["2.4 Chapter 2 The discovery of negatively-charged electron was later followed by the experiment conducted by Robert Millikan in 1909 to determine the quantity of charge on an electron. Millikan\u2019s Oil Drop Experiment Some fine oil droplets were allowed to be sprayed into the chamber by an atomiser. The air in the chamber was subjected to ionisation by X-rays. The electrons produced by the ionisation of air attached themselves to the oil drops. When sufficient amount of electric field is applied which can balance the gravitational force acting on an oil drop, the drop remains suspended in the air. From this experiment, Millikan observed that the smallest charge found on them was approximately 1.59 \u00d7 10\u201319 coulombs and the charge on each drop was always an integral multiple of that value. On the basis of this observation, he concluded that 1.59 \u00d7 10\u201319 coulombs is the smallest possible charge and considered that value as the charge of the electron. Oil spray Atomiser to produce (+) oil droplets X-rays produce Microscope charges on the oil drops Electrically charged plates (\u2013) FIGURE 2.3\u2002\u0007A schematic representation of the apparatus used by Millikan to determine the charge of an electron DISCOVERY OF PROTONS The presence of positively-charged particles in an atom has been predicted by Goldstein based on the electrical neutrality of an atom. The discovery of proton by Goldstein was done on the basis of the cathode ray experiment conducted by using a perforated cathode. Just like the cathode rays, some rays were found to emanate from an anode. These are called anode rays or canal rays. Anode rays were found as a stream of positively-charged particles in contrast to the cathode rays. When hydrogen gas is taken in a discharge tube, these positively-charged particles were found to be protons. Positive rays Cathode rays Positive Red glow from anode from cathode rays ++ \u2013 \u2013 + \u2013 + + + + Anode \u2013 Perforated cathode + \u2013 +\u2013 HV source FIGURE 2.4 Discovery of Protons Properties of Anode Rays 1.\t Anode rays travel in straight lines. 2.\t A\u0007 node rays possess positive charge since they were found to deflect towards negatively-charged electrodes.","Atomic Structure 2.5 3.\t The properties of anode rays depend upon the nature of the gas taken in the discharge tube. 4.\t The mass of the particles was same as the atomic mass of the gas inside the discharge tube. The discovery of electrons and protons as subatomic particles inside the atom lead to the conception of atomic models which depict the arrangement of fundamental particles in an atom. Various atomic models have been proposed by different scientists, like, J.J. Thomson, Rutherford, Bohr and Sommerfeld. Thomson\u2019s Atomic Model J.J. Thomson proposed his atomic model soon after his discovery of electrons as listed hereunder. 1.\t A\u0007 n atom contains negatively-charged particles called electrons embedded uniformly throughout a thinly spread positively-charged mass. 2.\t \u0007Since the atom is electrically neutral, the total negative charge of electrons is balanced by the total positive charge. Electrons Positively- charged sphere FIGURE 2.5\u2002 Thomson\u2019s atomic model Thomson\u2019s model of an atom is popularly known as \u2018plum pudding model\u2019 or \u2018apple pie model\u2019 or \u2018watermelon model\u2019. Validity of Thomson\u2019s Model Thomson\u2019s model could successfully explain the electrical neutrality of atom. However, it failed to explain how the positively-charged particles are shielded from the negatively-charged electrons without getting neutralised. EXAMPLE Write different isotopes of oxygen, carbon and chlorine. SOLUTION The isotopes of oxygen are: 8 O16 , 8 O17 , 8 O18 . The isotopes of chlorine are: 17 Cl35 , 17 Cl37 . The isotopes of carbon are: 6 C12 , 6 C13 , 6 C14 .","2.6 Chapter 2 EXAMPLE What was the basis for the proposal of Dalton\u2019s atomic theory? SOLUTION Laws of chemical combination, such as, the law of conservation of mass, the law of definite proportions and the law of multiple proportions, were the basis for the proposal of Dalton\u2019s atomic theory. EXAMPLE \t(i)\t What are A, B, C, D, and E in the given figure? A EB \t(ii)\t C \t(iii)\t What is the purpose of C? D E\u0007 xplain the role of D in the phenomenon taking place in the discharge tube. SOLUTION \t(i)\t A is cathode, B is perforated anode, C is suction pump, D is zinc sulphide screen and E is cathode ray. \t(ii)\t C is suction pump which can help in reducing the pressure in the discharge tube. \t(iii)\t D is zinc sulphide screen. Zinc sulphide is a fluorescent material. When cathode rays strike the zinc sulphide screen, bright spots are formed on the screen. EXAMPLE \u0007An atom of an element is represented as ZXA. After the emission of a \u03b2-particle, another element Y is formed. Represent Y with atomic number and mass number. SOLUTION z XA \u2192 z+1YA + \u03b2 particle. EXAMPLE C\u0007 alculate the specific charges (e\/m) of the following particles and then arrange the particles in the ascending order of their specific charges. (a)\tElectron\t (b)\tProton\t (c)\t \u03b1-particle SOLUTION Specific charge of an electron = 1.6\u00d710\u221219 coulomb\/kg 9.1\u00d710\u221231 \t\t\t\t \u2009= 0.176 \u00d7 1012 = 17.6 \u00d7 1010 coulomb\/kg Specific charge of a proton = 1.6\u00d710\u221219 coulomb\/kg = 0.96 \u00d7 108 coulomb\/kg 1.67\u00d710\u221227 Specific charge of \u03b1 - particle = 2\u00d71.6\u00d710\u221219 = 1.6\u00d7108 = 0.472 \u00d7 108 coulomb\/kg 2\u00d710\u221227(1.67 + 1.72) 3.39 Hence, ascending order of specific charges of electron, proton and \u03b1-particles is \u03b1-particle < proton < electron.","Atomic Structure 2.7 EXAMPLE C\u0007 alculate the mass of a charged particle in CGS units if its charge is x coulomb and specific charge is y coulomb\/g. SOLUTION The mass of the particle in CGS units is e g= x g. e y m EXAMPLE T\u0007 he isotopes of an element have mass numbers: A, A + 1, A + 2. The ratio of abundance of these isotopes is 3 : 2 : 4. Calculate the average atomic mass of the element. SOLUTION Average atomic mass = A\u00d73 + (A + 1)2 + (A + 2)4 = 3A + 2A +2 + 4A + 8 =A+ 10 9 9 9 Rutherford\u2019s \u03b1-ray Scattering Experiments In order to test the validity of Thomson\u2019s atomic model, Rutherford conducted \u03b1-ray scattering experiment. In this experiment, \u03b1-particles were allowed to pass through a pair of positively-charged parallel plates and the resultant narrow beam of \u03b1-particles was allowed to strike the gold foil which was surrounded by zinc sulphide screen. The observations or results of this experiment completely disproved Thomson\u2019s model. Perforation \u03b1-particles Thin gold foil + Zinc sulphide screen + Bright spot Spherical screen Lead Sources of container \u03b1-particles FIGURE 2.6 \u03b1-ray Scattering Experiment TABLE 2.2\u2002 Observations and conclusions of \u03b1-ray scattering experiment Observations Conclusions Most of the particles passed straight through the gold Presence of large empty space in an atom foil without any deflection Very few \u03b1-particles completely rebounded and few Presence of a central, positively-charged core known as nucleus \u03b1-particles showed large deflection Rutherford\u2019s Atomic Model The atom is mostly composed of empty space. The entire positive charge and mass of the atom is concentrated in the small, central part known as the nucleus. The size of the nucleus is so small that its diameter is 105 times less than that of the atom. The diameter of the nucleus has been estimated by","2.8 Chapter 2 Rutherford as 10\u201313 cm in contrast to that of the atom to be 10\u20138 cm. Nucleus The electrons present outside the nucleus revolve round the nucleus Electron with high velocities. The electrons revolve round the nucleus with high velocities to counterbalance the electrostatic forces of attraction between protons and electrons. Rutherford\u2019s atomic model resembles the planetary motion in solar system. Therefore, Rutherford\u2019s model of an atom is also called planetary model. Validity of Rutherford\u2019s Atomic Model FIGURE 2.7 \u0007The solar system Rutherford\u2019s atomic model could very well explain the presence of positively-charged nucleus and presence of electrons outside the nucleus in the atom. However, the failure of Rutherford\u2019s theory stemmed from two major objections. \t1.\tT\u0007 his model is in contradiction to the principle of classical electrodynamics. According to this, any charged particle in a circular motion radiates energy continuously. The electron being a charged particle in the circular motion loses energy. This should ultimately result in its spiral path towards nucleus and the atom should then collapse. \t2.\tThe second major objection to Rutherford\u2019s model came from the pattern of atomic spectra. When light passes through a prism, it gets split up into its components of different wave lengths, like, visible, ultraviolet, infrared light, etc. The arrangement of component light energies according to their wavelengths is called spectrum and spectroscope is the instrument designed to observe the spectra. Since white light is composed of lights of different wavelengths, a continuous band of different wave lengths is obtained which is called continuous spectrum. Light from the sun or incandescent bulb gives such type of spectra. Continuous spectrum (+) V I B G Y O R Electric arc (\u2013) (white light house) Slit Prism Detector (photographic plate) (a) High Detector Voltage (photographic plate) Hydrogen Prism gas Arc Slit (b) FIGURE 2.8 (a) Spectrum produced by the white light \u2003 . (b) Atomic spectra of hydrogen But when the spectrum is taken for the atoms of the gas present in the discharge tube, it is found to consist of discrete lines of different colours. This is called line spectrum, which is a discontinuous spectrum. Figure 2.8 (b) shows the atomic spectra of hydrogen gas. According to Rutherford\u2019s atomic model, electrons revolving around the nucleus should lose energy continuously. Hence, the spectra of the atom should be a continuous spectrum, whereas the observed atomic spectrum was a line spectrum.","Atomic Structure 2.9 In 1913, Danish scientist, Neils Bohr could overcome the limitation of Rutherford\u2019s atomic model successfully based on the quantum theory of radiation proposed by Max Planck. Quantum Theory of Radiation At the end of the 19th century, physicists had an idea that matter and energy are distinctly different. Matter consists of particles which have mass and have specific positions in space. Energy is the form of electromagnetic radiation which has no mass and does not have any specific position in space. Matter can absorb or emit any quantity of energy. But in the beginning of the 20th century, these ideas were proved to be incorrect on the basis of some experimental results. German physicist, Max Planck, in 1901 carried out the first important experiment by studying the radiation emitted by solid bodies heated to incandescence. He concluded from his experimental observation that energy can be absorbed or radiated by a body in the form of small packets of energy called quanta which are whole number multiples of the quantity hv where h = Planck\u2019s constant = 6.625 \u00d7 10\u201334 J.s v = frequency of the radiation EXAMPLE Following conclusions are drawn by observing \u03b1-ray scattering experiment. Write the respective observations based on which these conclusions are drawn. \t(i)\t Non-uniform distribution of positive charge. \t(ii)\t Presence of positively-charged core or nucleus. \t(iii)\tPresence of large empty space in an atom. SOLUTION Observations Conclusions (i) \u0007Non-uniform distribution of positive The angle of deflections of various \u03b1-particles were charge different (ii) P\u0007 resence of positively-charged core or Very few \u03b1-particles completely rebounded and few nucleus \u03b1-particles showed large deflection (iii) P\u0007 resence of large empty space in an Most of the \u03b1-particles passed through the gold foil atom without any deflection EXAMPLE C\u0007 ompare Thomson\u2019s atomic model with Rutherford\u2019s atomic model. SOLUTION Thomson\u2019s atomic model Rutherford\u2019s atomic model (a) In Thomson\u2019s model positively-charged In Rutherford\u2019s model, positively-charged particles particles are thinly spread throughout the atom. are concentrated in a small central part called nucleus. (b) Electrons are embedded in the thin positively- Electrons revolve around the nucleus with a high charged mass. velocity.","2.10 Chapter 2 EXAMPLE The wavelength of particular radiation is 700 nm (1 nm = 10\u22129 m). Find its frequency (\u03bd). SOLUTION \u03bb = 700 nm = 700 \u00d7 10\u20139 m = 7 \u00d7 10\u20137m. Also, c = velocity of light = 3 \u00d7 108 m\/s So, \u03bd = c = 3\u00d7108 = 0.42 \u00d7 1015\/s \u03bb 7\u00d710\u22127 This theory proves the particle nature of energy. On the basis of this theory, Bohr proposed his atom model. Bohr\u2019s Model of an Atom \t1.\t Electrons revolve around the nucleus in specified circular paths called orbits or shells. \t2.\tE\u0007 ach orbit or shell is associated with a definite amount of energy. Hence, these are also called energy levels and are designated K, L, M and N, respectively. \t3.\tT\u0007 he energy associated with a certain energy level increases with the increase of its distance from the nucleus. Hence, if the energy associated with the K, L, M and N shells are E1, E2, E3\u2026\u2026\u2026 respectively, then E1 < E2 < E3 \u2026\u2026\u2026., etc. \t4.\t\u0007As long as the electron revolves in a particular orbit, the electron does not lose its energy. Therefore, these orbits are called stationary orbits and the electrons are said to be in stationary energy states. \t\u20095.\t A\u0007 n electron jumps from a lower-energy level to a higher-energy level, by absorbing energy, but when it jumps from a higher- to lower-energy level, the energy is emitted in the form of electromagnetic radiation. The energy emitted or absorbed (\u2206E) is an integral multiple of \u2018h\u03bd.\u2019 \t6.\tT\u0007 he electron can revolve only in the orbit in which the angular momentum of the electron (mvr) is quantised, i.e., mvr is a whole number multiple of h\/2\u03c0.This is known as the principle of quantisation of angular momentum. \t\tThe angular momentum is written as \t\tmvr = nh , 2\u03c0 \t\twhere, n is an integer (n = 1,2,3,4,\u2026\u2026.) and is called principal quantum number. \t\tm = mass of the electron \t\tv = velocity of an electron in its orbit \t\tr = distance of the electron from the nucleus n=6 n=5 n=4 n=3 n=2 n=1 + K L M NOP FIGURE 2.S9tatSiotnaatriyoonrbaitrsyofoarnbaittosmof an atom","Atomic Structure 2.11 By applying the concept of quantisation of energy, Bohr calculated the radius and energy of the nth orbit of hydrogen atom. rn = n2h2 , En = \u22122\u03c0 2me 4 4 \u03c0 2me 2 n2h2 With the help of these expressions, Bohr gave a satisfactory explanation for the spectra of hydrogen and hydrogen-like species (ions having one electron, e.g., He+, Li+2, Be+3). Limitations of Bohr\u2019s Atomic Model The following are the limitations of Bohr\u2019s atomic model: \t1.\t Bohr could not explain the spectral series for the multi-electron atoms. \t 2.