Atomic Structure Chemistry Class 11 Notes Chapter2 | KPK textbook

Atomic Structure Chemistry Class 11 Notes Chapter2 | KPK textbook board, elements, worksheet, atom solutions, neutrons, Thomson, and Chemistry city fsc part 1 notes 2021.

Atomic Structure Notes for Chemistry Class 11 2021

How did J.J. Thomson discover the e/m ratio of the cathode rays?

J.J. Thomson performed an experiment using a special discharge tube, in which he made arrangements to apply both the electric field and magnetic field simultaneously. When both the fields are kept off, the electrons (cathode rays) strike at B on the florescent screen. When only the electric field is applied they strike at A, while in the presence of magnetic field alone, they strike at C. Both the fields are adjusted in such a way that the electrons again strike at B. In this manner the forces (i.e. electric force and magnetic force) are compared and balanced to evaluate the e/m of an electron, the final equation is:
e / m = E / B²r

Where E and B are the strengths of electric and magnetic fields respectively and r is the radius of the circular path of the electron. Using above equation, the value of e/m for electron is 1.7588 x 10¹¹ C/kg.

Q.2 i FSc Notes 2021 Part-1 for KPK

Answer:
Properties of Cathode Rays
Properties of cathode rays are given below.

• Cathode rays are negatively charged rays and travel towards anode.
• Cathode rays travel in straight line away from the cathode and cast shadows of opaque objects when placed in their path.
• Cathode rays produce X-rays when they strike a metal target i.e. anode.
• Cathode rays can ionize gases and also can cause a chemical change since they have a reducing effect.

Q.2 iii) Calculate the distance (Å) between the nucleus and an electron in the 5th orbit of an excited hydrogen atom.

Answer:
We know that the radius is calculated from the formula,
r_n = n² (0.529) Å
For 5^th orbit, n = 5
r₅ = (5)² (0.529) = 13.225 Å
So, the distance between the nucleus and an electron in the 5th orbit of an excited hydrogen atom is 13.225 Å.

Q.2 iv) The centripetal force, provided by the columbic force of attraction is balanced by the centrifugal force, mv²/r.

Answer:
As we know that, the equation to calculate wave number is:

Q.2 v) How will you differentiate between a continuous and a line spectrum?

Answer:

Q.2 vi) How did Moseley discover that the atomic number (Z) is the fundamental property of an element?

Answer:
Moseley performed a number of experiments using different X-ray tubes with anodes of different materials. He took spectrum of X-rays in each case, by following them to fall on a photographic plate. It was observed that the wavelengths of X-rays were the characteristics of each element, used as anode. He showed that the square root of frequency of a spectral line is strictly related with the atomic number (Z).
√v ∝ a (Z-b)
So, it was proved that atomic number is more fundamental property of an element than atomic mass.

Q.2 vii) What is the frequency of a radiation with wave number (⊽) equal to 0.5 x 10⁸ m^-1?

Answer:
As we know that, the formula to calculate frequency (v) is:
v = c⊽
Putting the values, c = 2.99 x 10⁸ m/s, ⊽ = 0.5 x 10⁸ m^-1
v = c⊽ = (2.99 x 10⁸) (0.5 x 10⁸) = 1.495 x 10¹⁶ s^-1

Q.2 viii) Calculate the wave length of an electron when it moves with the velocity of light.

Answer:
Wavelength of electron can be calculated from de Broglie’s relation,
λ = h / mv
h = Planck’s constant = 6.6262 x 10^-34 Js
v = speed of light = 2.99 x 10⁸ m/s
m = mass of electron = 9.1 x 10^-31 kg
Putting the values in equation,

Long Questions fsc notes Part-1 for kpk 2021

Q.3 b) Prove that E = hc⊽ , where E = energy, h = planck’s constant. c = velocity of light and ⊽ wave number.

Answer:
According to Planck’s equation,
E = hv            (1)
Where h is plank’s constant and v is the frequency (number of waves passing through a point in one second). The frequency of a photon is inversely proportional to its wavelength, i.e.
v ∝ 1 / λ
v = c / λ
Where c, the proportionality constant, is called the velocity of light. Its value is 2.99 x 108 m/s.
Putting the value in equation (1).
E = h = c / λ        (2)
The number of waves per unit length is called wave number (⊽). It is reciprocal of the wavelength.
⊽ = 1 / λ m-1 or cm-1
Inserting the relationship of wavenumber in equation (2).
E = hc⊽
Thus the energy of photon is related to the frequency, wavelength and wavenumber.

