Important Questions for CBSE Class 11 Chemistry Chapter 4 - Chemical Bonding and Molecular Structure

Very Short Answer Questions                                                                  1 Marks

1. Define a chemical bond.

Ans:  Chemical bond is defined as the attractive force which is responsible for the holding or we can say which holds the different constituents of atoms like electrons, protons or neutrons or any other ions together in various chemical species.
 

2. Give the main feature of Lewis' approach of chemical bonding.   

Ans: Lewis mainly approached that the atoms attain the stable octet when they are linked with the help of chemical bonds. He also discovered that atoms contain a positively charged center called nucleus and the outer shell of the atom can attain a maximum of eight electrons. These electrons are present at the corners of a cube and revolve around the center. Lewis also introduced Lewis symbols which are simple notations that represent valence electrons in an atom.

3. Write electron dot structure (Lewis structure) of $Na,\text{ }Ca,\text{ }B,\text{ }Br,\text{ }Xe,\text{ }As,\text{ }Ge,\text{ }{{N}^{3-}}$.                                                       

Ans: Lewis symbols are simple notations that represent valence electrons in an atom with the help of a dot so they are also known as electron dot structure or lewis structure.

4. Give the octet rule in short.

Ans: Octet word corresponds to the number eight so it gives us information about the atoms which tend to adjust the arrangement of their electrons in such a way that they achieve eight electrons in their outermost shell. Except hydrogen and helium as they will need only 2 electrons to complete their octet.

5. Define an ionic bonding.                                                                          

Ans: An ionic bond which is also known as electrovalent bond can be defined as the bond which is formed by the complete transference of one or more of outermost electrons from the atom of any metal to that of a non – metal.

6. Which one of the following has the highest bond order? ${{N}_{2}},N_{2}^{+}+or\text{ }N_{2}^{-}$. 

Ans: Bond order is generally defined as the number of bonds present between two atoms and in this case ${{N}_{2}}$ has the highest bond order as compared to others.

7. Define bond order. 

Ans:  Bond order is defined as the number of bonds between two atoms in a molecule. It can be calculated by the difference between bonding and antibonding number of electrons. 

8. What type of bond is formed when atoms have a high difference of electronegativity?

Ans:  When atoms have a high difference of electronegativity and bond formed is known as electrovalent or ionic bond.     

9. Define dipole moment.                                                                    

Ans: Dipole moment is defined as the product of the magnitude of the charge and the distance between the centers of positive and negative charge present in any compound.

10. Give the mathematical expression of the dipole moment.                        

Ans: Mathematically dipole moment is product of magnitude of charge represented by the letter Q and distance represented by r whereas dipole moment is represented by M. Dipole moment is usually expressed in Debye units $\left(D\right)$. Expression can be written as follows:  

Dipole moment $\left( M \right)=charge\left( Q \right)\times $ distance of separation $\left( r \right)$. 

11. Why is the dipole moment of $C{{O}_{2}},B{{F}_{3}},CC{{l}_{4}}$ is zero?                                       

Ans: Dipole moment of $C{{O}_{2}},B{{F}_{3}},CC{{l}_{4}}$ is zero this can be explained on the basis of the shape of these molecules these all have symmetrical shapes which cancels the  dipoles effect of each other and their net dipole moment become zero.

12.  Why is $B{{F}_{3}}$ non-polar?                                                                               

Ans: $B{{F}_{3}}$ is said to be non-polar in nature; this can also be explained on the basis of its symmetrical shape due to which the net dipole moment becomes zero and it becomes non – polar in nature.

13. Write the resonating structure of the ${{O}_{3}}$ molecule.                                   

Ans: Resonating structure generally gives the transference of electrons which can be shown as follows: 

14.  What is a sigma bond?                                                                                       

Ans: Sigma bond is a type of covalent bond which is formed due to the overlapping of orbitals of the two atoms along their same orbital axis.

15. What is pi – bond?

Ans: Pi bond is also a type of covalent bond which formed between the two atoms due to their sideways overlap along their $p$ – orbitals.