\t \u0007Bohr\u2019s model could not give a satisfactory justification for the assumption that electrons can revolve in those orbits where their angular momentum (mvr) is a whole number multiple of nh\/2\u03c0, i.e., he could not justify quantisation of angular momentum. \t3.\t\u0007According to Heisenberg\u2019s uncertainty principle, it is impossible to determine simultaneously with certainty the exact position and the momentum of the particle. Bohr assumed that an electron of an atom is located at a definite distance from the nucleus and revolves around the nucleus with a definite velocity, i.e., the momentum of the electron is fixed which is against Heisenberg\u2019s uncertainty principle. \t3.\t T\u0007 he atomic spectral lines split into a number of closely packed lines in the presence of a magnetic field and an electric field. These effects are called Zeeman effect and Stark effect, respectively. Bohr failed to explain these effects. \t4.\t\u0007When the hydrogen spectrum was observed with the spectroscope of high resolving power, it was found that the individual lines in the spectrum consisted of several fine lines lying close to each other. This is called fine spectrum and he failed to explain the fine structure of the spectrum. DISCOVERY OF NEUTRONS The electrons being particles having negligible mass and massive protons concentrated inside the nucleus, it could be predicted that the mass of an atom has to be equal to the mass of the total number of protons present in the atom. However, this was found to be true only in case of a hydrogen atom. The difference in the predicted mass and actual mass of the atom has been found to be equal to the mass of the proton or multiples of the mass of a proton. These particles are supposed to have neutral charge since the atom is electrically neutral. They were called neutrons and they were discovered by James Chadwick by an experiment involving the bombardment of beryllium nucleus with \u03b1-particles. The discovery of fundamental particles has ultimately resulted in the establishment of a basic atomic model. The basic model of an atom comprises small positively-charged nucleus at the centre of the atom and the electrons revolving round the nucleus in orbits.","2.12 Chapter 2 TABLE 2.3\u2002 Characteristics of fundamental particles S. No. Fundamental Particles Charges Masses Relative Charges 1. Electron (e) \u20131.6 \u00d7 10\u201319C 9.1 \u00d7 10\u201331 kg \u20131 \u20134.8 \u00d7 10\u201310 esu (or) +1 0. 00055 amu 0 2. Proton (p) + 1.6 \u00d7 10\u201319 C 1.67 \u00d7 10\u201327 kg + 4.8 \u00d7 10\u201310 esu (or) 1.0078 amu 3. Neutron (n) 0 1.72 \u00d7 10\u201327 kg (or) 1.0083 amu TABLE 2.4\u2002 Atomic number and mass number Representations Definitions Examples The number of protons in an atom Atomic number Z Cl atom has 17 protons in its atom, The total number of nucleons, i.e., Z = 17 Mass number A number of protons and neutrons in an atom Cl atom has 17 protons and 18 neutrons in its nucleus. \u2234 A = 17 + 18 = 35 An element when represented along with its atomic number and mass number is represented as ZXA. Elements are found to exist in their isotopic forms. Isotopes are the atoms of the same element having different mass number. Based on their percentage abundance of each isotopic form, average atomic mass of an element is calculated. Examples: 1H1, 1H2, 1H3, etc. It is also found that atoms of different elements have same mass number. These are called isobars. Examples: 18Ar40, 20Ca40, etc. Electronic Configuration The systematic arrangement of electrons in the various shells or orbits in an atom is called electronic configuration. The electrons are arranged in an atom in the various shells around the nucleus. The Shell 2n2 last shell or the outermost shell from the nucleus with electrons is called valence shell. K2 The shell inner to this is called penultimate shell and the one inner to the penultimate L8 shell is called anti-penultimate shell. M 18 M 32 The filling of electrons in various shells can be done according to Bohr\u2013Bury scheme. O 50 According to this, the maximum number of electrons that can be accommodated in any shell is given by 2n2, where n represents the number of the shell. The maximum number of electrons that can be filled in the valence shell is 8, that in the penultimate shell is 18 and the anti-penultimate shell has a maximum capacity of 32 electrons. The filling of electrons till atomic number 30 follows the following pattern:","Atomic Structure 2.13 K L MN 2 28 28 8 2 8 82 2 8 18 2 EXAMPLE A\u0007 n element has protons whose mass is equal to 23,881 times that of an electron. Identify the element and write its electronic configuration. SOLUTION Mass of a proton is 1837-times than that of an electron. Number of protons = 23,881 = 13 1837 Atomic number of the element is 13. So, the element is aluminium and its electronic configuration is 2, 8, 3. EXAMPLE W\u0007 rite the electronic configuration and the atomic number of the atom which becomes stable by gaining 3 electrons in fifth shell. SOLUTION The electronic configuration of the stable species is 2, 8, 18, 18, 8 Hence, the electronic configuration of the atom which becomes stable after gaining 3 electrons in the fifth shell is 2, 8, 18, 18, 5 and the atomic number is 51. EXAMPLE W\u0007 hat is the ratio of the amount of energy required to remove an electron from hydrogen and He+ ion? SOLUTION 11e22le=ct\u2212ro1n3.s6peeVcie;sE) nEenr=gy\u2212req1u3.i6rne\u00d72d Z2 to Energy of electron (single eV ; this electron = +13.6 eV For hydrogen \u21d2 \u2013 13.6 \u00d7 remove For He+ \u21d2 \u2013 13.6 \u00d7 22 =\u2212 54.4 eV ; Energy required to remove this electron = +54.4 eV 12 Ratio of energy required = 13.6 = 1 = 1 : 4. 54.4 4","2.14 Chapter 2 EXAMPLE \u0007Complete the following table: Valence Electronic Number of shells in Charges on configurations electrons in Number of atom stable ions of element penultimate shell core electrons M +3 +1 8 18 N +2 18 2, 8, 6 Note: Core electrons are inner electrons which exclude valence electrons. SOLUTION Valence shells Charges on Electronic Number of Number of core in atom stable ion configurations electrons in electrons M penultimate shell 10 N +3 of element 18 N +1 2, 8, 3 8 18 M +2 10 \u22122 2, 8, 8, 1 8 2, 8, 8, 2 8 2, 8, 6 8","Atomic Structure 2.15 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t Which postulate of Dalton\u2019s atomic theory is \t15.\t What is Heisenberg\u2019s uncertainty principle? PRACTICE QUESTIONS \u00adconsidered to be correct even today? \t16.\t According to _______, the charges in an atom are \t2.\t Like atoms are identical in all respects. This arranged like the pulp and seeds of a watermelon. s\u00adtatement of Dalton\u2019s atomic theory is contradicted. Which discovery contradicts this? \t17.\t Which theory supported the particle nature of an electron? \t3.\t The value of the Planck\u2019s constant \u2018h\u2019 in erg-s is ______. \t18.\t Give the value of Planck\u2019s constant in \t\t(1)\u2002erg-s \t4.\t Why was a gas at low pressure taken by Thomson \t\t(2)\u2002joule-s while conducting the experiment? \t19.\t Give equations to calculate the following: \t5.\t Why is Rutherford\u2019s model called nuclear model? \t\t (a)\u2002 the radius of the nth orbit of hydrogen atom \t\t (b)\u2002 energy of the nth orbit of hydrogen atom \t6.\t Mass of the electron is calculated from ____ and ____ values of electron. \t20.\t What happens when an electron jumps from a lower energy level to a higher energy level? \t7.\t Give the mass and charge of fundamental particles of an atom. \t21.\t What is Zeeman effect? \t8.\t What is an atomic model? \t22.\t What is Stark effect? \t9.\t The equation for the calculation of energy of nth \t23.\t With the increase in the radius of the orbit, the orbit of hydrogen atom derived by Bohr is ______. energy of an electron _____________. \t10.\t Who discovered protons? Based on what experi- \t24.\t What is an \u03b1-particle? ment was he able to discover these protons? \t25.\t What is a continuous spectrum? \t11.\t What was the mathematical equation given by Max Planck? \t26.\t The circular paths in which electrons revolve are called ______. \t12.\t Who discovered neutrons? How was the discovery made? \t27.\t Why are light rays known as electromagnetic waves? \t13.\t Neutrons were discovered by bombarding b\u00ad eryllium \t28.\t The ______ consists of well-defined lines of d\u00ad efinite with _____ particles. frequencies. \t14.\t What name did Max Planck give to energy packets? \t29.\t In the formula E = hv, E is ______ and v is ______. Short Answer Type Questions \t30.\t \u2018Electrons jump from one orbit to another orbit.\u2019 electron from a hydrogen atom to produce an H+ Justify this statement on the basis of Bohr\u2019s theory. ion? Explain. \t31.\t On what basis did Bohr propose his atomic model? \t35.\t If Rutherford\u2019s atomic model is correct, then the atom should collapse. Why? \t32.\t What are orbits and why are they called stationary orbits? \t36.\t Describe Millikan\u2019s oil drop experiment in brief. \t33.\t Mention the properties of anode rays. \t37.\t According to Rutherford\u2019s atomic model, where are the protons and electrons located in an atom? \t34.\t What is the amount of energy needed to remove an","2.16 Chapter 2 \t38.\t Why was the presence of neutrons in an atom \t42.\t The wavelength of a particular radiation is 700 nm \u00adpredicted? How were neutrons discovered? (1 nm = 10\u20139 m). Find its frequency ( \u03bd ). \t39.\t Describe J.J. Thomson\u2019s atomic model. \t43.\t Distinguish between continuous spectrum and discontinuous spectrum. Give some examples of \t40.\t Define angular momentum. In the relation sources for these spectra. mvr = nh , what do m, v, r and h denote? \t44.\t An electron revolving round in an orbit has angular 2\u03c0 \t41.\t On what basis did Bohr assume the concept of h momentum equal to 2\u03c0 . Can it lose energy? \u00adstationary orbits for an electron? Essay Type Questions \t47.\t What are the drawbacks of Rutherford\u2019s atomic model? \t45.\t Explain Bohr\u2019s atomic model. \t48.\t Describe Rutherford\u2019s atomic model. \t46.\t What are the observations and conclusions drawn by J.J. Thomson while conducting experiments \t49.\t State the limitations of Bohr\u2019s model. with a discharge tube for studying the properties of cathode rays? For Answer key, Hints and Explanations, please visit: www.pearsoned.co.in\/IITFoundationSeries CONCEPT APPLICATION Level 1 PRACTICE QUESTIONS Direction for questions from 1 to 7: \t9.\t Some of the \u03b1-rays deflect in acute and obtuse State whether the following statements are true angles due to the presence of the ______ in the or false. centre of the atom. \t1.\t According to Thomson\u2019s atomic model, electrons \t10.\t According to classical electrodynamics, if an electri- revolve round the nucleus. cally charged particle revolves in a circular path, it continuously ______ energy. \t2.\t In a discharge tube, anode rays originate when electrons collide with gas molecules. \t11.\t The energy of an electron present in the first orbit of an atom is ________ than the energy of electron \t3.\t 8O16 and 8O18 are isotopes while 20Ca40 and 18Ar40 in the other orbits. are isobars. \t12.\t Splitting of spectral lines in the presence \t4.\t Energy is absorbed when the electron jumps from of m\u00adagnetic field is known as _____ K to L energy shells. effect. \t5.\t \u03b1-ray scattering experiment proved that the posi- \t13.\t The kinetic energy of an electron present in the tive particles are present in the extra nuclear part of first orbit of an atom is _______ than that of the an atom. electron in the last orbit. \t6.\t Characteristic spectra of atoms are line spectra. \t14.\t The spectra produced by the deexcitation of an electron is called _______________. \t7.\t An electron in the excited state of an atom is highly unstable. Direction for questions from 8 to 14: Fill in the blanks. \t8.\t Anode rays are deflected towards the negative plate in the presence of an electric field because they consist of _____ particles.","Atomic Structure 2.17 Direction for question 15: \t\t(a)\u2002protons\t (b)\u2002neutrons\t Match the entries given in Column A with \t\t(c)\u2002electrons\t (d)\u2002nucleus appropriate ones from Column B. \t19.\t In which of the following pairs of shells, energy dif- \t15. Column A Column B ference between two adjacent orbits is minimum? \t A.\t\u0007e value varies (\u2002)\t a.\thv m \t\t (a)\u2002 K, L\t (b)\u2002 L, M\t with the nature of (\u2002)\t b.\tR\u0007 utherford\u2019s atomic gas model \t\t (c)\u2002 M, N\t (d)\u2002 N, O B.\tP\u0007lum pudding (\u2002)\t c.\t\u0007Sun rays \t20.\t Assertion A: An electron in the inner orbit is model more tightly bound to the nucleus. (\u2002)\t d.\tB\u0007 ohr\u2019s stationary orbit C.\tM\u0007ass of the atom (\u2002)\t e.\u2002T\u0007 homson\u2019s atomic \t\tReason B: The greater the absolute value of is concentrated at energy of an electron the more tightly the electron the centre of atom model is bound to the nucleus. (\u2002)\t f.\t\u0007Anode rays D.\tC\u0007ontinuous \t\t (a)\u2002\u0007Both A and B are true but B is not the appro- spectrum priate reason for A. E.\t mvr = nh \t\t (b)\u2002\u0007Both A and B are individually correct and B is 2\u03c0 the correct reason for A. D.\t Quantum \t\t (c)\u2002 A is correct but B is not correct. \t\t (d)\u2002 Both A and B are not correct. \t21.\t The electron revolves only in the orbits in which Direction for questions from 16 to 45: \t\t(a)\u2002mvr > nh \t(b)\u2002mvr \u2265 2n\u03c0h For each of the questions, four choices have been 2\u03c0 provided. Select the correct alternative. \t\t(c)\u2002mvr = nh \t(d)\u2002mvr < 2n\u03c0h \t16.\t Which of the following concepts was not consid- 2\u03c0 ered in Rutherford\u2019s atomic model? \t22.\t Which among the following pairs are having differ- ent number of valence electrons? \t\t (a)\u2002 the electrical neutrality of atom \t\t(a)\u2002Na+, Al+3\t (b)\u2002P\u20133, Ar \t\t (b)\u2002 the quantisation of energy \t\t (c)\u2002\u0007electrons revolve around nucleus at very high \t\t(c)\u2002Mg+2, Ar\t (d)\u2002 O\u20132, F speeds \t23.\t If two naturally occurring isotopes of an element PRACTICE QUESTIONS \t\t (d)\u2002\u0007existence of nuclear forces of attraction on the are 7X15, 7X11; what will be the percentage compo- electrons sition of each isotope of X occurring, respectively, 1\t 7.\tWhen alpha particles are sent through a thin metal if the average atomic weight accounts to 14? foil, only one out of ten thousand of them rebounded. This observation led to the conclusion that \t\t (a)\u2002 95, 5\t (b)\u2002 80, 20\t \t\t (a)\u2002\u0007positively-charged particles are concentrated at \t\t (c)\u2002 75, 25\t (d)\u2002 65, 35 the centre of the atom \t24.\t According to quantum theory of radiation, which \t\t (b)\u2002\u0007more number of electrons is revolving around is false? the nucleus of the atom \t\t (a)\u2002\u0007radiations are associated with energy \t\t (c)\u2002 unit positive charge is only present in an atom \t\t (d)\u2002\u0007a massive sphere with more negative charge and \t\t(b)\u2002r\u0007adiation is neither emitted nor absorbed discontinuously unit positive charge is present at the centre of the atom \t\t(c)\u2002t\u0007he magnitude of energy associated with a quantum is dependent on frequency \t18.