NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 Structure of Atom gives an in-depth knowledge of the topics, various theories and models associated with it, like – Rutherford’s Atomic model, Thomson’s Model Bohr’s model, Quantum Mechanical Model of an atom and so on. One can refer to the NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 for detailed explanations and the complete steps required while solving numerical questions related to wavelength, frequency and the energy associated with electromagnetic radiation.
NCERT exemplar class 11 Chemistry solutions chapter 2: MCQ (Type 1)
Which of the following conclusions could not be derived from Rutherford's α-particle scattering experiment?
(a) Most of the space in the atom is empty.
(b) The radius of the atom is about 10 -10 m while that of nucleus is 10 -15 m.
(c) Electrons move in a circular path of fixed energy called orbits.
(d) Electrons and the nucleus are held together by electrostatic forces of attraction.
The answer is the option (c)
Electrons move in a circular path of fixed energy called orbits because it does not explain anything about an atom’s stability, it’s electronic structures, their energies, and distribution around the nucleus.
Which of the following statements is not correct about the characteristics of cathode rays?
(a) They start from the cathode and move towards the anode.
(b) They travel in straight line in the absence of an external electrical or magnetic field.
(c) Characteristics of cathode rays do not depend upon the material of electrodes in cathode ray tube.
(d) Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube.
The answer is the option (d)
Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube because we know that nature of the gas in the cathode rays and characteristics of the cathode rays have no bearing on the material of the electrodes.
Which of the following statements about the electron is incorrect?
(a) It is a negatively charged particle.
(b) The mass of electron is equal to the mass of neutron.
(c) It is a basic constituent of all atoms.
(d) It is a constituent of cathode rays.
The answer is the option (b)
The mass of electron is equal to the mass of neutron because the mass of a neutron is kg and that of an electron is kg.
Two atoms are said to be isobars if
(a) They have same atomic number but different mass number.
(b) They have same number of electrons but different number of neutrons.
(c) They are same number of neutrons but different number of electrons.
(d) Sum of the number of protons and neutrons is same but the number of protons is different.
The answer is the option (d)
Sum of the number of protons and neutrons is same, but the number of protons is different because isobars have a different atomic number.
Number of angular nodes for 4d orbital is __________ .
(a) 4 (b) 3 (c) 2 (d) 1
The answer is the option (c)
2 because we know that for dxy orbitals, there are two angular nodes (represented as “I”) that pass through the origin and bisect the xy plane that contains the z-axis.
Orbital angular momentum depends on
(b) n and l
(c) n and m
(d) m and s
The answer is the option (a)
I, because “I” defines the 3-dimensional shape of the orbital and is commonly known as the subsidiary quantum number or the orbital angular momentum. For any given value of “n”, the “I” can have n values that range from 0 to n-1.
For the electrons of oxygen atom, which of the following statements is correct?
(a) Zeff for an electron in a 2s orbital is the same as Zeff for an electron in a 2p
(b) An electron in the 2s orbital has the same energy as an electron in the 2p
(c) Zeff for an electron in Is orbital is the same as Zeff for an electron in a 2s orbital.
(d) The two electrons present in the 2s orbital have spin quantum numbers ms but of opposite sign.
The answer is the option (d)
The two electrons present in the 2s orbital have spin quantum numbers ms but of opposite sign. This is so because in the same orbital plane, only two electrons can exist, and they must have opposite signs.
NCERT exemplar class 11 Chemistry solutions chapter 2: MCQ (Type 2)
Identify the pairs which are not of isotopes?
The answer is the options (iii) and (iv) because isotopes must have different atomic mass number but an identical atomic number.
Which of the following sets of quantum numbers are correct?
n l m1
(i) 1 1 +2
(ii) 2 1 +1
(iii) 3 2 -2
(iv) 3 4 -2
The answer is the options (ii) and (iii)
As, in the case n = 1, then I ≠ 1, therefore option (i) is incorrect.
In case n = 3, I ≠ 4, therefore option (iv) is also incorrect.
In case, n = 2, I = 0, 1. When l = 1, then m = -1, 0, +1. Therefore option (ii) is correct.
In case n = 3, l= 0, 1, 2. When l = 2, m = -2, -1, 0, +1, +2. Therefore, option (iii) is also correct.
Which of the following statements concerning the quantum numbers are correct?
(i) Angular quantum number determines the three dimensional shape of the orbital.
(ii) The principal quantum number determines the orientation and energy of the orbital.
(iii) Magnetic quantum number determines the size of the orbital.