16.  How many $s$- and $\pi $- bonds are there in a molecule of ${{C}_{2}}{{H}_{4}}$ (ethane).

Ans: In a molecule of ethane which is represented by the molecular formula ${{C}_{2}}{{H}_{4}}$ generally contains $5s$- bonds out of which one is between $C-C$ and four are between $C-H$ along with this it contains one $\pi $- bond.

17. How many $s$- and $\pi $- bonds are there in a molecule of $C{{H}_{2}}=CH-CH=C{{H}_{2}}$.                                                                                                                         

Ans: There are $9s$- bonds in a molecule containing molecular formula of $C{{H}_{2}}=CH-CH=C{{H}_{2}}$ out of  $9s$ three are between $C-C$ and  $6$ are between $C-H$ along with this it also contains $2\pi$- bonds.

18. What type of bond exists in multiple bonds (double / triple)?               

Ans:  pi – bond is always present in molecules which contain multiple bonding.

19. What type of bond is formed due to orbital overlap?                                

Ans:  Covalent bonds are generally formed due to the overlapping of orbitals which are able to move freely in space. The overlapping may be straight sides of sideways which form sigma and pi bonds respectively.

20. How do covalent bonds form due to orbital overlapping?                        

Ans:  Orbital overlap concept gives us the information about the formation of a covalent bond between two atoms which are formed as results of pairing of electrons present in the valence shell which attains opposite spins. 

21. Define hybridization.                                                                                          

Ans: Hybridization is defined as the process of intermixing of orbitals which are different to each other in their energies therefore to redistribute their energies they are able to form new sets of orbitals of equivalent energies and shape.

22. State the hybrid orbital is associated with $B$ in $BC{{l}_{3}}$ and $C$ in ${{C}_{2}}{{H}_{4}}$.     

Ans:   The hybrid orbitals associated with 

$B$ in $BC{{l}_{3}}$ shows $S{{p}^{2}}$ Hybridization while in

$C$ in ${{C}_{2}}{{H}_{4}}$ also shows $S{{p}^{2}}$ Hybridization.

23. What is the state of hybridization of carbon atoms in diamond and graphite?

Ans: Hybridization of carbon in diamond is $S{{p}^{3}}$ and in graphite it shows $S{{p}^{2}}$ hybridization.

24. What type of hybridization takes place in 

(i) $p$ in $PC{{l}_{5}}$

(ii) $S$ in $S{{F}_{6}}$?

Ans:   Hybridization of (i) phosphorus in $PC{{l}_{5}}$ is $S{{p}^{3}}d$

While hybridization of (ii) sulphur in $S{{F}_{6}}$ is $S{{p}^{3}}{{d}^{2}}$.

25. Define bonding molecular orbital. 

Ans:  This can be defined as the molecular orbital which is formed by the addition of atomic orbitals is called bonding molecular orbital. This can be represented as follows: $\sigma =\Psi A+\Psi B$

26. Define anti-bonding molecular orbital.

Ans: This can be defined as the molecular orbital which is formed by the subtraction of atomic orbitals is called anti-bonding molecular orbital. This is represented as: ${{\sigma }^{+}}=\Psi A-\Psi B$

27. Explain diagrammatically the formation of molecular orbital by LCAO. 

Ans: The molecular orbital is generally formed by the subtraction of the atomic orbital and is called an anti-bonding molecular orbital. Diagram can be shown as follows:

28. Which one $O_{2}^{-}$ and $O_{2}^{2-}$ may exhibit is paramagnetic?

Ans: Paramagnetic substances which contain one or more than one unpaired electron in its molecular orbital configuration here $O_{2}^{-}$ contain one unpaired electron in its Mo configuration and are said to be paramagnetic in nature.

29. Why are bonding molecular orbitals more stable than antibonding molecular orbitals?                                                                                                                

Ans. Bonding molecular orbital is said to be more stable as compared to its anti-bonding molecular orbital; this can be explained due to its lower energy and greater stability as compared to its corresponding anti-bonding molecular orbital. 

