybridization changes from $s{{p}^{2}}$ to $s{{p}^{3}}$. As a result, the shape gets changed to tetrahedral.


26. Is there any change in the hybridization of B and N atoms as a result of the following reaction? \[\mathbf{B{{F}_{3}}+N{{H}_{3}}\to {{F}_{3}}B.N{{H}_{3}}}\] 

Ans: Boron atoms in $B{{F}_{3}}$ is $s{{p}^{2}}$ hybridized. The orbital picture of boron in the excited state can be shown as:


seo images

Nitrogen atom in $N{{H}_{3}}$  is $s{{p}^{3}}$  hybridized. The orbital picture of nitrogen can be represented as: 


seo images

After the reaction has occurred, an adduct ${{F}_{3}}B.N{{H}_{3}}$  is formed as hybridization of ‘B’ changes to $s{{p}^{3}}$ . However, the hybridization of ‘N’ remains intact.


27. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in $\mathbf{{{C}_{2}}{{H}_{2}}}$ and $\mathbf{{{C}_{2}}{{H}_{4}}}$ molecules.

Ans: 

$\mathbf{{{C}_{2}}{{H}_{4}}}$ :

The electronic configuration of C-atom in the excited state is:

\[_{6}C=1{{s}^{2}}2{{s}^{1}}2p_{x}^{1}2p_{y}^{1}2p_{z}^{1}\]

In the formation of an ethane molecule (${{C}_{2}}{{H}_{4}}$), one sp2 hybrid orbital of carbon overlaps a $s{{p}^{2}}$  hybridized orbital of another carbon atom, thereby forming a C-C sigma bond. The remaining two $s{{p}^{2}}$ orbitals of each carbon atom from a $s{{p}^{2}}-s$ sigma bond with two hydrogen atoms. The unhybridized orbital of one carbon atom undergoes sidewise overlap with the orbital of a similar kind present on another carbon atom to form a weak n-bond.


seo images

$\mathbf{{{C}_{2}}{{H}_{2}}}$:

In the formation of ${{C}_{2}}{{H}_{2}}$ molecule, each C-atom is sp hybridized with two 2p-orbitals in an unhybridized state. 

One sp orbital of each carbon atom overlaps with the other along the internuclear axis forming a C-C sigma bond. The second sp orbital if each C-atom overlaps a half-filled 1s-orbital to form a $\sigma $ bond. 

The two unhybridized 2p-orbitals of the first carbon undergo sidewise overlap with the 2p orbital of another carbon atom, thereby forming two pi (n) bonds between carbon atoms. Hence, the triple bond between two carbon atoms is made up of one sigma and two n-bonds.


seo images

28. What is the total number of sigma and pi bonds in the following molecules?

  1. $\mathbf{{{C}_{2}}{{H}_{2}}}$

Ans: A single bond is a result of that axial overlap of bonding orbitals. Hence, it contributes a sigma bond. A multiple bond (double or triple bond) is always formed as a result of the sidewise overlap of orbitals. A pi-bond is always present in it. A triple bond is a combination of two pi-bonds and one sigma bond.

Structure of ${{C}_{2}}{{H}_{2}}$ can be represented as:


seo images

Hence, there are three sigma bonds and two pi-bonds in ${{C}_{2}}{{H}_{2}}$.

  1. $\mathbf{{{C}_{2}}{{H}_{4}}}$

The structure of ${{C}_{2}}{{H}_{4}}$ can be represented as:


seo images

Hence, there are five sigma bonds and one pi-bond in ${{C}_{2}}{{H}_{4}}$ .


2
9. Considering x-axis as the inter-nuclear which out of the following will not form a sigma bond and why? 

  1. 1s and 1s 

  2. 1s and $2{{p}_{x}}$ 

  3. $2{{p}_{y}}$ and $2{{p}_{z}}$ 

  4. 1s and 2s.

