Preview Physical Chemistry for the JEE and Other Engineering Entrance Examinations by K. Rama Rao, S. V. V. Satyanarayana (2013)

Preview Physical Chemistry for the JEE and Other Engineering Entrance Examinations by K. Rama Rao, S. V. V. Satyanarayana (2013) Preview Physical Chemistry for the JEE and Other Engineering Entrance Examinations by K. Rama Rao, S. V. V. Satyanarayana (2013) Preview Physical Chemistry for the JEE and Other Engineering Entrance Examinations by K. Rama Rao, S. V. V. Satyanarayana (2013) Preview Physical Chemistry for the JEE and Other Engineering Entrance Examinations by K. Rama Rao, S. V. V. Satyanarayana (2013)

Physical Chemistry for the JEE and Other Engineering Entrance Examinations K Rama Rao S.V.V Satyanarayana Delhi Ÿ Chennai FM.indd 4/2/2013 12:16:06 PM Copyright © 2013 Dorling Kindersley (India) Pvt Ltd Licensees of Pearson Education in South Asia No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent This eBook may or may not include all assets that were part of the print version The publisher reserves the right to remove any material in this eBook at any time ISBN 9788131787618 eISBN 9789332516366 Head Office: A-8(A), Sector 62, Knowledge Boulevard, 7th Floor, NOIDA 201 309, India Registered Office: 11 Local Shopping Centre, Panchsheel Park, New Delhi 110 017, India FM.indd 4/2/2013 12:16:06 PM Contents Preface Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 FM.indd iv StruCture of atom BaSiC ConCeptS of ChemiStry the StateS of matter Solid State SolutionS thermodynamiCS thermoChemiStry ChemiCal KinetiCS ChemiCal equiliBrium ioniC equiliBrium redox reaCtionS eleCtroChemiStry StoiChiometry SurfaCe ChemiStry nuClear ChemiStry 1.1–1.86 2.1–2.38 3.1–3.114 4.1–4.90 5.1–5.96 6.1–6.80 7.1–7.46 8.1–8.98 9.1–9.84 10.1–10.112 11.1–11.48 12.1–12.108 13.1–13.86 14.1–14.70 15.1–15.64 4/2/2013 12:16:06 PM Preface Physical Chemistry is the coordination of Physics and Chemistry, of which Oswald laid the foundation Without the knowledge of those physical methods, which raised Chemistry from being a mere collection of facts to a science with a rational basis, it is simply impossible to study Chemistry, as Robert Bunsen pointed out, “A chemist who is no physicist is almost valueless.” Although certain branches of Chemistry, such as Organic Chemistry continue to retain traces of individuality, it is not possible to study one branch of chemistry in isolation (It must be noted here that recent studies conducted on organometallic compounds has significantly blurred the line between Organic Chemistry and Inorganic Chemistry.) It is true that the study of neither Organic Chemistry nor Inorganic Chemistry is possible without a clear understanding of Physical Chemistry This book, targeted at students of classes XI and XII preparing for various competitive examinations, presents a logical and modern approach to the basic concepts of Physical Chemistry Although the historical approach is often believed to be the best way of inculcating a research outlook—ideally, the basis of all science teaching—we have purposely retained only those ideas and obsolete experimental procedures that provide a historical introduction or establish the experimental basis for current theories Physical Chemistry involves the theoretical interpretation of experimental observations However, all too frequently, students are unaware of the methods used in making these observations So, throughout the book, sections have been devoted to brief descriptions of the apparatus employed and discussions of the fundamental principles concerned in the measurement of various physical properties This will help students appreciate the logic of the various interpretations From experience, we know that combining the theory with its numerical application enables students to grasp the subject more easily Hence, many solved examples and problems for practice have been included in the text, with a clear emphasis on the questions that have appeared in several competitive examinations The intent is to give students a clear idea of the level of numerical questions that appear in competitive examinations Solutions to numerical questions are attached as an aid to the understanding of the principles of Physical Chemistry Physical Chemistry naturally starts with the basic concepts such as development of structure of matter, element, compound, mixtures, development of chemical combinations These are dealt with in the chapter ‘Basic Concepts of Chemistry’ The study of atom is one of the most outstanding advances in Physical Chemistry that brought about great changes in the presentation of facts of Inorganic as well as Organic Chemistry All the topics usually dealt with under the heading of Physical Chemistry will be found in this book It has been made up-to-date as much as possible by the inclusion of recent developments in the theories of several works K Rama Rao S V V Satyanarayana FM.indd 4/2/2013 12:16:06 PM Chapter Structure of Atom 1.1 iNTRODUCTION The introduction of atomic theory by John Dalton early in the nineteenth century marks the inception of a modern era in chemical thinking The virtue of Dalton’s theory was not that it was new or original, for theories of atoms are older than the science of chemistry, but that it represented the first attempt to place the corpuscular concept of matter upon a quantitative basis The theory of the atomic constitution of matter dates back at least 2,500 years to the scholars of ancient Greece and early Indian philosophers who were of the view that atoms are fundamental building blocks of matter According to them, the continued subdivision of matter would ultimately yield atoms which would not be further divisible The word ‘atom’ has been derived from the Greek word ‘a-tomio’ which means ‘uncuttable’ or non-divisible Thus, we might say that as far as atomic theory is concerned Dalton added nothing new He simply displayed a unique ability to crystallize and correlate the nebulous notions of the atomic constitution prevalent during the early nineteenth century into a few simple quantitative concepts 1.2 Atomic Theory The essentials of Dalton’s atomic theory may be summarized in the following postulates: All matter is composed of very small particles called atoms Atoms are indestructible They cannot be subdivided, created or destroyed Atoms of the same element are similar to one another and equal in weight Atoms of different elements have different properties and different weights Chemical combination results from the union of atoms in simple numerical proportions Chapter 01.indd T he universe is a concourse of atoms Marcus Aurelius John Dalton was born in England in 1766 His family was poor, and his formal education stopped when he was eleven years old He became a school teacher He was color blind His appearance and manners were awkward, he spoke with difficulty in public As an experiment he was clumsy and slow He had few, if any outward marks of genius From 1808, Dalton published his celebrated New Systems of Chemical Philosophy in a series of publications, in which he developed his conception of atoms as the fundamental building blocks of all matter It ranks among two greatest of all monuments to human intelligence No scientific discovery in history has had a more profund affect on the development of knowledge Dalton died in 1844 His stature as one of the greatest scientists of all time continues to grow Thus, to Dalton, the atoms were solid, hard, impenetrable particles as well as separate, unalterable individuals Dalton’s ideas of the structure of matter were borne out of considerable amount of subsequent experimental evidence as to the relative masses of substances entering into chemical combination Among the experimental results and relationship supporting this atomic theory were Gay-Lussac’s law of combination of gases by volume, Dalton’s law of multiple proportions, Avagadro’s hypothesis that equal volumes of gases under the same conditions contain the same number of molecules, Faraday’s laws relating to electrolysis and Berzelius painstaking determination of atomic weights Modern Atomic Theory Dalton’s atomic theory assumed that the atoms of elements were indivisible and that no particles smaller than atoms 4/2/2013 11:06:22 AM 1.2  Structure of Atom exist As a result of brilliant era in experimental physics which began towards the end of the nineteenth century extended into the 1930s paved the way to our present modern atomic theory These refinements established that atoms can be divisible into sub-atomic particles, i.e., electrons, protons and neutrons—a concept very different from that to Dalton The major problems before the scientists at that time were (i) How the sub-atomic particles are arranged within the atom and why the atoms are stable? (ii) Why the atoms of one element differ from the atoms of another elements in their physical and chemical properties? (iii) How and why the different atoms combine to form molecules? (iv) What is the origin and nature of the characteristics of electromagnetic radiation absorbed or emitted by atoms? 1.3.1 Discovery of Electron The term ‘electron’ was given to the smallest particle that could carry a negative charge equal in magnitude to the charge necessary to deposit one atom of a 1-valent element by Stoney in 1891 In 1879, Crookes discovered that when a high voltage is applied to a gas at low pressure streams of particles, which could communicate momentum, moved from the cathode to the anode It did not seem to matter what gas was used and there was strong evidence to suppose that the particles were common to all elements in a very high vacuum they could not be detected The cathode ray discharge tube is shown in Fig 1.1 1.3 Sub-Atomic Particles We know that the atom is composed of three basic subatomic particles namely the electron, the proton and the neutron The characteristics of these particles are given in Table 1.1 Table 1.1  The three main sub-atomic particles Particle Symbol Electron e Proton p Neutron n Mass Charge 1/1837 of H-atom or 9.109 ×10–28 g or 9.1×10–31 kg 1.008 amu or 1.672×10–24 g or 1.672×10–27 kg (1 unit) 1.0086 amu or 1.675×10–24 g or 1.675×10–27 kg (1 unit) – 4.8 × 10–10 esu or –1.602 × 10–19 coulombs (–1 unit) + 4.8 × 10–10 esu or + 1.602 × 10–19 coulombs (+ unit) No charge It is now known that many more sub-atomic particles exist, e.g., the positron, the neutrino, the meson, the hyperon etc, but in chemistry only those listed in Table 1.1 generally need to be considered The discovery of these particles and the way in which the structure of atom was worked out are discussed in this chapter Chapter 01.indd Fig 1.1  Cathode ray discharge tube The properties of these cathode rays are given below: (i) When a solid metal object is placed in a discharge tube in their path, a sharp shadow is cast on the end of the discharge tube, showing that they travel in straight lines (ii) They can be deflected by magnetic and electric fields, the direction of deflection showing that they are negatively charged (iii) A freely moving paddle wheel, placed in their path, is set in motion showing that they possess momentum, i.e., particle nature (iv) They cause many substances to fluoresce, e.g., the familiar zinc sulphide coated television tube (v) They can penetrate thin sheets of metal J.J Thomson (1897) extended these experiments and determined the velocity of these particles and their charge/ mass ratio as follows The particles from the cathode were made to pass through a slit in the anode and then through a second slit They then passed between two aluminium plates spaced about cm apart and eventually fell onto the end of the tube, producing a well-defined spot The position of the spot was noted and the magnetic field was then switched on, causing the electron beam to move in a circular arc while under the influence of this field (Fig 1.2.) 