30. Define bond order.                                                                                              

Ans: Bond order which is generally represented by B.O. is defined as one half of the differences between the number of electrons present in the bonding and the anti- bonding orbitals. The formula can be written as:

Bond Order (B.O) $=\frac{1}{2}({{N}_{a}}-{{N}_{b}})$

If ${{N}_{b}}>{{N}_{a}}$, molecule is stable and

If ${{N}_{b}}<{{N}_{a}}$, the molecule is unstable.

31. Define hydrogen bonding.

Ans: Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with other electronegative atom like $\left( F,\text{ }O\text{ }or\text{ }N \right)$ of another molecule. 

32. What are the types of $H$- bonding? Which of them is stronger?              

Ans: Hydrogen bonding is generally of two types known by the name

a. Intermolecular $H$- bonding and

b. Intramolecular $H$- bonding. 

Out of these two intermolecular H-bonding is stronger as compared to intramolecular H-bonding.

33. $N{{H}_{3}}$ has a higher boiling point than $P{{H}_{3}}$.Give reason.                                

Ans:  Boiling point generally depends upon the hydrogen bonding like in $N{{H}_{3}}$ there is hydrogen bonding whereas in $P{{H}_{3}}$ there is no hydrogen bonding.

34. Define electrovalent bond.

Ans: Electro covalent bond is defined as the formation of bonds due to the result of electrostatic attraction between their positive and negative ions.

Short Answer Questions                                                                               2 Marks

1. Give the main feature of Kossel’s explanation of chemical bonding. 

Ans:  Kossel gives explanation towards the chemical bonding have gives us the following features:

1. Highly electronegative halogens and highly electropositive alkali metals present in the periodic table are separated by noble gases. 

2. Formation of a negative ion from a halogen atom and a positive ion from an alkali metal atom is generally associated with a gain and loss of an electron by their respective atoms. 

3. The negative and positive ions formed generally attain stable noble gas electronic configurations. Noble gases are those gases which have eight electrons in their outermost shell and have a general configuration of $n{{s}^{2}}n{{p}^{6}}$. The $ve$ and $+ve$ ions are stabilized by electrostatic attraction.

2. How can you explain the formation of $NaCl$ according to the kossel concept? 

Ans: The formation of $NaCl$ from sodium and chlorine can be explained according to Kossel concept as follows:

Sodium will lose one electron as: $Na\to N{{a}^{+}}+{{e}^{-}}$

Configuration shown by $N{{a}^{+}}$in term of neon is $[Ne]3{{s}^{1}}$

Now chlorine will gain one electron reaction can be shown as: $Cl+{{e}^{-}}\to C{{l}^{-}}$

Configuration shown by $C{{l}^{-}}$ in term of neon is $[Ne]3{{s}^{2}}3{{p}^{6}}$this is corresponding to $[Ar]$.

Hence the reaction between $N{{a}^{+}}$and $C{{l}^{-}}$will be given as: $N{{a}^{+}}+C{{l}^{-}}\to N{{a}^{+}}C{{l}^{-}}\cong NaCl$.

3. Write the significance of octet rule.

Ans: The main significance of octet rule is it is useful for giving the proper concept about the structures of most of the organic compounds. Octet rule mainly applies towards the second period elements of the periodic table. 

4. Write the Lewis structure for $CO$ molecule.

Ans: Lewis’s structure of $CO$ can be written by knowing the outer or we can say valence shell configurations of carbon and oxygen atoms which can be written as:

Carbon: $(6)-1{{s}^{2}}2{{s}^{2}}2{{p}^{2}}$, Oxygen: $(8)-1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$

This corresponds the total number of valence electrons $(4+6)=10$

But it is not completing the octet due to which multiple bonds is exhibited which can be shown as:

Or 

5. Give the Lewis dot structure of $HN{{O}_{3}}$.                                                   

Ans: Lewis dot structure of $HN{{O}_{3}}$ can be shown as follows:

6. What changes are observed in atoms undergoing ionic bonding?    

Ans: Ionic bonding is formed due to the transference of electrons and it will give following changes: 

1. Both the atoms will acquire stable noble gas configuration. 

2. The atom which loses electrons becomes a $+vely$ charged ion known as cation whereas the other atom which gains electrons becomes a $-vely$ charged ion called anion. 