Ans: $2{{p}_{y}}$ and $2{{p}_{z}}$ orbitals will not form a sigma bond. Taking x-axis as the internuclear axis, $2{{p}_{y}}$ and $2{{p}_{z}}$ orbitals will undergo lateral overlapping, thereby is forming a pi (n) bond. Option (c)


30. Which hybrid orbitals are used by carbon atoms in the following molecules?

  1. $\mathbf{C{{H}_{3}}-C{{H}_{3}}}$ 

Ans: 


seo images

Both ${{C}_{1}}$ and ${{C}_{2}}$ are $s{{p}^{3}}$ hybridized

  1. $\mathbf{C{{H}_{3}}-CH=C{{H}_{2}}}$ 

Ans: 

seo images


${{C}_{1}}$ is $s{{p}^{3}}$ hybridized, ${{C}_{2}}$and ${{C}_{3}}$ are $s{{p}^{2}}$ hybridized

  1. $\mathbf{C{{H}_{3}}-C{{H}_{2}}-OH}$ 

Ans: 

seo images

Both ${{C}_{1}}$ and ${{C}_{2}}$ are $s{{p}^{3}}$ hybridized

  1. $\mathbf{C{{H}_{3}}CHO}$ 

Ans: 

${{C}_{1}}$ is $s{{p}^{3}}$ hybridized, ${{C}_{2}}$ is $s{{p}^{2}}$ hybridized. As the second carbon is present as an aldehyde group which contains double bond with oxygen atom and single bond with hydrogen atom.

  1. $\mathbf{C{{H}_{3}}COOH}$ 

Ans: 

seo images

${{C}_{1}}$ is $s{{p}^{3}}$ hybridized, ${{C}_{2}}$ is $s{{p}^{2}}$ hybridized


31. What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.

Ans: When two atoms combine by sharing their one or more valence electrons, a covalent bond is formed between them. 

The shared pairs of electrons present between the bonded atoms are called bond pairs. 

All valence electrons may not participate in bonding. The electron pairs that do not participate in bonding are called lone pairs of electrons. 

For example, in ${{C}_{2}}{{H}_{6}}$  (ethane), there are seven bond pairs but no lone pair present.


seo images

In ${{H}_{2}}O$ , there are two bond pairs and two lone pairs on the central atom (oxygen).


seo images

32. Distinguish between a sigma bond and a pi bond.

Ans: The following are the differences between sigma and pi bonds

s.no

Sigma ($\mathbf{\sigma} $) Bond

Pi ($\mathbf{\pi} $ ) Bond

1

It is formed by the end to end overlap of orbitals

It is formed by the lateral overlap of orbitals

2

The orbitals involved in the overlapping are s-s, s-p or p-p

These bonds are formed by the overlap of p-p orbitals only

3

It is a strong bond

It is a weak bond

4

The electron cloud is symmetrical about a line joining two nuclei

The electron cloud is not symmetrical

5

It consists of one electron cloud which is symmetrical about the internuclear axis.

There are two electron clouds lying above the plane of atomic nuclei.

6

Free rotation about 

$\sigma $-bonds is possible

Rotation restricted in case of $\pi $-bonds


33. Explain the formation of H2 molecules on the basis of valence bond theory.

Ans: Let us assume the two hydrogen atoms (A and B) with nuclei (${{N}_{A}}$  and ${{N}_{B}}$ ) and electrons (${{e}_{A}}$  and ${{e}_{B}}$ ) are taken to undergo a reaction to form a hydrogen molecule. 

When A and B are at a large distance, there is no interaction between them. As they begin to approach each other, the attractive and repulsive forces start operating. 

Attractive Force Arises Between:

  1. Nucleus of one atom and its own electron i.e., ${{N}_{A}}$ – ${{e}_{A}}$ and ${{N}_{B}}$- ${{e}_{B}}$. 

  2. The nucleus of one atom and electron of another atom i.e., ${{N}_{A}}$ – ${{e}_{B}}$ and ${{N}_{B}}$ –${{e}_{A}}$ . 

Repulsive Force Arises Between: 

  1. Electrons of two atoms i.e., eA – eB.

  2.  Nuclei of two atoms i.e., NA – NB. 

The force of attraction brings the two atoms together, whereas the force of repulsion tends to push them apart.


seo images

The magnitude of the attractive forces is more than that of the repulsive forces. Hence, the two atoms approach each other. As a result, the potential energy decreases. Finally, a state is reached when the attractive forces balance the repulsive forces and the system acquires minimum energy. This leads to the formation of a dihydrogen molecule.


34. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Ans: The given conditions should be satisfied by atomic orbitals of form molecular orbitals:

  1. The combining atomic orbitals must have an equivalent or nearly an equivalent energy. This means that during a homonuclear molecule, the 1s-atomic orbital of an atom can combine with the 1s-atomic orbital of another atom, and not with the 2s-orbital.

  2. The combining atomic orbitals must have proper orientations to make sure that the overlap is maximum.

  3. The extent of overlapping should be large


35. Use molecular orbital theory to explain why the $\mathbf{B{{e}_{2}}}$  molecule does not exist.

Ans: The electronic configuration of Beryllium is $1{{s}^{2}}2{{s}^{2}}$ 

The molecular orbital electronic configuration for $B{{e}_{2}}$ molecule can be written as:

\[\sigma _{1s}^{2}\sigma _{1s}^{*2}\sigma _{2s}^{2}\sigma _{2s}^{*2}\] 

Hence, the bond order for $B{{e}_{2}}$is $\frac{1}{2}({{N}_{b}}-{{N}_{a}})$ 

Where, ${{N}_{b}}$ = number of electrons in bonding orbitals

${{N}_{a}}$ = number of electrons in non-bonding orbitals

$\therefore $ Bond order of $B{{e}_{2}}=\frac{1}{2}(4-4)=0$ 

A negative or zero bond order means that the molecule is unstable. Hence, Be2 molecule does not exist.


36. Compare the relative stability of the following species and indicate their magnetic properties;

\[\mathbf{{{O}_{2}},O_{2}^{+},O_{2}^{2-}(\text{superoxide)},O_{2}^{2-}(peroxide)}\] 

Ans: There are 16 electrons in a molecule of dioxygen, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

\[{{[\sigma (1s)]}^{2}}{{[{{\sigma }^{*}}(1s)]}^{2}}{{[\sigma (2s)]}^{2}}{{[\sigma *(2s)]}^{2}}{{[\sigma (1{{p}_{z}})]}^{2}}{{[\pi (2{{p}_{x}})]}^{2}}{{[\pi (2{{p}_{y}})]}^{2}}{{[{{\pi }^{*}}(2{{p}_{x}})]}^{1}}{{[{{\pi }^{*}}(2{{p}_{y}})]}^{1}}\] 

Since the 1s orbital of each oxygen atom is not involved in bonding, the number of bonding electrons = 8 = ${{N}_{b}}$  and the number of antibonding orbitals = 4 = ${{N}_{a}}$ .

Bond order = $\frac{1}{2}({{N}_{b}}-{{N}_{a}})=\frac{1}{2}(8-4)=2$

Similarly the electronic configuration of $O_{2}^{+}$ can be written as:

\[KK{{[\sigma (2s)]}^{2}}{{[{{\sigma }^{*}}(2s)]}^{2}}{{[\sigma (2{{p}_{z}})]}^{2}}{{[\pi (2{{p}_{x}})]}^{2}}{{[\pi (2{{p}_{y}})]}^{2}}{{[{{\pi }^{*}}(2{{p}_{x}})]}^{1}}\] 

${{N}_{b}}$=8

${{N}_{a}}$ = 3

Bond order of $O_{2}^{+}$= $\frac{1}{2}(8-3)=2.5$ 

Similarly the electronic configuration of $O_{2}^{-}$ can be written as:

\[KK{{[\sigma (2s)]}^{2}}{{[{{\sigma }^{*}}(2s)]}^{2}}{{[\sigma (2{{p}_{z}})]}^{2}}{{[\pi (2{{p}_{x}})]}^{2}}{{[\pi (2{{p}_{y}})]}^{2}}{{[{{\pi }^{*}}(2{{p}_{x}})]}^{2}}{{[{{\pi }^{*}}(2{{p}_{y}})]}^{1}}\] 

${{N}_{b}}$=8

${{N}_{a}}$ = 5

Bond order of $O_{2}^{-}$= $\frac{1}{2}(8-5)=1.5$

Similarly the electronic configuration of $O_{2}^{2-}$  can be written as:

\[KK{{[\sigma (2s)]}^{2}}{{[{{\sigma }^{*}}(2s)]}^{2}}{{[\sigma (2{{p}_{z}})]}^{2}}{{[\pi (2{{p}_{x}})]}^{2}}{{[\pi (2{{p}_{y}})]}^{2}}{{[{{\pi }^{*}}(2{{p}_{x}})]}^{2}}{{[{{\pi }^{*}}(2{{p}_{y}})]}^{2}}\]