4/2/2013 11:06:22 AM Structure of Atom  1.3 Thomson proposed that the amount of deviation of the particles from their path in the presence of electrical and magnetic field depends on (i) Greater the magnitude of the charge on the particle, greater is the interaction with the electric and magnetic fields, and thus greater is the deflection (ii) Lighter the particle, greater the deflection (iii) The deflection of electrons from its original path increase in the voltage across the electrodes, or the strength of the magnetic field By careful and quantitative determination of the magnetic and electric fields on the motion of the cathode rays, Thomson was able to determine the value of charge to mass ratio as Fig 1.3  Millikan’s apparatus for determining the value of the electronic charge e = 1.758820 × 1011 C kg -1 (1.1) me me is the mass of the electron in kg and e is the magnitude of the charge on the electron in Coulomb 1.3.2 Charge on the Electron Thomson’s experiments show electrons to be negatively charged particles Evidence that electrons were discrete particles was obtained by Millikan by his well known oil drop experiment during the years 1910-14 By a series of very careful experiments Millikan was able to determine the value – electronic charge, and the mass Millikan found the charge on the electron to be –1.6 × 10–19 C The present-day accepted value for the charge on the electron is 1.602 × 10–19 C When this value for ‘e’ is compared with the most modern value of e/m, the mass of the electron can be calculated Small droplets of oil from an atomiser are blown into a still thermostated airspace between parallel plates, and the rate of fall of one of these droplets under gravity is observed, from which its weight can be calculated The airspace is now ionized with an X-ray beam, enabling the droplets to pick up charge by collision with the ionized air molecules By applying a potential of several thousand volts across the parallel metal plates, the oil droplet can either be speeded up or made to rise, depending upon the direction of the electric field Since, the speed of the droplet can be related to its weight, the magnitude of the electric field, and the charge it picks up, the value of the charge can be determined Anode (+) Cathode (–) Gas at low pressure me = 1.6022 × 10-19 C e (1.2) = e / me 1.758820 × 1011 C kg -1 = 9.1094 × 10–31 kg + (1.3) Spot of light when top plate is positive Spot of light when plate is not charged Deflecting plates – Fluorescent screen Fig 1.2  Thomson’s apparatus for determining e/m for the electron Chapter 01.indd 4/2/2013 11:06:23 AM 1.4  Structure of Atom 1.3.3 Discovery of Proton If the conduction of electricity, through gases is due to particles, which are similar to those involved during electrolysis, it was to be expected that positive as well as negative ones should be involved, and that they would be drawn to the cathode By using a discharge tube containing a perforated cathode, Goldstein (1886) had observed the formation of rays (shown to the right of the cathode in Fig 1.4.) noticed a particle of great penetrating power which was unaffected by magnetic and electric fields It was found to have approximately the same mass as the proton (hydrogen ion) The reaction is represented as Be + 42 He → 12 C +0 n Where the superscript refers to the atomic mass and the subscript refers to the atomic number (the number of protons in the nucleus) Notice that a new element, carbon, emerges from this reaction 1.4 Atomic Models Fig 1.4  J.J Thomson (1910) measured their charge/mass ratio from which he was able to deduce that the particles were positive ions, formed by the loss of electrons from the residual gas in the discharge tube The proton is the smallest positively charged particle equal in magnitude to that on the electron and is formed from the hydrogen atom by the loss of an electron The discovery that atoms contained electrons caused some consternation Left to themselves, atoms were known to be electrically neutral So, the negative charge of the electrons had to be balanced by an equal amount of positive charge The puzzle was to work out how the two types of charges were arranged To explain this, different atomic models were proposed Two models proposed by J.J Thomson and Earnest Rutherford are discussed here though they cannot explain about the stability of atoms 1.4.1 Thomson Model of Atom “A theory is a fool and not a creed.” J.J Thomson Sir Joseph John Thomson 1856-1940 Thomson’s researches on the discharge of electricity through gases led to the discovery of the electron and isotopes H → H+ + e– Unlike cathode rays, the characteristics of posi tively charged particles depend upon the nature of the gas present in the cathode ray tube These are positively charged ions The charge to mass ratio of these parti cles is found to be dependent upon the gas from which these originate Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge The behaviour of these particles in the magnetic or electric field is opposite to that observed for electron or cathode rays In 1898, Sir J.J Thomson proposed that the electrons are embedded in a ball of positive charge (Fig 1.5) This model of the atom was given the name plum pudding or raisin pudding or watermelon According to this model 1.3.4 Discovery of Neutron The neutron proved to be a very elusive particles to track down and its existence, predicted by Rutherford in 1920, was first noticed by Chadwick in 1932 Chadwick was bombarding the element beryllium with α-particles and Chapter 01.indd Fig 1.5  The Thomson model of atom The positive charge was imagined as being spread over the entire atom and the electrons were put in this background 4/5/2013 12:29:16 PM Structure of Atom  1.5 we can assume that just like the seeds of a watermelon are embedded within the reddish juicy material, the electrons are embedded in a ball of positive charge It is important to note that in Thomson’s model, the mass of the atom is uniformly distributed over the atom Though this model could explain the overall neutrality of the atom, but could not explain the results of later experiments The Discovery of X-rays and Radioactivity In 1895, Rontgen noticed when electrons strike a material in the cathode ray tubes, produces a penetrating radiation emitted from the discharge tubes, and it appeared to originate from the anode The radiation had the following properties: (i) It blackened the wrapped photographic film (ii) It ionized gases, so allowing them to conduct electricity (iii) It made certain substances fluoresce, e.g., zinc sulphide Furthermore the radiation was shown to carry no charge since it could not be deflected by magnetic or electric fields Since Rontgen did not know