2.3 Developments Leading to the that when electrically charged particle moves Bohr’s Model of Atom under accelaration, alternating electrical and magnetic fields are produced and transmitted.Historically, results observed from the studies These fields are transmitted in the formsof interactions of radiations with matter have of waves called electromagnetic waves orprovided immense information regarding electromagnetic radiation.the structure of atoms and molecules. Neils Light is the form of radiation known fromBohr utilised these results to improve upon early days and speculation about its naturethe model proposed by Rutherford. Two dates back to remote ancient times. In earlierdevelopments played a major role in the days (Newton) light was supposed to be madeformulation of Bohr’s model of atom. These of particles (corpuscules). It was only in thewere: 19th century when wave nature of light was(i) Dual character of the electromagnetic established. radiation which means that radiations Maxwell was again the first to reveal that possess both wave like and particle like light waves are associated with oscillating properties, and electric and magnetic character (Fig. 2.6).(ii) Experimental results regarding atomic spectra. First, we will discuss about the duel nature of electromagnetic radiations. Experimental results regarding atomic spectra will be discussed in Section 2.4. 2.3.1 Wave Nature of Electromagnetic Radiation In the mid-nineteenth century, physicists actively studied absorption and emission of radiation by heated objects. These are called Fig.2.6 The electric and magnetic field thermal radiations. They tried to find out of components of an electromagnetic what the thermal radiation is made. It is now wave. These components have the a well-known fact that thermal radiations same wavelength, frequency, speed consist of electromagnetic waves of various and amplitude, but they vibrate in two frequencies or wavelengths. It is based on mutually perpendicular planes. a number of modern concepts, which were Although electromagnetic wave motion isunknown in the mid-nineteenth century. complex in nature, we will consider here onlyFirst active study of thermal radiation laws a few simple properties.occured in the 1850’s and the theory of (i) The oscillating electric and magneticelectromagnetic waves and the emission of fields produced by oscillating chargedsuch waves by accelerating charged particles Reprint 2025-26 38 chemistry particles are perpendicular to each (iv) Different kinds of units are used to other and both are perpendicular to the represent electromagnetic radiation. direction of propagation of the wave. These radiations are characterised by Simplified picture of electromagnetic wave is shown in Fig. 2.6. the properties, namely, frequency (ν) and wavelength (λ).(ii) Unlike sound waves or waves produced in water, electromagnetic waves do The SI unit for frequency (ν) is hertz not require medium and can move in (Hz, s–1), after Heinrich Hertz. It is defined as vacuum. the number of waves that pass a given point (iii) It is now well established that there in one second. are many types of electromagnetic Wavelength should have the units of radiations, which differ from one length and as you know that the SI units of another in wavelength (or frequency). These constitute what is called length is meter (m). Since electromagnetic electromagnetic spectrum (Fig. 2.7). radiation consists of different kinds of waves Different regions of the spectrum are of much smaller wavelengths, smaller units identified by different names. Some are used. Fig. 2.7 shows various types of examples are: radio frequency region electro-magnetic radiations which differ from around 106 Hz, used for broadcasting; one another in wavelengths and frequencies. microwave region around 1010 Hz used for radar; infrared region around 1013 In vaccum all types of electromagnetic Hz used for heating; ultraviolet region radiations, regardless of wavelength, travel at around 1016Hz a component of sun’s the same speed, i.e., 3.0 × 108 m s–1 (2.997925 radiation. The small portion around 1015 × 108 ms–1, to be precise). This is called speed Hz, is what is ordinarily called visible of light and is given the symbol ‘c’. The light. It is only this part which our eyes frequency (ν ), wavelength (λ) and velocity of can see (or detect). Special instruments light (c) are related by the equation (2.5). are required to detect non-visible radiation. c = ν λ (2.5) (a) (b) Fig. 2.7 (a) The spectrum of electromagnetic radiation. (b) Visible spectrum. The visible region is only a small part of the entire spectrum. Reprint 2025-26 structure of atom 39 The other commonly used quantity Frequency of red light specially in spectroscopy, is the wavenumber ( ). It is defined as the number of wavelengths per unit length. Its units are reciprocal of ν = = 4.00 × 1014 Hz wavelength unit, i.e., m–1. However commonly The range of visible spectrum is fromused unit is cm–1 (not SI unit). 4.0 × 1014 to 7.5 × 1014 Hz in terms of frequency units. Problem 2.3 The Vividh Bharati station of All India Problem 2.5 Radio, Delhi, broadcasts on a frequency Calculate (a) wavenumber and (b) of 1,368 kHz (kilo hertz). Calculate frequency of yellow radiation having the wavelength of the electromagnetic wavelength 5800 Å. radiation emitted by transmitter. Which part of the electromagnetic spectrum Solution does it belong to? (a) Calculation of wavenumber ( ) Solution λ=5800Å = 5800 × 10–8 cm = 5800 × 10–10 m The wavelength, λ, is equal to c/ν, where c is the speed of electromagnetic radiation in vacuum and ν is the frequency. Substituting the given values, we have c v (b) Calculation of the frequency (ν ) 2.3.2 Particle Nature of Electromagnetic Radiation: Planck’s Quantum Theory This is a characteristic radiowave wavelength. Some of the experimental phenomenon such as diffraction* and interference** can Problem 2.4 be explained by the wave nature of the The wavelength range of the visible electromagnetic radiation. However, following spectrum extends from violet (400 nm) to are some of the observations which could red (750 nm). Express these wavelengths not be explained with the help of even the in frequencies (Hz). (1nm = 10–9 m) electromagentic theory of 19th century Solution physics (known as classical physics): Using equation 2.5, frequency of violet (i) the nature of emission of radiation from light hot bodies (black-body radiation) (ii) ejection of electrons from metal surface when radiation strikes it (photoelectric effect) = 7.50 × 1014 Hz (iii) variation of heat capacity of solids as a function of temperature * Diffraction is the bending of wave around an obstacle. ** Interference is the combination of two waves of the same or different frequencies to give a wave whose distribution at each point in space is the algebraic or vector sum of disturbances at that point resulting from each interfering wave. Reprint 2025-26 40 chemistry (iv) Line spectra of atoms with special entering the hole will be reflected by the cavity reference to hydrogen. walls and will be eventually absorbed by the walls. A black body is also a perfect radiator of These phenomena indicate that the system radiant energy. Furthermore, a black body iscan take energy only in discrete amounts. in thermal equilibrium with its surroundings.All possible energies cannot be taken up or It radiates same amount of energy per unitradiated. area as it absorbs from its surrounding in It is noteworthy that the first concrete any given time. The amount of light emitted explanation for the phenomenon of the black (intensity of radiation) from a black body body radiation mentioned above was given and its spectral distribution depends only by Max Planck in 1900. Let us first try to on its temperature. At a given temperature, understand this phenomenon, which is given intensity of radiation emitted increases below: with the increase of wavelength, reaches a Hot objects emit electromagnetic maximum value at a given wavelength and radiations over a wide range of wavelengths. then starts decreasing with further increase of At high temperatures, an appreciable wavelength, as shown in Fig. 2.8. Also, as the proportion of radiation is in the visible temperature increases, maxima of the curve region of the spectrum. As the temperature shifts to short wavelength. Several attempts is raised, a higher proportion of short were made to predict the intensity of radiation wavelength (blue light) is generated. For as a function of wavelength. example, when an iron rod is heated in a But the results of the above experiment furnace, it first turns to dull red and then could not be explained satisfactorily on progressively becomes more and more red the basis of the wave theory of light. Max as the temperature increases. As this is Planck arrived at a satisfactory relationship heated further, the radiation emitted becomes white and then becomes blue as the temperature becomes very high. This means that red radiation is most intense at a particular temperature and the blue radiation is more intense at another temperature. This means intensities of radiations