${{N}_{b}}$=8

${{N}_{a}}$ = 6

Bond order of $O_{2}^{2-}$= $\frac{1}{2}(8-6)=1$ 

Bond dissociation energy is directly proportional to bond order. Thus, the higher the bond order, the greater will be the stability. On this basis, the order of stability is

\[O_{2}^{+} > {{O}_{2}} > O_{2}^{-} > O_{2}^{2-}\] 


37. Write the significance of a plus and a minus sign shown in representing the orbitals.

Ans: Molecular orbitals are represented by wave functions. A plus sign in an orbital indicates a positive wave function while a minus sign in an orbital represents a negative wave function.


38. Describe the hybridization in case of $\mathbf{PC{{l}_{5}}}$. Why are the axial bonds longer as compared to equatorial bonds?

Ans: the ground state and excited state outer electronic configuration of phosphorus (Z=15) are:


seo images

Phosphorus atom is $s{{p}^{3}}d$  hybridized in the excited state. These orbitals are filled by the electron pairs donated by five Cl atoms as:


seo images

The five $s{{p}^{3}}d$ hybrid orbitals are directed towards the five corners of the trigonal bipyramidal. Hence, the geometry of $PC{{l}_{5}}$  can be represented as


seo images

There are five P-Cl sigma bonds in $PC{{l}_{5}}$. Three P-Cl bonds lie in one plane and make an angle of ${{120}^{0}}$  with each other. These bonds are called equatorial bonds. 

The remaining two P-Cl bonds lie above and below the equatorial plane and make an angle of ${{90}^{0}}$  with the plane. These bonds are called axial bonds. 

As the axial bond pairs suffer more repulsion from the equatorial bond pairs, axial bonds are slightly longer than equatorial bonds.


39. Define hydrogen bonds. Is it weaker or stronger than the van der Waals forces?

Ans: A hydrogen bond is defined as an attractive force acting between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule (may be of the same kind). 

Due to a difference between electro-negativities, the bond pair between hydrogen and the electronegative atom gets drifted away from the hydrogen atom. As a result, a hydrogen atom becomes electropositive with respect to the other atom and acquires a positive charge.

${{H}^{\delta +}}-{{X}^{\delta -}}$ ……..${{H}^{\delta +}}-{{X}^{\delta -}}$……. ${{H}^{\delta +}}-{{X}^{\delta -}}$

The magnitude of H-bonding is maximum in the solid state and minimum in the gaseous state. 

There are two types of H-Bonds: 

  1. Intermolecular H-bond e.g., $HF,{{H}_{2}}O$  etc.

  2. Intramolecular H-bond e.g., o-nitrophenol 


seo images

Hydrogen bonds are stronger than Vander Waals forces since hydrogen bonds are regarded as an extreme form of dipole-dipole interaction.


40. What is meant by the term bond order? Calculate the bond order of:

$\mathbf{{{N}_{2}},{{O}_{2}},O_{2}^{+},O_{2}^{-}}$ 

Ans: Bond order is defined as one half of the difference between the number of electrons present in the bonding and antibonding orbitals of a molecule. 

If  ${{N}_{a}}$ is equal to the number of electrons in an antibonding orbital, then  ${{N}_{b}}$ is equal to the number of electrons in a bonding orbital.

Bond order = $\dfrac{1}{2}({{N}_{b}}-{{N}_{a}})$

If ${{N}_{b}}>{{N}_{a}}$ , then the molecule is said to be stable.

However, ${{N}_{b}}\le {{N}_{a}}$ , then the molecule is considered to be unstable. 

Bond order of ${{N}_{2}}$  can be calculated from its electronic configuration as:

\[{{[\sigma (1s)]}^{2}}{{[{{\sigma }^{*}}(1s)]}^{2}}{{[\sigma (2s)]}^{2}}{{[\sigma *(2s)]}^{2}}{{[\pi (2{{p}_{x}})]}^{2}}{{[\pi (2{{p}_{y}})]}^{2}}{{[\pi (2{{p}_{z}})]}^{2}}\]