\t Canal ray experiment lead to the discovery of ______. \t\t (d)\u2002 photons are quanta of radiation \t25.\t Select True\/False among the following statements: \t\t\t(i)\t\u0007Bohr\u2019s theory successfully explained stability of the atom.","2.18 Chapter 2 \t\t(ii)\t Atoms give line spectra. \t33.\t 7X15, 7X11 are two naturally occurring isotopes \t\t(iii)\tV\u0007 elocity of electromagnetic waves depends on of an element X. What is the percentage of each the frequency. i\u00adsotope of X if the average atomic mass is 14? \t\t(iv)\tBohr introduced the concept of orbital. \t\t (a)\u2002 (i) T, (ii) T, (iii) F, (iv) F \t\t (a)\u2002 95, 5\t (b)\u2002 80, 20\t \t\t (b) (i) T, (ii) F, (iii) T, (iv) T \t\t (c)\u2002 (i) F, (ii) T, (iii) F, (iv) F \t\t (c)\u2002 75, 25\t (d)\u2002 65, 35 \t\t (d)\u2002 (i) F, (ii) F, (iii) T, (iv) T \t34.\t A trinegative ion of an element has 8 electrons in its M shell. The atomic number of the element is \t\t(a)\u200215\t (b)\u200218 \t26.\t Which of the following particles do not produce \t\t(c)\u200220\t (d)\u200216 electronic spectra? \t35.\t Arrange the following statements given by various scientists in chronological order: \t\t(a)\u2002Li+2\t (b)\u2002He+2\t \t\t(c)\u2002Be+2\t (d)\u2002Na+ \t\t(1)\t calculation of energy and radius of orbit \t27.\t An element has two isotopes with mass numbers \t\t(2)\t\u0007atoms of the same elements are identical in all 16 and 18. The average atomic weight is 16.5. The percentage abundance of these isotopes is ______ respects and ______, respectively. \t\t(3)\t\u0007calculation of diameter of the nucleus and the atom \t\t (a)\u2002 75, 25\t (b)\u2002 25, 75\t \t\t(4)\ta\u0007ssumption of thinly spread positively-charged mass \t\t (c)\u2002 50, 50\t (d)\u2002 33.33, 66.67 \t28.\t Which among the following are isobars? \t\t (a)\u2002 4 3 1 2\t (b)\u2002 4 2 3 1\t \t\t(a)\u2002bXa and bYa+1\t(b)\u2002bXa and cYb \t\t (c)\u2002 2 4 3 1\t (d)\u2002 3 4 2 1 \t\t(c)\u2002bXa and b+1Ya\t (d)\u2002 bXa and b\u20131Ya\u20131 \t36.\t What is the ratio of radii of the first successive odd orbits of hydrogen atom? \t29.\t Some of the elements have fractional atomic masses. The reason for this could be \t\t (a)\u2002 9 : 1\t (b)\u2002 1 : 9 \t\t (a)\u2002 the existence of isobars \t\t (c)\u2002 1 : 3\t (d)\u2002 3 : 1 \t\t (b)\u2002 the existence of isotopes \t37.\t An electron revolves round the nucleus in the 3rd \t\t (c)\u2002 the nuclear reactions orbit and jumped to a higher orbit X showing a PRACTICE QUESTIONS \t\t (d)\u2002 the presence of neutrons in the nucleus difference in angular momentum equal to h . The \u03c0 \t30.\t Which of these pairs has almost similar masses? value of \u2018X\u2019 could be \t\t(a)\u2002proton\u2013electron\t (b)\u2002neutron\u2013electron \t\t(a)\u20024\t (b)\u20026 \t\t(c)\u2002electron\u20131H1\t (d)\u2002neutron\u22121H1 \t\t(c)\u20025\t (d)\u20027 \t31.\t The energy of an electron revolving in the 3rd orbit \t38.\t Rutherford\u2019s \u03b1-particle scattering experiment of Be+3 ion is _________ ev eventually led to the conclusion that \t\t(a)\u2002\u201310.2\t (b)\u2002\u201313.6 \t\t (a)\u2002 mass and energy are related \t\t(c)\u2002\u201324.2\t (d)\u2002\u201318.1 \t\t(b)\u2002\u0007the point of impact with matter can be pre- cisely determined \t32.\t Which of the following concepts, was not consid- ered in Rutherford\u2019s atomic model? \t\t (c)\u2002 neutrons are buried deep in the nucleus \t\t (a)\u2002 the electrical neutrality of atom \t\t (d)\u2002\u0007electrons are distributed in a large space around the nucleus \t\t (b)\u2002 the quantisation of energy \t39.\t Arrange the following steps which are carried out \t\t(c)\u2002\u0007electrons revolve around the nucleus at very in \u03bc-ray experiment in the correct sequence: high speeds \t\t(1)\t Passage of \u03bc-particles through a slit \t\t (d)\u2002\u0007existence of nuclear forces of attraction on the electrons \t\t(2)\t bombardment of \u03bc-particles with a gold foil","Atomic Structure 2.19 \t\t(3)\t deflection of \u03bc-particles \t42.\t The ratio of atomic numbers of two elements A \t\t(4)\t production of \u03bc-particles and B is 1 : 2. The number of electrons present in the valence shell (3rd) of A is equal to the \u00addifference \t\t (a)\u2002 4 1 2 3\t (b)\u2002 4 1 3 2\t in the number of electrons present in the other two \t\t (c)\u2002 1 4 3 2\t (d)\u2002 1 4 2 3 shells. Steps involved for the calculation of ratio of \t40.\t Which among the following pairs are having differ- number of electrons present in a penultimate shell ent number of total electrons? to anti-penultimate shell of B are given below. \t\t(a)\u2002Na+ and Al+3\t (b)\u2002P\u20133 and Ar Arrange them in the correct sequence: \t\t(c)\u2002Mg+2 and Ar\t (d)\u2002 O\u20132 and F\u2013 \t\t(1)\t calculation of atomic number of B \t41.\t The postulates of Bohr\u2019s atomic model are given \t\t(2)\t calculation of valence electrons present in A below. Arrange them in the correct sequence: \t\t(3)\t calculation of atomic number of A \t\t(1)\t\u0007As long as the electron revolves in a particular orbit, \t\t(4)\tc\u0007alculation of number of electrons present in the electron does not lose its energy. Therefore, the penultimate and anti-penultimate shells of B these orbits are called stationary orbits and the electrons are said to be in stationary energy states. \t\t(5)\t writing electronic configuration of B \t\t (a)\u2002 2 3 4 1 5\t (b)\u2002 2 3 1 5 4 \t\t(2)\t\u0007Electrons revolve round the nucleus in specified \t\t (c)\u2002 4 5 2 3 1\t (d)\u2002 4 2 1 3 5 circular paths called orbits or shells. \t43.\t The equation given by Bohr to calculate radius of \t\t(3)\tT\u0007 he energy associated with a certain energy nth orbit of hydrogen atom is level increases with the increase of its distance from the nucleus. \t\t(a)\u2002 rn = n2h2 \t(b)\u2002rn = 4n\u03c022hm2 e 4\u03c02me \t\t(4)\t\u0007An electron jumps from a lower energy level to a higher energy level by absorbing energy. But \t \t(c)\u2002 rn = nh 2 \t(d)\u2002rn = 4\u03c0n22hm22e when it jumps from a higher to lower energy 4\u03c0 2me level, energy is emitted in the form of electro- magnetic radiation. \t44.\t The number of electrons present in the valence \t\t(5)\t\u0007Each orbit or shell is associated with a definite shell of an atom with atomic number 38 is amount of energy. Hence, these are also called energy levels and are designated as K, L, M and \t\t(a)\u20022\t (b)\u200210 N, respectively. \t\t(c)\u20021\t (d)\u20028 \t\t (a)\u2002 1 3 4 5 2 \t (b)\u2002 2 3 5 1 4 \t45.\t The mass number of an atom whose unipositive ion PRACTICE QUESTIONS has 10 electrons and 12 neutrons is \t\t (c)\u2002 2 5 3 1 4\t (d)\u2002 2 1 4 3 5 \t\t(a)\u200222\t (b)\u200223 \t\t(c)\u200221\t (d)\u200220 Level 2 \t1.\t When the same isotopic gas is taken in two dis- \t5.\t If the energy released when an electron jumped charge tubes, the angle of deflection is found to be from the 4th orbit to the 3rd orbit of hydrogen is \u2018x,\u2019 different though the strength of the external elec- then what would be the energy difference when it tric field applied is the same. Explain. jumps from the 3rd orbit to the 2nd orbit? \t2.\t In a canal ray experiment, different gases were \t6.\t Electronic spectra can distinguish isobars but not found to produce canal rays with the same specific isotopes. Justify. charge. Explain. \t7.\t If the energy difference between the orbits when \t3.\t When canal rays experiment is conducted with an electron in H atom gets excited to higher energy orbit from its ground state is 12.1 eV\/atom, cal- hydrogen gas, smceienvatilsutes sw. Jeurestiffoyu. nd to give particles culate the frequency of radiation emitted ( l eV with different = 1.602 \u00d7 10\u201319J) when electron comes back to \u00adsecond energy level. \t4.\t Energy of the electron in the atom is negative. Explain. \t\t[Hint: Energy of a free electron is taken as zero.]","PRACTICE QUESTIONS 2.20 Chapter 2 \t17.\t Why is the source of \u03b1-particles kept inside the lead block? \t8.\t Is the energy difference between successive orbits the same for all orbits? Justify your answer. \t18.\t If Thomson\u2019s model is considered to be correct, what would be the observation of Rutherford\u2019s \t9.\t Though there is only one electron in a hydrogen \u03b1-ray scattering experiment? atom, the spectrum of hydrogen contains a number of lines. How do you explain this? \t19.\t The ratio of the atomic numbers of two elements A and B is 2 : 3. A is an inert gas with the first 3 orbits \t10.\t What is the ratio of the radius of the 1st orbit to 2nd completely filled. Identify A and B and write their orbit, if the velocity of the electron in the 1st orbit electronic configurations. is twice that of the 2nd orbit. \t20.\t A stable dipositive ion and a dinegative ion are isoelec- \t11.\t A particular atom has the 4th shell as its valence tronic with an octet configuration in the second shell shell. If the difference between the number of elec- of their atoms. Identify the preceding and succeeding trons between K and N shells and L and M shells elements and write their electronic configurations. is zero, find the atomic number of the element and electronic configuration of its stable ion. \t21.\t Predict the possible atomic number(s) of an atom in which the third shell is incompletely filled and maxi- \t12.\t A stable unipositive ion of an element contains mum 3 more electrons can be added in that shell. three fully filled orbits. What is the atomic number of the element? \t22.\t The radius of nth orbit of a single electron species is 0.132 n2 A\u00b0. Identify the element. \t13. \tExplain why a blackened platinum strip when placed at the radius of curvature turns red hot, only \t23. \tWhat is the frequency of light emitted when an when the cathode taken has concave shape. electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit? \t14.\t The average atomic mass of two isotopes with mass numbers A and A + 2 is A + 0.25. Calculate the \t24.\t An electron having an angular momentum of percentage abundance of the isotopes. 1.05 \u00d7 10\u201334 joules jumps to another orbit such that it has an angular momentum of 4.20 \u00d7 10\u201334 joules. \t15.\t Spectral line given by an atom is a kind of signature Explain the possible transitions. of the respective atom. Comment on this statement. \t25.\t The mass number of a particular element which has \t\t[Hint: The nuclear charges of different atoms are equal number of protons and neutrons is 32. What different.] is the electronic configuration of the atom and its stable ion? Directions for questions from 16 to 25: Application-Based Questions \t16.\t Why was a spherical sulphide screen used in \u03b1-ray scattering experiment? Level 3 \t1.\t In Millikan\u2019s oil drop experiment, the distance \t3.\t Is the velocity of an electron in all orbits the same between the metal plates, A and B to which an for an atom of a particular element? How does electric potential is applied such that A is positive it vary for different single electron species? Give and B is negative is 5 mm. An oil drop is found to \u00adreasons in support of your answer. be suspended at a distance of 2 mm from B. Predict the change in the position of the oil drop when \t4.\t What is the ratio of distance between successive there is a sudden drop or rise in potential. Justify. orbits of 1 and 2 to 2 and 3 of hydrogen atom? \t2.\t Different gases in the discharge tube produce dif- \t\t[Hint: radius of nth orbit in hydrogen is 0.529 ferent colours under suitable conditions of pressure \u00d7 n2 A\u00b0] and voltage. Explain. \t5.\t If Ay +1 or By\u22122 +1 were to be used instead of \t\t[Hint: Each element has its own characteristic atomic spectrum.] x x \u22121 \u03b1-particles in Rutherford\u2019s experiment, which would be better and why?","Atomic Structure 2.21 Directions for questions from 6 to 10: duced show the same deflection under the external Application-Based Questions electric field? Give reasons to support your answer. \t6.\t Draw a comparison between the potential energy \t10.\t Anode Cathode and kinetic energy of electrons in the 1st orbits of hydrogen and He+ ion. Also comment on the total + Fluorescent energy of the electrons in the above cases. material (ZnS) \t7.\t Though the kinetic energy of electrons decreases with an increase in the distance from the nucleus, +\u2212 \u2212 Bright the potential energy of the electron increases. How H.V. Schematic diagram spot do you account for this? \t\tIf the given schematic diagram represents Thomson\u2019s \t8.\t Why is high voltage and low pressure maintained in experiment and the corresponding observation, the discharge tube? what would be his atomic model? \t9.\t If canal ray experiments are conducted with differ- ent isotopes of hydrogen gas, do the canal rays pro- PRACTICE QUESTIONS","2.22 Chapter 2 CONCEPT APPLICATION Level 1 True or false \t1.\t False \t2.\t True \t3.\t True \t4.\t True \t5.\t False \t6.\t True \t7.\t True Fill in the blanks \t10.\t loses \t12.\t Zeeman \t14.\t emission spectrum \t11.\t less \t13.\t more \t8.\t positively-charged \t9.\t positive charge Match the following B : e\t C:b \t15.\t A : f\t E : d\t F:a \t\tD :\t Multiple choice questions \t16.\t b \t20.\t a \t24.\t b \t28.\t c HINTS AND EXPLANATION \t17.\t a \t21.\t c \t25.\t a \t29.\t b \t18.\t a \t22.\t c \t26. \tb \t30.\t d \t19.\t d \t23.\t c \t27.\t a \t31.\t Energy, En = \u221213.6Z2 eV 13 6 \t\t\u21d21400 = 4x + 1100 \u21d2 4x = 300 n2 \t\tx = 300 =75 4 \u221213.6 16 \t\tE3 = 9 \u00d7 eV= \u2212 24.2 eV \t\t\u2234 The percentage of 7X11 = 100 \u2013 x \t32.\t According to Rutherford\u2019s theory, an atom is elec- \t\t= 100 \u2013 75 = 25 and that of 7X15 is 75 trically neutral and electrons revolve around the \t34.\t Since electronic configuration of the trinegative ion is 2, 8, 8 the electronic configuration of the neutral nucleus. Nuclear forces of attraction exist between atom is 2, 8, 5 and its atomic number is 15. the nucleus and electrons. The only assumption that Rutherford did not consider is quantisation of energy. \t35.\t \u2002 (i)\t\u0007Atoms of the same elements are identical in all respects. \t33.\t Let the percentage of 7X15 is x. \t\t\u2234 The percentage composition of 7X11 is 100 \u2013 x \t\t(ii)\tA\u0007ssumption of thinly spread positively- charged mass \t\tAverage atomic weight = 14 = x(15) + (100 \u2212 x)11x 100 \t\t(iii)\t\u0007Calculation of the diameters of the nucleus and the atom \t\t\u21d21400 = 15x + 1100 \u2013 11x \t\t(iv)\t Calculation of energy and radius of orbit","Atomic Structure 2.23 \t36.\t rn = 0.529 \u00d7 n2 A\u00b0 \t41.\t \u2002(i)\t \u0007Electrons revolve around the nucleus in speci- \t\tSuccessive first odd orbits are 1 and 3 fied circular paths called orbits or shells. r1 = 0.529 \u00d7 12 = 1 = 1:9 \t\t(ii)\t\u0007Each orbit or shell is associated with a definite r3 0.529 \u00d7 32 9 \b amount of energy. Hence, these are also called \t energy levels and are designated as K, L, M \t37.