3. These cations and anions are held together by the Coulombic forces of attraction which further form an ionic bond. 

7. Mention the factors that influence the formation of an Ionic bond.  

Ans: Formation of Ionic bond mainly depends upon the following three factors: 

1. Low ionization energy: Those elements which have low ionization enthalpy have greater tendency to form an ionic bond. 

2. High electron gain enthalpy: Those compounds which have high negative value of electron gain enthalpy generally favors the formation of ionic bonds. 

3. Lattice energy: Compounds having high lattice energy value favors ionic bond formation.

8. Give a reason why $H_{2}^{+}$ ions are more stable than $H_{2}^{-}$ though they have the same bond Order.                                                                                                     

Ans: $H_{2}^{+}$ ions are more stable than $H_{2}^{-}$ though they have the same bond Order as in $H_{2}^{-}$ ion one electron is present in anti-bonding orbital due to this destabilizing effect is more in this its stability becomes less than that of $H_{2}^{+}$ ion.

9. How would the bond lengths vary in the following species? ${{C}_{2}},C_{2}^{-},C_{2}^{2-}$.      

Ans: The order of bond lengths in ${{C}_{2}},C_{2}^{-},C_{2}^{2-}$ is defined as ${{C}_{2}}>C_{2}^{-}>C_{2}^{2-}$, Higher the number of orbitals lower is the bond length.

10. Out of covalent and hydrogen bonds, which is stronger.                    

Ans:  Out of covalent and hydrogen bonds, covalent bonds are said to be stronger.

11. Define covalent radius.                                                                                

Ans: The covalent radius is defined as the radius or we can say distance between the one atom to its corresponding atom or radius of an atom's core which is in contact with the core of an adjacent atom in a bonded situation. This can be shown as follows:

12. Why does $N{{H}_{3}}$ have a higher dipole moment than $N{{F}_{3}}$ though both are pyramidal? 

Ans:  $N{{H}_{3}}$ has high dipole moment than $N{{F}_{3}}$ as in case of $N{{H}_{3}}$ the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the $N-H$ bonds but on the other hand in $N{{F}_{3}}$ the orbital dipole is in the direction opposite to the resultant dipole moment of three $N-F$ bonds. Hence the orbital dipole becoming a lone pair decreases which results in the low dipole moment.

13. Draw the resonating structure of $NO_{3}^{-}$.                            

Ans: The resonating structure generally involves the movement of electrons and resonating structures of $NO_{3}^{-}$ can be shown as:

14. On which factor does dipole moment depend in case of polyatomic molecules? 

Ans: The dipole moment of the polyatomic molecule generally depends on the individual dipole moments of bonds and also on the spatial arrangement of various bonds present in the molecule. 

15. Dipole moment of $Be{{F}_{2}}$ is zero. Give a reason.

Ans: In $Be{{F}_{2}}$ the dipole moment is zero this can be explained on the basis of existence of two equal bond dipoles which are present in opposite directions and cancel the effect of each other so it form zero dipole moment which can be shown as:

16. Give the main features of VSEPR Theory.

Ans: The main postulates of VSEPR theory can be explained as follows: 

1. The shape of any molecule generally depends upon the number of valence shell electron pairs around the central atom. 

2. Pairs of electrons in the valence shell repel each other, that’s why their electron clouds are negatively charged. 

3. The pairs of electrons tend to occupy that position in space which minimize repulsion and maximize the distance between them. 

4. The valence shell is considered as a sphere with the electron pairs localizing on the sphere at maximum distance from one another. 

5. A multiple bond is treated as a single electron pair and two or three electron pairs of a multiple bond is treated as a super pair. 

6. When two or more resonance structures can represent a molecule, the VSEPR nodal is applicable to any type of structure. 

17. What's the difference between lone pair and bonded pair of electrons?  

Ans:  The main difference between lone pair and bond pair is that lone pair electrons do not take part in bond formation whereas bond pair electrons take part in bond formation.