${{N}_{b}}$=10

${{N}_{a}}$ = 4

Bond order of nitrogen molecule = $\frac{1}{2}(10-4)=3$

There are 16 electrons in a molecule of dioxygen, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

\[{{[\sigma (1s)]}^{2}}{{[{{\sigma }^{*}}(1s)]}^{2}}{{[\sigma (2s)]}^{2}}{{[\sigma *(2s)]}^{2}}{{[\sigma (1{{p}_{z}})]}^{2}}{{[\pi (2{{p}_{x}})]}^{2}}{{[\pi (2{{p}_{y}})]}^{2}}{{[{{\pi }^{*}}(2{{p}_{x}})]}^{1}}{{[{{\pi }^{*}}(2{{p}_{y}})]}^{1}}\] 

Since the 1s orbital of each oxygen atom is not involved in bonding, the number of bonding electrons = 8 = ${{N}_{b}}$  and the number of antibonding orbitals = 4 = ${{N}_{a}}$ .

Bond order = $\frac{1}{2}({{N}_{b}}-{{N}_{a}})=\frac{1}{2}(8-4)=2$

Hence, the bond order of oxygen molecule is 2

Similarly the electronic configuration of $O_{2}^{+}$ can be written as:

\[KK{{[\sigma (2s)]}^{2}}{{[{{\sigma }^{*}}(2s)]}^{2}}{{[\sigma (2{{p}_{z}})]}^{2}}{{[\pi (2{{p}_{x}})]}^{2}}{{[\pi (2{{p}_{y}})]}^{2}}{{[{{\pi }^{*}}(2{{p}_{x}})]}^{1}}\] 

${{N}_{b}}$=8

${{N}_{a}}$ = 3

Bond order of $O_{2}^{+}$= $\frac{1}{2}(8-3)=2.5$ 

Similarly the electronic configuration of $O_{2}^{-}$ can be written as:

\[KK{{[\sigma (2s)]}^{2}}{{[{{\sigma }^{*}}(2s)]}^{2}}{{[\sigma (2{{p}_{z}})]}^{2}}{{[\pi (2{{p}_{x}})]}^{2}}{{[\pi (2{{p}_{y}})]}^{2}}{{[{{\pi }^{*}}(2{{p}_{x}})]}^{2}}{{[{{\pi }^{*}}(2{{p}_{y}})]}^{1}}\] 

${{N}_{b}}$=8

${{N}_{a}}$ = 5

Bond order of $O_{2}^{-}$= $\frac{1}{2}(8-5)=1.5$


Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure Quick Overview of Topics

Class 11 Chemistry Chapter 4 NCERT Solutions -Quick Overview of Detailed Structure of Topics and Subtopics Covered in Chemical Bonding and Molecular Structure.


Topic

Subtopics

Chemical Bonding

- Types of chemical bonds

- Ionic, covalent, and metallic bonds

- Lewis structure and octet rule

- Formal charge and resonance

Molecular Structure

- Molecular geometry and shape

- VSEPR theory and electron pair repulsion

- Hybridization of atomic orbitals

- Molecular orbital theory and bond order

Intermolecular Forces

- Van der Waals forces

- Hydrogen bonding and dipole-dipole interactions

- London dispersion forces

Ionic Solids

- Crystal lattice structure

- Properties and applications

Covalent Solids

- Network and molecular solids


- Properties and examples

Metallic Solids

- Metallic bonding and electron sea model

- Properties and applications


Some Important Concepts for Chemical Bonding and Molecular Structure

Class 11 NCERT solutions help the students to go through the formulas and concepts easily. Here find the Important formulas and concepts of Chapter 4 - Chemical Bonding and Molecular Structure to crack your exams.


  1. Lewis Dot Structures:

    • Used to represent the valence electrons of atoms and their bonding patterns in molecules.

    • Octet Rule: Atoms tend to gain, lose, or share electrons to achieve a noble gas electron configuration.


  1. Ionic Bonding:

    • Forms between a metal and a nonmetal.

    • Electrostatic attraction between oppositely charged ions.


  1. Covalent Bonding:

    • Forms between nonmetal atoms by sharing of electrons.

    • Polar Covalent Bond: Unequal sharing of electrons due to differences in electronegativity.

    • Non-polar Covalent Bond: Equal sharing of electrons.


  1. Valence Shell Electron Pair Repulsion (VSEPR) Theory:

    • Predicts the geometry of molecules based on the repulsion between electron pairs.