\t Angular momentum in third orbit is given by and N, respectively. \t\tmvr = 3h \t \u21d2\b \t\t(iii)\t\u0007The energy associated with a certain energy 2\u03c0 level increases with the increase of its distance from the nucleus. \t\tAngular momentum in X orbit is given by \t\t(iv)\t\u0007As long as the electron revolves in a particular orbit, mvr = Xh \u2192 (2) the electron does not lose its energy. Therefore, 2\u03c0 these orbits are called stationary orbits and the electrons are said to be in stationary energy states. Xh \u2212 3h = h 2\u03c0 2\u03c0 \u03c0 \t\t\u2002 (v)\t\u0007An electron jumps from a lower energy level to a higher energy level by absorbing energy. (X\u2212 3) h = h \u21d2 X=5 But when it jumps from a higher to lower \t\t 2\u03c0 \u03c0 \b energy level, energy is emitted in the form of electromagnetic radiation. \t38.\t The \u03b1-ray scattering experiment led to the discov- \t42.\t \u2002 (i)\t calculation of valence electrons present in A ery of nucleus which in turn helped him conclude \t\t(ii)\t calculation of atomic number of A that electrons revolve around the nucleus to over- \t\t(iii)\t calculation of atomic number of B come the strong electrostatic force of attraction. \t\t(iv)\t writing electronic configuration of B \t39.\t \u2002 (i)\t production of \u03b1-particles HINTS AND EXPLANATION \t\t(ii)\t production of a narrow beam of \u03b1-particles \t\t\u2009(v)\t\u0007calculation of number of electrons present in \t\t(iii)\t bombardment of \u03b1-particles with gold foil the penultimate and anti-penultimate shells of \t\t(iv)\t deflection of \u03b1-particles B \t40.\t The electronic configuration of Mg+2 is 2, 8 \t43.\t The equation given by Bohr to calculate radii of nth \t\t\u2234 Total no. of electrons = 10 orbit of hydrogen atom is \t\tThe electronic configuration of Ar is 2, 8, 8 \t\t\u2234 Total no. of electrons = 18. rn = n2h2 \t\tHence, Mg+2 and Ar are having different number 4\u03c0 2me2 of total electrons. \t44.\t Z = 38, electronic configuration = 2, 8, 18, 8, 2 \t\t\u2234 Two valence electrons \t45.\t Mass number = 11 + 12 = 23 Level 2 \t3.\t \u2002 (i)\t\u0007e\/m depends on number of protons and neutrons \t\t(ii)\t existence of isotopes \t1.\t (i)\t\u0007factors which affect angle of deflection in an \t\t(iii)\t variation in e\/m for isotopes electric field \t4.\t \u2002 (i)\t\u0007comparing energy of free electron and energy \t\t(ii)\t\u0007conditions for changing the factors which affect of electron in an atom the angle of deflection \t\t(ii)\t\u0007change in energy when an electron is brought \t2.\t (i)\t factors affecting specific charge closer to the atom \t\t(ii)\tc\u0007onditions where different gases can have the same specific charge","2.24 Chapter 2 \t14.\t \u2002(i)\t (n1 \u00d7 A) + (n2 \u00d7 (A + 2)) = (n1 + n2) \t\t\t (A + 0.25) form \t\t(iii)\t\u0007reason for the change in the energy of an \t\t(ii)\t 87.5% and 12.5% electron \t15.\t \u2002 (i)\t\u0007calculation of energy of electrons in He+ and \t6.\t \u2002 (i)\t\u0007fundamental particle responsible for the spectra Li+2 ions \t\t(ii)\t\u0007relation between fundamental particle and \t\t(ii)\t 4 : 9 structure of spectra \t\t(iii)\t\u0007difference in number of the above particles \t16.\t to observe scintillations even if the \u03b1-rays get deflected at large angles between isobars and isotopes \t\t(iv)\t effect of this on structure of spectra \t17.\t\u03b1-particles cannot penetrate lead, but \u03b2- and \u03b3-rays can. In order to screen \u03b1-particles from \u03b2- and \t7.\t \u2002 (i)\t relation between energy and n \u03b3-rays, lead block was used. \t\t(ii)\t\u0007calculation of n2 value from the difference in \t18.\t If Thomson\u2019s model is correct, all the \u03b1-particles energy would have penetrated the gold foil and their angle \t\t(iii)\t calculation of frequency from energy of deflection would be insignificant. \t\t(iv)\t\u0007calculation of energy difference based on n \t19.\t A \u2192 electronic configuration \u2013 2, 8, 18, 8. Thus, values the atomic number = 36. \t\t(v)\t calculation of frequency \t\t(vi)\t \u03bd = 0.456 \u00d7 1015 s\u20131 \t\tA : B = 2 : 3 \t8.\t \u2002 (i)\t factors affecting deflection \t\t\u2234 B = 36 \u00d7 3 = 54 \t\t(ii)\t the effect of atomic number on deflection 2 \t\t(iii)\t the effect of kinetic energy on deflection \t\tThe electronic configuration of \t9.\t \u2002 (i)\t Bohr\u2019s model of atom HINTS AND EXPLANATION \t\t(ii)\t\u0007relation between energy absorbed and excitation \t\tB = 2, 8, 18, 18, 8. \t\t(iii)\t\u0007relation between the path of electron during \t20.\t Let the dipositive ion be X+2 and dinegative ion is deexcitation and energy emitted Y\u20132. \t\t(iv)\t relation between energy emitted and spectrum \t\tThe octet in the second shell \u2192 2, 8 \t10.\t \u2002(i)\t \u0007comparing the angular momentum of elec- tron in the orbits \t\tThe number of electrons in X = 10 + 2 = 12. \t\t(ii)\t comparison of radius \t\tThe number of electrons in Y = 10 \u2013 2 = 8. \t\t(iii)\t r1 : r2 = 1 : 4 \t\tX is magnesium with an electronic configuration 2, \t11.\t \u2002(i)\t\u0007number of electrons in K and L shells when N 8, 2. Y is oxygen with a configuration 2, 6. shell is the valence shell \t\tThe element preceding magnesium is sodium and \t\t(ii)\t\u0007calculation of number of electrons in K, L, M the succeeding one is aluminium. They have elec- and N shells tronic configurations 2, 8, 1 and 2, 8, 3, respec- tively. The elements preceding and succeeding for \t\t(iii)\t calculation of atomic number oxygen are nitrogen and fluorine which have elec- \t\t(iv)\t\u0007number of electrons to be lost to form stable ion tronic configurations 2, 5 and 2, 7, respectively. \t\t(v)\t atomic number = 20 \t21.\t Electronic configuration of the atom which can \t12.\t\u2002 (i)\t\u0007maximum number of electrons present in an accommodate three more electrons in the 3rd shell orbit (Bohr\u2013Bury scheme). could be 2, 8, 5 and 2, 8, 15, 2. Hence, the prob- able atomic numbers are 15 and 27. \t\t(ii)\t electronic configuration of neutral atom \t\t(iii)\t37 \t13.\t \u2002 (i)\t factors responsible for the strip to turn red hot \t\t(ii)\t\u0007The path in which the electrons travel from concave cathode.","Atomic Structure 2.25 \t22.\t rn Kn2 \t\tInitial angular momentum = 1.05 \u00d7 10\u201334 joules Z = n \u00d7 6.625 \u00d7 10\u221234 2 \u00d7 3.14 1.05 \u00d7 10\u221234 = ,n = 1 \t\t \u22340.132n2 = 0.529 \u00d7 n2 \t\tThe electron is present in the 1st orbit originally. \t\t Z \t\tWhen the electron gets excited, the angular momentum 0.529 \u2234Z = 0.132 = 4 \t\t= 4.20 \u00d7 10\u201334 joules. \t\t \t\tSince the atomic number is 4, the element is \u00d7 6.625 \u00d7 10\u221234 4.20 \u00d7 10\u221234 = n 2 \u00d7 3.14 ,n = 4 \t beryllium. \t\t 23.\t E3 \u2212 E2 = \u221221.72 \u00d7 10\u221219 \uf8eb\uf8ec\uf8ed 1 \u2212 1 \uf8f8\uf8f7\uf8f6 \t\tAn electron can lose energy when it is present in 9 4 the 4th orbit and not from the 1st orbit. \t\t= 21.72 \u00d7 10\u221219 \u00d7 5 \t\tThe possible transitions are 4 \u2192 3, 4 \u2192 2, 4 \u2192 1, 36 3 \u2192 2, 3 \u2192 1 and 2 \u2192 1. \t\t\u2234 Difference in energy, E3 \u2013 E2 = 3.01 \u00d7 10\u201319 J \t25.\t Mass number = 32 \t\tNo. of protons = No. of neutrons = 16 \t\tAccording to Planck\u2019s equation, \u0394E = hv \t\t\u2234 Electronic configuration = 2, 8, 6 \t\tElement is sulphur and stable ion is S\u22122 \t\t3.01 \u00d7 10\u201319 = 6.625 \u00d7 10\u201334 \u03bd \t\tElectronic configuration of S\u22122 is 2, 8, 8 \t\tFrequency, (v) = 3.01 \u00d7 13\u221219 = 4.5 \u00d7 1014 s\u22121 6.625 \u00d7 10\u221234 \t24.\t Angular momentum of an orbit \t\tmvr = nh 2\u03c0 HINTS AND EXPLANATION Level 3 \t1.\t \u2002 (i)\t charge acquired by oil drop \t\t(iv)\t\u0007comparison of nuclear charge in different sin- gle electron species \t\t(ii)\t\u0007different forces acting on the charged oil drop when it is at a distance of 2 mm from B. \t4.\t \u2002 (i)\t\u0007effects of high voltage and low pressure on the \t\t(iii)\t\u0007relation between position of oil drop and dif- gas molecules in the discharge tube ferent forces \t\t(ii)\t the effect of velocity of electrons on ionisation \t\t(iii)\t\u0007relation between the voltage and the velocity \t\t(iv)\t\u0007effect of a particular force on the position of oil drop of electrons \t\t(iv)\t\u0007relation between the pressure and the number \t\t(v)\t\u0007change in position of oil drop with change in potential of gas molecules \t\t(v)\t\u0007the effect of the number of gas molecules on \t2.\t \u2002 (i)\tenergy of electron the impact of collision \t\t\t Excitation and deexcitation \t\t(ii)\tfactors affecting the energy of electron \t5.\t \u2002 (i)\t the factors affecting angle of deflection \t\t(iii)\tc\u0007omparison of the energy emitted during the \t\t(ii)\t\u0007the characteristics in which the two particles deexcitation of electron in different atoms differ \t3.\t \u2002 (i)\t forces acting on moving electron \t6.\t The electrons revolve round the nucleus with high \t\t(ii)\t position of the electron in the orbit velocities to counter balance the nuclear force of \t\t(iii)\tr\u0007elationship between position, forces and attraction. As nuclear force of attraction in the 1st orbit of He+ is more than that in H atom. velocity","2.26 Chapter 2 ber of molecules is very less, the collisions between the electrons which move towards the anode with a Kinetic energy of electron in He+ is more (due to high velocity and the gas molecules become effec- greater velocity) than in H atom. When an elec- tive. These collisions lead to the dislodgement of tron approaches towards an atom, it loses poten- electrons from gaseous molecules and formation of tial energy because it works towards the force of cathode rays takes place. Moreover, high voltage a\u00adttraction. The greater the force of attraction, the increases the kinetic energy of the electrons which more is the loss of potential energy. Hence, the in turn increases the probability of removal of elec- electron in He+ has lesser potential energy than the trons from the gaseous molecules. electron in H atom. But the loss of PE is more sig- nificant than the change in KE. Hence, total energy \t9.\t Protium (1H1), deuterium (1H2) and tritium (1H3) of helium is less than that of hydrogen. are naturally occurring isotopes of hydrogen. Unipositive ions of protium, deuterium and tritium \t7.\t The kinetic energy of an electron is proportional differ in their mass. to its velocity. With the increase in distance from the nucleus, the velocity of the electron decreases \t\tThus, their e\/m ratio is different. Therefore, they as the electron has to overcome a lesser nuclear deflect at different angles in an external electric force of attraction. An electron loses its potential field. energy when it approaches towards an atom that is a nucleus because it works towards the force of \t10.\t Since in the given experiments, it is observed that attraction during this process. Hence, the poten- positively-charged particles are detached from the tial energy of the electron decreases with decrease molecules under low pressure and high voltage, in distance between the nucleus and the electron Thomson\u2019s model would be the other way round, and increases with increase in distance between the i.e., positively-charged particles would be embed- nucleus and electron. ded in a thinly spread negatively-charged mass. \t8.\t Low pressure means that less number of gas mol- ecules is present in the discharge tube. If the num- HINTS AND EXPLANATION","13CChhaapptteerr PNeruimodbiecr CSlayssteifmicsation of Elements Remember Before beginning this chapter, you should be able to: \u2022\u2002\u0007review the basic concepts of physical and chemical properties of matter. \u2022\u2002\u0007understand classification of matter. Key Ideas After completing this chapter, you should be able to: \u2022\u2002\u0007learn earlier versions of classification of elements, like, Newland\u2019s law of octaves and Dobereiner\u2019s Triads with emphasis on Mendeleev\u2019s periodic table. \u2022\u2002\u0007understand the placement of elements on modern periodic table, its merits and defects. \u2022\u2002s\u0007tudy about the similarities and gradual changes in the properties, such as, atomic size, ionisation energy, electron affinity, electro-positivity, oxidizing, reducing property, etc. F I G U R E 1 . 1 \u2002 Figure Caption","3.2 Chapter 3 INTRODUCTION Elements such as gold, silver, tin, copper, lead and mercury have been known since antiquity. The first scientific discovery of element occurred in 1649 and it was the discovery of phosphorus by Henning Brand. The discovery of elements then continued and the scientists started recognizing the similarities and dissimilarities among the elements in order to make the study of the elements easy and systematic. A series of efforts for the classification of elements depending on the pattern of their properties eventually led to the construction of the periodic table, presently comprising of 118 elements. The periodic table is a tabular arrangement of elements, which depicts the trend of their physical behaviour and chemical properties. The German Chemist, Johann Dobereiner made the first significant effort in the development of the periodic table. In 1817, he noticed that a group of three elements (like, Ca, Sr and Ba) showed similar chemical properties and there was a regular trend in their physical properties when these elements are arranged in an increasing order of atomic weight. The atomic weight of strontium is approximately equal to the average of the atomic weights of calcium and barium. Later on, in 1829, he discovered the halogen triad and alkali metal triad consisting of (Cl, Br and I) and (Li, Na and K) and proposed that nature contained triads of elements with similar chemical properties and the atomic weight of the second element of the triad is the average of the other two. In 1864, the English Chemist, John Newland arranged the then known 56 elements in an increasing order of their atomic weights. He noticed that atomic weights of many pairs of elements which differed by eight or a multiple of eight exhibited similarities in their chemical properties. Based on this observation, he proposed the law of octaves. NEWLAND\u2019S ARRANGEMENT OF ELEMENTS IN OCTAVES TABLE 3.1\u2002 Newland\u2019s arrangement of elements HF Cl Co\/Ni Br Pd I Pt\/Ir Li Na K Cu Rb Ag Cs Tl Be Mg Ca Zn Sr Cd Ba\/V Pb B Al Cr Y Ce\/La U Ta Th C Si Ti In Zr Sn W Hg NP Mn As Di\/Mo Sb Nb Bi OS Fe Se Ro\/Ru Te Au Os The law states that the eighth element starting from a given one is a kind of repetition of the first like the eighth note of an octave of music. Newland\u2019s system of classification failed to explain similarities in higher elements. Dobereiner, Newland and many other scientists contributed significantly to the systematic classification of elements. Their efforts ultimately enabled the Russian chemist Mendeleev to construct his periodic table in 1869. He stated in his famous periodic law that physical and chemical properties of elements are periodic functions of their atomic weights. Like Newland, he also arranged the then known 63 elements in the increasing order of their atomic weights in his periodic table in 8 vertical columns and 6 horizontal rows called groups and periods, respectively.","Periodic Classification of Elements 3.3 After the discovery of noble gases and radioactive elements another version of Mendeleev\u2019s periodic table was published in 1905. In this version \u20180\u2019 group was added to accommodate the noble gases and the seventh period for the radioactive elements of 1905. Mendeleev\u2019s Periodic Table Previous attempts of classification show the similarities only in small units like a triad or in a small group, but this table represents the relationship in the entire network, i.e., in vertical columns and in horizontal rows. Merits of Mendeleev\u2019s Periodic Table \t1.