18. $C{{O}_{2}}$ is linear whereas $S{{O}_{2}}$ is bent – shaped. Give a reason.                     

Ans:  $C{{O}_{2}}$ is said to be linear as in $C{{O}_{2}}$ the bond electrons are furthest away from each other which further forms the angle of ${{180}^{\circ }}$. 

$S{{O}_{2}}$ contains a bend shape as in the compound number of bonding pairs is $4$ and it also contains one lone pair of electrons which does not participate in bond formation due to which we can experience repulsive strain in the molecule. Shape of $S{{O}_{2}}$ can be shown as follows:

19. Why does ${{H}_{2}}O$ have a bent structure?                                                           

Ans:  In water molecules there are two bonding pairs and two lone pairs of electrons. The shape would be said to be tetrahedral if there were all bonding pair electrons but in this case two lone pairs are present due to which its shape gets distorted to an angular shape. As $lp\text{ }\text{ }lp$ repulsion is more than $bp\text{ }\text{ b}p$ repulsion.

20. For the molecule,

Why is structure (b) more stable than structure (a)?                                       

Ans:  In case of structure 

(a) the lone pair is present at axial position which suggests that there are three $lp\text{ }\text{ }lp$ repulsions at ${{90}^{\circ }}$ . 

Whereas in case of (b) the lone pair is present at an equatorial position and there are two $lp\text{ }\text{ }lp$ repulsions. Which represents that structure (b) is more stable as compared to structure (a).

21. How would you attribute the structure of the $P{{H}_{3}}$ molecule using the VSEPR model? 

Ans:  The outermost shell of the Phosphorus atom contains $5$ electrons. In this case $H$– atoms contribute one electron with each to make a total of $8$ electron around the phosphorus atom. In this way $4$ pairs of electrons are distributed in a tetrahedral manner around the central atom. Out of which three pairs from three $P\text{ }\text{ }H$ bonds while the fourth pair remains unused. Due to repulsion between the bond pair of these three and lone pair of the remaining one the shape is not of tetrahedral nature but becomes trigonal pyramidal. 

22. In the $S{{F}_{4}}$ molecule, the $lp$ electrons occupy an equatorial position in the trigonal bipyramidal arrangement to an axial position. Give a reason.                    

Ans:  In $S{{F}_{4}}$ molecule the $lp$ electrons occupy an equatorial position because the repulsion of $lp\text{ }lp$ electrons is minimum. 

23. How is VBT different from Lewis' concept?                                                    

Ans: In Lewis' concept the bond formation is formed by the sharing of electron pairs according to the Octet rule and in VBT bond formation is described in terms of hybridization and overlap of the orbitals. 

24. $S$ – orbital does not show any preference for direction. Why?                

Ans: $S$ – Orbital does not show any preference for direction because it is spherically symmetrical in nature only those orbitals will show preference to any direction which are not symmetrical in nature.

25. Why is $s$ – bond stronger than $\pi $- bond?                                                            

Ans:  $s$ – bond stronger than $\pi $- bond as orbitals in $s$ – bond can overlap to a greater extent due to axial orientation on the other hand in $\pi $- bond sideways overlapping is there which is as much appreciable because presence of $s$ - bond in this restricts the distance between the involved atoms. 

26. What are the different types of $s$ - bond formation?                                    

Ans:  $s$ - bond can be formed by any of the following types of combinations of atomic orbitals.

• $S\text{ }\text{ }S$ - overlapping: In this case overlapping is between two half – filled $S$- orbitals along their internuclear axis which can be shown as:

• $S-P$ Overlapping: In this case overlapping occurs between half – filled $s$-orbitals of one atom and half-filled $p$ -orbitals of another atom. This can be shown as:

• $P-P$ Overlapping: In this type of overlapping the overlap takes place between half-filled $p$ -orbitals of the two approaching atoms which can be shown as: 

27. What is zero overlap?                                                                                        

Ans:  The unsymmetrical overlap of orbitals resulting in zero overlap, this type of overlapping will be seen between $px-s$ and $px-py$ orbital

28. The features of hybridization.                                                                         

Ans: The main features of hybridization can be explained as follows:

1. The number of hybrid orbitals are always equal to the number of the atomic orbitals in which they get hybridized. 

2. The hybridized orbitals are always equivalent in energy and shape. 

3. The hybrid orbitals are said to be more effective in forming stable bonds as compared to pure atomic orbitals. 

4. The hybrid orbitals revolve in such a manner that minimum repulsion will occur in any particular geometrical shape of the compound.