the nature of the radiation, he named them as X-rays and the name is still carried on The true nature of X-rays was not discovered until 1912, when it became apparent that its properties could be explained by assuming to be wave like in character, i.e., similar to light but is much smaller wavelength It is now known that X-rays are produced whenever fastmoving electrons are stopped in their tracks by impinging on a target, the excess energy appearing mainly in the form of X-radiation The year after Rontgen discovered X-rays, Henry Becquerel observed that uranium salts emitted a radiation with properties similar to those possessed by X-rays The Curies followed up this work and discov ered that the ore pitchblend was more radioactive than purified uranium oxide; this suggested that something more intensely radioactive than uranium was responsible for this increased activity and eventually the Curies succeeded in isolating two new elements called polonium and radium, which were responsible for this increased activity In 1889, Becquerel reported that the radiation from the element radium could be deflected by a magnetic field and in the same year, Rutherford noticed that the radiation from uranium was composed of at least two distinct types Subsequently, it was shown that the radiation from both sources contained three distinct components and are named as a – b and g-rays Rutherford found that a-rays consist of high energy particles carrying two units of Chapter 01.indd positive charge and four units of atomic mass He concluded that a-particles are helium nuclei as when a-particles combined with two electrons yielded helium gas b-rays are negatively charged particles similar to electrons The g-rays are high energy radiations like X-rays, are neutral in nature and not consist of particles As regards penetrating power, a-particles have the least followed by b-rays (100 times that of a-particles) and g-rays (1000 times that of a-particles) 1.4.2 Rutherford’s Nuclear Model of Atom Ernest Rutherford 1871-1937 Rutherford was born in New Zealand in 1871 He was educated at the university of New Zealand and at Cambridge university He taught at McGill University, and at the university of Manchester In 1919, he became director of the celebrated Cavendish laboratory at Cambridge In 1908, he received Nobel prize in chemistry Rutherford made many of the basic discoveries in the field of radioactivity With Bohr and others, he elaborated a theory of atomic structure In 1919, he produced the first artificial transmutation of an element (that of nitrogen into oxygen) For many years he was a vigorous leader in laying the foundations of the greatest developments in atomic science, which he did not live to see He died in 1937 In 1911, Earnest (later Lord) Rutherford demonstrated a classic experiment for testing the Thomson’s model Rutherford, Geiger and Harsden studied in detail the effect of bombarding a thin gold foil by high speed a-particle (positively charged helium particles) A thin parallel beam of a-particles was directed onto a thin strip of gold and the subsequent path of the particles was determined Since the a-particles were very energetic, it was thought they would go right through the metal foils To their surprise they observed some unexpected results which are summarized as follows: (i) It was observed that 99% of the alpha particles passed straight through the foil and struck the screen at the centre (ii) A few of the alpha particles deflected from their original path through varying angles (iii) Hardly one out of the 20, 000 a-particles was bounced back on its path The results of Rutherford’s experiment is represented more explicitly in Fig 1.6 4/5/2013 12:29:17 PM 1.72  Structure of Atom Matching Type Questions (a) s (a) rs (a) qs (a) p (a) qs (a) q (a) rs (a) qsrt (a) qrs 10 (a) s (b) p (b) ps (b) pq (b) pqrs (b) r (b) p (b) pqr (b) qst (b) p (b) q Integer Type Questions (c) q (c) ps (c) pqrs (c) p (c) pq (c) s (c) pqr (c) pqs (c) ps (c) pqs (d) r (d) qs (d) qr (d) pq (d) q (d) r (d) r (d) s (d) qr (d) prs Assertion (A) and Reason (R) Type Questions a a a Chapter 01.indd 72 b c b b b b a 11 a 12 a 10 625 547 10 11 12 13 14 15 5 17 18 19 20 21 Previous Years’ IIT Questions a b d c a a c d d 10 c 11 (a) r (b) q (c) p (d) s 12 (a) qr (b) pqrs (c) pqr (d) pq 13 14 15 16 17 13 c 14 c 15 b 4/2/2013 11:07:31 AM Structure of Atom  1.73 HINTS and solutions Multiple Choice Questions with Only One Answer Level I Mass of electron is less when compared to mass of positive rays r ø = e ao y Probability density means y2 10 λë = 52 Dχ = Dp h 4p Dp = \DV = h 2KEm 12 Angular speed = V r 16 Orbital angular momentum = l (l +1) h 2p h v1 = h vo + x 19 n = 3, l = So orbital is 3d v1 - vo = v2 - vo k kv1 - v2 k -1 -13.6 -13.6 E2 - E1 = 27 = 20 E3 - E2 -13.6 -13.6 21 l = 2πr = nλ 2πx = λ In third orbit 2π(9x) = 3λ \ λ = 6πx 66 When reaches nucleus kinetic energy converted to potential energy h v2 = hvo + kx vo = h m 4p h mv = p p v= m P2 \ mV = 2m 2 p Ze \ = m r \ If momentum is P KEm 6.625 × 10 3.31 × 10 -29 = -34 2( KE ) × 0.1 × 10 -5 -4 \ KE = × 10 J = 1.25×10-13×(64)1/3 =5×10-13cm π(5×10-13 )3 =1.25×10-13 Fraction of volume = -8 π (10 ) 31 Radius of nucleus 32 The probability of finding election in all directions are same P Ze = 2m r1 \4 Ze Ze = r r1 \r1 = r 80 The time taken by one revolution = \ Ratio = \1: 2p r V 2p r 2p (4r ) : -8 2.188 × 10 2.188 × 158 / 38 V = l h ml h \l = m \l = Chapter 01.indd 73 4/2/2013 11:07:33 AM 1.74  Structure of Atom Multiple Choice Questions with Only One Answer Level II Orbital angular momentum = l (l + 1) h For n 2p = 5l values are 0, 1, 2, 3, and these cannot satisfy the given condition (i) The no of peaks in a radial probability curve is equal to n – l So, no of peaks in 1s and 2s are not equal (ii) No of nodal planes equals to l, so it is same for 2p, 3p and 4p orbitals (iii) Ti2+ and Ni2+ both have same no of unpaired e– and have same magnetic moments (iv) Any orbital of any sub shell can accommodate only two e– +x M and M+y both have same magnetic moment 5.916 which means that both have same no (5) of unpaired e– It is possible only if x and y are +3 and +1 Electron configuration corresponding