of different wavelengths emitted by hot body depend upon its temperature. By late 1850’s it was known that objects made of different material and kept at different temperatures emit different amount of radiation. Also, when the surface of an object is irradiated with light (electromagnetic radiation), a part of radiant energy is generally reflected as such, a part Fig. 2.8 Wavelength-intensity relationship is absorbed and a part of it is transmitted. The reason for incomplete absorption is that ordinary objects are as a rule imperfect absorbers of radiation. An ideal body, which emits and absorbs radiations of all frequencies uniformly, is called a black body and the radiation emitted by such a body is called black body radiation. In practice, no such body exists. Carbon black approximates fairly closely to black body. A good physical approximation to a black body is a cavity with a tiny hole, which has no other opening. Any ray Fig. 2.8(a) Black body Reprint 2025-26 structure of atom 41 by making an assumption that absorption and emmission of radiation arises from Max Planck (1858–1947)oscillator i.e., atoms in the wall of black Max Planck, a German physicist,body. Their frequency of oscillation is received his Ph.D in theoreticalchanged by interaction with oscilators of physics from the University ofelectromagnetic radiation. Planck assumed Munich in 1879. In 1888, he that radiation could be sub-divided into was appointed Director of the discrete chunks of energy. He suggested that Institute of Theoretical Physics atoms and molecules could emit or absorb at the University of Berlin. energy only in discrete quantities and not Planck was awarded the Nobel Prize in Physics in a continuous manner. He gave the name in 1918 for his quantum theory. Planck also made quantum to the smallest quantity of energy significant contributions in thermodynamics and that can be emitted or absorbed in the form other areas of physics. of electromagnetic radiation. The energy (E) of a quantum of radiation is proportional Photoelectric Effect to its frequency (ν) and is expressed by In 1887, H. Hertz performed a very interesting equation (2.6). experiment in which electrons (or electric E = hυ (2.6) current) were ejected when certain metals (for example potassium, rubidium, caesium The proportionality constant, ‘h’ is known etc.) were exposed to a beam of light as shownas Planck’s constant and has the value in Fig. 2.9. The phenomenon is called6.626×10–34 J s. Photoelectric effect. The results observed With this theory, Planck was able to explain in this experiment were: the distribution of intensity in the radiation (i) The electrons are ejected from the metalfrom black body as a function of frequency or surface as soon as the beam of lightwavelength at different temperatures. strikes the surface, i.e., there is no time Quantisation has been compared to lag between the striking of light beam andstanding on a staircase. A person can stand the ejection of electrons from the metalon any step of a staircase, but it is not possible surface.for him/her to stand in between the two steps. The energy can take any one of the values (ii) The number of electrons ejected is from the following set, but cannot take on any proportional to the intensity or brightness values between them. of light. E = 0, hυ, 2hυ , 3hυ....nhυ..... (iii) For each metal, there is a characteristic minimum frequency, ν0 (also known as threshold frequency) below which photoelectric effect is not observed. At a frequency ν >ν0, the ejected electrons come out with certain kinetic energy. The kinetic energies of these electrons increase with the increase of frequency of the light used. All the above results could not be explained on the basis of laws of classical physics. According to latter, the energy content of theFig.2.9 Equipment for studying the photoelectric effect. Light of a particular frequency beam of light depends upon the brightness of strikes a clean metal surface inside a the light. In other words, number of electrons vacuum chamber. Electrons are ejected ejected and kinetic energy associated with from the metal and are counted by a detector that measures their kinetic them should depend on the brightness of light. energy. It has been observed that though the number Reprint 2025-26 42 chemistry Table 2.2 Values of Work Function (W0) for a Few Metals Metal Li Na K Mg Cu Ag W0 /eV 2.42 2.3 2.25 3.7 4.8 4.3 of electrons ejected does depend upon the the minimum energy required to eject the brightness of light, the kinetic energy of the electron is hν0 (also called work function, ejected electrons does not. For example, red W0; Table 2.2), then the difference in energy light [ν = (4.3 to 4.6) × 1014 Hz] of any brightness (hν – hν0 ) is transferred as the kinetic energy of (intensity) may shine on a piece of potassium the photoelectron. Following the conservation metal for hours but no photoelectrons are of energy principle, the kinetic energy of the ejected. But, as soon as even a very weak ejected electron is given by the equation 2.7. yellow light (ν = 5.1–5.2 × 1014 Hz) shines on the potassium metal, the photoelectric effect (2.7) is observed. The threshold frequency (ν0) for where me is the mass of the electron and v is the potassium metal is 5.0×1014 Hz. velocity associated with the ejected electron. Einstein (1905) was able to explain the Lastly, a more intense beam of light consists photoelectric effect using Planck’s quantum of larger number of photons, consequently the theory of electromagnetic radiation as a number of electrons ejected is also larger as starting point. compared to that in an experiment in which a beam of weaker intensity of light is employed. Albert Einstein, a German born American physicist, is regarded Dual Behaviour of Electromagnetic by many as one of the two great Radiation physicists the world has known (the other is Isaac Newton). His The particle nature of light posed a dilemma for three research papers (on special scientists. On the one hand, it could explain relativity, Brownian motion and the black body radiation and photoelectric the photoelectric effect) which Albert Einstein effect satisfactorily but on the other hand, he published in 1905, while he (1879–1955) was employed as a technical it was not consistent with the known wave assistant in a Swiss patent office in Berne have behaviour of light which could account for the profoundly influenced the development of physics. phenomena of interference and diffraction. He received the Nobel Prize in Physics in 1921 for The only way to resolve the dilemma was his explanation of the photoelectric effect. to accept the idea that light possesses both particle and wave-like properties, i.e., Shining a beam of light on to a metal light has dual behaviour. Depending onsurface can, therefore, be viewed as shooting the experiment, we find that light behavesa beam of particles, the photons. When a either as a wave or as a stream of particles.photon of sufficient energy strikes an electron Whenever radiation interacts with matter, itin the atom of the metal, it transfers its energy displays particle like properties in contrastinstantaneously to the electron during the to the wavelike properties (interferencecollision and the electron is ejected without and diffraction), which it exhibits when itany time lag or delay. Greater the energy possessed by the photon, greater will be propagates. This concept was totally alien to transfer of energy to the electron and greater the way the scientists thought about matter the kinetic energy of the ejected electron. In and radiation and it took them a long time to other words, kinetic energy of the ejected become convinced of its validity. It turns out, electron is proportional to the frequency as you shall see later, that some microscopic of the electromagnetic radiation. Since the particles like electrons also exhibit this wave- striking photon has energy equal to hν and particle duality. Reprint 2025-26 structure of atom 43 Problem 2.6 Solution Calculate energy of one mole of photons The energy (E) of a 300 nm photon is of radiation whose frequency is 5 ×1014 given by Hz. Solution Energy (E) of one photon is given by the expression = 6.626 × 10–19 J E = hν The energy of one mole of photons h = 6.626 ×10–34 J s = 6.626 ×10–19 J × 6.022 × 1023 mol–1 ν = 5×1014 s–1 (given) = 3.99 × 105 J mol–1 E = (6.626 ×10–34 J s) × (5 ×1014 s–1) The minimum energy needed to remove = 3.313 ×10–19 J one mole of electrons from sodium Energy of one mole of photons = (3.99 –1.68) 105 J mol–1 = 2.31 × 105 J mol–1 = (3.313 ×10–19 J) × (6.022 × 1023 mol–1) The minimum energy for one electron = 199.51 kJ mol–1 Problem 2.7 A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second This corresponds to the wavelength by the bulb. h c = Solution E 34 8 1 6.626 10 J s 3.0 10 m s = 19 Power of the bulb = 100 watt 3.84 10 J = 100 J s–1 = 517 nm Energy of one photon E = hν = hc/λ (This corresponds to green light) 6.626  10 34 J s  3  10 8 m s 1 Problem 2.9 =  400  10 9m The threshold frequency ν0 for a metal is 7.0 ×1014 s–1. Calculate the kinetic energy = 4.969 × 10–19 J of an electron emitted when radiation of Number of photons emitted frequency ν =1.0 ×1015 s–1 hits the metal. 