(iv) Spin quantum number of an electron determines the orientation of the spin of electron relative to the chosen axis.
The answer is the options (i) and (iv) as only these two statements are correct. In contrast, the principal quantum number determines the orbit's size while the orientation of the electron cloud within a sub-shell is determined by the magnetic quantum number. Hence, statements (ii) and (iii) are incorrect.
Which of the following orbitals are degenerate?
Orbitals with the same shells and sub-shells are known as degenerate orbitals. From the given options, the options that represent degenerate orbitals due to the same main shell, i.e. n=3 are
Moreover, represent degenerate orbitals due to the same value of n = 4.
The arrangement of orbitals on the basis of energy is based upon their (n+l) value. Lower the value of (n+l), lower is the energy. For orbitals having same values of (n+l), the orbital with lower value of n will have lower energy.
Based upon the above information, arrange the following orbitals in the increasing order of energy.
(a) 1s, 2s, 3s, 2p
(b) 4s, 3s, 3p, 4d
(c) 5p, 4d, 5d, 4f, 6s
(d) 5f, 6d, 7s, 7p
Based upon the above information, solve the questions given below:
(a) Which of the following orbitals has the lowest energy?
4d, 4f, 5s, 5p
(b) Which of the following orbitals has the highest energy?
5p, 5d, 5f, 6s, 6p
I. (a) 1s<2s<2p<3s
(d) 7s < 5f < 6d < 7p
II. (a) 5s
What is the difference between the terms orbit and orbital?
Orbits are circular and disc-like.
Different orbitals have different shapes, i.e. s orbitals are spherically symmetrical, p-orbitals are dumb-bell shaped and so on.
It represents the planar motion of an electron around the nucleus.
It represents the three-dimensional motion of an electron around the nucleus.
Orbit is a well-defined circular path around the nucleus in which the electrons revolve.
An orbital is a three-dimensional space around the nucleus within which the probability of finding an electron is maximum.
Hydrogen atom has only one electron, so mutual repulsion between electrons is absent. However, in multielectron atoms mutual repulsion between the electrons is significant. How does this affect the energy of an electron in the orbitals of the same principal quantum number in multielectron atoms?
The energy of an electron in a hydrogen atom is determined by the principal quantum number. Thus, the energy of the orbitals increases as: 1s< 2s= 2p< 3s= 3p= 3d< 4s= 4p < 4d= 4f.
However, the energy of an electron in a multielectron atom is dependent on not just the principal quantum number (shell) as well as also on the azimuthal quantum number (subshell) which is quite unlike that of an hydrogen atom. This implies that, for a given principal quantum number, s, p, d, f, all will have different energies.
NCERT Exemplar Class 11 Chemistry Solutions Chapter 2: Long Answer Type
What is photoelectric effect? State the result of photoelectric effect experiment that could not be explained on the basis of laws of classical physics. Explain this effect on the basis of quantum theory of electromagnetic radiations.
Photoelectric effect states that when specific metals are exposed to a light beam, the metal ejects the electrons. This phenomenon is known as the photoelectric effect. The electrons ejected are called photoelectrons.
The results of this experiment were:
i) The electrons are ejected from the metal surface only as the beam of light strikes the surface.
ii) The number of electrons is proportional to the intensity of radiation.
iii) For each metal; there is a minimum frequency, below which photoelectric effect is not observed.
iv) K.E. of electrons is proportional to the frequency of light.
When a photon of adequate energy strikes an electron in the atom of the metal, it transfers its energy to the electron, and the electron is expelled without any delay. Greater is the energy of the photon, greater will also be the kinetic energy of the ejected electron.
Threshold frequency, is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency was allowed to hit a metal surface, an electron having of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.
As per the quantum theory,
Where h represents the plank's constant, i.e.
As per the information provided in the question, we must calculate which represents the threshold frequency of photons.
v = frequency of incident photons. As per the information provided,
electrons kinetic energy =
(which is, in fact, the threshold frequency)
The relation between frequency and wavelength can be represented by the equation,
, where λ represents the wavelength and c represents the speed of light, i.e.
Here, in this case of the threshold frequency
Thus, the photon's maximum wavelength can be represented as,
Now, we know that which is certainly greater than the calculated value of the . Thus, the electron will not be emitted in case a photon which has a wavelength that is equal to 600 nm hits the surface of the metal.
When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula
What points of Bohr's model of an atom can be used to arrive at this formula? Based on these points derive the above formula giving description of each step and each term.