29. What are the important consolations for hybridization?

Ans: The important consolations for hybridization can be explained as follows:

1. The orbitals present in the valence shell of the atom are said to be hybridized orbital. 

2. Hybridized orbitals should have almost the same energy. 

3. It is not necessary that all electrons get promoted prior to hybridization.

4. It is also not necessary that only half-filled orbitals participate in hybridization while filled orbitals can also take part in hybridization.

30. Describe the shape of $sp, s{{p}^{2}} and s{{p}^{3}}$ hybrid orbital.                               

Ans: The shape of $sp,s{{p}^{2}} and s{{p}^{3}}$ hybrid orbital is as follows:

1. $sp$-hybrid orbital is oriented to an angle ${{180}^{\circ }}$ and is said to have a linear shape.

2. $s{{p}^{2}}$-hybrid orbital lie in a plane and is directed towards the corners of equilateral triangle making an angle of ${{120}^{\circ }}$ and trigonal symmetric in shape.

3. $s{{p}^{3}}$-hybrid orbitals are directed towards the four corners of tetrahedron making an angle of ${{109}^{\circ }}28'$ and have shape of tetrahedron.

31. Ethylene is a planar molecule whereas acetylene is a linear molecule. Give a reason.                                                                                                              

Ans: Ethylene which is represented by the molecular formula ${{C}_{2}}{{H}_{4}}$ generally shows $s{{p}^{2}}$ hybridization in which the four hydrogen atoms are placed in four corners of a plane having bond angle of ${{120}^{\circ }}$ and said to be planar in nature.

While in acetylene carbon shows $sp$ hybridization which shares an angle of ${{180}^{\circ }}$ and shows linear shape which can be shown as:

32. In ${{H}_{2}}O,{{H}_{2}}S,{{H}_{2}}Se,{{H}_{2}}Te$ the bond angle decreases though all have the same bent shape. Why?                                                                                          

Ans: In all the four cases the molecules will undergo $S{{p}^{3}}$ hybridization which further forms four hybrid orbitals out of which two are occupied by lone pairs of electrons and two by bond pair electrons. In this way they are expected to have ${{109}^{\circ }}28'$ angle but in actuality it will not show this angle. As in case of ${{H}_{2}}O$ molecule, oxygen is small in size and said to be highly electronegative in nature due to which the bond pairs are closer and this will be subjected to larger repulsion called $\left( bp-bp \right)$. In case of  ${{H}_{2}}S$ as $S$ atom is larger than $O$, $bp-bp$ repulsion is less as compared to ${{H}_{2}}O$ and it is also true for ${{H}_{2}}Se$ and ${{H}_{2}}Te$ as well.

33. Out of $p$-orbital and $sp$-hybrid orbital which has greater directional character and Why?                                                                                                      

Ans: $sp$-hybrid orbital has greater directional character as compared to $p$-orbital. This can be explained on the basis that the two lobes of $p$-orbitals are equal in size and equal electron density is distributed whereas in $sp$-hybrid orbital electron density is greater on one side.

34. $H{{e}_{2}}$ does not exist. Explain in terms of LCAO.                                      

Ans: The electronic configuration of the helium atom is $1{{s}^{2}}$. Each helium atom contains $2$ electrons therefore we can say that in $H{{e}_{2}}$molecule there would be $4$ electrons. These electrons will accommodate in $\sigma 1s$ and ${{\sigma }^{*}}1s$ and the molecular orbitals electronic configuration can be shown as:

$H{{e}_{2}}:{{(\sigma 1s)}^{2}}{{(\sigma *1s)}^{2}}$

Now bond Order of $H{{e}_{2}} is \frac{1}{2}(2-2)=0$

This suggests that the $H{{e}_{2}}$ molecule is unstable in nature and does not exist.

