Sample Chemistry Assignment
Transcription
Sample Chemistry Assignment
Sample Chemistry Assignment Multiple Choice Questions 1. The incorrect statement about Anode Rays is: a) Anode Rays are deflected in electrical or magnetic fields b) Their e / m values depend on the gas present in the discharge tube c) The e / m value of Anode Rays is a constant d) Anode Rays are generated by the ionization of the gas in the discharge tube Answer . (c) The e / m values of Anode Rays are not the same but depends on the particular gas present in the discharge tube. 2. The velocity of electrons in various Bohr orbits of hydrogen atom can be expressed by: 2πe 2 2πe 3 2π 2 e 2 2πe 2 a) b) c) d) nh nh nh nh 3 2 Ze . Answer : (a) The velocity of an electron in a Bohr orbit v = 2πnh In case of hydrogen, Z = 1. Therefore, 2πe 2 nh 3. The species on which the Bohr’s Theory is applicable: b) H+ c) Li a) He+ d) Li+ Answer : (a) Bohr’s Atom Model is applicable to only single-electron systems (atoms and ions). The number of electrons in He+ H+ Li and Li+ are, respectively, 1 , 0 , 3 , 2. Therefore, among the given examples, Bohr’s Atom Model is applicable to only He+. 4 . If there are 5 electrons in a d orbital, the electron distribution according to Hund’s Rule will be: a) ↑↓ ↑↓ ↑ b) ↑↓ ↑ c) ↑ ↑ ↑ ↑ ↑ ↑ d) ↑↑ ↑↑ ↑ Answer : (c) According to Hund’s Rule of maximum multiplicity, the degenerate orbitals will be first singly occupied by electrons of same spin, until all the orbitals are so occupied. 1 5. The correct order of energies of orbitals is: b) 1s > 4 f > 4 p > 3d > 2 s a) 4d > 6 s > 4 p > 4 s > 3d c) 4d > 5s > 4 p > 3d > 4 s d) 3 p > 4d > 2 p > 2 s > 1s Answer : (c) Between two orbitals, the one with a lower value of (n + l ) has a lower energy. If two orbitals have the same value of (n + l ) , then the one with a lower value of n has a lower energy. For the 4d orbital, (n + l ) = 4 + 2 = 6 For the 5s orbital, (n + l ) = 5 + 0 = 5 For the 4 p orbital, (n + l ) = 4 + 1 = 5 For the 3d orbital, (n + l ) = 3 + 2 = 5 For the 4 s orbital, (n + l ) = 4 + 0 = 4 Since the 5s , 4 p and 3d orbitals have the same value of (n + l ) , the order of their energies are determined by the values of n . 6 . Two electrons in the same orbital can be distinguished by their a) Principle Quantum Number b) Azimuthal Quantum Number c) Magnetic Quantum Number d) Spin Quantum Number (d) Two electrons in the same orbital have the same values for each of n , l and m , but different values of s + 12 and − 12 . 7. Bohr’s Atom Model can explain: a) The molecular spectrum of hydrogen b) The atomic spectrum of hydrogen only c) The solar spectrum d) Spectra of atoms or ions with only one electron Answer : (d) Bohr’s Atom Model can explain the emission spectra of only single-electron systems (atoms and ions), e.g., H , He+ , Li2+. ----------------------------------------------------------------------------------------------------------------------8 . The number of unpaired electrons in Mn 2+ , Fe 3+ and Ni 2+ ions are, respectively, a) 5, 5, 1 b) 4, 5, 2 c) 5, 4, 2 d) 5, 5, 2 Answer: (d) Mn 2 + → [Ar ]3d 5 or [ Ar ] Number of unpaired electrons 5 Fe 3 + → [ Ar ]3d 5 Ni 2 + → [ Ar ]3d 8 or or [Ar ] ↑ ↑ ↑ ↑ ↑ Number of unpaired electrons 5 [Ar ] ↑↓ ↑↓ ↑↓ ↑ ↑ Number of unpaired electrons 2 2 9. When an electron is promoted from the first to the third orbit, the radius of the orbit increases by a factor of : a) 4 b) 2 c) 9 d) 10 Answer: (c) The radius of the n th Bohr orbit of hydrogen atom rn = n2h2 4π 2 me 2 or, rn ∝ n 2 [ h, π , m, e are constants] . Therefore r3 : r1 = 32 : 12 or, r3 = 9r1 10. Pick the incorrect statement: a) Electronic configuration of Cr is [ Ar ] 3d 5 4 s1 b) Magnetic quantum numbers can have negative values c) An f orbital is possible in the fourth principal quantum level d) NO + ion is paramagnetic Answer: (d) NO + does not have any unpaired electron. Therefore, it is not paramagnetic. Short Answer Type Questions 1. Which experimental observation proved that electrons are universal constituents of all matter? Answer : Determination of the e/m value of electrons in cathode rays by J. J. Thomson (1897) showed that this characteristic of the cathode rays remains the same no matter what the nature of the gas in the cathode ray tube. Since it was recognized that the electrons are produced by the ionization of the gas in the discharge tube, this proved that electrons are universal constituents of all matter. 2. Briefly describe Rutherford’s -particle scattering experiment and explain the conclusions drawn from the results. Answer : Rutherford (1911) directed a stream of -particles from radioactive material on to a very thin gold foil and placed a Zn sulphide screen just behind it with the objective of detecting the -particles by the scintillations they produce. Through a careful count of the number of scintillations with and without the gold foil in place, he found that only 1 in approximately 20,000 -particles is scattered through a large angle, while the rest mostly went right through the gold foil. He correctly explained scattering through such large angle due to a close encounter with a positively charged massive body, subsequently called the nucleus, in a gold atom. Since such large deflections were so rare, he concluded that most of the space inside an atom must be empty. 3 3. Define “Wavelength” and “Wave Number”, and deduce their relationships with frequency of an electromagnetic radiation. Answer : Wavelength ( λ ) is the distance between the two successive crests or troughs of any wave motion, and hence has the unit of length (m or cm). Wave Number (ν ) is the number of waves per unit length, is equal to the reciprocal of wavelength, and has the unit m −1 or cm −1 . 1 Velocity of an electromagnetic wave, c = νλ = ν × or, ν = cν , where is the frequency. ν 4. What is the significance of the energy of an electron being negative in a stable atomic orbit? Answer : Energy of an electron is, by definition, zero at infinite distance from the nucleus of an atom, where the energy is highest and the electron most unstable. At any point nearer to the nucleus, therefore, the energy of an electron is less than zero, i.e., negative. This also signifies that energy must be added to remove an electron to an infinite distance from the nucleus and thus ionize an atom, since it is by addition to only a negative quantity that one can get zero. 5. Name the Quantum Numbers required to fully define an electron in an atomic orbit, and briefly explain their roles. Answer : The Quantum Numbers required to fully define an electron in an atomic orbit are Principal (n), Azimuthal (l), magnetic (m) and spin (s) quantum numbers. Principal Quantum Number defines the size of the orbit, Azimuthal Quantum Number defines the shape of the orbit, the Magnetic Quantum Number defines the orientation of the orbits in a magnetic field, and the Spin Quantum Number distinguishes two electrons in the same orbital. 4