Q.3 c) What will be the energy of a radiation with λ = 2 x 10^-8 m?

Answer:
Using the relation modified from planck’s equation,
E = h (c / λ)
h = 6.6262 x 10^-34 Js
c = 2.99 x 10⁸ m/s
λ = 2 x 10⁸ m
Putting the values in equation

Q.4 a) What are the postulates of Bohr’s atomic model?

Answer:
Postulates of Bohr’s Atomic Model
The main postulates of Bohr’s atomic model are following:

• Electron revolves around the nucleus in fixed circular paths called the orbits (or shells). Each orbit is associated with a definite amount of energy.
• As long as electron remains in one of the circular orbits it does not radiate energy. The electron emits or absorbs energy only, when it jumps from one orbit to another.
• When the electron jumps from higher to lower orbit, it loses energy and when it jumps from lower to higher orbit, it absorbs energy. The energy change, ΔE is given by:
  ΔE = E₂ – E₁ = hv
  Where E₁ is the energy of lower orbit and E₂ is that of higher orbit, h is the planck’s constant and v is the frequency of radiation emitted or absorbed by the electron.
• The angular momentum (mvr) of the electron in the hydrogen atom is quantized and can have values which are the whole multiple of  h/2π, that is,
  mvr = nh / 2π
  h = Planck’s constant, n(orbit number)=1,2,3, …….etc., m being the mass of the electron, v the velocity of electron and r is the radius of the orbit.

Q.4 b) How can Bohr’s model of atom be applied to hydrogen atom to calculate the radius of nth shell?

Answer:
Calculation of the Radius of nth Shell
Bohr assumed that proton, being 1836 times heavier remains stationary with respect to electron which revolves around the nucleus in the hydrogen atom. Let ‘e’ be the charge of the revolving electron of mass m which moves on the circular path (orbit) of radius r with velocity v. If nuclear charge is +Ze, Z being (proton number). Then according to Coulomb’s law, the electrostatic force of attraction (centripetal force) between the electron and the nucleus is given by:

Where the proportionality constant K is equal to 1 / 4πε_o and ε_o is the vacuum permittivity and is a measure of how easy it is for electrostatic forces to pass through the vacuum (free space). Its value is 8.85 x 10^-12 C² J^-1 m^-1. Thus,

The centripetal force, provided by the columbic force of attraction is balanced by the centrifugal force, mv²/r.

For a stable orbit

Rearranging

Above equation shows that radius is inversely proportional to the square of the velocity of the electron. That is to say the electron moves faster in an orbit of smaller radius.
From Bohr’s postulate,
mvr = nh / 2π
Separating ‘v’ and taking its square on both sides gives the equation,

Q.4 c) Derive expression using Bohr’s model, for the energy difference (ΔE), frequency (v) and wave number in hydrogen atom.

Answer:
Energy Difference between Two Orbits
The general equation for the energy of nth orbit for hydrogen atom is,

Let E₁ and E₂ be the energies of lower and higher orbits respectively, then above equation for two orbits can be written as;

Taking the difference of equation (A) and equation (B),

Q.4 d) How does Bohr’s model explain the hydrogen spectrum?

Answer:
Hydrogen Spectrum
Bohr’s model can successfully explain the hydrogen spectrum. When an electron jumps from one energy level (orbit) to another energy level (orbit), the associated frequencies, wavelengths and wavenumbers are calculated from Bohr’s theory.
Then these calculated values are compared with the experimental values obtained from the line spectrum of the hydrogen atom. These calculated values are in close agreement with the values obtained experimentally from the hydrogen spectrum. Thus Bohr’s model is able to explain the hydrogen spectrum.

Q.4 e) What are the shortcomings of Bohr’s atomic model?

Answer:
Shortcomings of Bohr’s atomic model:
Bohr’s atomic model can explain hydrogen and hydrogen like ions He⁺, Li⁺⁺, Be⁺⁺⁺, but fails to explain the following.

1. It is unable to explain the spectrum of more complicated atoms (multi electron system).
2. It cannot explain the multiplicity of the spectral lines (fine structure of atom) observed under a high resolving power spectrometer.
3. Bohr’s theory cannot explain the effect of magnetic field (Zeeman effect) and electric field (Stark effect) on the spectra of atoms.
4. Bohr’s model of atom goes against the Heisenberg’s uncertainty principle.