to these two oxidation states are M+x → 3s2 3p6 3d5 M+y → 4s1 3d6 So, M+x is more stable due to the half-filled configuration According to photoelectric effect eq hc hc + E where E = KE = λ λ0  l -l  E = hc    ll  de Broglie’s wavelength h , where M = mass of e– and E = KE 2mE or   h  l -l  2mhc    ll  = 1/ h (ll ) 2mhc(l - l)   hll =    2mhc(l - l)  From the observations no of nodal regions for that orbital n – l – = xy is a nodal plane So orbital is dyz Chapter 01.indd 74 Along x, y axis the shape of orbital is as shown below: (i) No of nodal planes = l = (ii) de Broglie’s wavelength l = h h = mv p (iii) The principal quantum no n indicates the size and energy of orbit (iv) p has electronic configuration 3s2 3p3 The length of a series depends on the difference between wavelength of successive lines It is more in Lyman series 13.6 × = 3.4 eV 42 Work function = 1.4 eV The additional energy supplied due to acceleration is eV So, net resultant KE = (3.4 – 1.4) + = eV 10 Energy given to the e– = 12.27 de Broglie’s wavelength = 11 Angular momentum of e– = V Å= 12.27 5Å nh 2p So the transition occurs from → 2; that means the transition lies in Balmer series and will be observed in the visible region 12 de Broglie’s wavelength = h 2meV where m is mass of particle, e is charge and V is stopping potential h k : l :l = So, p a 2m × e × × 4m × 2e × v p or l p : l a = p =2 :1 : 13 Energy of emitted photon 13.6   13.6 =  × - ×  eV   = × 13.6 eV \ KE of e– = [3 × 13.6 – 13.6] = 27.2 eV 4/2/2013 11:07:34 AM Structure of Atom  1.75 14 DX × DV = \ DV = h 4pm 6.625 × 10-34 × × 3.142 × 0.1505 10-10 @ 3.4 × 10-24 m/s 15 Six nodes are as shown: 35 So, it has 2.5 waves, so wavelength = = 14 cm 2.5 No of m values = 2l + So, no of m values for l = is This value is possible for m in case of f and higher sub shells only 17 T  o have the same de Broglie’s wavelength they must have the same momentum ⇒ Pp = Pe ⇒ mp × vp = me ve ⇒ 1837 me × vp = me ve So, vp = ve 1837 18 Let us suppose that ‘na’ is the no of quanta absorbed and ‘ne’ is the no of quanta emitted na and ne were related as ne = 0.53 na hc So, the absorbed energy Ea = na × 4530 Emitted energy Ee = ne × ⇒ hc 5080 ne hc 4530 0.53na ⋅ × na 5080 na hc = 0.4726 19 From the data no of e– in both cases is (n – 1) So, x has +1 charge and y has +2 charge 20 Energy involved in transition E1 = 2E – E = E = hc l 4E E –E= = Energy involved in transition E1 = 3 E1 hc / l1 l E \ = = ⇒ =3 E2 E / hc / l l1 Chapter 01.indd 75 21 From the photoelectric effect h(x) = hv0 + y(1) h(2x) = hv0 + 3y(2) from Eq (1) → y = hx – hv0 ⇒ h (2x) = hv0 + 3[h(x) – h(v0)] x ⇒ v0 = 22 From the data the difference between radii of ground state and excited state orbit radius ⇒ rn – r1 = 0.7935 × 10–9 m ⇒ rn – (0.0529 × 10–9 m) = 0.7935 × 10–9 m ⇒ rn = 0.8464 × 10–9 m = 0.0529 × 10–9 × n2 ⇒ n = So, the maximum spectral lines formed = (n2 - n1 )(n2 - n1 + 1) =6 23 F  or Balmer series n1 = and for the least wavelength n2 = ∞ and for the highest wavelength n2 = So, wave no for spectral line with least wavelength,  R 1 v1 = RH  -  = H ∞  2 wave no for spectral line highest wavelength  5R 1 v2 = RH  -  = H 36  2 So, v1 : v2 = RH RH : =9:5 36 24 No of nodal planes = l = No of Radial nodes = n – l – = – – = No of peaks = n – l = – = 25 No of revolutions made by one e– per sec = velocity of e - z @ × 6.66 × 1015 2prn n 26 From Rydberg’s equation 1  = R  -  l 1 n  ⇒ lR 1 = 1⇒ n2 = lR - lR n2 28 At radial node i.e., at r = r0, y2 =   1/  r0  - r0 / a0   =0 ⇒   2 - e a0   2p  a0    r ⇒ - = ⇒ r0 2a0 a0 4/2/2013 11:07:35 AM 1.76  Structure of Atom 29 In case of unielectron systems energy can be determined only from n values i.e., orbital of same shell have same energy 30 The energy of an e– in a multielectronic atom can be determined from (n + l) value 31 (i) Carbonates with thermal stability can be produced by alkali metals So, — V (ii) Colored compounds will be formed by transition elements So, — X (iii) Largest atomic radius is possible with maximum no of shells So, — Y (iv) Acidic oxides will be formed by non-metals So, — W 32 No of l values possible for first shell = (0, 1, 2) So, the total possible no of m values are nm = 9(1 + + 5) [No of m values for given l =2l +1] So, the no of e– that can be fitted = × n.18e– 33 According to the given rules Zr (40) has electronic configuration as follows; 1s3 2s3 2p9 3s3 3p9 3d13 35 l  is the wavelength corresponding to initial kinetic energy E1 l2 is the wavelength corresponding to final kinetic energy E2 and l2 = 0.9l1 l1 = ⇒ hc 2mE1 : l2 = l1 = = l 0.9 hc 2mE2 E2 E1 ⇒ E2 = 1.23456 E1 So, E2 is 23.456% is more than that of E1 nh h = Angular momentum = 2p 11 ⇒  = n = 0.529 × n Radius of nth orbit rn = ⇒  rn = 2.82 Å 37 T  he data indicates that the transition is only possible with Lyman, Balmer, Paschen series that means transition to 4th orbit is not possible which means that transition from 4th orbit occurs Chapter 01.indd 76 Maximum no of spectral lines possible = Σ (n – 1) = 15 ⇒  n = So, from Rydberg’s equation v= 1  = RH  -  l 1  l= 36 × 912 Å 35   = 912 Å    RH  = 938.05 n3 Z2 T Z n3 ⇒ = 22 × 13 T2 Z1 n2 [z is the same] 23 ⇒ = : 27 40 Velocity of an e– in nth orbit 39 Time period ∝ 2.17 × 106 m/s n From the data 2.17 × 106 × 108 = n ⇒ n = No of lines in Balmer series is – = = 41 l = h 2meV0 42 Wavelength of first line in Lyman series l1 = ⋅ × RH Z Wavelength corresponding to Balmer series l2 = 36 × × RH Z ⇒  l - l1 = ⇒  91.2 × Z2 1 ⋅ RH Z  36   - 4    129   20  = 33.4   ⇒  Z @ 43 From the data h (3.2 × 1016) = hv0 + 2E(i) h (2.5 × 1016) = hv0 + E(ii) By solving Eqs (i) and (ii) v0 = 18 × 1015 Hz 4/2/2013 11:07:36 AM Structure of Atom  1.77 44  hc = E2 – E1 l 6.625 × 10-34 × × 108 ⇒ l 53 Moseley’s equation is v = a (Z – b) So, from the data 100 = (Z – 1) ⇒ Z = 11  -13.786  -13.786   -  = 1.602 × 10–19     55 B  y the given condition, each primary shell should consist of (n + 1) subshells So, configuration will be 1s2 1p6 2s2 2p6 3s2 2d6 = l = 1.199 × 10–7 m = 1200 Å – th 45 Energy of e in n orbit of an H like system, En = -13.6 × Z eV/atom n2 l value for g sub shell is (i) Maximum no of m values = 2l + So, for g = (ii) Maximum no of unpaired e–1 = 2l + So, for g = (iii) The