100 J s 1 20 1 Solution 19  2 .012  10 s 4 .969  10 J According to Einstein’s equation Kinetic energy = ½ mev2=h(ν – ν0 ) Problem 2.8 = (6.626 × 10–34 J s) (1.0 × 1015 s–1 – 7.0 When electromagnetic radiation of ×1014 s–1) wavelength 300 nm falls on the surface of sodium, electrons are emitted with a = (6.626 × 10–34 J s) (10.0 × 1014 s–1 – 7.0 kinetic energy of 1.68 ×105 J mol–1. What ×1014 s–1) is the minimum energy needed to remove = (6.626 × 10–34 J s) × (3.0 × 1014 s–1) an electron from sodium? What is the maximum wavelength that will cause a = 1.988 × 10–19 J photoelectron to be emitted? Reprint 2025-26 44 chemistry 2.3.3 Evidence for the quantized* spectrum. A continuum of radiation is passed Electronic Energy Levels: Atomic through a sample which absorbs radiation of spectra certain wavelengths. The missing wavelength The speed of light depends upon the nature which corresponds to the radiation absorbed of the medium through which it passes. As a by the matter, leave dark spaces in the bright result, the beam of light is deviated or refracted continuous spectrum. from its original path as it passes from one The study of emission or absorption medium to another. It is observed that when spectra is referred to as spectroscopy. The a ray of white light is passed through a prism, spectrum of the visible light, as discussed the wave with shorter wavelength bends more above, was continuous as all wavelengths (red than the one with a longer wavelength. Since to violet) of the visible light are represented in ordinary white light consists of waves with the spectra. The emission spectra of atoms in all the wavelengths in the visible range, a ray the gas phase, on the other hand, do not show of white light is spread out into a series of a continuous spread of wavelength from red coloured bands called spectrum. The light of to violet, rather they emit light only at specific wavelengths with dark spaces between them.red colour which has longest wavelength is Such spectra are called line spectra ordeviated the least while the violet light, which has shortest wavelength is deviated the most. atomic spectra because the emitted radiation is identified by the appearance of bright linesThe spectrum of white light, that we can in the spectra (Fig. 2.10 page 45).see, ranges from violet at 7.50 × 1014 Hz to red at 4×1014 Hz. Such a spectrum is called Line emission spectra are of great continuous spectrum. Continuous because interest in the study of electronic structure. violet merges into blue, blue into green and Each element has a unique line emission so on. A similar spectrum is produced when spectrum. The characteristic lines in atomic a rainbow forms in the sky. Remember that spectra can be used in chemical analysis to visible light is just a small portion of the identify unknown atoms in the same way electromagnetic radiation (Fig.2.7). When as fingerprints are used to identify people. electromagnetic radiation interacts with The exact matching of lines of the emission matter, atoms and molecules may absorb spectrum of the atoms of a known element energy and reach to a higher energy state. With with the lines from an unknown sample higher energy, these are in an unstable state. quickly establishes the identity of the latter, For returning to their normal (more stable, German chemist, Robert Bunsen (1811-1899) lower energy states) energy state, the atoms was one of the first investigators to use line and molecules emit radiations in various spectra to identify elements. Elements like rubidium (Rb), caesium (Cs)regions of the electromagnetic spectrum. thallium (Tl), indium (In), gallium (Ga) and Emission and Absorption Spectra scandium (Sc) were discovered when their The spectrum of radiation emitted by a minerals were analysed by spectroscopic substance that has absorbed energy is called methods. The element helium (He) was an emission spectrum. Atoms, molecules or discovered in the sun by spectroscopic method. ions that have absorbed radiation are said Line Spectrum of Hydrogen to be “excited”. To produce an emission When an electric discharge is passed through spectrum, energy is supplied to a sample by gaseous hydrogen, the H2 molecules dissociate heating it or irradiating it and the wavelength and the energetically excited hydrogen atoms (or frequency) of the radiation emitted, as produced emit electromagnetic radiation of the sample gives up the absorbed energy, is discrete frequencies. The hydrogen spectrum recorded. consists of several series of lines named after An absorption spectrum is like the their discoverers. Balmer showed in 1885 photographic negative of an emission on the basis of experimental observations * The restriction of any property to discrete values is called quantization. Reprint 2025-26 structure of atom 45 (a) (b) Fig. 2.10 (a) Atomic emission. The light emitted by a sample of excited hydrogen atoms (or any other element) can be passed through a prism and separated into certain discrete wavelengths. Thus an emission spectrum, which is a photographic recording of the separated wavelengths is called as line spectrum. Any sample of reasonable size contains an enormous number of atoms. Although a single atom can be in only one excited state at a time, the collection of atoms contains all possible excited states. The light emitted as these atoms fall to lower energy states is responsible for the spectrum. (b) Atomic absorption. When white light is passed through unexcited atomic hydrogen and then through a slit and prism, the transmitted light is lacking in intensity at the same wavelengths as are emitted in (a) The recorded absorption spectrum is also a line spectrum and the photographic negative of the emission spectrum. that if spectral lines are expressed in terms The value 109,677 cm–1 is called the of wavenumber ( ), then the visible lines of Rydberg constant for hydrogen. The first five the hydrogen spectrum obey the following series of lines that correspond to n1 = 1, 2, 3, formula: 4, 5 are known as Lyman, Balmer, Paschen, Bracket and Pfund series, respectively, (2.8) Table 2.3 shows these series of transitions in the hydrogen spectrum. Fig. 2.11 (page, 46)where n is an integer equal to or greater than shows the Lyman, Balmer and Paschen series3 (i.e., n = 3,4,5,....) of transitions for hydrogen atom. The series of lines described by this formula Of all the elements, hydrogen atom hasare called the Balmer series. The Balmer the simplest line spectrum. Line spectrumseries of lines are the only lines in the hydrogen spectrum which appear in the visible region Table 2.3 The Spectral Lines for Atomicof the electromagnetic spectrum. The Swedish Hydrogenspectroscopist, Johannes Rydberg, noted that all series of lines in the hydrogen Series n1 n2 Spectral Regionspectrum could be described by the following expression : Lyman 1 2,3.... Ultraviolet Balmer 2 3,4.... Visible (2.9) Paschen 3 4,5.... Infrared where n1=1,2........ Brackett 4 5,6.... Infrared n2 = n1 + 1, n1 + 2...... Pfund 5 6,7.... Infrared Reprint 2025-26 46 chemistry atomic structure and spectra. Bohr’s model for hydrogen atom is based on the following postulates: i) The electron in the hydrogen atom can move around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states. These orbits are arranged concentrically around the nucleus. ii) The energy of an electron in the orbit does not change with time. However, the electron will move from a lower stationary state to a higher stationary state when required amount of energy is absorbed by the electron or energy is emitted when electron moves from higher stationary state to lower stationary state (equation 2.16). The energy change does not take place in a continuous manner. Angular Momentum Fig. 2.11 Transitions of the electron in the Just as linear momentum is the product hydrogen atom (The diagram shows the of mass (m) and linear velocity (v), angular Lyman, Balmer and Paschen series of momentum is the product of moment of transitions) inertia (I) and angular velocity (ω). For an electron of mass me, moving in a circular becomes more and more complex for heavier path of radius r around the nucleus, atom. There are, however, certain features angular momentum = I × ω which are common to all line spectra, i.e., Since I = mer2, and ω = v/r where v is the(i) line spectrum of element is unique and linear velocity, (ii) there is regularity in the line spectrum of ∴angular momentum = mer2 × v/r = mevreach element. The questions which arise are: What are the reasons for these similarities? iii) The frequency of radiation absorbed orIs it something to do with the electronic emitted when transition occurs betweenstructure of atoms? These are the questions two stationary states that differ inneed to be answered. We shall find later that energy by ∆E, is given by: the answers to these questions provide the key in understanding electronic structure of (2.10) these elements. Where E1 and E2 are the energies of the2.4 Bohr’s Model for Hydrogen lower and higher allowed energy states Atom respectively. This