Class 11 Chemistry · 55 questions · 48 with answers
Which of the following conclusions could not be derived from Rutherford’s α -particle scattering experiement?
(iii) Electrons move in a circular path of fixed energy called orbits.
Which of the following options does not represent ground state electronic configuration of an atom?
(ii) 1s2 2s2 2p6 3s2 3p6 3d9 4s2
The probability density plots of 1s and 2s orbitals are given in Fig. 2.1: 1s 2s Fig. 2.1 The density of dots in a region represents the probability density of finding electrons in the region. On the basis of above diagram which of the following statements is incorrect?
This question refers to a figure in the original PDF.
(iv) The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases.
Which of the following statement is not correct about the characteristics of cathode rays?
(iv) Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube.
Which of the following statements about the electron is incorrect?
(ii) The mass of electron is equal to the mass of neutron.
Which of the following properties of atom could be explained correctly by Thomson Model of atom?
(i) Overall neutrality of atom.
Two atoms are said to be isobars if.
(iv) sum of the number of protons and neutrons is same but the number of protons is different.
The number of radial nodes for 3p orbital is __________.
(iv) 1 15 Structure of Atom
Number of angular nodes for 4d orbital is __________.
(iii) 2
Which of the following is responsible to rule out the existence of definite paths or trajectories of electrons?
(ii) Heisenberg’s uncertainty principle.
Total number of orbitals associated with third shell will be __________.
(iii) 9
Orbital angular momentum depends on __________.
(i) l
Chlorine exists in two isotopic forms, Cl-37 and Cl-35 but its atomic mass is 35.5. This indicates the ratio of Cl-37 and Cl-35 is approximately
(iii) 1:3
The pair of ions having same electronic configuration is __________.
(ii) Fe3+, Mn2+
For the electrons of oxygen atom, which of the following statements is correct?
(iv) The two electrons present in the 2s orbital have spin quantum numbers ms but of opposite sign.
If travelling at same speeds, which of the following matter waves have the shortest wavelength?
(ii) Alpha particle (He2+)
Identify the pairs which are not of isotopes? 12 13
(iii) 6 X, 7 8 8
(iv) 4 X, 5 Y
Out of the following pairs of electrons, identify the pairs of electrons present in degenerate orbitals :
(i) (a) n = 3, l = 2, ml = –2, ms = − (b) n = 3, l = 2, ml = –1, ms = −
(iv) (a) n = 3, l = 2, ml = +2, ms = − (b) n = 3, l = 2, ml = +2, ms = +
Which of the following sets of quantum numbers are correct? n l ml
(ii) 2 1 +1
(iii) 3 2 –2
In which of the following pairs, the ions are iso-electronic?
(i) Na+, Mg2+
(iii) Na+, O2– –
Which of the following statements concerning the quantum numbers are correct?
(i) Angular quantum number determines the three dimensional shape of the orbital.
(iv) Spin quantum number of an electron determines the orientation of the spin of electron relative to the chosen axis.
Arrange s, p and d sub-shells of a shell in the increasing order of effective nuclear charge (Zeff) experienced by the electron present in them.
d < p < s
Show the distribution of electrons in oxygen atom (atomic number 8) using orbital diagram.
Nickel atom can lose two electrons to form Ni2+ ion. The atomic number of nickel is 28. From which orbital will nickel lose two electrons.
4s
Which of the following orbitals are degenerate? 3d xy ,4d xy 3d 2 ,3dyz ,4dyz ,4d 2 z z
3dxy, 3d 2 , 3dyz and 4dxy , 4dyz, 4d 2 z z
Calculate the total number of angular nodes and radial nodes present in 3p orbital.
For 3p orbital n = 3, l = 1 Number of angular nodes = l = 1 Number of radial nodes = n – l – 1 = 3–1–1=1
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. I. 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 II. 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
Which of the following will not show deflection from the path on passing through an electric field? Proton, cathode rays, electron, neutron.
neutron
An atom having atomic mass number 13 has 7 neutrons. What is the atomic number of the atom?
A = 13, A – Z = 7 Z = 6 atomic number = 6
Wavelengths of different radiations are given below : λ( A) = 300 nm λ(B) = 300 μ m λ (C) = 3 nm λ(D) = 30 A Arrange these radiations in the increasing order of their energies.
B<A<C=D Hint : E α λ
The electronic configuration of valence shell of Cu is 3d104s1 and not 3d 94s2. How is this configuration explained?
Completely filled and half filled orbitals have extra stability. In 3d104s1, d orbital is completely filled and s is half filled. So it is more stable configuration. 1 1
The Balmer series in the hydrogen spectrum corresponds to the transition from n1 = 2 to n 2 = 3,4,......... . This series lies in the visible region. Calculate the wave number of line associated with the transition in Balmer series when the electron moves to n = 4 orbit. (RH = 109677 cm–1)
109677 2 2 cm 1 n i n f For ni = 2 to nf = 4 transition in Balmer series. ∴ 109677 12 12 cm 1 2 4 1 1 1 109677 cm = 20564.44 cm–1 4 16
According to de Broglie, matter should exhibit dual behaviour, that is both particle and wave like properties. However, a cricket ball of mass 100 g does not move like a wave when it is thrown by a bowler at a speed of 100 km/h. Calculate the wavelength of the ball and explain why it does not show wave nature. 19 Structure of Atom
m = 100 g = 0.1 kg. 100 1000 m 000 1 v = 100 km/hr 60 60 s h = 6.626 × 10–34 Js 34 6.626 10 Js 36 1 6.626 10 36 m = 238.5 × 10–36m–1 1000 1 0.1 kg ms Since the wavelength is very small, the wave nature cannot be detected.