    • Shapes include linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral, among others.


  1. Molecular Orbital Theory:

    • Describes the behaviour of electrons in molecules using molecular orbitals formed by the overlap of atomic orbitals.

    • Bonding Molecular Orbital: Formed by in-phase overlap of atomic orbitals.

    • Antibonding Molecular Orbital: Formed by out-of-phase overlap of atomic orbitals.


Benefits of Referring to Vedantu’s NCERT Solutions for class 11 Chemistry Chapter 4

By referring to Vedantu’s NCERT Solutions for Class 11 Chemistry Chapter 4: Chemical Bonding and Molecular Structure, students can build a strong foundation, excel in their exams, and develop a deeper understanding of the essential principles of chemical bonding. Here are few Benefits of Referring to Vedantu’s NCERT Solutions for class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure:


  • Vedantu’s NCERT solutions provide thorough explanations for all topics in Chapter 4, ensuring a deep understanding of chemical bonding and molecular structure.

  • Includes detailed discussions on ionic bonds, covalent bonds, and coordinate bonds.

  • Covers important concepts such as bond parameters, bond length, bond angle, bond enthalpy, and bond order.

  • Each concept is explained in a clear and concise manner, using precise scientific terminology.

  • Detailed analysis of VSEPR theory, valence bond theory, and molecular orbital theory.

  • Aligned with CBSE Syllabus: Solutions are prepared to the latest CBSE syllabus and exam pattern.

  • Important Topics: Detailed explanation of hybridization, molecular shapes, and the concept of resonance.

  • Solutions are available for free download, allowing easy offline access.


Related Study Materials for Class 11 Chemistry Chapter 4  


Conclusion

Chemical bonding is the core concept of Chemistry taught at the Higher Secondary level of education. Learning how atoms bond with each other to form molecules is the basis of studying this subject. Hence, this chapter is of utmost importance for the development of a strong conceptual foundation of chemistry among students.


Students will study the concepts and chemical principles explained in this chapter and will proceed to solve the exercise questions. This is where the solutions for these exercises will come in very handy. The experts have solved all the questions included in each exercise of Class 11 Chemistry Chemical Bonding and Molecular Structure and provided accurate answers to learn and follow.


All the answers have been formulated by following the CBSE Class 11 standards so that students can easily comprehend the context of the questions. Studying these answers directly will be ideal to follow how to attempt such questions during an exam.


NCERT Solutions Class 11 Chemistry | Chapter-wise Links 

Access Vedantu’s chapter-wise NCERT Chemistry Class 11 Solutions PDFs below for all other chapters.



NCERT Solutions Class 11 Chemistry - Related Links

FAQs on NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

1. Give a brief overview of the chapter chemical bonding and molecular structure in NCERT Class 11.

Chemical bonding and molecular structure in NCERT Class 11 deals with the core concepts of molecular structures along with the chemical bonding. This chapter will introduce the concepts of an ionic bond, valence electrons, covalent bonds, etc. Furthermore, students will also learn about the Lewis structure, bond parameters, polar character of covalent bonds, valence bond theory, the geometry of covalent molecules, the covalent character of an ionic bond, etc. You will also read about balanced chemical equations, types of chemical equations, combination reaction, decomposition reaction, displacement reaction, corrosion and rancidity. Further, the students will also be introduced to the concept of hybridization, s,  p and d orbitals, VSEPR theory, the molecular orbital theory of homonuclear diatomic molecules hydrogen bond, shapes of simple molecules, etc.

2. What are the topics that are covered in chapter 4 chemical bonding and molecular structure?

  • Octet Rule

    • Covalent Bond

    • Lewis Representation of Simple Molecules (The Lewis Structures)

    • Formal Charge

    • Limitations of the Octet Rule

  • Ionic or Electrovalent Bond

    • Lattice Enthalpy

  • Bond Parameters

    • Bond Length

    • Bond Angle

    • Bond Enthalpy

    • Bond Order

    • Resonance Structures

    • 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

  • Hybridisation

    • Types of Hybridisation

    • Other Examples of Sp3, Sp2 and Sp Hybridisation

    • The Hybridisation of Elements Involving D Orbitals

  • 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

  • Hydrogen Bonding

    • Cause of Formation of Hydrogen Bond

    • Types of H-bonds.