minimum principal quantum no of gshell = × = 1: 16 47 Electronic configuration in second excited state = 1s2 2s2 2p6 3s2 3p3 3d2 So, max (n + l + m) value for this configuration = [[(3 + +(–1)] + (3 + + 0) + (3 + + 1)] + (3 + + 2) + (3 + + 1)] = 25 48 In any orbit the no of waves by an e– equals to its principal quantum no and total length of all the waves must equal n circumference of the orbit ⇒ nl = 2prn ⇒ nl = 2pr (n2) – 51 Energy of e in ground state En = -2p2 Z me n2 n2 RH = Rydberg constant = 2p2 nZ e Ch3 hc l ⇒ l = hc 6.625 × 10-34 × × 108 = DE ( DEe.V ) × 1.602 × 10-19 = 1.2406 × 10–10 m Chapter 01.indd 77 59 Dx ⋅ Dv = h 4pm ⇒ Dx = 6.625 × 10-27 5.272 × 10-28 × = cm × 3.142 × 2 @ 2.64 × 10–30 m 60 From the data hc = hv0 + E1 l1 hc = hv0 + 2E1 l2 By substitution and on solving 1/4 and 3/4 ⇒  v0 = 1.19 × 1015 s–1 So, energy of e– in ground state configuration –RH × ch 52 DE = n -1 I = ∑ 2(2l + 1) i =0 E Z n2 So, = 12 × 22 E2 Z n1 = 57 N  o of e– corresponding for a given l value equals to (2l + 1) For a given n value the possible ‘l’ values are to n – So, total no of e– in a given orbit From the data n1 + n2 = and n1 – n2 = ⇒ n1 = and n2 = So, from Rydberg’s constant 1 1 = RH ⋅ Z  -  l  n1 n2  ⇒ 1  = 109677 ×  -  l 1  ⇒ l = 114 Å = 1.14 × 10–6 cm 4/2/2013 11:07:36 AM 1.78  Structure of Atom 62 If na is the number of absorbed quanta and ne is the no of emitted quanta, then n e = 0.5 na Emitted energy Ee is 47% of the absorbed energy Ea hc ne ⋅ X = 0.47 So,   hc na 4700 ⇒   4700 ne × = 0.47 X na ⇒   X = 4700 × 0.5 = 5000 Å 0.47 63 Before the excitation the molecule should undergo the homolytic cleavage So, energy required for dissociation Ed = 0.04 × 400 = 16 KJ On dissociation the sample consists of 0.08 moles of H gas atoms Energy required for excitation  -13.6  -13.6   = 0.08 × 6.023 × 1023  -     = 4.914768 × 1023 eV = 4.914768 × 1023 × 1.602 × 10-19 KJ 103 = 78.734 KJ So, total energy required = 16 + 78.734 = 94.7345 KJ 64 Angular momentum of e mvr = or – nh = 4.217 × 10-34 kg.m2/s 2p 65 From the data 1 1 RH × ×  -  = RH   By comparison 2   4 =n     2 ⇒ n2 = 2 6 de Broglie’s wavelength in nth orbit is nl = 2prn ⇒ nl = 2pr1n [for hydrogen] ⇒ l = 2pr1n = 2pr (x)3 ⇒ l = 6px 67 If w is angular velocity and r is radius of orbit, v is velocity of e– v = rw v ⇒ w = ∞ r n 68 We know that the no of waves made by an e– in an orbit equals to its principle quantum number ‘n’ So, from the figure n = No of revolutions made by an e– per sec (or) Frequency = = Z2 × 6.66 × 1015 s–1 n3 × 6.66 × 1015 = 1× 1014 s -1 (4)3 69 In case of an elliptical orbit, we know that length of major axis n = length of minor axis k And length of major axis is double the radius of that circular orbit nh 2p ⇒ n = 3.999 @ The no of spectral lines in visible region = n – = – 2=2 Chapter 01.indd 78 2   = n ⇒ n1 =   1 1  - 2  n1 n2  ⇒ × 16 × 0.529 = l ⇒ l = 8.464 Å 70 Ionization potential in H-like system means, it is the energy of e– in ground state E So, En = (0.01) E1 = 21 n ⇒ n = 10 73 DE = 12.1 = En2 - En1 4/2/2013 11:07:37 AM Structure of Atom  1.79 -13.6 - (-13.6) n2 So, 12.1 = ⇒ n = Therefore,  Dn = change in angular momentum = Magnetic moment m = h 2p -34 2p @ 2.11 × 10–34 J s 74 If E is KE of a particle, de Broglie wavelength h 2mE So, if KE doubles, l becomes times 1 1  - 2  n1 n2  – As e has double the mass in hypothetical system Rydberg’s constant in the equality will be equal to 2RH 75  = RH l 1 1 = R  -  [transition is to the first l   excited level i.e., to 2n orbit] So, ⇒ l = 18 5R 77 For, Hb line of Balmer series n1 = and n2 = E E So, hv = 21 - 21 n2 n1 [Here, E1 is –ve of ionization Energy]  -2 × 10-18  -2 × 10-18   ⇒ 6.000 × 10-34 × v =  -     16 ⇒ v = 6.25 × 1014 = 625 × 1012 Hz Radius of nth orbit rn = 0.529 × n2 Å ⇒ 8.464 Å = 0.529 × n2 ⇒ n = Velocity of e– in nth orbit = @ n(n + 2) = 4(4 + 2) = 4.89 BM ( - 1) × 6.625 ×10 l= below: Cr → 1s2 2s2 2p6 3s2 3p6 4s1 3d5 Cr+2 → 1s2 2s2 2p6 3s2 3p6 3d4 So, no of unpaired e– = 2.18 × 106 m/s n 2pZe nh ⇒ Velocity of e–1 in 4th orbit = 545 × 103 m/s Electronic configuration of Fe2+ = 1s2 2s2 2p6 3s2 3p6 3d6 So, no of d e– in Fe2+ = Configuration of Ne — 1s2 2s2 2p6 ” ” Mg — 1s2 2s2 2p6 3s2 ” ” Cl– — 1s2 2s2 2p6 3s2 3p6 de Broglie e’s wavelength ln = 2pr = nl So, l1 2pr1 = l 2pr1 × l = 4l1 = 2pr4 ⇒ Circumference of the fourth orbit = 4l4   = 16l1 83  Shortest wavelength is possible for any series if n2 = ∞ Longest wavelength is possible for any series if n2 = n1 + So, for hydrogen 11  1 = RH  -  Xx 1 ∞  ⇒ RH = (i) x For helium 1 1 = RH ⋅ Z  -  (ii) l   From Eqs (i) and (ii), 9x l= (a) In H-like systems, energy of electron in an orbital increases with increase in ‘n’ So, energy of 3d is less than energy of 4s (b) In d x2 - y orbitals cloud is along x and y-axes So, it lies in xy plane (c) Electronic configuration of Cr — 1s2 2s2 2p6 3s2 79 Electronic configuration of Cr and Cr2+ are as given Chapter 01.indd 79 4/2/2013 11:07:38 AM 1.80  Structure of Atom 3p6 4s1 3d5 (d) Except the spin quantum no the remaining Quantum number are obtained from Schrodinger equation 86 From the data = Velocity of e -1 = Velocity of light 275 ⇒ V = 1090909 m/s Velocity of e– in nth orbit Vn = n=2 87 If three spin quantum no were possible, then each orbital can accommodate 3e– So, electronic configuration of K is 1s3 2s3 2p9 3s3 3p1 From the data r(n + 1) – rn = r(n – 1) ⇒ [(0.529) (n + 1)2] – [0.529 × n2] = (0.529)(n – 1)2 ⇒ n2 + 2n + – n2 = n2 – 2n + ⇒ n2 = 4n ⇒ n = 95 At nodal surface, y2 value must be equal to zero By equating y2= we will get the expresion as (2 – Zr) = 89 For the line in ultraviolet region, the energy must be more than that corresponding to E Energy of A > Energy of E 90 In the given sub shell, half of the max no of e– in sub shell will have the same spin quantum number 91 To find the distance at which finding probability d y = dr 2 of e– is maximum, we have to count ⇒ e - r / a0 = ⇒ – In (r / a0 ) = – In ⇒ r = a0 = 6.625 × 10-34 × × 108 – (2 × 1.602 × 10–19) 310 × 10-9 h 2mE So, lH : lHe : l CH4 = Chapter 01.indd 80 1 Equation for photoelectric effect hv = hv0 + KE ⇒ v = v0 × ⋅ KE h or KE = hv – hv0 y = mx – c So, slope of the curve m = h h 2meV 98 Velocity of e– = 2.18 × 106 m/s