expression is commonly Neils Bohr (1913) was the first to explain known as Bohr’s frequency rule. quantitatively the general features of the iv) The angular momentum of an electronstructure of hydrogen atom and its spectrum. is quantised. In a given stationary stateHe used Planck’s concept of quantisation it can be expressed as in equation (2.11)of energy. Though the theory is not the hmodern quantum mechanics, it can still n = 1,2,3..... (2.11) m e v r nbe used to rationalize many points in the .2 Reprint 2025-26 structure of atom 47 Where me is the mass of electron, v is the Niels Bohrvelocity and r is the radius of the orbit in (1885–1962)which electron is moving. Niels Bohr, a Danish physicist Thus an electron can move only in those received his Ph.D. from the orbits for which its angular momentum is University of Copenhagen in integral multiple of h/2π. That means angular 1911. He then spent a year momentum is quantised. Radiation is emitted with J.J. Thomson and Ernest or obsorbed only when transition of electron Rutherford in England. In 1913, takes place from one quantised value of he returned to Copenhagen where he remained for angular momentum to another. Therefore, the rest of his life. In 1920 he was named Director of the Institute of theoretical Physics. After firstMaxwell’s electromagnetic theory does not World War, Bohr worked energetically for peacefulapply here that is why only certain fixed orbits uses of atomic energy. He received the first Atomsare allowed. for Peace award in 1957. Bohr was awarded the The details regarding the derivation of Nobel Prize in Physics in 1922. energies of the stationary states used by Bohr, are quite complicated and will be discussed in higher classes. However, according to Bohr’s Fig. 2.11 depicts the energies of different theory for hydrogen atom: stationary states or energy levels of hydrogen a) The stationary states for electron are atom. This representation is called an energy numbered n = 1,2,3.......... These integral level diagram. numbers (Section 2.6.2) are known as When the electron is free from the influence Principal quantum numbers. of nucleus, the energy is taken as zero. The b) The radii of the stationary states are electron in this situation is associated with the expressed as: stationary state of Principal Quantum number = n = ∞ and is called as ionized hydrogen atom. rn = n2 a0 (2.12) When the electron is attracted by the nucleus where a0 = 52.9 pm. Thus the radius of and is present in orbit n, the energy is emitted the first stationary state, called the Bohr and its energy is lowered. That is the reason orbit, is 52.9 pm. Normally the electron in the hydrogen atom is found in this What does the negative electronic orbit (that is n=1). As n increases the energy (En) for hydrogen atom mean? value of r will increase. In other words the electron will be present away from The energy of the electron in a hydrogen the nucleus. atom has a negative sign for all possible orbits (eq. 2.13). What does this negativec) The most important property associated sign convey? This negative sign means with the electron, is the energy of its that the energy of the electron in the stationary state. It is given by the atom is lower than the energy of a free expression. electron at rest. A free electron at rest  1  is an electron that is infinitely far away 2 from the nucleus and is assigned the  n = 1,2,3.... (2.13)E n  R H  n energy value of zero. Mathematically, thiswhere RH is called Rydberg constant and corresponds to setting n equal to infinityits value is 2.18×10–18 J. The energy of the in the equation (2.13) so that E∞=0. As thelowest state, also called as the ground state, is electron gets closer to the nucleus (as n 1 decreases), En becomes larger in absoluteE1 = –2.18×10–18 ( ) = –2.18×10–18 J. The 12 value and more and more negative. The energy of the stationary state for n = 2, will most negative energy value is given by n=1 which corresponds to the most stable 1 orbit. We call this the ground state. ) = –0.545×10–18be : E2 = –2.18×10–18J ( J. 22 Reprint 2025-26 48 chemistry for the presence of negative sign in equation  R H   R H (2.13) and depicts its stability relative to the E  2 2  (where ni and nf  n f  n ireference state of zero energy and n = ∞. d) Bohr’s theory can also be applied to stand for initial orbit and final orbits) the ions containing only one electron,  1 1  18  1 1  similar to that present in hydrogen ∆E  R H 2  2 2 .18  10 J 2  2  n i n f   n i n f  atom. For example, He+ Li2+, Be3+ and so on. The energies of the stationary states associated with these kinds of ions (also (2.17) known as hydrogen like species) are The frequency (ν) associated with the given by the expression. absorption and emission of the photon can 2 be evaluated by using equation (2.18)   Z (2.14) J E n  2 .18  10 18 2  n  and radii by the expression 52 .9 (n 2 ) 18 rn = pm (2.15) 2 .18  10 J  1 1  Z  34 2  2 (2.18) 6 .626  10 Js  n i n f where Z is the atomic number and has values 2,3 for the helium and lithium atoms 15  1 1 respectively. From the above equations, it is  3 .29  10 2  2 Hz (2.19)evident that the value of energy becomes more  n i n f  negative and that of radius becomes smaller and in terms of wavenumbers ( )with increase of Z. This means that electron will be tightly bound to the nucleus. (2.20)e) It is also possible to calculate the velocities of electrons moving in these orbits. Although the precise equation 3 .29  1015 s 1  1 1  = 8  s 2  2 is not given here, qualitatively the 3  10 m s  n i n f  magnitude of velocity of electron increases with increase of positive 7  1 1  1 charge on the nucleus and decreases = 1 .09677  10 2  2 m (2.21)  n i n f  with increase of principal quantum number. In case of absorption spectrum, nf > ni and the term in the parenthesis is positive and 2.4.1 Explanation of Line Spectrum of energy is absorbed. On the other hand in case ∆ E is negative Hydrogen of emission spectrum ni > nf , Line spectrum observed in case of hydrogen and energy is released. atom, as mentioned in section 2.3.3, can be The expression (2.17) is similar to that explained quantitatively using Bohr’s model. used by Rydberg (2.9) derived empirically According to assumption 2, radiation (energy) using the experimental data available at that is absorbed if the electron moves from the time. Further, each spectral line, whether orbit of smaller Principal quantum number to in absorption or emission spectrum, can the orbit of higher Principal quantum number, be associated to the particular transition in whereas the radiation (energy) is emitted if hydrogen atom. In case of large number of the electron moves from higher orbit to lower hydrogen atoms, different possible transitions orbit. The energy gap between the two orbits can be observed and thus leading to large is given by equation (2.16) number of spectral lines. The brightness or ∆E = Ef – Ei (2.16) intensity of spectral lines depends upon the number of photons of same wavelength or Combining equations (2.13) and (2.16) frequency absorbed or emitted. Reprint 2025-26 structure of atom 49 2.4.2 Limitations of Bohr’s Model Problem 2.10 Bohr’s model of the hydrogen atom was no What are the frequency and wavelength doubt an improvement over Rutherford’s of a photon emitted during a transition nuclear model, as it could account for the from n = 5 state to the n = 2 state in the stability and line spectra of hydrogen atom hydrogen atom? and hydrogen like ions (for example, He+, Li2+, Solution Be3+, and so on). However, Bohr’s model was too simple to account for the following points. Since ni = 5 and nf = 2, this transition i) It fails to account for the finer details gives rise to a spectral line in the visible (doublet, that is two closely spaced lines) region of the Balmer series. From of the hydrogen atom spectrum observed equation (2.17) by using sophisticated spectroscopic 18  1 1  techniques. This model is also unable E = 2 .18  10 J 2  2  5 2  to explain the spectrum of atoms 19 other than hydrogen, for example, =  4 .58  10 J helium atom which possesses only two It is an emission energy electrons. Further, Bohr’s theory was also unable to explain the splitting The frequency of the photon (taking of spectral lines in the presence of energy in terms of magnitude) is given by magnetic field (Zeeman effect) or an electric field (Stark effect). ii) It could not explain the ability of atoms to form molecules by chemical bonds. In other words, taking into account the points mentioned above, one needs a better = 6.91×1014 Hz theory which can explain the salient features of the structure of complex atoms. 