\tM\u0007 endeleev left some gaps in his periodic table for some of the then unknown elements based on the properties of the other elements present in the same group. He named these missing elements as Eka boron, Eka aluminium and Eka silicon. Later on, these elements were discovered and named as scandium, gallium and germanium, respectively. \t2.\t\u0007Mendeleev corrected the doubtful atomic weights of some elements. For example, the atomic weight of indium, i.e., 76 was based on the assumption that indium oxide had the formula InO. This atomic weight placed indium, which has metallic properties, among the non-metals. Mendeleev corrected the formula of indium oxide from InO to In2O3. Based on this corrected formula, the atomic weight of indium is 113. He also corrected the atomic weights of the elements beryllium and uranium. Limitations of Mendeleev\u2019s Periodic Table \t1.\tA\u0007 nomalous pairs: In Mendeleev\u2019s periodic table, there are some pairs of elements in which the element with a higher atomic weight is placed before that of the lower atomic weight element in order to maintain the similarity in chemical properties of the elements belonging to the same group. \t\tThe pairs are\t (i) Co Ni (ii) Te I (iii) Ar K 58.9 58.6 127.6 126.9 39.94 39.10 \t2.\tPosition of hydrogen: The position of hydrogen is not justified. \t3.\t\u0007Positions of isotopes: The isotopes of an element have different atomic weights. But there is no position for the isotopes in Mendeleev\u2019s periodic table. \t4.\t P\u0007 osition of lanthanides and actinides: Two sets of 14 elements each (lanthanides and actinides, respectively), which follow lanthanum and actinium in the order of atomic weights, have not been provided regular or separate places in the periodic table. They have just been put together along with lanthanum and actinium, respectively. \t5.\t P\u0007 osition of elements of group VIII: Three triads of the transition elements are placed in group VIII of the periodic table. They should have been given separate positions. \t6.\tAnomalous placement of the transition and coinage metals: \t\t(a)\t Transition elements were placed along with typical elements under the same group. \t\t(b)\tCoinage metals were placed with alkali metals. Owing to these limitations and the later discovery of atomic structure, Mendeleev\u2019s periodic table has no practical significance at present. Nevertheless, its significance lies in the fact that the concept of periodicity in properties and the resultant arrangement of elements in a periodic table originated from Mendeleev\u2019s classification of elements.","TABLE 3.2 \u2002 Mendeleev\u2019s periodic table 3.4 Chapter 3 SERIES GROUPS OF ELEMENTS S 0 I II III IV V VI VII VIII 1 \u2014 Hydrogen \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 2 H 1.008 Nickel 3 Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Iron Cobalt Ni 59 (Cu) 4 He 4 Li 7.03 Be 9.1 B 11.0 C 12.0 N 14.04 O 16.00 F 19.0 Fe 55.9 Co 59 Palladium (Ag) 5 Neon Sodium Magnesium Aluminium Silicon Phosphorus Sulphur Chlorine Pd 106.5 6 Ne 19.9 Na 23.05 Mg 24.3 Al 27.0 Si 28.4 P 31.0 S 32.06 Cl 35.45 Ruthenium Rhodium (Au) 7 Argon Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Ru 101.7 Rh 103.0 \u2014 Ar 38 K 39.1 Ca 40.1 Sc 44.1 Ti 48.1 V 51.4 Cr 52.1 Mn 55.0 Platinum 8 Copper Zinc Gallium Germanium Arsenic Selenium Bromine \u2014 \u2014 Pt 194.9 9 Krypton Cu 63.6 Zn 65.4 Ga 70.0 Ge 72.3 As 75 Se 79 Br 79.95 10 Kr 81.8 Rubidium Strontium Yttrium Zirconium Niobium Molybdenum \u2013\u2013 \u2014 Iridium RO4 11 Silver Rb 85.4 Sr 87.6 Y 89.0 Zr 90.6 Nb 94.0 Mo 96.0 Osmium Ir 193 12 Ag Cadmium Indium Tin Antimony Tellurium Iodine Os 191 107.9 Cd 112.4 In 114.0 Sn 119.0 Sb 120.0 Te 127 I 127 \u2014 Xenon Xe 128 Caesium Barium Lanthanum Cerium \u2014 \u2014 \u2014 Cs 132.9 Ba 137.4 \u2014 \u2014 \u2014 La 139 Ce 140 \u2014 \u2014 \u2014 \u2014\u2014\u2014\u2014 \u2014 \u2014 Gold R Au 197.2 Ytterbium \u2014 Tantalum Tungsten \u2014 R2O Radium Rd 224 Yb 173 Ta 183 W 184 RO Mercury Thallium Lead Bismuth \u2014 Hgf 200.0 TI 204.1 Pb 206.9 Bi 208 \u2014 Thorium \u2014 Uranium Th 232 U 239 HIGHER SALINE OXIDES R2O3 RO2 R2O5 RO3 R2O7 HIGHER GASEOUS HYDROGEN COMPOUNDS RH4 RH3 RH2 RH","Periodic Classification of Elements 3.5 EXAMPLE Is the Dobereiner\u2019s law of triads applicable for phosphorus, arsenic and antimony? Explain. SOLUTION weigh Yes, P, As and Sb also follow Dobereiner\u2019s law of triad: tomic 31+121 = 76 2 Atomic weight of P Atomic weight of As Atomic weight of Sb 31 75 121 EXAMPLE Consider a Dobereiner\u2019s triad with elements X, Y and Z. The sum of atomic weights of extreme elements, X and Z is 46 and the difference of atomic weights of Z and Y is two times the atomic number of oxygen. Identify X, Y and Z. SOLUTION Given, X + Z = 46 \u2192 (1) \t Z \u2212 Y = 16 \u2192 (2) As X +Z =Y , (Dobereiner\u2019s law of triads). 2 \u2234Y= 46 = 23 \u2192 (3) 2 From equation (2) and (3), Z \u2013 23 = 16 \u21d2 Z = 39 From equation (1), X + Z = 46 \u21d2 X + 39 = 46 \u21d2 X = 7 So, X, Y and Z are Li, Na and K, respectively. Modern Periodic Table\u2014the Long Form of the Periodic Table In1913, H.G.J. Moseley worked on X-ray spectra of elements and established that the atomic number is equal to the total nuclear charge and it is a more fundamental characteristic of an element than atomic weight. Moreover, it was found that some of the major defects of Mendeleev\u2019s periodic table like the anomalous position of some pairs of elements (Co\u2013Ni, Te\u2013I, Ar\u2013K), placements of isotopes, etc., would automatically disappear, if atomic number, i.e., the total nuclear charge is used as the basis of the compilation of the periodic table instead of the atomic weight. Based on these observations, Moseley proposed the modern periodic law which states that \u2018the physical and chemical properties of elements are the periodic functions of their atomic numbers.\u2019 Thus, according to the modern periodic law, if the elements are arranged in the order of their increasing atomic numbers, the elements with similar properties are repeated after certain regular intervals. Description of modern periodic table The long form of the periodic table (also called Bohr\u2019s table) consists of horizontal rows called periods and vertical columns called groups or families. There are 7 periods and 16 groups in the periodic table. The groups from I to VII are divided into subgroups \u2018A\u2019 and \u2018B.\u2019 VIII group consists of 3 vertical columns. All noble gases are kept in the zero group. According to the recent recommendation of IUPAC, the groups are numbered from 1 to 18 (18 groups) instead of subdivision of groups into A and B subgroups.","3.6 Chapter 3 Different Portions of Long Form of Periodic Table Long Form of Periodic Table Left portion Middle portion Right portion Group I A Groups: III B, IV B Groups: III A, IV A, Group II A V A, VI A, VII A, 0 (extremely V B, VI B, VII B (consists of metals, non- electropositive metals) VIII, I B, II B metals, metalloids and noble gases \u2013including extremely Transition elements electronegative halogens) Inner transition elements First transition Second transition Third transition Fourth transition Actinides series consists of series consists of Lanthanides comprise 10 elements 10 elements Y(39) series consists of series consists of comprise 14 elements Sc (21) to 10 elements La 10 elements 14 elements to Cd (48) and they (57) and Hf (72) to Ac(89) and Zn (30), and they belong to the fifth belong to the fourth Hg (80) and they Rf (104) to Rg (kept separately below the main body of the periodic table) period period belong to (111) and they the sixth period. belong to the seventh period FIGURE 3.1 Division of periodic table into different portions TABLE 3.3\u2002 Periods and their characteristics Periods Characteristics First This period has only two elements, hydrogen and helium which have only one energy level Second or shell. This period is called very-short period Third It has 8 elements: Li 3 to Ne 10. Elements have two shells. The elements of this period are called bridge elements Fourth Fifth This period also has 8 elements: Na 11 to Ar 18. The second and third periods are called short Sixth periods. These elements have 3 shells. The elements of this period are called typical elements Seventh This period has 18 elements: K 19 to Kr 36. They have 4 energy levels or 4 shells This period also has 18 elements: Rb 37 to Xe 54. They have 5 energy levels or 5 shells This period has 32 elements: Cs 55 to Rn 86. They have six energy levels or 6 shells. The fourth, fifth and sixth periods are called long periods This period has 32 elements. All these elements are radioactive. Out of these, the naturally occurring radioactive elements are Fr 87, Ra 88, Ac 89 Th 90, Pa 91, and U 92 while the remaining elements after uranium, i.e., Np 93 to Db 105 are artificially prepared radioactive elements and these are called transuranic elements. Atoms of these elements have 7 energy levels or 7 shells Groups \t1.\tG\u0007 roups IA, IIA, IIIA, IVA, VA, VIA, VIIA contain metals, non-metals and metalloids. These elements are called representative elements because their valence electrons represent their group numbers. \t2.\t\u0007Groups IB, IIB, IIIB (only Sc, Y, La and Ac), IV B, V B, VI B, VII B and VIII contain only transition metals. \t3.\t\u0007In group III B except Sc, Y, La and Ac, the other elements are called inner-transition metals and are placed separately in two rows under the periodic table. They belong to the sixth period (Ce to Lu) and seventh period (Th to Lr). \t4.\tThe zero\/0 group contains noble gases.","Periodic Classification of Elements 3.7 Merits of the Long Form of the Periodic Table \t1.\tT\u0007 he classification is based on the atomic number of elements, which is more fundamental than the atomic weight. \t2.\tI\u0007 n this periodic table, the position of an element is related to the electronic configuration of the atom. Elements belonging to the same group possess same number of electrons in their valence shells and show similar chemical characteristics. Examples: TABLE 3.4\u2002 Elements of the same group and their electronic configurations Alkali metals Element Atomic No. of Electronic No. of Electrons in Halogens Numbers Shells Configurations Valence Shell Li 2, 1 1 Na 3 2 2, 8, 1 1 K 11 3 2, 8, 8, 1 1 Rb 19 4 2, 8, 18, 8, 1 1 Cs 37 5 2, 8, 18, 18, 8, 1 1 Fr 55 6 2, 8, 18, 32, 18, 8, 1 1 F 87 7 2, 7 7 Cl 9 2 2, 8, 7 7 Br 17 3 2, 8, 18, 7 7 I 35 4 2, 8, 18, 18, 7 7 At 53 5 2, 8, 18, 32, 18, 7 7 85 6 \t3.\t\u0007Variation of chemical properties along a period is correlated with gradual filling of electrons in a particular shell in the period. \t4.\t\u0007In this table, inert gases are placed at the end of each period which is very logical because valence shells have an octet configuration. So, their inertness can also be explained in terms of the electronic configuration. \t5.\tD\u0007 ue to the separation of the two subgroups, dissimilar elements (e.g., alkali and coinage metals) do not fall together. \t6.\t\u0007Different types of elements, such as, (a) active metals, (b) transition metals, (c) non-metals and metalloids (d) noble gases and (e) inner transition elements (lanthanides and actinides) have their separate locations in this periodic table. Defects of the Long Form of the Periodic Table \t1.\tE\u0007 ven after the compilation of the modern periodic table, the position of hydrogen remained controversial. This is because hydrogen with one electron in its valence shell shows similarities with both alkali metals and halogens. Therefore, placing hydrogen in IA group is not completely justifiable. \t2.\tT\u0007 he elements lanthanides and actinides could not be placed in the main body of the modern periodic table. \t3.\t\u0007This periodic table does not reflect the exact distribution of electrons of some of the transition and inner transition elements.","TABLE 3.5\u2002 Modern periodic table 3.8 Chapter 3 Modern Periodic table 1 1 Atomic number, Z 18 Element symbol 1 H Relative atomic mass, A 2 H 1.008 r He 1.008 2 13 14 15 16 17 4.00 3 4 5 6 7 8 9 10 Li Be B C N O F Ne 6.94 9.01 10.81 12.01 14.01 16.00 19.00 20.18 11 12 3 4 567 8 9 10 11 12 13 14 15 16 17 18 Na Mg 21 23 26 Al Si P S Cl Ar 22.99 24.31 Sc V Fe 26.98 28.09 30.97 32.06 35.45 39.95 19 20 44.96 22 50.94 24 25 55.85 27 28 29 30 31 32 33 34 35 36 39 K Ca Ti 41 Cr Mn 44 Co Ni Cu Zn Ga Ge As Se Br Kr Y 39.10 40.08 47.90 Nb 52.01 54.94 Ru 58.93 58.69 63.54 65.41 69.72 72.59 74.92 78.96 79.91 83.80 88.91 37 38 40 92.91 42 43 101.07 45 46 47 48 49 50 51 52 53 54 La\u2013Lu Rb Sr Zr 73 Mo Tc 76 Rh Pd Ag Cd In Sn Sb Te I Xe Ac\u2013Lr 85.47 87.62 91.22 Ta 95.94 98.91 Os 102.91 106.42 107.87 112.40 114.82 118.71 121.75 127.60 126.90 131.30 55 56 72 180.95 74 75 190.23 77 78 79 80 81 82 83 84 85 86 Cs Ba Hf 105 W Re 108 Ir Pt Au Hg Tl Pb Bi Po At Rn 132.91 137.34 178.49 Db 183.85 186.21 Hs 192.22 195.08 196.97 200.59 204.37 207.19 208.98 210 210 222 87 88 104 [262] 106 107 [277] 109 110 111 112 Fr Ra Rf Sg Bh Mt Ds Rg Uub 223 226.03 [261] [266] [264] [268] [271] [272] [285] Lanthanoids 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Actinoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 138.91 140.12 140.91 144.24 146.92 150.36 151.96 157.25 158.92 162.50 164.93 167.26 168.93 173.04 174.97 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 227.03 232.04 231.04 238.03 237.05 239.05 241.06 244.07 249.08 252.08 252.09 257.10 258.10 259 262","Periodic Classification of Elements 3.9 EXAMPLE W\u0007 hy is the atomic number more important than the atomic weight in predicting chemical properties of elements SOLUTION Chemical properties of elements depend on the number of electrons present in an atom and their arrangement in different orbits. The number of electrons in an atom is equal to that of protons and the number of protons is known as atomic number which is unique for each element. Hence, atomic number is a more fundamental property than atomic weight. Atomic weights of two elements can be almost same. For example, atomic weights of calcium and argon are 10.1 and 39.9, respectively. But there is no similarity between these two elements. EXAMPLE T\u0007 he position of hydrogen is not justified even in modern periodic table and it is called rogue element. Why? SOLUTION Hydrogen has only one electron in the first shell. Therefore, it becomes stable either by losing an electron like alkali metals or by gaining an electrons like halogens. Since it has similarities in chemical properties with alkali metals as well as halogens, its position is not justified even in modern periodic table and it is known as rogue element. PERIODICITY The regular gradation in the properties of elements and their repetition at regular intervals is called periodicity. The properties which show such regular trends are called periodic properties. Some of the periodic properties include the following: (\ta)\u2002 atomic size\t (b)\u2002 ionisation potential\t (c)\u2002 electron affinity (\td)\u2002electronegativity\t (e)\u2002electropositive character\t (f)\u2002oxidizing\/reducing capacity \t(g)\u2002 metallic\/non-metallic