When through the hydrogen gas, an electric discharge is passed, the molecules produce the excited hydrogen atoms by disassociating. Electromagnetic radiation is emitted by these excited atoms of discrete frequencies. It can be represented by the following formula :
2 basic postulates of Bohr's model of an atom can be used to reach this interpretation of the spectrum of atomic hydrogen:
1. The following principle is followed by the criterion for selecting the electron's stationary orbits, "The angular momentum of the electron should be an integral multiple of"
If we assume that at an orbit of radius r, the mass and velocity of the electron is m and v, then its angular momentum can be given by the following formula: -
where n is a non-zero positive integer.
2. In accordance with the quantum theory of radiation, when an electron changes its orbit from one orbit to another, there will be a difference of energy between the two energy levels which will be either absorbed or emitted.
Therefore, when from an orbit with energy E2, an electron jumps to an orbit of energy E1 (E2>E1), the difference in energy is given out in the form of quantized radiation. If we assume the frequency of the radiation to be ϑ, then the energy will be given by :
With these two points, Rydberg's constant can be reached –
In a Bohr type system orbit, consider 2 electron orbits that have quantum numbers n1 and n2, in such a way that n2>n1. The corresponding energies are En1 and En2 and En1<En2.
Therefore, during the transfer of electrons from n2 to n1 orbit, the energy emitted will be: -
, wherein, ϑ is the frequency of radiation.
As per the Bohr atomic model for the hydrogen atom (Z=1):
Now, on substituting the values form, e, and h
NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 pdf download can also be used by the students to get in-depth knowledge of topics like quantum numbers and their combinations, writing electronic configurations. These topics are of value from the examination point of view.
NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 Main Subtopics
NCERT Exemplar Solutions for Class 11 Chemistry Chapter 2 Structure of Atom include the following topics:
2.1 Subatomic Particles
2.1.1 Discovery of Electron
2.1.2 Charge to Mass Ratio of Electron
2.1.3 Charge on Electron
2.1.4 Discovery of Protons and Neutrons
2.2 Atomic Models
2.2.1 Thomson Model of Atom
2.2.2 Rutherford’s Nuclear Model of Atom
2.2.3 Atomic Number and Mass Number
2.2.4 Isobars and Isotopes
2.2.5 Drawbacks of Rutherford’s Model
2.3 Development Leading to Bohr’s Model of Atom
2.3.1 Wave Nature of Electromagnetic Radiation
2.3.2 Planck’s Quantum Theory
2.3.3 Atomic Spectra: Evidence of Quantized Energy Levels
2.4 Bohr’s Model for Hydrogen Atom
2.4.1 Line Spectrum of Hydrogen
2.4.2 Drawbacks of Bohr’s Model
2.5 Towards Quantum Mechanical Model of Atom
2.5.1 Dual Behaviour of Matter
2.5.2 Heisenberg’s Uncertainty Principle
2.6 Quantum Mechanical Model of Atom
2.6.1 Orbitals and Quantum Numbers
2.6.2 Shapes of Atomic Orbitals
2.6.3 Energies of Orbitals
2.6.4 Filling of Orbitals in Atom
2.6.5 Electronic Configuration of atoms
2.6.6 Stability of filled and half-filled subshells
What will students learn from NCERT Exemplar Class 11 Chemistry Solutions Chapter 2?
Through the use of NCERT Exemplar Solutions Class 11 Chemistry Chapter 2 students will be able to understand the concepts of the Quantum Mechanical Model of Atom in which they’ll learn about the shapes and energies of atomic orbitals and how the filling of orbitals takes place in an atom.
NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 Structure of Atom provides detailed knowledge on topics including the Bohr’s Model for Hydrogen Atom and its drawbacks. The learners will be able to comprehend and solve numerical value questions regarding subatomic particles. The discovery of the subatomic particles and their properties are also discussed.
Chemistry Class 11 Chapter Wise Links
Important Topics to cover from NCERT Exemplar Class 11 Chemistry Solutions Chapter 2
Some of the important topics from the chapter that must be specifically learned by the students are:
Class 11 Chemistry NCERT Exemplar Solutions Chapter 2 talks about the subatomic particles and also about cathode and anode rays, their origin and properties. Apart from this one can learn about different models of atoms and their drawbacks.
The students will get to learn about electromagnetic radiations and spectrum. It’s properties and reasons for failure have also been discussed with the help of suitable graphs and diagrams. One can also learn about topics like black body radiation and the photoelectric effect.
Topics about the transition of electrons in the Hydrogen atom and its line spectrum that includes Lyman, Balmer, and Paschen series which have been discussed in the NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 and its related questions and solutions are included to offer a better understanding to the students.
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