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

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NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure is explained in detail in this article. Some of the important topics in Chemical bonding and Molecular structure includes:

Bond Enthalpy
Covalent Bond
Ionic bond
Nuclear bond
Complex bonding
Hybridisation
Lewis' dot structure
Molecular Orbital Theory
Valence Bond Theory
Chemical Bonding and Molecular Structure

Expected no of questions: 2 to 4 questions of total 7 marks

Download PDF: NCERT Solutions for Class 11 Chemistry Chapter 4 pdf

NCERT Solutions for Class 11 Chemistry Chapter 4

NCERT Solutions for Class 11 Chemistry Chapter 4 is given below.

Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure – Important Topics

Chemical bonding exists between molecules of the same or different elements. It also exists between the subatomic particles. Some of the different types of bonding are:

Ionic/ Electrovalent Bond: This bond is established by the exchange of oppositely charged ion commonly taking place in a metal and non metal (eg: Sodium and Chlorine).
Covalent Bond: Covalent bonding is where electrons are shared between the participating molecules.
Nuclear Bond: This is the bonding force that keeps the protons and neutrons together inside the nucleon.

Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions.

Types of bonds can be characterized based on the bond parameters. Some of the bond parameters are:

Bond length
Bond angle
Bond enthalpy
Bond order
Resonance structures

Hybridisation is the process of intermixing the orbitals of slightly different energies to redistribute their energies resulting in the formation of new sets of orbitals of equivalent energies and shapes. The various hybridisation are:

sp
sp²
sp³
sp³d
sp³d²

Read More:

Bond Energy	Sigma and Pi Bond	Valence Electrons
VSEPR Theory	Electron Affinity	PCl5 Hybridisation
Bond Order formula	Lattice Energy Formula	Polar Compounds

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CBSE CLASS XII Related Questions

1.
Complete each synthesis by giving missing starting material, reagent, or products:
View Solution
2.
Hoffmann Bromamide degradation reaction is given by:
- ArNO$_2$
- ArNH$_2$
- ArCONH$_2$
- ArCH$_2$NH$_2$
View Solution
3.
Calculate emf of the following cell at 25°C: \[ \text{Zn}(s) \, | \, \text{Zn}^{2+} (0.1M) \, || \, \text{Cd}^{2+} (0.01M) \, | \, \text{Cd}(s) \] Given: \[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.40 \, \text{V}, \quad E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.76 \, \text{V}. \]
View Solution
4.
Complete the following reactions by writing the structural formulae of ‘A’ and ‘B’:
(i) CH$_3$CH=CH$_2$ + HBr $\xrightarrow{\text{Peroxide}}$ ‘A’ $\xrightarrow{KOH}$ ‘B’
(ii) CH$_3$CH$_2$CHCl + alc. KOH $\xrightarrow{\Delta}$ ‘A’ $\xrightarrow{\text{H}_2\text{O}}$ ‘B’ (Main product)
View Solution
5.
Assertion (A): n-Butyl chloride has higher boiling point than n-Butyl bromide.
Reason (R): C—Cl bond is more polar than C—Br bond.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution
6.
Assertion (A): Electrolysis of aqueous NaCl gives H$_2$ at cathode and Cl$_2$ at anode.
Reason (R): Chlorine has higher oxidation potential than H$_2$O.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution

View All

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

Download PDF: NCERT Solutions for Class 11 Chemistry Chapter 4 pdf

NCERT Solutions for Class 11 Chemistry Chapter 4

Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure – Important Topics

CBSE CLASS XII Related Questions

1.
Complete each synthesis by giving missing starting material, reagent, or products:

2.
Hoffmann Bromamide degradation reaction is given by:

3.
Calculate emf of the following cell at 25°C: \[ \text{Zn}(s) \, | \, \text{Zn}^{2+} (0.1M) \, || \, \text{Cd}^{2+} (0.01M) \, | \, \text{Cd}(s) \] Given: \[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.40 \, \text{V}, \quad E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.76 \, \text{V}. \]

5.
Assertion (A): n-Butyl chloride has higher boiling point than n-Butyl bromide.
Reason (R): C—Cl bond is more polar than C—Br bond.

6.
Assertion (A): Electrolysis of aqueous NaCl gives H$_2$ at cathode and Cl$_2$ at anode.
Reason (R): Chlorine has higher oxidation potential than H$_2$O.

Similar Chemistry Concepts

Comments

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

Download PDF: NCERT Solutions for Class 11 Chemistry Chapter 4 pdf

NCERT Solutions for Class 11 Chemistry Chapter 4

Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure – Important Topics

CBSE CLASS XII Related Questions

1.Complete each synthesis by giving missing starting material, reagent, or products:

2.Hoffmann Bromamide degradation reaction is given by:

3.Calculate emf of the following cell at 25°C: \[ \text{Zn}(s) \, | \, \text{Zn}^{2+} (0.1M) \, || \, \text{Cd}^{2+} (0.01M) \, | \, \text{Cd}(s) \] Given: \[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.40 \, \text{V}, \quad E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.76 \, \text{V}. \]

5.Assertion (A): n-Butyl chloride has higher boiling point than n-Butyl bromide. Reason (R): C—Cl bond is more polar than C—Br bond.

6.Assertion (A): Electrolysis of aqueous NaCl gives H$_2$ at cathode and Cl$_2$ at anode. Reason (R): Chlorine has higher oxidation potential than H$_2$O.

Similar Chemistry Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Complete each synthesis by giving missing starting material, reagent, or products:

2.
Hoffmann Bromamide degradation reaction is given by:

3.
Calculate emf of the following cell at 25°C: \[ \text{Zn}(s) \, | \, \text{Zn}^{2+} (0.1M) \, || \, \text{Cd}^{2+} (0.01M) \, | \, \text{Cd}(s) \] Given: \[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.40 \, \text{V}, \quad E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.76 \, \text{V}. \]

5.
Assertion (A): n-Butyl chloride has higher boiling point than n-Butyl bromide.
Reason (R): C—Cl bond is more polar than C—Br bond.

6.
Assertion (A): Electrolysis of aqueous NaCl gives H$_2$ at cathode and Cl$_2$ at anode.
Reason (R): Chlorine has higher oxidation potential than H$_2$O.