NCERT Solutions for class 12 Chemistry Chapter 1: The Solid State

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NCERT Solutions for class 12 Chemistry Chapter 1 The Solid State are provided in this article. Some of the important concepts covered in this topic include:

As this chapter has been omitted in the new CBSE Class 12 Syllabus, it does not have any weightage but is useful in preparing for competitive exams.

Download PDF: NCERT Solutions for Class 12 Chemistry Chapter 1 pdf

NCERT Solutions for Class 12 Chemistry Chapter 1

NCERT Solutions for Class 12 Chemistry Chapter 1 Solid State is given below.

Class 12 Chemistry Chapter 1: Solid State – Important Topics

Some of the important topics in Class 12 Chemistry Chapter 1: Solid State is given below

Types of Solids – Crystalline and Amorphous Solids

Properties of Crystalline Solids:

They have fixed geometrical shapes.
Crystalline solids have their constituent particles arranged in long-range order.
They have a fixed melting point.
They have fixed heat of fusion.
They are anisotropic.
They are considered as true solids.

Properties of Amorphous Solids:

They have an irregular shape.
Amorphous solids have their constituent particles arranged in short-range order.
When heated, they soften and do not have a sharp melting point.
They do not have a sharp melting point.
They are isotropic.
They are considered as pseudo solids or super cooled liquids.

Solid State

Solid State

Types of Bonding – Ionic Bond, Covalent Bond

Ionic Bond: Ionic bond is formed by the exchange of electrons between two reacting molecules. It usually occurs between metals and non metals (eg: Sodium and Chlorine) where there is an interchange of oppositely charged particles.

Covalent Bond: Covalent Bond is the sharing of electrons between two molecules to form a new product. This types of bonding is not as strong as Ionic bonds.

Some of the factors which determine the type of Bonding are called as Bond parameters. They include:

Bond Length
Bond Angle
Bond Enthalpy
Bond Order
Resonance Structures

Lattice Enthalpy: It is the amount of energy required to separate one mole of crystalline ionic compounds from its constituent gaseous ions.

Also Read:

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CBSE CLASS XII Related Questions

1.
Predict the alkene formed by dehydrohalogenation of 1-Bromo-1-methylcyclohexane.

View Solution

2.
What happens when acidic solution of potassium permanganate is allowed to stand for sometime ? Give the equation involved. What is this type of reaction called ?

View Solution

3.
61 g benzoic acid (M = 122 g mol$^{-1}$) dissolved in 500 g benzene. Vapour pressure of pure benzene = 66 torr. Assume complete dimerisation. Calculate vapour pressure of solution.

View Solution

4.
Predict the alkene that would be formed by dehydrohalogenation of 1-Bromo-1-methylcyclohexane.

View Solution

5.
Though chlorine shows strong $-I$ effect, why is it ortho/para directing?

View Solution

6.
For decomposition of $H_2O_2$ by $I^-$: Step I: $H_2O_2 + I^- \rightarrow H_2O + IO^-$ (slow). Step II: $H_2O_2 + IO^- \rightarrow H_2O + I^- + O_2$ (fast). (a) Write rate law. (b) Determine order w.r.t. $H_2O_2$ and $I^-$ and overall order. (c) Molecularity of Step II.

View Solution

NCERT Solutions for class 12 Chemistry Chapter 1: The Solid State

Download PDF: NCERT Solutions for Class 12 Chemistry Chapter 1 pdf

NCERT Solutions for Class 12 Chemistry Chapter 1

Class 12 Chemistry Chapter 1: Solid State – Important Topics

CBSE CLASS XII Related Questions

1.
Predict the alkene formed by dehydrohalogenation of 1-Bromo-1-methylcyclohexane.

2.
What happens when acidic solution of potassium permanganate is allowed to stand for sometime ? Give the equation involved. What is this type of reaction called ?

3.
61 g benzoic acid (M = 122 g mol$^{-1}$) dissolved in 500 g benzene. Vapour pressure of pure benzene = 66 torr. Assume complete dimerisation. Calculate vapour pressure of solution.

4.
Predict the alkene that would be formed by dehydrohalogenation of 1-Bromo-1-methylcyclohexane.

5.
Though chlorine shows strong $-I$ effect, why is it ortho/para directing?

6.
For decomposition of $H_2O_2$ by $I^-$: Step I: $H_2O_2 + I^- \rightarrow H_2O + IO^-$ (slow). Step II: $H_2O_2 + IO^- \rightarrow H_2O + I^- + O_2$ (fast). (a) Write rate law. (b) Determine order w.r.t. $H_2O_2$ and $I^-$ and overall order. (c) Molecularity of Step II.

Similar Chemistry Concepts

Comments

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NCERT Solutions for class 12 Chemistry Chapter 1: The Solid State

Download PDF: NCERT Solutions for Class 12 Chemistry Chapter 1 pdf

NCERT Solutions for Class 12 Chemistry Chapter 1

Class 12 Chemistry Chapter 1: Solid State – Important Topics

CBSE CLASS XII Related Questions

1.Predict the alkene formed by dehydrohalogenation of 1-Bromo-1-methylcyclohexane.

2.What happens when acidic solution of potassium permanganate is allowed to stand for sometime ? Give the equation involved. What is this type of reaction called ?

3.61 g benzoic acid (M = 122 g mol$^{-1}$) dissolved in 500 g benzene. Vapour pressure of pure benzene = 66 torr. Assume complete dimerisation. Calculate vapour pressure of solution.

4.Predict the alkene that would be formed by dehydrohalogenation of 1-Bromo-1-methylcyclohexane.

5.Though chlorine shows strong $-I$ effect, why is it ortho/para directing?

6.For decomposition of $H_2O_2$ by $I^-$: Step I: $H_2O_2 + I^- \rightarrow H_2O + IO^-$ (slow). Step II: $H_2O_2 + IO^- \rightarrow H_2O + I^- + O_2$ (fast). (a) Write rate law. (b) Determine order w.r.t. $H_2O_2$ and $I^-$ and overall order. (c) Molecularity of Step II.

Similar Chemistry Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Predict the alkene formed by dehydrohalogenation of 1-Bromo-1-methylcyclohexane.

2.
What happens when acidic solution of potassium permanganate is allowed to stand for sometime ? Give the equation involved. What is this type of reaction called ?

3.
61 g benzoic acid (M = 122 g mol$^{-1}$) dissolved in 500 g benzene. Vapour pressure of pure benzene = 66 torr. Assume complete dimerisation. Calculate vapour pressure of solution.

4.
Predict the alkene that would be formed by dehydrohalogenation of 1-Bromo-1-methylcyclohexane.

5.
Though chlorine shows strong $-I$ effect, why is it ortho/para directing?

6.
For decomposition of $H_2O_2$ by $I^-$: Step I: $H_2O_2 + I^- \rightarrow H_2O + IO^-$ (slow). Step II: $H_2O_2 + IO^- \rightarrow H_2O + I^- + O_2$ (fast). (a) Write rate law. (b) Determine order w.r.t. $H_2O_2$ and $I^-$ and overall order. (c) Molecularity of Step II.