n So, graphic representation must be a hyperbola curve 99  v = a (z – b) -mv mv mv :+ : 2r r 2r l= 2 = =1Å Z from the curve, we can say that at any point lB > lA, means that mB < mA hc – Work function l : –1 : If magnitude is only considered, ratio, : : (iii) No of maximum (or) no of peaks = n – l = – = (iv) In terms of KE (E) de Broglie’s wavelength ⇒ r = 97 de Broglie’ wavelength l = = (2 × 10–19) (ii) Total energy : Kinetic energy : Potential energy 1 : =4:2:1 94 Orbital can be represented in terms of wave function as ‘ynlm’ and we know that in general, value for m can be assigned for orbital along with z axis 2.18 × 106 m/s ⇒ n 92  (i) KE = 1: : : v = az – ab a = slope of the curve = tan 45° = ab = Intercept on the axis = ⇒ v = 52 – ⇒ v = 2601 s–1 Multiple Choice Questions with One or More Than One Answer (a, c) From the data we can write the expression as 16 4/2/2013 11:07:38 AM Structure of Atom  1.81 follows: 4.25 = WA + TA (i) 4.20 = WB + TB (ii) TB = TA – 1.5 (iii) h 2h = 2mTB 2mTA ⇒  TA = 4TB (iv) So, from above equations TB = 0.5 eV TA = eV WA = 2.25 eV WB = 4.15 eV hv = hv0 + KE and stopping potential is equal to K.E ⇒ hc hc = + v0 l l0 v0 = v0( eV ) = y = mx – c hc - hc ⋅ l l0 Passage II hc hc ⋅ - ⋅ e l e l0 So, slope = m = 6.625 × 10-34 × × 108 1.602 × 10-19 Work function for metal = 0.24 eV hc = w + v0 l ⇒ v0 = -34 6.625 × 10 × × 10 100 × 10-9 –0.24 × 1.602 × 10–19 = 1.949 × 10–18 J = 12.1 eV The work function for metal II is more than metal I So, if both metals were exposed to same light, KE i.e., stopping potential for metal II is less than metal I (a, c) For the same magnetic moment ions (a) atoms should contain same no of unpaired e– 16 BE = Ionization energy of H-like system = Based on the given conditions, No of sub shells = n + 1, and No of orbitals in a sub shell = 2l + That means 1st orbit consists of s and p sub shells with and orbitals and 2nd orbit consists of s, p and d sub shells with 3, and orbitals, respectively EC with atomic no 25 1s2 1p10 2s6 2p3 EC corresponding to atomic No 18 is 1s6 1p10 2s2, so it has two unpaired e– which causes a magnetic moment of BM After 3s, 3p and 2d have the same (n + l) values So, because of lower n value e– enters into 2d If three spin quantum no were possible each orbital can accommodate electrons, So, configuration of atom with atomic no is 1s9 1p15 2s6 13.6 13.6 × Z = × Z = 13.6 n2 Because of half of the mass of –ve particle in the hypothetical system Energy, velocity, radius, expression were as follows: -2p2 Z me E E= , So Ehypothetical = 2 nh V= 2pZe , So Vhypothetical = V nh r= n2 h2 , So rh = 2r 4p2 me Rydberg’s constant, R = Longest wavelength is possible if transition occurs from the immediate orbit So, the given transition is → So, = = Passage I Chapter 01.indd 81 1 1 = Rn  -  l   R 9 - 4 72 R ⇒ l=  ×  Vn = V = ⇒ Z = Comprehensive Type Questions 2p2 mZ e R ; So, Rh = ch3 2.188 × 106 m/s n 2.188 × 106 = 1.094 × 106 m/s r = rn × n2 = (2r) n2 = 0.529 × × = 9.522 Å  We know that KE and total energy are with 4/2/2013 11:07:39 AM 1.82  Structure of Atom opposite sign and are of equal magnitude So, KEn = – En = E 13.6 = = 6.8 eV/atom 2 Passage III Passage VI We know that E3 – E2 = -13.6 × 22  -13.6  - ×Z2    1 1 So, 47.2 = 13.6 ×  -  ⇒ Z = 4 9  hc = (–13.6 × 25 × 1.602 × 10–19) l ⇒ l = 3.6489 × 10–9 m KE = – E = – (13.6 × 1.602 × 10–12) ergs = 5.4468 × 10–9 ergs Passage IV From the data A is the 1st orbit and B is nth orbit After absorption of 2.55 eV, e–1 excites to n1 orbit The possible no of spectral lines for de-excitation is Σ (n1 – 1) = Which mass that n1 = and n may be (or) From the given data En1 - En = 2.55 -13.6  -13.6  × Z2 –   n2 = 2.55 i.e., 16  n  The above expression is possible only if n = and Z = Minimum energy is possible due to transition of e– from 4th orbit to 3rd orbit E=- 13.6  -13.6  -  = 0.6611 V   Passage V hc  = W + KE l E = hc 6.625 × 10-34 × × 102 = 6.625 × 10–19 J = l 3000 × 10-80 Passage VII de Broglie wavelength of a cricket ball = h 6.625 × 10-34 = = 1.1 × 10–34 m mV 0.2 × 30 DP =  6.625 × 10 × × 10  × eV ⇒  -10 -19  4000 × 10 1.602 × 10   ⇒ + KE ⇒ KE = 1.10 eV KE1 = hv1 – hv0 = h (10 × 1014 – × 1014) = h (5 × 1014) In second case KE2 = 2KE1 ⇒ h (10 × 1014) = hv1 – × 1014 × h ⇒ v1 = 15 × 1014 s–1 h 6.625 × 10-34 = 5.272 × 10–2 = 4p ⋅ Dx 4p× 10-10 Dx = DP = m × DV So, according to Heisenberg’s principle h Dx × DP = m2 (DV)2 = 4p h ⇒ DV = 2m 2p Passage VIII Longest wavelength corresponding to the transition (n + 1) → n, so in case of Lyman series the transition is → So, 1 1 = RH ⋅ Z  -  l  n1 n2  = RH l ⇒l = -34 Chapter 01.indd 82 Stopping potential is the –ve potential that must be applied on collection plate to stop photoelectrons It will be equal to KE of e– in (eV.) So, V0 = 25 – = 18 eV 1  ⋅ -  1 a  912 = Å = 22.8 nm RH  Shortest wavelength corresponds to the transition of ∞ → n So, 1  R 1 = RH  -  = H l1 2 ∞  For the longest wavelength  1  3R = RH  -  = H l 1  4/2/2013 11:07:39 AM Structure of Atom  1.83 ⇒ l = l1 = 0.33 l1 Passage IX The data can be represented in the form of a diagram as follows: r n2 Z = 12 × r2 n2 Z1 V Z n = × V2 Z n1 T n3 Z = 13 × 22 T2 n2 Z1 ⇒  K1 n22 = K n12 Assertion (A) and Reason (R) Type Questions So,  En – E2 = (10.2 + 17) ⇒ -13.6  -13.6  × Z -   Z = 27.2 n2   Similarly, -13.6  -13.6  × Z -   Z = 10.2 n   By solving Eqs (i) and (ii), n = 6, Z = Passage X Radius r = n2 h2 , So for hypothetical, it can be 4p2 mZe modified as rm = n2 h2 0.529 = Å 414 4p2 (207 m) (2e) Ionization energy is the energy of e– in first orbit with +ve sign E= -2p2 Z me for hypothetical n2 h2 Em = = 828 × 13.6 eV -2p2 Z (207 m) (2e) n2 h2 0.529 2 0.529 3 -  = r3 – r1 = ×8 Å 414  414 Matching Type Questions Note: Molecules at very high temperature are monoatomic gases that gives a line-spectrum We know that Chapter 01.indd 83 Peaks in radial distribution curve = n – l radial nodes = n – l – Which means that the radial probability curves depend only on ‘n’ and ‘l’ and not on ‘m’ An electron can never be found in the nucleus and this can be explained by uncertainty principle Any object can not move with velocity of light Reason is the consequence of assertion and it can not explain assertion The shape of orbital