character Atomic Size The size of an atom depends on the distance between the valence shell and the nucleus, i.e., the atomic radius. Variation of Atomic Size Along a Period and a Group in the Periodic Table Period: The atomic radii generally decrease from left to right across a period. When we proceed along a period, electrons are added to the subshells of the same main energy level. With the addition of each electron, the nuclear charge increases by 1. Hence, the increased nuclear charge attracts the electrons more strongly and the size of the atom decreases. Group:\u2002 On moving down a group, the atomic radii of the elements increase with an increase in the atomic number. On proceeding downward in a group, the electrons are added to the next main energy level which are farther from the nucleus. This effect decreases the electrostatic force of attraction between the nucleus and the valence shell electrons. Consequently, the atomic radii of the atoms of the elements increase along a group from top to bottom.","3.10 Chapter 3 Ionisation Energy Ionisation energy of an element is the minimum amount of energy required to remove the most loosely held electron from the outermost shell of an isolated neutral gaseous atom of that element in its lowest-energy state to produce a cation. Successive Ionisation Potentials The amount of energy required to remove the first electron from a neutral atom is called first ionisation energy. The energy required to remove the electron from a unipositive charged ion is called second ionisation energy. Similarly, the third and the fourth ionisation energy or ionisation potentials can be defined in the same way. If the first ionisation energy is considered as I1 and the second is I2 and so on, then I1 < I2 < I3 < I4 < \u2026\u2026 The successive increase in these values is due to the fact that it is relatively more difficult to remove an electron from a cation having higher positive charge than from a cation having lower positive charge than from a neutral atom. Variation of Ionisation Potential Along a Period and a Group in the Periodic Table Period:\u2002 Ionisation energy generally increases from left to right along a period because the atomic size of the elements gradually decreases due to the increase in the atomic number. Group:\u2002 Ionisation energy decreases from top to bottom in a group because of the increase in the atomic radius which is due to the increase in the number of shells. Electron Affinity The amount of energy released when an electron is added to an isolated neutral gaseous atom in its lowest-energy state to produce an anion is called its electron affinity. Factors influencing the magnitude of electron affinity are the following: (a)\u2002 atomic size\t \t (b)\u2002 effective nuclear charge. Variation of Electron Affinity Along a Period and Along a Group of the Periodic Table Period:\u2002 Electron affinity values generally increase on moving from left to right along a period due to the decrease in the atomic size and the increase in its effective nuclear charge. Group:\u2002 Electron affinity gradually decreases from top to bottom in a group. This is due to the steady increase in the atomic radius and the decrease in the effective nuclear charge of the elements. Electronegativity Electronegativity of an element is defined as the relative tendency or ability of the atom of an element present in a molecule to attract the shared electron pair towards itself.","Periodic Classification of Elements 3.11 Variation of Electronegativity Along a Group and a Period of the Periodic Table Period:\u2002 Electronegativity increases on moving along a period from left to right. This is due to the increase in nuclear charge and decrease in atomic size, as a result of which shared electron pair can be attracted more towards itself. Group:\u2002 Electronegativity decreases from top to bottom in a group due to an increase in the atomic size, as a result of which the shared electron pair is attracted less towards itself. (OHFWURSRVLWLYH\u0003&KDUDFWHU Electropositive character of an element is its ability to lose one or more electrons to form a positively charged ion. Variation of Electropositivity Along a Period and a Group in a Periodic Table Period:\u2002 Electropositive character decreases from left to right along a period due to the increase in ionisation potential. Group:\u2002 Electropositive character increases from top to bottom along a group due to the decrease in ionisation potential. Oxidizing and Reducing Property The higher the value of electron affinity of the element, the greater would be its capacity to accept an electron, i.e., the element behaves as a strong oxidizing agent. The lower the value of the ionisation potential of an element, the greater would be its capacity to lose electrons, i.e., the element behaves as a strong reducing agent. Variation of Oxidizing and Reducing Property Along a Period and a Group in a Periodic Table Period:\u2002 Due to the increase in electron affinity from left to right of a period, the oxidizing capacity of the elements increases and the reducing capacity decreases from left to right along a period. Group:\u2002 Due to the decrease in ionisation potential as well as electron affinity from top to bottom of a group, oxidizing capacity of the element decreases, whereas its reducing capacity increases on descending a group. Metallic and Non-metallic Property Non-metals generally have high electronegativity and metals have low electronegativity. Variation of Metallic and Non-metallic Character Along a Period and a Group of the Periodic Table Period:\u2002 Electronegativity increases from left to right along a period in the periodic table. Consequently, the metallic character along a period decreases from left to right and the non-metallic character increases. Group:\u2002 Since electronegativity decreases from top to bottom along a group in the periodic table, the metallic character of an element increases on descending a group and non-metallic character decreases.","3.12 Chapter 3 This trend is very much prominent in the case of elements of IVA and VA groups. These groups begin with non-metals C and N, respectively, and end with metals Pb and Bi, respectively. EXAMPLE Why do the chemical properties along a period vary? SOLUTION Chemical property of an element is due to electrons present in the valence shell. Elements of the same period possess different number of valence electrons. Hence, their chemical properties vary along a period. EXAMPLE How does the magnitude of effective nuclear charge influence the electron affinity of an element? SOLUTION The nucleus has positive charge, and hence, it attracts the electrons present in the atom. As the distance of electrons from the nucleus increases, the attraction experienced by electrons because of nuclear charge decreases.The net nuclear force of attraction experienced by the electron is called effective nuclear charge. The greater the effective nuclear charge, the higher is the electron affinity of an atom. EXAMPLE \u0007Predict the position and properties like metallic\/non-metallic character and oxidizing\/reducing capacity of an element with atomic number 35 in the periodic table. SOLUTION Element with atomic number 35 is bromine and its electronic configuration is 2, 8, 18, 7. Bromine belongs to VIIA (17) group (halogens). Due to this electronic configuration, bromine has strong electron affinity. Hence, it is a reducing agent. Since bromine has seven valence electrons, it is a non-metal. EXAMPLE \u0007Write the electronic configuration of the most stable cation of the fourth period element which has the lowest-ionisation potential value. Justify your answer. SOLUTION The elements belonging to group IA or 1 have the least ionisation potential value in a period. Since it has only one valence electron, the monopositive cation is the most stable cation. The element belongs to fourth period, hence, the electronic configuration of the cation is 2, 8, 8.","Periodic Classification of Elements 3.13 EXAMPLE \u0007How is electropositive character related to reducing property of an element? How does it vary in the group? SOLUTION Electropositive character is defined as the tendency of the element to lose electrons. Reducing property of an element is its ability to give up electron(s) and by virtue of this ability, it reduces the other element. The greater the electropositive character, the more is its reducing property. In a group, reducing property increases from top to bottom as electropositive character increases from top to bottom.","3.14 Chapter 3 TEST YOUR CONCEPTS Very Short Answer Type Questions PRACTICE QUESTIONS \t1.\t What was the need to classify elements? \t17.\t What do you mean by representative elements? Give example. \t2.\t The element in between lithium and potassium in Dobereiner\u2019s classification is ________. \t18.\t Where are the extremely electropositive metals positioned in the periodic table? \t3.\t What was the basis of Dobereiner\u2019s classification of the elements? \t19.\t What type of elements does the right portion of the periodic table contain? \t4.\t In Newland\u2019s classification, elements are arranged in an increasing order of their ______. \t20.\t What is meant by atomic radius? \t5.\t Mendeleev placed coinage metals with _________ \t21.\t An increase in atomic radius results in ________ in metals. the electron affinity of an element. \t6.\t What is Newland\u2019s law of octaves? \t22.\t Which group contains the noble gases and what are the special characteristics of these gases? \t7.\t The Mendeleev\u2019s periodic table, Ar and ________ form the anomalous pair. \t23.\t Which elements are called transuranic elements? \t8.\t What is Mendeleev\u2019s periodic law? \t24.\t _______ group elements have a greater tendency to form cations. \t9.\t Name the element whose atomic weight was \u00adcorrected by Mendeleev. \t25.\t Define ionisation potential. \t10.\t Name some elements whose properties were \t26.\t Describe the trend of the metallic and non-metallic \u00adpredicted by Mendeleev prior to their discovery. character of the elements in a group. \t11.\t Lanthanides and actinides are called ________ \t27.\t Define electronegativity. What are its units? elements. \t28.\t Give the number of valence electrons of (a) alkali \t12.\t Which group contains the inner transition elements? elements and (b) halogens. \t13.\t What is the modern periodic law? \t29.\t How does ionisation potential vary in a group and in a period? \t14.\t What is the basis of the modern periodic table? \t30.\t Describe the trend of the metallic and non-metallic \t15.\t Which periods are called short periods and which character of the elements of a period. periods are called long periods? Why? \t16.\t The elements in the very-short period are __________ and _________. Short Answer Type Questions \t31.\t What are the advantages of the classification of \t35.\t Give brief description of Mendeleev \u2019s periodic elements? table. \t32.\t Give some examples of Dobereiner\u2019s triads. Explain \t36.\t What is the basic difference between Mendeleev \u2019s how they are grouped as triads. periodic table and the modern periodic table? \t33.\t What are the achievements and the limitations of \t37.\t Why are the lanthanides and the actinides given a Newland\u2019s classification? separate position in the modern periodic table? \t34.\t Give the limitations of Dobereiner\u2019s classification. \t38.\t Why is there an increase in successive IP values?","39.\t Why do the elements of the same group show Periodic Classification of Elements 3.15 i\u00addentical property? Explain. \t43.\t Why is the modern periodic table also referred to as \t40.\t How does electronegativity vary along a group and Bohr\u2019s table? along a period? Give reasons and explain with examples. \t44.\t How do the oxidizing and reducing capacity of an \t41.\t How does the atomic size vary along a group and a element depend on the ionisation potential and period? Give reasons and explain with the help of electron affinity? an example. \t42.\t What is the difference between electronegativity and electropositivity? Essay Type Questions \t45.\t What are the merits and the limitations of \t48.\t What are the merits and limitations of the long Mendeleev \u2019s periodic table? form of periodic table? \t46.\t Explain in detail about the different periods of the \t49.\t How does ionisation potential, electron affinity, modern periodic table. electropositivity, oxidizing and reducing capacity vary along a group and a period? Give reasons. \t47.\t What is meant by periodicity? What are the various periodic properties? Explain how and why atomic radius varies in a group and period. Give an example. For Answer key, Hints and Explanations, please visit: www.pearsoned.co.in\/IITFoundationSeries CONCEPT APPLICATION Level 1 Direction for questions from 1 to 7: Direction for questions from 8 to 14: PRACTICE QUESTIONS State whether the following statements are true Fill in the blanks. or false. \t8.\t In the modified Mendeleev \u2019s periodic table, \t1.\t In Dobereiner\u2019s classification, the atomic weight ________ groups and ________ periods are pres- and the properties of the first element are almost ent, whereas in modern periodic table, _____ equal to the average of those of the 2nd and 3rd groups and _____ periods are present. elements of a triad. \t9.\t If three elements X, Y and Z form a triad and \t2.\t Atomic numbers of the isotopes of an element are atomic weights of X and Z are 9 and 40, respec- different. tively, then the atomic weight of an element Y is _____. \t3.\t Differentiating electrons enter anti-penultimate shell in transition elements. \t10.\t Mendeleev named certain undiscovered elements as eka boron and eka aluminium which when later \t4.\t Along a period, atomic volume of the elements discovered were called ______and________. gradually increases from left to right due to an increase in the number of valence electrons. \t11.\t According to Newland\u2019s classification, the prop- erties of the 3rd element will find similarity with \t5.\t The element with atomic number 17 has the least those of the______ element. size among all the elements of that period. \t12.\t Extremely electropositive metals are present in \t6.\t The lower is the value of ionisation potential of an ________ and ________ groups. element, the greater is its reducing power. \t13.\t The atomic number of an element that belongs to \t7.\t Among all the groups, the maximum number of the IVA group and 4th period is ___________. elements is found in IIIB group.","3.16 Chapter 3 \t14.\t Good reducing agents are found in _____ and \t\t (c)\u2002\u0007In fourth period elements differentiating elec- _____ groups. trons enter either the 4th shell or the third shell. Direction for question 15: \t\t (d)\u2002\u00071st transition series elements are present in the Match the entries given in Column A with fourth period. appropriate ones in Column B. \t20.\t The values of the second electron affinities of \t15.\t Column A Column B e\u00ad lements are positive. Which of the following could be the appropriate reason? A.\t Calcium (\u2002)\t a.\tModern periodic table \t\t (a)\u2002\u0007Work has to be done against the force of \u00adrepulsion B.\t Sodium (\u2002)\t b.\tN\u0007 aturally occurring of the valence electron of the \u00aduninegative ion. radioactive element \t\t(b)\u2002W\u0007 ork has to be done against the force of C.