2.5 Towards Quantum Mechanical Problem 2.11 Model of the Atom Calculate the energy associated with the In view of the shortcoming of the Bohr’s first orbit of He+. What is the radius of model, attempts were made to develop a more this orbit? suitable and general model for atoms. Two important developments which contributed Solution significantly in the formulation of such a ( 2 .18  10 18 J )Z 2 model were: E n  2 atom–1 n 1. Dual behaviour of matter, For He+, n = 1, Z = 2 2. Heisenberg u ( 2 .18  10 18 J )( 2 2 ) 18 E1  2 8 .72  10 J 2.5.1 Dual Behaviour of Matter 1 The French physicist, de Broglie, in 1924 The radius of the orbit is given by proposed that matter, like radiation, should equation (2.15) also exhibit dual behaviour i.e., both particle ( 0 .0529 nm )n 2 and wavelike properties. This means that rn = just as the photon has momentum as well Z as wavelength, electrons should also have Since n = 1, and Z = 2 momentum as well as wavelength, de Broglie, ( 0 .0529 nm )12 from this analogy, gave the following relation rn = = 0 .02645 nm between wavelength (λ) and momentum (p) of 2 a material particle. Reprint 2025-26 50 chemistry Louis de Broglie Solution (1892 – 1987) Louis de Broglie, a French According to de Brogile equation (2.22) physicist, studied history 34 h ( 6 .626  10 Js ) as an undergraduate in the   1 early 1910’s. His interest mv ( 0 .1 kg )(10 m s ) turned to science as a result of his assignment to radio = 6.626 × 10–34 m (J = kg m2 s–2) communications in World Problem 2.13 War I. He received his Dr. Sc. from the University of Paris in 1924. The mass of an electron is 9.1×10–31 kg. He was professor of theoretical physics at If its K.E. is 3.0×10–25 J, calculate its the University of Paris from 1932 untill his wavelength. retirement in 1962. He was awarded the Solution Nobel Prize in Physics in 1929. Since K.E. = ½ mv2 1/ 2 25 2 2 1/ 2 h h  2 K .E.   2  3 .0  10 kg m s    (2.22) v =  m  =  9 .1  10 31 kg  p mv where m is the mass of the particle, v its = 812 m s–1velocity and p its momentum. de Broglie’s prediction was confirmed experimentally when h 6 .626 10 34 Js it was found that an electron beam undergoes m v ( 9. 1 10 31 kg )( 812 m s 1 ) diffraction, a phenomenon characteristic of waves. This fact has been put to use in making = 8967 × 10–10 m = 896.7 nm an electron microscope, which is based on Problem 2.14 the wavelike behaviour of electrons just as an Calculate the mass of a photon withordinary microscope utilises the wave nature wavelength 3.6 Å.of light. An electron microscope is a powerful tool in modern scientific research because it Solution achieves a magnification of about 15 million λ = 3.6 Å = 3.6 × 10–10 m times. Velocity of photon = velocity of light It needs to be noted that according to de 34 Broglie, every object in motion has a wave character. The wavelengths associated with 10 8 1 ordinary objects are so short (because of their = 6.135 × 10–29 kglarge masses) that their wave properties cannot be detected. The wavelengths associated with electrons and other subatomic particles (with very small mass) can however be detected 2.5.2 Heisenberg’s Uexperimentally. Results obtained from Werner Heisenberg a German physicist in the following problems prove these points 1927, stated uqualitatively. the consequence of dual behaviour of matter and radiation. It states that it is impossible to determine simultaneously, the exact Problem 2.12 position and exact momentum (or velocity) What will be the wavelength of a ball of of an electron. mass 0.1 kg moving with a velocity of Mathematically, it can be given as in 10 m s–1 ? equation (2.23). Reprint 2025-26 structure of atom 51 (2.23)  h  momentum photons of such light  p =   would change the energy of electrons by collisions. In this process we, no doubt, would be able to calculate the position of the electron, but we would know very little about where ∆x is the u∆px (or ∆vx) is the u(or velocity) of the particle. If the position of One of the important implications of thethe electron is known with high degree of Heisenberg Uelectron will be uthe other hand, if the velocity of the electron particles. The trajectory of an object is is known precisely (∆(vx ) is small), then the determined by its location and velocity at position of the electron will be u(∆x will be large). Thus, if we carry out some is at a particular instant and if we also know physical measurements on the electron’s its velocity and the forces acting on it at that position or velocity, the outcome will always instant, we can tell where the body would depict a fuzzy or blur picture. be sometime later. We, therefore, conclude that the position of an object and its velocity The u fix its trajectory. Since for a sub-atomicunderstood with the help of an example. object such as an electron, it is not possibleSuppose you are asked to measure the simultaneously to determine the position and thickness of a sheet of paper with an velocity at any given instant to an arbitrary unmarked metrestick. Obviously, the results degree of precision, it is not possible to talk obtained would be extremely inaccurate of the trajectory of an electron. and meaningless. In order to obtain any The effect of Heisenberg U Principle is significant only for motion ofgraduated in units smaller than the thickness microscopic objects and is negligible forof a sheet of the paper. Analogously, in order that of macroscopic objects. This can beto determine the position of an electron, we seen from the following examples. must use a meterstick calibrated in units of If u object of mass, say about a milligram (10–6 kg),in mind that an electron is considered as a thenpoint charge and is therefore, dimensionless). To observe an electron, we can illuminate it with “light” or electromagnetic radiation. The “light” used must have a wavelength smaller than the dimensions of an electron. The high Werner Heisenberg (1901 – 1976) Werner Heisenberg (1901 – 1976) received his Ph.D. in physics from the University of Munich in 1923. He then spent a year working with Max Born at Gottingen and three years with Niels Bohr in Copenhagen. He was professor of physics at the University of Leipzig from 1927 to 1941. During World War II, Heisenberg was in charge of German research on the atomic bomb. After the war he was named director of Max Planck Institute for physics in Gottingen. He was also accomplished mountain climber. Heisenberg was awarded the Nobel Prize in Physics in 1932. Reprint 2025-26 52 chemistry The value of ∆v∆x obtained is extremely small and is insignificant. Therefore, one may say that in dealing with milligram- sized or heavier objects, the associated uconsequence. In the case of a microscopic object like an = 0.579×107 m s–1 (1J = 1 kg m2 s–2) electron on the other hand. ∆v.∆x obtained is = 5.79×106 m s–1 much larger and such u Problem 2.16consequence. For example, for an electron whose mass is 9.11×10–31 kg., according to A golf ball has a mass of 40g, and a speed Heisenberg u within accuracy of 2%, calculate the u Solution The u Using the equation (2.22) It, therefore, means that if one tries to find the exact location of the electron, say to an uu 10 4 m 2 s 1 4 1 8  10 ms 10 m = 1.46×10–33 m This is nearly ~ 1018 times smaller which is so large that the classical picture than the diameter of a typical atomic of electrons moving in Bohr’s orbits (fixed) nucleus. As mentioned earlier for large cannot hold good. It, therefore, means that particles, the uthe precise statements of the position no meaningful limit to the precision of and momentum of electrons have to be measurements. replaced by the statements of probability, that the electron has at a given position Reasons for the Failure of the Bohr Modeland momentum. This is what happens in the quantum mechanical model of atom. One can now understand the reasons for the failure of the Bohr model. In Bohr model, an electron is regarded as a charged particle Problem 2.15 moving in well defined circular orbits about the nucleus. The wave character of the A microscope using suitable photons is electron is not considered in Bohr model. employed to locate an electron in an atom within a distance of 0.1 Å. What is the Further, an orbit is a clearly defined path u of its velocity? only if both the position and the velocity of the electron are known exactly at the same Solution time. This is not possible according to the ∆ x ∆p = or ∆ x m ∆ v Heisenberg u of the hydrogen atom, therefore, not only ignores dual behaviour of matter but also contradicts Heisenberg u Reprint 2025-26 structure of atom 53 these objects obey. When quantum mechanics Erwin Schrödinger, an is applied to macroscopic objects (for which Austrian physicist received wave like properties are insignificant) the his Ph.D. in theoretical results are the same as those from the physics from the University classical mechanics. of Vienna in 1910. In 1927 Schrödinger succeeded Max Quantum mechanics was developed Planck at the University of independently in 1926 by Werner Heisenberg Berlin at Planck’s request. and Erwin Schrödinger. Here, however, we In 1933, Schrödinger left shall be discussing the quantum mechanics Berlin because of his which is based on the ideas of wave motion. Erwin Schrödinger opposition to Hitler and (1887–1961) The fundamental equation of quantum Nazi policies and returned mechanics was developed by Schrödinger to Austria in 1936. After the invasion of Austria and it won him the Nobel Prize in Physics in by Germany, Schrödinger was forcibly removed 1933. This equation which incorporates wave- from his professorship. He then moved to Dublin, particle duality of matter as proposed