What is the experimental evidence in support of the idea that electronic energies in an atom are quantized?
Out of electron and proton which one will have, a higher velocity to produce matter waves of the same wavelength? Explain it.
Being lighter particles, electrons will have higher velocity. h Hint : mv
A hypothetical electromagnetic wave is shown in Fig. 2.2. Find out the wavelength of the radiation. Fig. 2.2
This question refers to a figure in the original PDF.
Wavelength is the distance between two successive peaks or two successive troughs of a wave. So 4 2.16 pm = 8.64 pm 8 –1 c 3.0 10 ms –6
Chlorophyll present in green leaves of plants absorbs light at 4.620 × 1014 Hz. Calculate the wavelength of radiation in nanometer. Which part of the electromagnetic spectrum does it belong to?
= = 0.6494 10 m = 649.4 nm; Visible light. 4.620 10 90 4 360 –1
What is the difference between the terms orbit and orbital?
Table-tennis ball has a mass 10 g and a speed of 90 m/s. If speed can be measured within an accuracy of 4% what will be the uncertainty in speed and position?
Uncertainty in the speed of ball = 3.6 ms 100 100 Uncertainty in position = 4 mv 25 Structure of Atom –34 6.626 10 Js = –3 –1 –1 4 3.14 10 10 kg g 3.6 ms = 1.46 × 10–33 m
The effect of uncertainty principle is significant only for motion of microscopic particles and is negligible for the macroscopic particles. Justify the statement with the help of a suitable example.
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 electron is determined by the value of n in hydrogen atom and by n + l in multielectron atom. So for a given principal quantum number electrons of s, p, d and f orbitals have different energy. IV. Matching Type
Match the following species with their corresponding ground state electronic configuration. Atom / Ion Electronic configuration (i) Cu
(a) 1s2 2s2 2p6 3s2 3p6 3d10 (ii) Cu2+
(c) 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (iv) Cr3+
(b) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 (iii) Zn2+
(d) 1s2 2s2 2p6 3s2 3p6 3d 9 (e) 1s2 2s2 2p6 3s2 3p6 3d 3
Match the quantum numbers with the information provided by these. Quantum number Information provided (i) Principal quantum number
(a) orientation of the orbital (ii) Azimuthal quantum number
(b) energy and size of orbital (iii) Magnetic quantum number
(d) shape of the orbital
(c) spin of electron (iv) Spin quantum number
Match the following rules with their statements : Rules Statements (i) Hund’s Rule
(a) No two electrons in an atom can have the same set of four quantum numbers.
(c) Pairing of electrons in the orbitals belonging to the same subshell does not take place until each orbital is singly (iii) Pauli Exclusion Principle occupied.
(b) Half-filled and completely filled orbitals have extra stablity. (ii) Aufbau Principle
(d) It is impossible to determine the exact position and exact momentum of a subatomic particle simultaneously. (iv) Heisenberg’s Uncertainty Principle (e) In the ground state of atoms, orbitals are filled in the order of their increasing energies.
Match the following (i) X-rays
(a) ν = 100 – 104 Hz (ii) UV
(d) ν = 1018 Hz 21 Structure of Atom
(b) ν = 1010 Hz (iii) Long radio waves
(c) ν = 1016 Hz (iv) Microwave
Match the following (i) Photon
(a) Value is 4 for N shell (ii) Electron
(d) Exhibits both momentum and wavelength
(b) Probability density ψ (iii)
(c) Always positive value (iv) Principal quantum number n
Match species given in Column I with the electronic configuration given in Column II. Column I Column II (i) Cr
(a) [Ar]3d84s0 (ii) Fe2+
(d) [Ar] 3d54s1 (e) [Ar]3d64s2
(b) [Ar]3d104s1 (iii) Ni2+
(c) [Ar]3d64s0 (iv) Cu
Assertion (A): All isotopes of a given element show the same type of chemical behaviour.
Reason (R): The chemical properties of an atom are controlled by the number of electrons in the atom. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) Both A and R are false.
(i) Both A and R are true and R is the correct explanation of A.
Assertion (A): Black body is an ideal body that emits and absorbs radiations of all frequencies.
Reason (R): The frequency of radiation emitted by a body goes from a lower frequency to higher frequency with an increase in temperature. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the explanation of A. (iii) A is true and R is false. (iv) Both A and R are false.
(ii) Both A and R are true but R is not the explanation of A.
Assertion (A): It is impossible to determine the exact position and exact momentum of an electron simultaneously.
Reason (R): The path of an electron in an atom is clearly defined. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true and R is not the correct explanation of A. (iii) A is true and R is false. (iv) Both A and R are false.
(iii) A is true and R is false.
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.
Threshold frequency, ν0 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 1.0×1015 s–1 was allowed to hit a metal surface, an electron having 1.988 × 10–19 J 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.
Hint : h h 0 mv
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 1 1 109677 2 2 n i n f 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.
Calculate the energy and frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom.
∆E = –3.052 × 10–19 J, ν = 4.606 × 1016 Hz
Why was a change in the Bohr Model of atom required? Due to which important development (s), concept of movement of an electron in an orbit was replaced by, the concept of probability of finding electron in an orbital? What is the name given to the changed model of atom? 23 Structure of Atom