\t\u0007Uranium (\u2002 )\t c.\t\u0007II A a\u00ad ttraction of the nucleus. D.\tL\u0007anthanides and (\u2002)\t d.\tAlkali metal \t\t(c)\u2002\u0007The electron loses its energy since it has to actinides work against the repulsive force. E.\t \u000718 groups and (\u2002)\t e.\t8 elements \t\t (d)\u2002\u0007In uninegative ions, the effective nuclear force 7 periods of attraction towards the valence electrons becomes more. F.\t\u00072nd and 3rd (\u2002)\t f.\t III B period in modern periodic table \t21.\t Why is the ionisation energy of sulphur less than that of phosphorous, though sulphur is next to Direction for questions from 16 to 45: phosphorous in the period? For each of the questions, four choices have been provided. Select the correct alternative. \t\t (a)\u2002\u0007Atomic radius of sulphur is greater than that of phosphorous. \t16.\t Atomic weights of three elements in a Doberenier traid are x, 81 and 127. Find the missing atomic \t\t (b)\u2002 Sulphur has half-filled electronic configuration. weight. \t\t (c)\u2002\u0007Phosphorushashalf-filledelectronicconfiguration. \t\t(a)\u2002104\t (b)\u200235\t \t\t(d)\u2002I\u0007n the third period the ionisation energy decreases along the period. \t\t(c)\u200246\t (d)\u200223 \t22.\t The element which accepts electrons readily to form anion belongs to PRACTICE QUESTIONS 1\t 7.\tThe properties which increase on going down the group are__________. \t\t (a)\u2002 VII A group and second period\t \t\t (b)\u2002 VII A group and third period \t\t (a)\u2002 ionisation energy and electro negativity \t\t (c)\u2002 O group and third period\t \t\t (b)\u2002 atomic size and ionisation energy \t\t (d)\u2002 VIA group and third period \t\t (c)\u2002 electro negativity and atomic size \t\t (d)\u2002 metallic character and reducing power \t23.\t Identify the oxide which forms the strongest base: \t18.\t Which of the following belong to the same group? \t\t(a)\u2002MgO\t (b)\u2002Al2O3\t \t\tA3+ (number of electrons = 10) \t\t(c)\u2002Na2O\t (d)\u2002CaO \t\tB2+ (number of electrons = 10) \t24.\t The formula of the oxide of an element M is M2O3. \t\t C (number of electrons = 5) The first four ionisation energies of the element M \t\t D (number of electrons = 31) can be in the order: \t\t(a)\u2002ABC\t (b)\u2002BCD \t\t (a)\u2002 120 kcal, 270 kcal, 400 kcal and 5098 kcal \t\t (b)\u2002 210 kcal, 150 kcal, 370 kcal and 590 kcal \t\t(c)\u2002ACD\t (d)\u2002ADB \t\t (c)\u2002 70 kcal, 100 kcal, 105 kcal and 120 kcal \t\t (d)\u2002 560 kcal, 410 kcal, 320 kcal and 290 kcal \t19.\t Which of the following is not true? \t25.\t The chemistry of alkali metals is essentially the \t\t(a)\u2002\u0007All the fourth period elements have the 4th chemistry of unipositive ions because shell as the valence shell. \t\t (a)\u2002 they have low ionisation energy values \t\t(b)\u2002I\u0007n all fourth period elements differentiating e\u00ad lectrons enter the 4th shell.","Periodic Classification of Elements 3.17 \t\t (b)\u2002\u0007they have strong tendency to lose single valence chemical properties just as the eighth note on a electron musical scale resembles the first one. \t\t (c)\u2002 second ionisation energy values are very high \t\t(a)\u20021432\t (b)\u20021342 \t\t (d)\u2002 all the above \t\t(c)\u20024213\t (d)\u20022431 \t26.\t Na+, F\u2013 and Mg+2 ions have the same \t32.\t The electron affinity increases on moving from left \t\t(a)\u2002size\t to right along a period. Arrange the reasons in a \t\t (b)\u2002 electronic configuration proper sequence. \t\t (c)\u2002 ionisation energy\t \t\t (d)\u2002 nuclear charge \t\t(1)\u2002T\u0007he amount of energy released during the addition of an electron increases from left to \t27.\t Gallium is more electronegative than aluminium right along a period. even though electronegativity of the elements decreases down the group. It is due to ______. \t\t(2)\u2002E\u0007ffective nuclear charge of the elements increases from left to right. \t\t (a)\u2002 increase in the atomic size\t \t\t (3)\u2002\u0007The atomic size of the elements decreases from \t\t (b)\u2002 increase in the ionisation energy left to right. \t\t (c)\u2002 increase of effective nuclear charge\t \t\t (4)\u2002\u0007The tendency to gain electrons and form anion increases from left to right. \t\t (d)\u2002 all the above \t\t(a)\u20021324\t (b)\u20023241 \t28.\t The total number of inner transition elements is \t\t(c)\u20023142\t (d)\u20024321 ________. \t33.\t X belongs to IA or first group and fifth period and \t\t(a)\u200214\t (b)\u200228 Y succeeds X in the group. Z succeeds Y in the period. Arrange the suitable statements in the cor- \t\t(c)\u200220\t (d)\u200215 \t29.\t Which one of the following doesn\u2019t come under rect sequence in order to arrange X, Y and Z in the zero group? increasing order of their atomic sizes. \t\t(a)\u2002He\t (b)\u2002K \t\t (a)\u2002\u0007effect of number of valence electrons and num- ber of shells on the atomic size \t\t(c)\u2002Ne\t (d)\u2002Ar \t\t (b)\u2002\u0007identification of the elements X, Y and Z \t30.\t The atomic numbers of 4 elements A, B, C and D are 8, 9, 10 and 11, respectively. The order of their \t\t(c)\u2002d\u0007etermination of the number of shells and the PRACTICE QUESTIONS atomic volume is number of valence electrons present in X, Y and Z \t\t (a)\u2002 A > B > C < D\t (b)\u2002 A < B < C > D \t\t (d)\u2002\u0007determination of the positions of Y and Z in \t\t (c)\u2002 A > B < C < D\t (d)\u2002 A < B > C > D the periodic table based on the position of X \t\t(a)\u2002dbca\t (b)\u2002bdca\t \t31.\t Arrange the following statements in chronological \t\t(c)\u2002cad\t (d)\u2002dca order: \t34.\t Among third period elements halogen and alkaline \t\t (1)\u2002\u0007Some elements be grouped in sets of three ele- earth metals are, respectively. ments in the increasing order of their atomic weights in which the atomic weight of the mid- \t\t (a)\u2002 Cl, Na\t (b)\u2002 S, Na dle element was found to be the arithmetic mean of the atomic weights of the other two elements. \t\t (c)\u2002 S, Mg\t (d)\u2002 Cl, Mg \t\t(2)\u2002\u0007The physical and chemical properties of ele- \t35.\t Which among the following pairs of elements have ments are periodic functions of their atomic maximum and minimum electron affinity values numbers. respectively? \t\t (3)\u2002\u0007The physical and chemical properties of elements \t\t (a)\u2002 Be, Cl\t (b)\u2002 He, K are periodic functions of their atomic weights. \t\t (c)\u2002 Be, S\t (d)\u2002 Cl, Cs \t\t(4)\u2002\u0007When elements are arranged in the increas- ing order of their atomic weights, the eighth \t36.\t Which of these triads could not be justified as \u00adelement resembles the first in physical and Dobereiner\u2019s triad? \t\t (a)\u2002 Li, Na, K\t (b)\u2002 Cl, Br, I \t\t (c)\u2002 C, N, O\t (d)\u2002 Ca, Sr, Ba","3.18 Chapter 3 \t37.\t An element belongs to IIIA group and forth period \t\t(a)\u2002Be\t (b)\u2002Mg in the modern periodic table. What could be the \t\t(c)\u2002Li\t (d)\u2002Ca probable atomic number of that element? \t42.\t Which of the following is not a transition element? \t\t(a)\u200223\t (b)\u200249\t \t\t(a)\u2002Mn\t (b)\u2002Fe \t\t(c)\u200231\t (d)\u200213 \t\t(c)\u2002Cu\t (d)\u2002K \t38.\t The formula of ion formed by an element A is A+2. \t43.\t Element X has 12 protons in its nucleus. To which The element A can probably belong to which of group in the periodic table would it belong? the following groups? \t\t (a)\u2002 IVA (14)\t (b)\u2002 IIA (2) \t\t(a)\u2002IIIA\t (b)\u2002VIIA\t \t\t (c)\u2002 IIIA (13)\t (d)\u2002 VIA (16) \t\t(c)\u2002IIA\t (d)\u2002IA \t39.\t The anomalous pairs in Mendeleev \u2019s periodic table \t44.\t Element X has 12 neutrons in its nucleus. To which is\/are _________. group in the periodic table would it belong? \t\t (a)\u2002 Co, Ni\t (b)\u2002 Te, I\t \t\t(a)\u20021\t (b)\u20022 \t\t (c)\u2002 Ar, K\t (d)\u2002 all the above \t\t (c)\u2002 3\t (d)\u2002 cannot be predicted \t40.\t Which of the following triads do not follow \t45.\t If I1 is the first ionisation potential, I2 is the second Dobereiner\u2019s law of triads? ionisation potential, I3 is the third ionisation poten- tial and I4 is the fourth ionisation potential of an \t\t (a)\u2002 Li, Na, K\t (b)\u2002 Ca, Sr, Ba\t element; then which of the following has the least \t\t (c)\u2002 Be, Mg, Ca\t (d)\u2002 Cu, Ag, Au value? \t41.\t Which of the following elements do not belong to \t\t(a)\u2002I1\t (b)\u2002I2 group IIA or second group? \t \t(c)\u2002I3\t (d)\u2002I4 PRACTICE QUESTIONS Level 2 largest cation and the smallest anion in the modern periodic table. \t1.\t Why is the positive ion always smaller than the cor- responding neutral atom? \t9.\t Why do transition metals show catalytic behaviour? \t2.\t Alkaline earth metals are denser than the corre- \t10.\t Elements with more non-metallic character are good sponding alkali metals. Comment on the statement. oxidizing agents and those with more electropositive character are good reducing agents. Explain. \t3.\t Why is the zero group present at the right-hand side of the periodic table? \t11.\t An element X belongs to group II A and fourth period in the periodic table. Find out the atomic \t4.\t Why is it not possible to form Na+2 ion? Explain number of X and the element which is placed just with respect to periodic properties. below X in the periodic table. \t5.\t If the atomic numbers of some elements in the \t12.\t Explain why the addition of an electron to a neutral modern periodic table are 8, 7, 11, 12, 13 and 9, atom is associated with release of energy. what type of ions do they form? Arrange the ions in the increasing order of their size and justify. \t13.\t The first four ionisation energies of an element R are 580, 640, 1000 and 2700 k cal\/mole, \u00adrespectively. \t6.\t \u2018The densities of transition metals are greater than Find the group number of R and write formulae of those of the alkali and alkaline earth metals.\u2019 Justify. its chloride and oxide. \t7.\t An element R belongs to IVA group and third \t14.\t B and Al belong to the same group, but the nature period in the periodic table. Arrange the elements of oxides formed by them are different, whereas Be, that are placed below R and towards right in the Al form same type of oxides although they belong periodic table in increasing order of atomic size and to different groups. Comment. ionisation potential. \t8.\t Predict the position of the elements which form the","Periodic Classification of Elements 3.19 Directions for questions from 15 to 24: \t\t (a)\u2002 identity the noble gas Application-Based Questions \t\t (b)\u2002 which of the following elements could D be? \t15.\t Chlorine, Y and iodine form a Dobereiner\u2019s triad. Identify the atomic weight of Y. \t\t\t (i)\u2002Ne\t\t (ii)\u2002Ca\t \t16.\t Three elements X, Y and Z form a Dobereiner triad. \t\t\t (iii)\u2002Na The ratio of the atomic weight of X to that of Z is 7 : 25. If the sum of the atomic weights of X and Z is \t\t (c)\u2002 Which element has the highest electron affinity? 160, find the atomic weights of X, Y and Z. \t\t(d)\u2002W\u0007 hich element has the strongest metallic \t17.\t The total number of electrons present in the first character? two and the last two shells is the same for an atom of an element X. The sum of the electrons pres- \t\t (e)\u2002 Write the unit of ionisation energy. ent in second, fourth and fifth shells is equal to the number of electrons present in the third shell and \t21.\t A physical science teacher said that work is directly the fifth shell is the valence shell. Identify X and proportional to force and displacement. One of the predict its position in the periodic table. students who attended the \u2018TIME IIT\u2019 foundation course said that work done in removing an elec- \t18.\t The atomic number of elements X, Y and Z are tron from a cation is directly proportional to the (A \u2013 2), A and (A + 2), respectively. Y is a noble gas charge on it. The teacher appreciated the student (not helium) then and explained the above concept. What was her explanation? \t\t (a)\u2002 predict the group to which X, Y and Z belong \t22.\t Elements A, B and C with atomic numbers z, z + \t\t (b)\u2002\u0007predict the formula of a molecule formed by X 1 and z + 2, respectively, form positive ions having and Z equal number of electrons. Compare and contrast the sizes of the respective ions. \t19.\t Consider the isotopes of carbon, i.e., C \u2013 12,\u00ad C \u2013 13, C \u2013 14, would you place them in same or \t23.\t Elements with higher electronegativity are good different slots in the periodic table? Give reason. oxidizing agents. Give reasons. \t20.\t Given below is the graph representing the ionisation \t24.\t If the atomic numbers of some elements in the energies of a few elements, A to H with s\u00aduccessive modern periodic table are 8, 7, 11, 12, 13 and 9, atomic numbers. what type of ions do they form? Arrange the ions in the increasing order of their size and justify. C B Ionisation energy H PRACTICE QUESTIONSG EF A D Atomic number Level 3 metallic elements. However, elements of 2nd period have lower electron affinity values than do the \t1.\t Write the electronic configuration of the most c\u00adorresponding elements of 3rd period.\u2019 Comment stable cation of the element having the lowest- on this statement. ionisation potential value and belonging to the 4th period. Justify your answer. \t3.\t Why do alkali metals tarnish on exposure to air? \t2.\t \u2018Generally, electron affinity values of elements decrease from top to bottom in a group of non-","3.20 Chapter 3 \t9.\t A, B, C and D are four 9th class students. Their chemistry teacher conducted a role play in the class \t4.\t Electron affinity values of noble gases are zero, which is as given below. She assumed that A is a whereas they are negative for alkaline earth metals. metal and B is a non-metal. C and D need to carry How do you account for this? out two separate experiments. For this purpose, C needs an oxidizing agent and D needs a reducing \t5.\t Explain how the density changes across a period agent. Accordingly they have to form two groups from K to Ni and compare it with the change down consisting of two students each. How will they form the group from K to Cs, respectively. the group? Explain this with appropriate reasons. Directions for questions from 6 to 10: Application- \t10.\t Three elements A, B and C have successive atomic Based Questions numbers in increasing order. A attains stability when an electron is added to the third shell which \t6.\t Which elements can be used in solar cells and why? is the valence shell. Identify A, B and C and pre- dict the respective elements possessing the maxi- \t7.\t Explain the energy changes involved in the forma- mum ionisation energy, oxidizing power, reducing tion of F\u2013 and O2\u2013 from their respective atoms. power, electron affinity and atomic size. \t8.\t \u2018Addition of an electron to a uninegative ion (except to VIA group elements) is an endothermic process, whereas the formation of uninegative ion from its neutral atom is an exothermic process.\u2019 Comment. PRACTICE QUESTIONS","Periodic Classification of Elements 3.21 CONCEPT APPLICATION Level 1 True or false \t2.\t False \t3.\t False \t4.\t False \t1.\t False \t6.\t True \t7.\t True \t5.