35. Dipole moment is a scalar or a vector quantity?                                  

Ans: Dipole moment is said to be a vector quantity and is depicted by a small arrow with tail on the $+ve~$ center and head pointing towards the negative center.

CBSE Class 11 Chemistry Chapter 4 Important Questions - Free PDF Download

Chemical Bonding Class 11 Chapter 4 - Chemistry

Chemical bonding class 11 covers several fundamental concepts in the field of chemistry. Various concepts covered in this section are as follows:

1. Lewis structure

2. Valence bond theory

3. VSEPR theory

4. The polar character of covalent bond

5. Hydrogen bonding

6. Orbital theory of homonuclear diatomic molecules

7. Hybridization

8. Octet rule 

9. Ionic and electrovalent bond

10. Lattice enthalpy

Chemical bond: The force that holds different atoms in a molecule is known as a chemical bond. Atoms are never present in a free state, rather they are present in a combined state as a single species. The group of atoms that live together in a single species is known as a molecule and the force that attracts them together is known as a chemical bond. 

Apart from the above listed important points of class 11 chemistry chapter 4, students will get to learn various other sections such as polar characters of covalent bonds, how to balance a chemical equation,s and different types of chemical equations.

One of the most important topics of class 11 chemistry chapter 4 is hybridization. It is considered as important as many past exam papers have this topic covered. Hybridization is the idea of mixing orbital into new hybrid orbitals with different energies, shapes. Hybrid orbitals are used in the explanation of molecular geometry and atomic bonding. They are symmetrically disposed of in space. These are usually formed by mixing atomic orbitals. The new orbitals that have been formed are known as hybrid orbitals. Below are the types of Hybridization discussed in brief for an easier understanding of the concept.

Class 11 Chapter 4 Chemistry - Chemical Bonding and Molecular Structure

Types of Hybridization:

1. Sp3 hybridization

2. Sp2 hybridization

3. Sp hybridization

4. Sp3 d hybridization

5. Sp3 d2 hybridization

6. Sp3 d3 hybridization

These are the types of orbits involved in mixing. As discussed earlier, it is considered as one of the most important discussed topics of class 11 chapter 4 chemistry - chemical bonding and molecular structure. The important questions are extracted in particular from the subject matter experts of Vedantu in free pdf form. Students can download the pdf form for a better understanding of the concerned topic. 

Features of Hybridization

Its concept was introduced by Pauling where he explained the shapes of polyatomic molecules. He found that hybrid orbitals are used to form bonds. He named this phenomenon hybridization. Its important features are discussed below:

1. Both hybrid orbitals and the number of atomic orbitals that get hybridized are the same.

2. Hybridized orbitals represent equal energy and shape

3. Electron density in hybrid orbitals is concentrated at one side

4. Both pure atomic orbital and hybrid orbital have different shapes in case of undergoing hybridization.

5. Hybrid orbitals minimize repulsion between electrons due to their orientation in space. Because of its orientation, it gains maximum symmetry.

6. Hybridization does not take place in the isolated gaseous atom.

7. The shape can be easily predicted if the hybridization of the molecule is known.

8. In the arrangement, the bigger lobe has a positive sign, while the smaller lobe placed on the opposite side always has a negative sign.

Conditions For Hybridization

1. Hybridization takes place with the valence shells orbitals

2. It also takes place between similar energy orbitals

3. In some cases not half-filled, but full-filled orbitals can also take part in hybridization.

4. For hybridization, it is not necessary for electrons to move from a lower energy level to a higher energy level.

Let us find in brief about the types of hybridization and how important questions are framed from this section.