3. Help me with the process of Chemistry preparation with Vedantu?

Preparing in a step by step and structured manner will help in understanding the concepts in Chemistry easier. First, consider the most scoring and easy concept to score higher marks. The weightage of this topic is also high. You may think of Qualitatives analysis towards the end, as it needs very less time.


chemical bonding class 11 NCERT pdf gives you an in-depth knowledge of the conceptual topics. NCERT Solutions have been drafted as per the latest CBSE Class 11 Science Syllabus. Students will feel that the solutions are in a simple language and can understand the difficult topics easily.

4. What is hybridization?

Redistribution of the energy of orbitals of individual atoms to give orbitals of equivalent energy happens when two atomic orbitals combine together to form hybrid orbital in a molecule. This process is called hybridization. The new orbitals which are formed are known as hybrid orbitals.


Types of Hybridization:

  • sp Hybridization

  • sp2 Hybridization

  • sp3 Hybridization

  • sp3d Hybridization

  • sp3d2 Hybridization

5. Give me a summary of NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure.

NCERT Solutions for Class 11 Chemistry Chapter deals primarily with a chemical bond and the attractive force that acts upon it. It talks about atoms, ions, and molecules that all come together to form any given compound. In addition to this, the formation of a  chemical bond has been explained via several theories, including the valence shell electron pair repulsion theory, the electronic theory, and the molecular orbital theory. 

6. Is it necessary to learn all the questions present in NCERT Solutions for Class 11 Chemistry Chapter 4?

In NCERT Solutions for Class 11 Chemistry Chapter 4, important questions are discussed with their solutions. If you study these questions, you should do fairly well in all your examinations. Thus, although it is not necessary per se to learn all the questions present in  chemical bonding NCERT Pdf it will help you in your studies if you do learn them.

7. What is the VSEPR theory?

The Valence Shell Electron Pair Repulsion (VSEPR) theory provides a simple procedure to predict the shapes of covalent molecules. Sidgwick and Powell in 1940 proposed this simple theory based on the repulsive interactions of the electron pairs in the valence shell of the atoms. It was further developed and redefined by Nyholm and Gillespie and stated that the shape of a molecule depends upon the number of valence shell electron pairs around the central atom. Chemical bonding class 11 NCERT pdf provides the best practice to get expertise in the chapter.

8. How can I understand the chapter Chemical Bonding and Molecular structure?

Vedantu provides an excellent repository of study materials for students. In case you are having difficulty understanding any of the chapters in your syllabus, you can check out Vedantu’s online classes or refer to their NCERT solutions. If you read chapters 3 to 4 times and solve questions using NCERT Solutions for Class 11 Chemistry Chapter 4, you will better understand the chapter Chemical Bonding and Molecular Structure. The solutions are free of cost and also available on the Vedantu Mobile app.

9. What are the directional properties of bonds?

Any covalent bond in nature is formed by the overlapping of atomic orbitals. The molecule of hydrogen is formed due to the overlap of 1s-orbitals of two hydrogen atoms. In the case of polyatomic molecules, the geometry of the molecules is also important. The VSEPR theory plays a role in dictating the molecule’s geometry, but the valence bond theory is better suited to explain the directional properties of bonds. Students can also refer to chemistry class 11 chapter 4.

10. What are the most important topics of chemical bonding class 11 NCERT solutions?

Chemical bonding class 11 NCERT solutions covers stoichiometry and chemical reactions, emphasising fundamental concepts and calculations.

11. What are important questions in Chemical Bonding and Molecular Structure class 11?

Chemical bonding Class 11 Solutions  revolves around key topics like Lewis structures, molecular geometry, and intermolecular forces.

12. What keeps chemical bonds together?

Electrostatic attraction is the primary force that keeps chemical bonds together, maintaining stability in molecules and compounds. For better understanding Students vedantu also provides chemical bonding class 11 solutions for better exam preparations.

13. What is the most important part in Chemical Bonding and Molecular Structure?

Understanding electron arrangement and the sharing or transfer of electrons is crucial for comprehending the essence of chemical bonding. Students can also refer to chapter 4 chemistry class 11 for better understanding.

14. What are the two main types of chemical bonding?

The two primary types of chemical bonding are ionic bonding, involving transfer of electrons, and covalent bonding, involving sharing of electrons between atoms.