\t True Fill in the blanks \t9.\t 24.5 \t10.\t scandium, gallium \t11.\t 10th \t8.\t 8, 7, 18, 7 \t13.\t 32 \t14.\t IA, IIA \t12.\t IA and IIA Match the following B : d\t C:b E : a\t F:e \t15.\t A : c\t \t\tD : f\t Multiple choice questions \t16.\t b \t20.\t a \t24.\t a \t28.\t b \t25.\t d \t29.\t b \t17.\t d \t21.\t c \t26.\t b \t30.\t c \t27.\t c \t18.\t c \t22.\t b \t19.\t b \t23.\t c \t31.\t \u2002 (i)\t\u0007Some elements be grouped in sets of three ele- \t\t(iii)\t\u0007The tendency to gain electrons and form HINTS AND EXPLANATION ments in the increasing order of their atomic anions increases from left to right. weights in which the atomic weight of the middle element was found to be the arithme- \t\t(iv)\t\u0007The amount of energy released during the tic mean of the atomic weights of the other addition of an electron increases from left to two elements. right along periods. \t\t(ii)\tW\u0007 hen the elements are arranged in the \t33.\t \u2002 (i)\t\u0007Determination of the positions of Y and Z in increasing order of their atomic weights, the the periodic table based on the position of X. eighth element resembles the first in physical and chemical properties just as the eighth note \t\t(ii)\t\u0007Determination of the number of shells and on a musical scale resembles the first one. the number of valence electrons present in X, Y and Z. \t\t(iii)\t\u0007The physical and chemical properties of ele- ments are periodic functions of their atomic \t\t(iii)\t\u0007Effect of the number of valence electrons and weights. number of shells on atomic size. \t\t(iv)\t\u0007The physical and chemical properties of ele- \t34.\t Among the third period elements halogen and ments are periodic functions of their atomic alkaline earth metal are chlorine and magnesium, numbers. respectively. \t32.\t \u2002(i)\t\u0007The atomic size of the elements decreases \t35.\t Inert gases have zero electron affinity due to their from left to right. stable octet configuration, whereas IA group e\u00ad lements have minimum electron affinity value due \t\t(ii)\tE\u0007ffective nuclear charge of the elements to large atomic size. Halogens have high electron increases from left to right. affinity due to small atomic size. Hence, Cl and Cs","3.22 Chapter 3 the mean of the atomic weights of Cu and Au, and hence, (Cu, Ag, Au) triad do not follow have maximum and minimum electron affinity val- Dobereiner\u2019s law of triads. ues, respectively. \t41.\t Lithium belongs to IA(1) group. \t36.\t According to Dobereiner\u2019s classification, the ele- ments of the triads (Li, Na, K; Cl, Br, I; Ca, Sr, \t42.\t Potassium belongs to IA(1) group called alkali Ba) have similar chemical properties. Elements C, metal, but not transition element. N and O do not show similar properties.\b \t43.\t The atomic number of X = 12. \t37.\t Elements in IIIA group are B, Al, Ga, In, Tl and in \t\tIts electronic configuration = 2, 8, 2 this B belongs to the second period. \t\t\u2234 X\u0007 belongs to IIA(2) group since there are two \t\t\u2234 The element in IIIA group and fourth period is electrons in the valence shell. \t\tGa (gallium), whose atomic number is 31. \t44.\t The position of the element in the periodic table \t38.\t The formula of ion formed by A is A+2 cannot be predicted based on the number of neu- \t\t\u21d2 Valency of A is 2 trons present in the nucleus. \t\t\u2234 A belongs to II A group \t45.\t Ionisation potential values increases in the follow- \t39.\t Anomalous pairs in Mendeleev \u2019s periodic table are ing order: Cu, Ni; Te, I; Ar, K. \t\tI1 < I2 < I4 < \u2026\u2026 \t40.\t Since the atomic weight of Ag is not equal to HINTS AND EXPLANATION Level 2 \t\t(ii)\t\u0007Tendency of an atom of an element to lose or gain based on number of valence electrons. \t1.\t \u2002 (i)\t difference between cation and neutral atom \t\t(ii)\t\u0007comparison of number of electrons in neutral \t\t(iii)\t type of ions formed \t\t(iv)\t variation of size of an ion with charge atom to cation \t\t(iii)\t\u0007comparison of number of protons in neutral \t6.\t \u2002 (i)\tT\u0007ransition metals come to the right of the respective alkali and alkaline earth metals in atom and cation the period. \t\t(iv)\t\u0007comparison of effect of nuclear charge in neu- \t\t(ii)\t factors that affect density tral atom and cation \t\t(iii)\t\u0007comparison of the variation in mass in alkali, \t2.\t \u2002(i)\t\u0007comparison of nuclear charge in alkali and alkaline and transition metals alkaline earth metals \t\t(iv)\t comparison of the variation in volume \t\t(ii)\tr\u0007elation and comparison between nuclear \t7.\t \u2002 (i)\t identification of R charge and atomic size of alkali and alkaline \t\t(ii)\t\u0007position of an element R in the periodic earth metals table \t\t(iii)\t\u0007comparison of atomic mass of corresponding \t\t(iii)\t\u0007comparison of effect of nuclear charge on elements of IA and IIA valence electrons across a period and down the \t\t(iv)\t relation between mass, size and density group \t\t(iv)\t\u0007Relation between nuclear charge, size and \t3.\t \u2002(i)\t\u0007comparison of electronic configuration of \u00adionisation potential zero group elements with elements of other group \t8.\t \u2002 (i)\t\u0007Cations are smaller than their corresponding atoms and anions are larger than their corre- \t\t(ii)\t characteristics of zero group elements sponding atoms \t4.\t \u2002 (i)\t electronic configuration of sodium \t\t(ii)\t\u0007trend of atomic size along a period and down \t\t(ii)\t position of sodium in the periodic table a group \t\t(iii)\t periodic properties of these elements \t5.\t \u2002(i)\ti\u0007dentification of the number of valence \u00adelectrons in each element","Periodic Classification of Elements 3.23 \t\t(iii)\t\u0007position of the largest atom which can form \t\tGiven that Zx + Zz = 160 cation \u2234 7 .Zz + Zz = 160 \u21d2 Zz = 125 \t\t(iv)\t\u0007position of the smallest atom which can form 25 anion \t\tFrom equation (1) \t9.\t \u2002 (i)\t requisite for catalytic behaviour Zx = 7 .125 = 35 25 \t\t(ii)\t electronic configuration of metals \t\tAtomic weight of Y is Zy = 35 + 125 = 80 . \t\t(iii)\t\u0007relation between electronic configuration and 2 bonding \t10.\t \u2002 (i)\t Nature of metal and non-metals \t\t[\u2234 X, Y and Z form Dobereiner triad] \t\t(ii)\t\u0007oxidizing agent and reducing agent in terms of \t18.\t Electronic configuration of element X is electronic concept \t K\t L\t M\tN\tO \t 2\t8\t18\t8\t2 \t\t(iii)\t\u0007relation between metals, non-metals, oxidiz- ing and reducing agents \t\tThe element X is strontium with atomic number 38. It belongs to the fifth period and 2nd group \t11.\t \u2002 (i)\t electronic configuration of X (IIA) or group 2. \t\t(ii)\ta\u0007tomic number of X based on periodic \t18.\t (a)\t\u0007As Y is a noble gas, the group of y is 18th group arrangement of elements or zero group. \t\t(iii)\t electronic configuration of element below X \t\t(iv)\t atomic number of element below X \t\t\t X = A \u2013 2; group of X is 16th group or VI A. \t\t\t Z = A + 2; group of Y is 2nd group or II A \t12.\t \u2002(i)\t\u0007type of forces present between an added e\u00ad lectron and neutral atom \t\t(b)\u2002X \u21d2 Valency = 2 (electronegative) \t\t(ii)\t\u0007the type of forces responsible for addition of \t\t\tZ \u21d2 Valency = 2 (electropositive) electrons \t\t\t Hence, the formula is ZX. HINTS AND EXPLANATION \t\t(iii)\t\u0007energy changes involved when electrons are added \t19.\t They should be placed in the same slot, i.e., they belong to IVA or 14 group and 2nd period as they \t13.\t \u2002(i)\ti\u0007dentification of valence shell electronic c\u00ad onfiguration from the given IP values have same atomic number. \t\t(ii)\ti\u0007dentification of respective group number \t20.\t (a)\t The noble gas is C. from the valence electrons \t\t(b)\tD is an alkali metal, i.e., Na. \t\t(c)\t B has the highest-electron affinity. \t\t(iii)\t writing the formulae of its chloride and oxide \t\t(d)\tD has the strongest-metallic character. \t\t(e)\t The unit of ionisation energy is kJ\/mole. \t14.\t \u2002 (i)\t\u0007position of B, Be, Al in the periodic table \t21.\t As the positive charge increases, the effective \t\t(ii)\t\u0007comparison of size and ionisation potential of nuclear charge on the outermost electron increases. B, Be and Al So, work done for removing an electron increases with the increase of charge on the cation. \t\t(iii)\t\u0007relation between size, IP and nature of the oxides formed \t15.\t Triad of chlorine, Y and iodine \t\tAtomic weights 35.5 \u2013 127 \t22.\t As the positive ions have equal number of electrons, \t\tMean atomic wt = 35.5 + 127 \u21d2 81.25 A, B and C must be isoelectronic and form uni- positive, dipositive and tripositive ions. With the 2 increase in nuclear charge, the effective nuclear \t\t\u2234 atomic weight of Y = 81.25 force of attraction on the outermost electrons increases. Therefore, the ionic size decreases from \t16.\t Let the atomic weights of X, Y and Z be Zx, Zy and A+, B+2 and C+3. Zz, respectively. \u2234 Zx = 7 \u21d2 Zx = 7 .Zz \u2192 (1) Zz 25 25","3.24 Chapter 3 \t24.\t The elements with atomic numbers 8, 7, 11, 12, 13 and 9 are oxygen, nitrogen, sodium, magne- \t23.\t Elements with higher electronegativity have greater sium, aluminium and fluorine, respectively. These tendency to attract the shared pair of electrons pres- e\u00ad lements belong to VIA (16), VA (15), IA (1), IIA ent in a compound. An atom which undergoes (2), IIIA (13) and VIIA (17) groups. Generally these reduction is an oxidizing agent. The greater the elements form O\u20132, N\u20133, Na+, Mg+2, Al3+ and electronegativity value, the more is its tendency to F\u2013 ions. The order of ionic size is undergo reduction. So, it acts as a good oxidizing agent. \t\tN\u20133 > O\u20132 >F\u2013> Na+ > Mg+2 > Al3+. HINTS AND EXPLANATION Level 3 \t6.\t The metals which belong to alkali metal groups, such as, K, Cs and Rb can be used in solar cells. \t1.\t \u2002 (i)\t trend of IP value along a period The reason is that these metals have very low first \t\t(ii)\t trend of ionisation potential along a period ionisation energy. Hence, they can lose electron \t\t(iii)\t\u0007the group that has elements with lowest- easily when exposed to sunlight. ionisation potential in their respective the \t7.\t Though F\u2013 and O\u20132 are isoelectronic species, periods the formation of F\u2013 is exothermic while O\u20132 is \t\t(iv)\t\u0007electronic configuration of the element and endothermic. Elements (or) ions having inert gas cation c\u00adonfiguration are highly stable. For attaining the nearest inert gas configuration fluorine has one \t2.\t \u2002(i)\t\u0007Comparison of atomic size of 2nd and 3rd electron less. So, it readily accepts electrons and period elements the formation of F\u2013 takes place. In this process, a large amount of energy is released. But in the case \t\t(ii)\t effect of numbers of electrons on size of the formation of O\u20132 repulsions exist between \t\t(iii)\t\u0007energy changes that take place by the addition the added electron and already existing O\u20131 ion. So, this process requires supply of energy to overcome of an electron in the III-shell and the II-shell this repulsion. Hence, the formation of O\u20132 is an endothermic process. \t3.\t \u2002 (i)\t reactivity of alkali metals \t\t(ii)\t electronic configuration of alkali metals \t8.\t Addition of an electron to uninegative ion involves \t\t(iii)\t ionisation potential of alkali metals repulsion between the electron and negatively- \t\t(iv)\t effect of ionisation potential on reactivity charged ion. So, energy is required to add an \t\t(v)\t relating reactivity and tarnishing electron to a uninegative ion. Hence, it is endo- thermic process. When an electron is added to a \t4.\t \u2002 (i)\t\u0007valence shell electronic configuration of noble neutral atom, it works towards the force of attrac- gases and alkaline earth metals tion exerted by the nucleus of the neutral atom, and hence, energy is released during this process. \t\t(ii)\t\u0007relation between valence shell electronic con- figuration and stability in both the cases \t9.\t Oxidizing agent is the substance which not only oxidises other substances but also undergoes reduc- \t\t(iii)\t\u0007types of forces existing between an electron tion by itself, i.e., it accepts electrons from the going to be added to a noble gas o\u00ad thers. The more the non-metallic character of an element, the more is its tendency to accept elec- \t\t(iv)\t\u0007energy change involved when an electron is trons, and thus acting as a good oxidizing agent. added to a noble gas element \t\tReducing agent is the substance which reduces \t5.\t \u2002 (i)\t\u0007comparison between different electrons enter- the other substances but itself undergoes oxidation. ing different shells That means it gives up its own electrons. \t\t(ii)\t\u0007variation of nuclear charge on the size of the elements from K to Ni \t\t(iii)\t\u0007variation of the nuclear charge on elements K to Cs down the group \t\t(iv)\t effect of nuclear charge on size \t\t(v)\t relation between mass, size and density","The more the metallic character, the more is the Periodic Classification of Elements 3.25 tendency of losing electrons. Hence, metals are good reducing agents. The pairing should be A and \t\tSince Cl has the highest-electron affinity, it gains D and B and C. an electron and attains stability, thus, it has greater oxidizing power. K which belongs to the IA(1) \t10.\t As A attains stability by gaining an electron, it must group and 4th period possesses greater atomic size, form a uninegative ion. The third shell being the lower ionisation potential and greater reducing valence shell of A, the electronic configuration of power. Atomic size gradually increases from Cl to A is 2, 8, 7. So the atomic number is 17. A, B and K. Ar which belongs to the 18th group possesses C have successive atomic numbers. So the atomic maximum ionisation potential due to a stable octet number of B is 18 and that of C is 19. A, B and C configuration. are chlorine, argon and potassium, respectively. Group VIIA(17) 0(18) IA(1) A B C Z Cl Ar K Electronic configuration 17 18 19 2, 8, 7 2, 8, 8 2, 8, 8, HINTS AND EXPLANATION","This page is intentionally left blank.","14CChhaapptteerr BCSNhoyuensmmdteiibnmcegasrl F I G U R E 1 . 1 \u2002 Figure Caption Remember Before beginning this chapter, you should be able to: \u2022\u2002\u0007know concepts of atom and subatomic particles. \u2022\u2002\u0007u\u0007 nderstand structure of atom and related theories. \u2022\u2002\u0007\u0007express chemical reactions. Key Ideas After completing this chapter, you should be able to: \u2022\u2002u\u0007n\u0007derstand the relation between electronic configuration and bond formation. \u2022\u2002\u0007s\u0007tudy the mode of formation of ionic and covalent bonds along with representation by Lewis dot structures and properties of ionic and covalent compounds \u2022\u2002\u0007u\u0007nderstand the mode of formation of coordinate covalent and metallic bonds, and to correlate with the properties of metals. \u2022\u2002\u0007\u0007compare the different intermolecular forces and their relation with properties of substances. \u2022\u2002\u0007u\u0007 nderstand the electronic concept of oxidation and reduction."]
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