Q.5 a) What are X-rays? How are these produced?

Answer:
X-rays
X-rays are electromagnetic radiations of very high frequency (shorter wavelengths). The wavelengths of the radiations, constituting x-rays, ranges from 10-2 Å to 10+2 Å (0.001 nm to 10 nm).
Production of X-rays
A German physicist W.C. Roentgen (1895) discovered x-rays accidentally while he was studying the properties of cathode rays. X-rays can be produced by several methods. For example,

1. Roentgen Method (gas tube)
2. Coolidge Method
3. By using betatron (an electron accelerating machine, devised by DW Kerst in 1941).

The Roentgen method is discussed below in detail.
Roentgen Method
The experimental setup for this method consists of a special type of discharge tube at a very low pressure of 0.001 mmHg. A very high potential difference of about 30,000 to 50,000 volts is required to be applied between the cathode and anode. At a pressure as low as 0.001 mmHg, some air molecules become ionized due to the influence of a very high electric field inside the tube. The heavier positive ions are attracted towards the cathode and due to their collisions electrons are emitted from the cathode.
These electrons are then attracted toward the anode and hit it with greater momentum (velocity) gained by acceleration, due to the high potential difference between the cathode and the anode. This causes the emission of X-rays from the surface of the anode. The cathode is usually concave shaped with its focus on the anode so that the electrons emitted from the cathode are focused on a small region of the anode and X-rays are emitted from this small region.

Q.5 b) Enlist some characteristics of X-rays.

Answer:
Characteristics of X-rays

• The X-rays travel in straight line like ordinary light.
• They are not deflected by electric or magnetic field, indicating that they are neutral.
• They have the ability to ionize gases. The ionizing power depends on the intensity of X-ray beam.
• X-rays produce fluorescence in many substances like rock salt, uranium, glass, compounds of calcium and barium.
• X-rays are reflected as well as refracted.
• X-rays can be diffracted by crystalline substances.
• X-rays can penetrate through many substances. Their penetration power is different in different substances.
• X-rays can blacken photographic plates. The extent of this property depends on the intensity of the X-rays, falling on the plate.

Q.6) What are quantum numbers? Discuss their significance in detail.

Answer:
Quantum Numbers
Quantum numbers are certain numbers (which are set of numerical values) that give information about the designation (energy, shape of orbital etc.) of an electron in an atom. There are four quantum numbers, three of which have been obtained by the solution of Schrodinger’s wave equation for hydrogen atom and the 4th one, which is called spin quantum number was discovered independently.

1. Principal quantum number
2. Azimuthal quantum number
3. Magnetic quantum number
4. Spin quantum number

1. Principal Quantum Number (n)
This quantum number represents the main energy level or shell in which the electron is present. It is represented by “n” and can have values;
n= 1, 2, 3, 4, . . . . .

Significance
Energy of the electron depends on the value of n, the lower the value of “n”, the lower will be the energy of the electron and vice versa. Letter designations are also used to denote the various n-values e.g. K, L, M, N, O, P, Q . . . . shells are used for n= 1, 2, 3, 4, 5, 6, 7 etc. values respectively. This quantum number is the same as the simplest whole number in the Bohr’s atomic model. This value of n is associated with the size of the shell. The maximum number of electrons in the principal quantum number is calculated by the formula 2n2, where n cannot have zero value.

2. Azimuthal Quantum Number (l)
This quantum number which is represented by “l”, accounts for the appearance of fine lines in hydrogen spectra. The spectral lines, associated with the main energy level (principal quantum number) is seen to have been split up into fine lines shown by a high resolving power spectrometer. These splittings are due to the presence of sub-shells or energy sub-levels in a shell. So a secondary quantum number called the azimuthal quantum number is also needed to represent the sub-shells. The “l” may have all possible whole number values from 0 to n-1. That is, for n=1, l=0 for n=2, l=0, 1 for n=3, l=0, 1, 2 and for n=4, l=0, 1, 2, 3

The value of “l” also determines the shape of the sub-shell. The shapes of sub-shells are due to revolution of electron around the nucleus. For example, when l=0 then it is s-orbital and is spherical, when l=1, the subshell is dumbbell shaped and is called p-subshell, when l=2, the subshell is sausage shaped and is called d-subshell. And when l=3, the subshell is even more complicated and is called the f-subshell.
The letters s, p, d and f have been taken from the old spectroscopic terms, sharp, principal, diffused and fundamental, respectively.