by de Ireland where he remained for seventeen years. Broglie is quite complex and knowledge of Schrödinger shared the Nobel Prize for Physics higher mathematics is needed to solve it. You with P.A.M. Dirac in 1933. will learn its solutions for different systems in higher classes. In view of these inherent weaknesses in the For a system (such as an atom or a Bohr model, there was no point in extending molecule whose energy does not change with Bohr model to other atoms. In fact an insight time) the Schrödinger equation is written into the structure of the atom was needed as where is a mathematical which could account for wave-particle duality operator called Hamiltonian. Schrödinger of matter and be consistent with Heisenberg gave a recipe of constructing this operator uadvent of quantum mechanics. the system. The total energy of the system takes into account the kinetic energies of all 2.6 Quantum Mechanical Model of the sub-atomic particles (electrons, nuclei), Atom attractive potential between the electrons Classical mechanics, based on Newton’s and nuclei and repulsive potential among the laws of motion, successfully describes the electrons and nuclei individually. Solution of motion of all macroscopic objects such as a this equation gives E and ψ. falling stone, orbiting planets etc., which have Hydrogen Atom and the Schrödingeressentially a particle-like behaviour as shown Equationin the previous section. However it fails when When Schrödinger equation is solved forapplied to microscopic objects like electrons, hydrogen atom, the solution gives the possibleatoms, molecules etc. This is mainly because energy levels the electron can occupy andof the fact that classical mechanics ignores the corresponding wave function(s) (ψ) ofthe concept of dual behaviour of matter the electron associated with each energyespecially for sub-atomic particles and the level. These quantized energy states andu corresponding wave functions which arethat takes into account this dual behaviour characterized by a set of three quantumof matter is called quantum mechanics. numbers (principal quantum number Quantum mechanics is a theoretical n, azimuthal quantum number l and science that deals with the study of the magnetic quantum number ml ) arise as a motions of the microscopic objects that have natural consequence in the solution of the both observable wave like and particle like Schrödinger equation. When an electron properties. It specifies the laws of motion that is in any energy state, the wave function Reprint 2025-26 54 chemistry corresponding to that energy state contains 2. The existence of quantised electronicall information about the electron. The wave energy levels is a direct result of thefunction is a mathematical function whose wave like properties of electrons andvalue depends upon the coordinates of the are allowed solutions of Schrödinger electron in the atom and does not carry any wave equation. physical meaning. Such wave functions of 3. Both the exact position and exact hydrogen or hydrogen like species with one velocity of an electron in an atom electron are called atomic orbitals. Such cannot be determined simultaneously wave functions pertaining to one-electron (Heisenberg uspecies are called one-electron systems. The path of an electron in an atom therefore, probability of finding an electron at a point can never be determined or known within an atom is proportional to the |ψ|2 at accurately. That is why, as you shall see later on, one talks of only probability ofthat point. The quantum mechanical results finding the electron at different points inof the hydrogen atom successfully predict an atom. all aspects of the hydrogen atom spectrum 4. An atomic orbital is the wave including some phenomena that could not be function ψfor an electron in an atom. explained by the Bohr model. Whenever an electron is described by a wave function, we say that the Application of Schrödinger equation to electron occupies that orbital. Sincemulti-electron atoms presents a difficulty: the many such wave functions are possible Schrödinger equation cannot be solved exactly for an electron, there are many atomic for a multi-electron atom. This difficulty can orbitals in an atom. These “one electron be overcome by using approximate methods. orbital wave functions” or orbitals form Such calculations with the aid of modern the basis of the electronic structure computers show that orbitals in atoms other of atoms. In each orbital, the electron than hydrogen do not differ in any radical has a definite energy. An orbital cannot way from the hydrogen orbitals discussed contain more than two electrons. In a above. The principal difference lies in the multi-electron atom, the electrons are filled in various orbitals in the order ofconsequence of increased nuclear charge. increasing energy. For each electronBecause of this all the orbitals are somewhat of a multi-electron atom, there shall,contracted. Further, as you shall see later (in therefore, be an orbital wave function subsections 2.6.3 and 2.6.4), unlike orbitals characteristic of the orbital it occupies. of hydrogen or hydrogen like species, whose All the information about the electron energies depend only on the quantum number in an atom is stored in its orbital wave n, the energies of the orbitals in multi-electron function ψ and quantum mechanics atoms depend on quantum numbers n and l. makes it possible to extract this information out of ψ. Important Features of the Quantum 5. The probability of finding an electron at Mechanical Model of Atom a point within an atom is proportional to the square of the orbital wave function Quantum mechanical model of atom is i.e., |ψ|2 at that point. |ψ|2 is known the picture of the structure of the atom, as probability density and is always which emerges from the application of positive. From the value of |ψ|2 at the Schrödinger equation to atoms. The different points within an atom, following are the important features of the it is possible to predict the region quantum-mechanical model of atom: around the nucleus where electron 1. The energy of electrons in atoms is will most probably be found. quantized (i.e., can only have certain specific values), for example when 2.6.1 Orbitals and Quantum Numbers electrons are bound to the nucleus in A large number of orbitals are possible in atoms. an atom. Qualitatively these orbitals can Reprint 2025-26 structure of atom 55 be distinguished by their size, shape and sub-shells in a principal shell is equal to the orientation. An orbital of smaller size means value of n. For example in the first shell (n = 1), there is more chance of finding the electron there is only one sub-shell which corresponds near the nucleus. Similarly shape and to l = 0. There are two sub-shells (l = 0, 1) in orientation mean that there is more probability the second shell (n = 2), three (l = 0, 1, 2) in of finding the electron along certain directions third shell (n = 3) and so on. Each sub-shell is than along others. Atomic orbitals are precisely assigned an azimuthal quantum number (l). distinguished by what are known as quantum Sub-shells corresponding to different values numbers. Each orbital is designated by three of l are represented by the following symbols. quantum numbers labelled as n, l and ml. Value for l : 0 1 2 3 4 5 ............ notation for s p d f g h ............ The principal quantum number ‘n’ is a positive integer with value of n = 1,2,3....... sub-shell The principal quantum number determines Table 2.4 shows the permissible values of the size and to large extent the energy of the ‘l ’ for a given principal quantum number and orbital. For hydrogen atom and hydrogen like the corresponding sub-shell notation. species (He+, Li2+, .... etc.) energy and size of the orbital depends only on ‘n’. Table 2.4 Subshell Notations The principal quantum number also n l Subshell notation identifies the shell. With the increase in the value of ‘n’, the number of allowed orbital 1 0 1s increases and are given by ‘n2’ All the 2 0 2sorbitals of a given value of ‘n’ constitute a single shell of atom and are represented 2 1 2p by the following letters 3 0 3s n = 1 2 3 4 ............ 3 1 3p Shell = K L M N ............ 3 2 3d Size of an orbital increases with increase of principal quantum number ‘n’. In other words 4 0 4s the electron will be located away from the 4 1 4p nucleus. Since energy is required in shifting away the negatively charged electron from the 4 2 4d positively charged nucleus, the energy of the 4 3 4f orbital will increase with increase of n. Azimuthal quantum number. ‘l’ is also Magnetic orbital quantum number. known as orbital angular momentum or ‘ml’ gives information about the spatial subsidiary quantum number. It defines the orientation of the orbital with respect to three-dimensional shape of the orbital. For a standard set of co-ordinate axis. For any given value of n, l can have n values ranging sub-shell (defined by ‘l’ value) 2l+1 values of from 0 to n – 1, that is, for a given value of n, ml are possible and these values are given by : the possible value of l are : l = 0, 1, 2, .......... ml = – l, – (l–1), – (l–2)... 