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NCERT Class 11 Chemistry Textbook Chapter 5 With Answer PDF Free Download
Chapter 5: States of Matter
5.1 Intermolecular Forces
Intermolecular forces are the forces of attraction and repulsion between interacting particles (atoms and molecules).
This term does not include the electrostatic forces that exist between the two oppositely charged ions and the forces that hold atoms of a molecule together i.e., covalent bonds.
Attractive intermolecular forces are known as van der Waals forces, in honor of Dutch scientist Johannes van der Waals (1837-1923), who explained the deviation of real gases from the ideal behavior through these forces. We will learn about this later in this unit.
van der Waals forces vary considerably in magnitude and include dispersion forces or London forces, dipole-dipole forces, and dipole-induced dipole forces.
A particularly strong type of dipole-dipole interaction is hydrogen bonding. Only a few elements can participate in hydrogen bond formation, therefore it is treated as a separate category.
We have already learned about this interaction in Unit 4. At this point, it is important to note that
attractive forces between an ion and a dipole are known as ion-dipole forces and these are not van der Waals forces. We will now learn about different types of van der Waals forces.
5.1.1 Dispersion Forces or London Forces
Atoms and nonpolar molecules are electrically symmetrical and have no dipole moment because their electronic charge cloud is symmetrically distributed. But a dipole may develop momentarily even in such atoms and molecules. This can be understood as follows.
Suppose we have two atoms ‘A’ and ‘B’ in the close vicinity of each other (Fig. 5.1a). It may so happen that momentarily electronic charge distribution in one of the atoms says ‘A’, becomes unsymmetrical i.e., the charge cloud is more on one side than the other (Fig. 5.1 b
and c).
This results in the development of instantaneous dipole on the atom ‘A’ for a very short time.
This instantaneous or transient dipole distorts the electron density of the other atom ‘B’, which is close to it and as a consequence, a dipole is induced in the atom ‘B’.
The temporary dipoles of atoms ‘A’ and ‘B’ attract each other. Similarly, temporary dipoles are induced in molecules also.
This force of attraction was first proposed by the German physicist Fritz London and for this reason force of attraction between two temporary dipoles is known as the London force.
Another name for this force is dispersion force. These forces are always attractive and interaction energy is inversely proportional to the sixth power of the distance between two interacting particles (i.e., 1/r 6 where r is the distance between two particles).
These forces are important only at short distances (~500 pm) and their magnitude depends on the polarisability of the particle.
| Author | NCERT |
| Language | English |
| No. of Pages | 22 |
| PDF Size | 1 MB |
| Category | Chemistry |
| Source/Credits | ncert.nic.in |
NCERT Solutions Class 11 Chemistry Chapter 5 States of Matter: Gases and Liquid
Q1. The critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C respectively. Which of these has stronger intermolecular forces and why?
Answer:
If the critical temperature of a gas is higher then it is easier to liquefy. That is the intermolecular forces of attraction among the molecules of gas are directly proportional to its critical temp.
Therefore, in CO2 intermolecular forces of attraction are stronger.
Q2. Explain the physical significance of Van der Waals parameters?
Answer:
After accounting for pressure and volume corrections, the van der Waals equation is(p+\frac{an^{2}}{V^{2}}) (V-nb) = nRT(p+V2an2)(V−nb)=nRT
The van der Waals constants or parameters are a and b.
The relevance of a and b is crucial here:
The magnitude of intermolecular attractive forces within the gas is measured by the value of ‘a,’ which is independent of temperature and pressure.
The volume occupied by the molecule is represented by ‘b,’ while the total volume occupied by the molecules is represented by ‘nb.’
Q3. Using the equation of state pV=nRT; show that at a given temperature density of a gas is proportional to gas pressure p.
Answer:
The equation of state is given by,
pV = nRT ……..(1)
Where, p = pressure
V = volume
N = number of moles
R = Gas constant
T = temp\frac{n}{V}Vn = \frac{p}{RT}RTp
Replace n with \frac{m}{M}Mm, therefore,\frac{m}{MV}MVm = \frac{p}{RT}RTp……..(2)
Where, m = mass
M = molar mass
But, \frac{m}{V}Vm = d
Where, d = density
Therefore, from equation (2), we get\frac{d}{M}Md = \frac{p}{RT}RTp
d = (\frac{M}{RT}RTM) p
d \propto∝ p
Therefore, at a given temp, the density of the gas (d) is proportional to its pressure (p).
Q4. At 0°C, the density of a certain oxide of a gas at 2 bar is the same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Answer:
Density (d) of the substance at temp (T) can be given by,
d = \frac{Mp}{RT}RTMp
Now, density of oxide (d1) is as given,d_{1}d1 = \frac{M_{1}p_{1}}{RT}RTM1p1
Where, M1 = mass of the oxide
p1 = pressure of the oxide
Density of dinitrogen gas (d2) is as given,d_{2}d2 = \frac{M_{1}p_{2}}{RT}RTM1p2
Where, M2 = mass of the oxide
p2 = pressure of the oxide
Acc to the question,
d1 = d2
Therefore, M_{1}p_{1} = M_{2}p_{2}M1p1=M2p2
Given:p_{1}p1 = 2 barp_{2}p2 = 5 bar
Molecular mass of nitrogen, M_{2}M2 = 28 g/mol
Now, M_{1}M1
= \frac{M_{ 2 }p_{2}}{p_{ 1 }}p1M2p2
= \frac{ 28 × 5 }{ 2 }228×5
= 70 g/mol
Therefore, the molecular mass of the oxide is 70 g/mol.
States of Matter: Gases and Liquid Textbook With Solutions PDF Free Download