1. Sp hybridization: Also named as linear or diagonal hybridization. Here, sp hybridized orbital is formed by intermixing of one s orbital and one p orbital. This arrangement leads to the formation of two equivalent sp hybridized orbital. Each hybrid orbitals have 50% s character and 50% p character

2. Sp2 hybridization: This arrangement is formed by intermixing of one s orbital and two p orbital. These hybrid orbitals are aligned in trigonal planar symmetry having a 120-degree angle between them. Here, talking about the percentage acquisition of each character. S character possess 33% and p character consist of 66.6%

3. Sp3 hybridization: This arrangement is formed by intermixing of one s orbital and three p orbitals. It further leads to the formation of four sp3 hybridized orbitals. These hybrid orbitals form 109.5 degrees angle between them. It consists of 25% s character and 75% p character

4. Sp3d hybridization: It involves the arrangement of three p orbitals and one d orbital to form sp3d hybridized orbitals. It possesses trigonal bipyramidal geometry. It lies in the horizontal plane at an angle of 120 degrees

5. Sp3d2 hybridization: It has one s, three p, and two d orbitals, it undergoes mixing and form six sp3d2 hybrid orbitals. They are directed towards the corners of the octahedron and are inclined at an angle of 90 degrees to each other.

The above section discusses in brief chemical bonding and some important topics related to class 11 chapter 4 - chemical bonding and molecular structure. These are some important topics as important questions are curated from this particular section. Students are provided with the important questions and topics so that it is easier for them to have a better understanding of exam and question patterns. These are provided by subject matter experts of Vedantu keeping in mind the general outline of the CBSE syllabus. 

Significance of CBSE Class 11 Chemistry Chapter 4 Important Questions

Chemical bonds are the core of matter around us. These bonds form between the same or different atoms to form molecules of elements and compounds respectively. The entire concept of the chemistry of matter rests on the topic of chemical bonding. Hence, this chapter is of utmost importance for the students of Class 11 to prepare a conceptual foundation for this subject.

Once the theoretical portion and exercises of this chapter are studied and solved respectively, students will need an evaluation platform to check their preparation level. They will need this platform to find out which part of this chapter needs more attention. This is where the important questions for CBSE Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure will come in handy.

Students can practise answering these questions and compare their answers with the given solution. They will easily find out which questions they have attempted correctly and do the needful. Hence, add these important questions to your practice sessions and make this chapter your strength.

Important Questions

Chemical bonding and molecular structure class 11 important questions are provided to students keeping in mind the outline of the CBSE syllabus and important topics as well. Some of the important questions include the following:

Q1. What is a Valence Bond Theory?

Ans. Electrons in a molecule occupy atomic orbitals rather than molecular orbitals. These atomic orbitals overlap on the bond which results in a larger overlap. The larger the lap, the stronger is the bond. The bonding is generally covalent in origin. It has a certain application of valence bond theory which is discussed below in brief:

1. The bond in an HF molecule is formed by the overlap of one s orbital of the hydrogen atom and two p orbital belonging to the fluorine atom. It is further explained in detail by the valence bond theory.

2. The difference in the length and strength of chemical bonds in H2 and F2 molecules is explained by overlapping orbitals in the molecules.

Looking at the application, it has certain limitations in its structure as well. This is discussed below in brief:

1. It fails to explain the tetravalency exhibited by carbon.

2. The assumption in theory that electrons are localized in specific areas

3. No insight offered on the energies.

4. No proper distinction between weak and strong ligands.

Q2. Explain the Formation of a Chemical Bond?

Ans. it is a force that attracts and bounds the constituents of species together. It has many theories for bond formation. Some of its theories are valence shell electron pair repulsion theory, electronic theory, valence bond theory, and molecular orbital theory. To achieve stability, the formation of a chemical bond is credited. Atoms combine with neighbours and finish their octet and duplets to achieve stability of closest inert gases. This stability is occurred either by sharing of electrons. The formed chemical bond is a result of sharing of electrons. This is known as a covalent bond

Conclusion:

Class 11 chemistry chapter 4 important questions with answers and important topics are extracted to help students get an understanding of chemical bonding. This is a very important chapter as it lays its foundation for all future higher studies. In order to understand the chapter thoroughly, students are advised to go through the pdf form of important questions prepared by Vedantu subject matter experts. Students will also get a deep understanding of important topics along with a detailed answer for each of the questions that are solved by subject experts in a simple language for better understandability.

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