0,1... (l–2), (l–1), l (n–1) Thus for l = 0, the only permitted value For example, when n = 1, value of l is only of ml = 0, [2(0)+1 = 1, one s orbital]. For l =0. For n = 2, the possible value of l can be 0 1, ml can be –1, 0 and +1 [2(1)+1 = 3, three p and 1. For n = 3, the possible l values are 0, orbitals]. For l = 2, ml = –2, –1, 0, +1 and +2, 1 and 2. [2(2)+1 = 5, five d orbitals]. It should be noted Each shell consists of one or more that the values of ml are derived from l and sub-shells or sub-levels. The number of that the value of l are derived from n. Reprint 2025-26 56 chemistry Each orbital in an atom, therefore, is angular momentum of the electron — a vector defined by a set of values for n, l and ml. An quantity, can have two orientations relative to orbital described by the quantum numbers the chosen axis. These two orientations are n = 2, l = 1, ml = 0 is an orbital in the p sub- distinguished by the spin quantum numbers shell of the second shell. The following chart ms which can take the values of +½ or –½. gives the relation between the subshell and These are called the two spin states of the the number of orbitals associated with it. electron and are normally represented by Value of l 0 1 2 3 4 5 two arrows, ↑ (spin up) and ↓ (spin down). Two electrons that have different ms values (one +½ Subshell notation s p d f g h and the other –½) are said to have opposite number of orbitals 1 3 5 7 9 11 spins. An orbital cannot hold more than two Electron spin ‘s’ : The three quantum electrons and these two electrons should have numbers labelling an atomic orbital can be opposite spins. used equally well to define its energy, shape To sum up, the four quantum numbers and orientation. But all these quantum provide the following information : numbers are not enough to explain the line i) n defines the shell, determines the size spectra observed in the case of multi-electron of the orbital and also to a large extent atoms, that is, some of the lines actually occur the energy of the orbital. in doublets (two lines closely spaced), triplets ii) There are n subshells in the nth shell. l(three lines, closely spaced) etc. This suggests identifies the subshell and determinesthe presence of a few more energy levels than the shape of the orbital (see sectionpredicted by the three quantum numbers. 2.6.2). There are (2l+1) orbitals of each In 1925, George Uhlenbeck and Samuel type in a subshell, that is, one s orbital Goudsmit proposed the presence of the fourth (l = 0), three p orbitals (l = 1) and five quantum number known as the electron d orbitals (l = 2) per subshell. To some spin quantum number (ms). An electron extent l also determines the energy of spins around its own axis, much in a similar the orbital in a multi-electron atom. way as earth spins around its own axis while iii) ml designates the orientation of therevolving around the sun. In other words, orbital. For a given value of l, ml has an electron has, besides charge and mass, (2l+1) values, the same as the number intrinsic spin angular quantum number. Spin of orbitals per subshell. It means that Orbit, orbital and its importance Orbit and orbital are not synonymous. An orbit, as proposed by Bohr, is a circular path around the nucleus in which an electron moves. A precise description of this path of the electron is impossible according to Heisenberg u their existence can never be demonstrated experimentally. An atomic orbital, on the other hand, is a quantum mechanical concept and refers to the one electron wave function ψ in an atom. It is characterized by three quantum numbers (n, l and ml) and its value depends upon the coordinates of the electron. ψhas, by itself, no physical meaning. It is the square of the wave function i.e., |ψ|2 which has a physical meaning. |ψ|2 at any point in an atom gives the value of probability density at that point. Probability density (|ψ|2) is the probability per unit volume and the product of |ψ|2 and a small volume (called a volume element) yields the probability of finding the electron in that volume (the reason for specifying a small volume element is that |ψ|2 varies from one region to another in space but its value can be assumed to be constant within a small volume element). The total probability of finding the electron in a given volume can then be calculated by the sum of all the products of |ψ|2 and the corresponding volume elements. It is thus possible to get the probable distribution of an electron in an orbital. Reprint 2025-26 structure of atom 57 the number of orbitals is equal to the number of ways in which they are oriented. iv) ms refers to orientation of the spin of the electron. Problem 2.17 What is the total number of orbitals associated with the principal quantum number n = 3 ? Solution For n = 3, the possible values of l are 0, 1 and 2. Thus there is one 3s orbital (n = 3, l = 0 and ml = 0); there are three 3p orbitals (n = 3, l = 1 and ml = –1, 0, +1); there are five 3d orbitals (n = 3, l = 2 and ml = –2, –1, 0, +1+, +2). Fig. 2.12 The plots of (a) the orbital wave Therefore, the total number of orbitals function ψ(r); (b) the variation of is 1+3+5 = 9 probability density ψ2(r) as a function of distance r of the electron from the The same value can also be obtained by nucleus for 1s and 2s orbitals. using the relation; number of orbitals = n2, i.e. 32 = 9. According to the German physicist, Problem 2.18 Max Born, the square of the wave function (i.e.,ψ2) at a point gives the probability density Using s, p, d, f notations, describe the of the electron at that point. The variation orbital with the following quantum 2 of ψ as a function of r for 1s and 2s orbitals numbers is given in Fig. 2.12(b). Here again, you may (a) n = 2, l = 1, (b) n = 4, l = 0, (c) n = 5, note that the curves for 1s and 2s orbitals l = 3, (d) n = 3, l = 2 are different. Solution It may be noted that for 1s orbital the n l orbital probability density is maximum at the nucleus and it decreases sharply as we move a) 2 1 2p away from it. On the other hand, for 2s b) 4 0 4s orbital the probability density first decreases c) 5 3 5f sharply to zero and again starts increasing. d) 3 2 3d After reaching a small maxima it decreases again and approaches zero as the value of r increases further. The region where this 2.6.2 Shapes of Atomic Orbitals probability density function reduces to zero The orbital wave function or ψfor an electron is called nodal surfaces or simply nodes. in an atom has no physical meaning. It In general, it has been found that ns-orbital is simply a mathematical function of the has (n – 1) nodes, that is, number of nodes coordinates of the electron. However, for increases with increase of principal quantum different orbitals the plots of corresponding number n. In other words, number of nodes wave functions as a function of r (the distance for 2s orbital is one, two for 3s and so on. from the nucleus) are different. Fig. 2.12(a), These probability density variation can be gives such plots for 1s (n = 1, l = 0) and 2s visualised in terms of charge cloud diagrams (n = 2, l = 0) orbitals. [Fig. 2.13(a)]. In these diagrams, the density Reprint 2025-26 58 chemistry of the dots in a region represents electron probability density in that region. Boundary surface diagrams of constant probability density for different orbitals give a fairly good representation of the shapes of the orbitals. In this representation, a boundary surface or contour surface is drawn in space for an orbital on which the value of probability density |ψ|2 is constant. In principle many such boundary surfaces may be possible. However, for a given orbital, only that boundary surface diagram of constant probability density* is taken to be good representation of the shape of the orbital which encloses a region or volume in Fig. 2.13 (a) Probability density plots of 1s and which the probability of finding the electron 2s atomic orbitals. The density of the is very high, say, 90%. The boundary surface dots represents the probability density diagram for 1s and 2s orbitals are given in of finding the electron in that region. Fig. 2.13(b). One may ask a question : Why (b) Boundary surface diagram for 1s do we not draw a boundary surface diagram, and 2s orbitals. which bounds a region in which the probability of finding the electron is, 100 %? The answer to this question is that the probability density |ψ|2 has always some value, howsoever small it may be, at any finite distance from the nucleus. It is therefore, not possible to draw a boundary surface diagram of a rigid size in which the probability of finding the electron is 100%. Boundary surface diagram for a s orbital is actually a sphere centred on the nucleus. In two dimensions, this sphere looks like a circle. It encloses a region in which probability of finding the electron is about 90%. Thus, we see that 1s and 2s orbitals are spherical in shape. In reality all the s-orbitals are spherically symmetric, that is, the probability of finding the electron at a given Fig. 2.14 Boundary surface diagrams of the distance is equal in all the directions. It is also three 2p orbitals. observed