is independent of principal quantum number Anode rays are +ve ions of different gases and they have different charges and masses So, e/m ratio for anode rays depends on nature of gas y2 is the square of amplitude of e– wave and we know that I ∝ a2 where I is intensity and a is amplitude The no of angular nodes = l and angular wave functions depend on ‘I’ and m Diamagnetic nature is due to pairing of e– 10 In the absence of electric and magnetic fields px, py and pz orbitals will have same energy and transition in between px and py can not give spectral line 11  In case of H like systems, energy of e– can be determined only from ‘n’ values and all the sub shells with same ‘n’ value will have same energy 12 The electron can revolve in a particular orbit if it’s wave is in phase only For the He, condition is nl = 2pr which can be modified as mvr = nh 2p 14 +  (or) – (ve) sign indicates that electrons are with opposite spins only 15 As the orbit moves towards the nucleus, the energy difference between successive orbits increases Integer Type Questions After the absorption of energy, energy of e– = –0.01 E -E So –0.01 E = n 4/2/2013 11:07:40 AM 1.84  Structure of Atom ⇒ n = 10 Hb line is due to the transition of e– from to -2 × 10-18  -2 × 10-18  -  16   v = 6.25 × 1014 or 625 Radius of orbit rn = 0.529 × n2 = 0.8464 ⇒ n = So, hvz = Velocity of e– = 2.188 × 106 m/s n = 547.00 m/s Velocity of e– 2.188 × 106 × 108 = n 275 ⇒ n = 1 1 = RH  -  l n  2 1 1  = 4814 912  n  ⇒ n = ⇒ If l is de Broglie’s wavelength and n is principal quantum no, nl = 2prn = 2p × 0.529 × n2 ⇒ l = 2p × 0.529 × n ⇒ n = Minimum value of magnetic quantum no = –l So, ‘l’ for the given orbital is No of angular nodes = l = 3rd excited state means 4th orbit So, spectral lines in Lyman series = n – = – = 10 Max no of spectral lines = Σ (n2 –n1) (6 – 1) = 11 hc = En2 - En1 l hc 1  = Z  -  × 2.178 × 10–8 l 1  ⇒ Minimum possible number for Z = 12 En2 - En1 = 1   = 6.625 × 10–34 × × 108  + -9 -9  108.5 10 30.4 10 × ×   ⇒ n = 13 hc = w + eV0 l ⇒ w = eV 14 Degeneracy means the total no of degenerate orbitals In H-like system all the orbitals present in a particular orbit will have the same energy So, no of degenerate orbitals (d) degeneracy n2 = 15 No of visible lines means, lines in Balmer series, =n–2=5–2=3 16 16.2 17 l = 2pr1n 6.64 =2 ⇒ n = 2p× 0.529 18 In second period 2s, 2p orbital are to be filled with three e– in each orbital So, total no of elements in second will be 12 19 No of revolutions (a) frequency f∝ z2 n3 So, f1 Z12 n23 = × f Z 22 n13 ⇒ 23 = × ⇒n=1 n2 From the data KE of e– = 13.6 × 0.5 eV ⇒ mv = 13.6 × 0.5 × 1.602 × 10–19 ⇒ v = 1.5473 × 106 So, the correct answer is 21 (6) Previous Years’ IIT Questions Smaller is the value of (n+l), lesser is the energy of sub shell When (n+l) values are same then energy of sub shell depends on the principal quantum number ‘n’; smaller will be the value of ‘n’ lesser is the energy of sub shell hc hc + l1 l  13.6  -13.6   × 1.602 × 10-19  ⇒  - -      n  Chapter 01.indd 84 4/2/2013 11:07:40 AM Structure of Atom  1.85 ( iv)  n = 3, l = < (ii)  n = 4, l = < (iii)  n = 3, l = < (i)  n = 4, l = px orbital has only one nodal plane i.e, yz-plane 1s2, 2s22p6; 3s23p63d5, 4s1 is ground state electronic configuration of Cr24 l = h 6.626 × 10-34 = mv 0.2 × / 3600 Vn =−2 Kn rn ∝ En−1 In the lowest orbit (1s), the angular momentum h = l (l + 1) 2π = ( for l = 0) rn = Beam of α-particles (nuclei of helium) are used in the Rutherford’s gold leaf experiment An orbital can accommodate maximum two electrons with opposite spin and no two electrons in the atom can have same values of all four quantum numbers r = n2 × 0.529 Å z For Be3+, n = 2, z = \ r = 0.529 Å Number of radial nodes = (n-l-1) Number of radial nodes for 3s = (3-0-1) = Number of radial nodes for 2p = (2-1-1) = 2p e m h e m p m = = e 4h 2h 2h h2 ao = 2 4h e m h4 \ e4 = 16p m a h2 \ KE = 32p ma 10  KE = Matching Type Questions 11 (a-r) (b-q) (c-p) (d-s) − KZe ; E n ( Total energy ) = 2r KZe Z2 K n ( Kinetic energy ) = − × 2.18 × 10−18 J = 2r n KZe Vn = potential energy = − r n2 rn = Radius of Orbit = × 0.5529 Å Z Vn =−2 Kn rn ∝ En−1 n2 × 0.529 Å Z ∝ Z y i.e , ∝ Z i.e., y = rn 12 Orbital angular momentum = l (l + 1) h cos θ 2p i.e., it depends on azimuthal quantum number, magnetic quantum number Shape, size and orientation of hydrogen like atomic orbitals is determined by principal, azimuthal and magnetic quantum numbers Probability density of electron at the nucleus is determined by the principal and azimuthal quantum number Comprehensive Type Questions 13 2s orbital is spherically symmetrical and it has one radial node Number of radial nodes of 2s orbital = n-l-1=2-0-1=1 Z2 32 × 13.6 eV = - × 13.6 eV = - × 13.6 eV n 12 EH = - × 13.6 eV ES1 \ = = 2.25 EH 14 ES1 = - 15 ES2 = EH Z2 × 13.6 eV = - 13.6 eV n2 Z = n2 - Z = n = for Li + Thus,S2 will be 3p because it has one radial node For p, n = 3, l = Number of radial node = n - l - = - - = In the lowest orbit (1s), the angular momentum h 2π = ( for l = 0) = l (l + 1) Chapter 01.indd 85 rn = n2 × 0.529 Å 4/2/2013 11:07:42 AM 1.86  Structure of Atom Integer Type Questions Total number of electrons with (n = 3) = 2n2 = 18 Out of these 18 electrons, electrons will have 1  anticlockwise spin  ms = -  2  Chapter 01.indd 86 17 Photoelectrons are ejected only when the energy of absorbed quantum is greater than the threshold energy or work function (Φ) of metal hc 6.626 × 10 -34 × × 108 Absorbed Energy, E = = l 300 × 10 -9 6.626 × 10 -19 = 6.626 × 10 -19 J = eV = 4.17eV 1.6 × 1019 Thus, four metals, i.e., Li, Na, K, Mg will show photoelectron emission 4/2/2013 11:07:42 AM .. .Physical Chemistry for the JEE and Other Engineering Entrance Examinations K Rama Rao S.V.V Satyanarayana Delhi Ÿ Chennai FM.indd 4/2/2013 12:16:06... spread along all the directions In other words, the distribution of charge is nonuniform or unsymmetrical On the other hand, the distribution of charge in the configurations p1x p1y p1z and px2 p y2... electrostatic forces of attraction Rutherford, therefore, contemplated the dynamical stability and imagined the electrons to be whirling about the nucleus, similar to the planetary motion Drawbacks of the