The maximum value of electrons in a sub-shell is given by the formula 2(2l+1). That is, s=2, p=6, d=10 and f=14.

Magnetic Quantum Number (m)
This quantum number, which is represented by m, explains the magnetic properties of the electron. The magnetic field created by the charged electron can interact with the external magnetic field, due to which the electrons in a given energy sub-level, orient themselves in certain specific regions of space around the nucleus, called orbitals. The orbitals of a given sub-shell are degenerate (same energy). The number allowed to m depends on the “l” values and ranges from –l through 0 to +l, making a total of (2l+1) values i.e. when l=0, m=0 (subshell is “s”) when l=1, m=-1, 0, +1 (subshell is p which is oriented in three directions x, y, z in space). That is to say the subshell p has three degenerate orbitals p_x, p_y and p_z, arranged in space along x, y and z axis.

When l=2, m= -2, -1, 0, +1, +2 (subshell is “d”, which implies that it has five space orientations due to five “m” values and are designated as d_xy, d_xz, d_yz, d_x²–_y² and d_z²).
When l=3, m= -3, -2, -1, 0, +1, +2, +3 (subshell is “f”, having orientations in seven different directions in space).

Spin Quantum Number (s)
This quantum number describes the spin of an electron on its axis. An electron may spin either in clockwise direction or in anti-clockwise direction. The direction of the spin can be found out by the application of an external magnetic field. Since an electron has equal probability to revolve clockwise and anti-clockwise direction around the nucleus on its own axis so the value for s may be +½ or –½ i.e.

Anti-clockwise spin is assigned positive sign while clockwise spin is assigned negative sign as a convention.
Two electrons, with the same spin, are said to have the parallel spins and are represented by (↑↑), while other are said to have antiparallel or pair up spins, and are represented as (↑↓) or (↓↑).

Q.7 a) What is an orbital? How does it differ from an orbit?

Answer:OrbitalA definite region in three dimensional space around the nucleus, where there is maximum probability of finding an electron of a specific energy “E”.Difference between Orbit and Orbital

Q.7 b) Discuss the shapes of s, p and d orbitals.

Answer:
Orbitals have no physical existence. These are in fact regions of space around the nucleus, where there is maximum probability of finding an electron, with a definite amount of energy. These regions have no strict boundaries. They can however, be pictured by looking at the relative locations of electrons in the principal quantum levels.
The type and shape of the orbital, depends on the value of the azimuthal quantum number, “l”.

Shapes of s orbitals
When l=0, the energy sublevel and the orbital are the same which is called “s” orbital and is spherically symmetrical around the nucleus. Thus all “s” orbitals (l=0) are spherical in shape, the size, however, increases with increasing “n” value. They are just like tennis balls.

Shapes of p orbitals
When l=1, the sub energy level is p which consists of three orbitals for the three m values (-1, 0, +1). Each p orbital has two lobes one on each side of the nucleus. The lobe of a p-orbital may be represented by a plane that contains nucleus and is perpendicular to the corresponding axis. Such a plane is known as nodal plane. For a pxorbital the Y and Z planes are the nodal planes. The lobes of the three different p orbitals can be thought of as directed along x, y and z axes of a set of coordinates.

Shapes of d orbitals
When l=2, the sub energy level is “d” which consists of five different orbitals (for five m values i.e. -2, -1, 0, +1, +2).They are designated as d_xy, d_xz, d_yz, d_x²–_y² and d_z². Their shapes are more complicated than those of p-orbitals. Three of the d orbitals have their lobes lying on the planes xy, xz and yz and two of them have their lobes along the three axis. One of the d orbital dz² resembles p orbital with an additional doughnut shaped space in the center.

Q.8 a) Calculate the wave number and wave length of a photon when the electron jumps from n₂ = 4 to n₁ = 1.

Answer:

Q.8 b) To which series of spectral lines this photon belongs?

Answer:
The photon belongs to Lyman series (U.V region), because electron is returning to its ground state i.e. n₁ = 1 from higher energy level (n₂ = 4).

Q.9 a) What will be the energy (kJ/mol) of an electron residing in n = 3 in a hydrogen atom?

Answer:

Q.9 b) How much energy is lost when an electron in a hydrogen atom jumps from n₂ = 3 to n₁ = 1?

Answer:
As we know that energy lost is equal to the difference in energies of the corresponding orbits, it can be calculated as,

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