that the size of the s orbital increases with increase in n, that is, 4s > 3s > 2s > 1s these diagrams, the nucleus is at the origin. and the electron is located further away from Here, unlike s-orbitals, the boundary surface the nucleus as the principal quantum number diagrams are not spherical. Instead each increases. p orbital consists of two sections called lobes Boundary surface diagrams for three that are on either side of the plane that passes 2p orbitals (l = 1) are shown in Fig. 2.14. In through the nucleus. The probability density * If probability density |ψ|2 is constant on a given surface, |ψ| is also constant over the surface. The boundary surface for |ψ|2 and |ψ| are identical. Reprint 2025-26 structure of atom 59 function is zero on the plane where the two it is sufficient to remember that, because lobes touch each other. The size, shape and there are three possible values of ml, there energy of the three orbitals are identical. are, therefore, three p orbitals whose axes They differ however, in the way the lobes are are mutually perpendicular. Like s orbitals, oriented. Since the lobes may be considered to p orbitals increase in size and energy with lie along the x, y or z axis, they are given the increase in the principal quantum number designations 2px, 2py, and 2pz. It should be and hence the order of the energy and size of understood, however, that there is no simple various p orbitals is 4p > 3p > 2p. Further, like relation between the values of ml (–1, 0 and +1) s orbitals, the probability density functions for and the x, y and z directions. For our purpose, p-orbital also pass through value zero, besides at zero and infinite distance, as the distance from the nucleus increases. The number of nodes are given by the n –2, that is number of radial node is 1 for 3p orbital, two for 4p orbital and so on. For l = 2, the orbital is known as d-orbital and the minimum value of principal quantum number (n) has to be 3. as the value of l cannot be greater than n–1. There are five ml values (–2, –1, 0, +1 and +2) for l = 2 and thus there are five d orbitals. The boundary surface diagram of d orbitals are shown in Fig. 2.15. The five d-orbitals are designated as dxy, dyz, dxz, dx2–y2 and dz2. The shapes of the first four d-orbitals are similar to each other, where as that of the fifth one, dz2, is different from others, but all five 3d orbitals are equivalent in energy. The d orbitals for which n is greater than 3 (4d, 5d...) also have shapes similar to 3d orbital, but differ in energy and size. Besides the radial nodes (i.e., probability density function is zero), the probability density functions for the np and nd orbitals are zero at the plane (s), passing through the nucleus (origin). For example, in case of pz orbital, xy-plane is a nodal plane, in case of dxy orbital, there are two nodal planes passing through the origin and bisecting the xy plane containing z-axis. These are called angular nodes and number of angular nodes are given by ‘l’, i.e., one angular node for p orbitals, two angular nodes for ‘d’ orbitals and so on. The total number of nodes are given by (n–1), i.e., sum of l angular nodes and (n – l – 1) radial nodes. 2.6.3 Energies of Orbitals Fig. 2.15 Boundary surface diagrams of the five The energy of an electron in a hydrogen atom 3d orbitals. is determined solely by the principal quantum Reprint 2025-26 60 chemistry number. Thus the energy of the orbitals in The energy of an electron in a multi- hydrogen atom increases as follows : electron atom, unlike that of the hydrogen atom, depends not only on its principal1s < 2s = 2p < 3s = 3p = 3d <4s = 4p = 4d quantum number (shell), but also on its= 4f < (2.23) azimuthal quantum number (subshell). Thatand is depicted in Fig. 2.16. Although the is, for a given principal quantum number, s,shapes of 2s and 2p orbitals are different, p, d, f ... all have different energies. Withinan electron has the same energy when it is a given principal quantum number, thein the 2s orbital as when it is present in 2p energy of orbitals increases in the orderorbital. The orbitals having the same energy s<p<d<f. For higher energy levels, theseare called degenerate. The 1s orbital in a differences are sufficiently pronounced andhydrogen atom, as said earlier, corresponds straggering of orbital energy may result,to the most stable condition and is called the e.g., 4s<3d and 6s<5d; 4f<6p. The mainground state and an electron residing in this reason for having different energies of theorbital is most strongly held by the nucleus. subshells is the mutual repulsion among theAn electron in the 2s, 2p or higher orbitals in electrons in multi-electron atoms. The onlya hydrogen atom is in excited state. electrical interaction present in hydrogen atom is the attraction between the negatively charged electron and the positively charged nucleus. In multi-electron atoms, besides the presence of attraction between the electron and nucleus, there are repulsion terms between every electron and other electrons present in the atom. Thus the stability of an electron in a multi-electron atom is because total attractive interactions are more than the repulsive interactions. In general, the repulsive interaction of the electrons in the outer shell with the electrons in the inner shell are more important. On the other hand, the attractive interactions of an electron increases with increase of positive charge (Ze) on the nucleus. Due to the presence of electrons in the inner shells, the electron in the outer shell will not experience the full positive charge of the nucleus (Ze). The effect will be lowered due to the partial screening of positive charge on the nucleus by the inner shell electrons. This is known as the shielding of the outer Fig. 2.16 Energy level diagrams for the few shell electrons from the nucleus by the electronic shells of (a) hydrogen atom inner shell electrons, and the net positive and (b) multi-electronic atoms. Note that charge experienced by the outer electrons is orbitals for the same value of principal known as effective nuclear charge (Zeff e). quantum number, have the same energies even for different azimuthal Despite the shielding of the outer electrons quantum number for hydrogen atom. from the nucleus by the inner shell electrons, In case of multi-electron atoms, orbitals the attractive force experienced by the outer with same principal quantum number shell electrons increases with increase of possess different energies for different nuclear charge. In other words, the energy of azimuthal quantum numbers. interaction between, the nucleus and electron Reprint 2025-26 structure of atom 61 (that is orbital energy) decreases (that is Table 2.5 Arrangement of Orbitals with more negative) with the increase of atomic Increasing Energy on the Basis number (Z ). of (n+l) Rule Both the attractive and repulsive interactions depend upon the shell and shape of the orbital in which the electron is present. For example electrons present in spherical shaped, s orbital shields the outer electrons from the nucleus more effectively as compared to electrons present in p orbital. Similarly electrons present in p orbitals shield the outer electrons from the nucleus more than the electrons present in d orbitals, even though all these orbitals are present in the same shell. Further within a shell, due to spherical shape of s orbital, the s orbital electron spends more time close to the nucleus in comparison to p orbital electron which spends more time in the vicinity of nucleus in comparison to d orbital electron. In other words, for a given shell (principal quantum number), the Zeff experienced by the electron decreases with increase of azimuthal quantum number (l), that is, the s orbital electron will be more tightly bound to the nucleus than p orbital electron which in turn will be better tightly bound than the d orbital electron. The energy of electrons in s orbital will be lower (more negative) than that of p orbital electron which will have less energy than that of d orbital electron and so on. Since the extent same energy. Lastly it may be mentioned here of shielding from the nucleus is different for that energies of the orbitals in the same electrons in different orbitals, it leads to the subshell decrease with increase in the splitting of energy levels within the same atomic number (Zeff). For example, energy of shell (or same principal quantum number), 2s orbital of hydrogen atom is greater than that is, energy of electron in an orbital, as that of 2s orbital of lithium and that of lithium mentioned earlier, depends upon the values is greater than that of sodium and so on, that of n and l. Mathematically, the dependence is, E2s(H) > E2s(Li) > E2s(Na) > E2s(K). of energies of the orbitals on n and l are quite 2.6.4 Filling of Orbitals in Atomcomplicated but one simple rule is that, the lower the value of (n + l) for an orbital, the The filling of electrons into the orbitals of lower is its energy. If two orbitals have different atoms takes place according to the same value of (n + l), the orbital with the aufbau principle which is based on the lower value of n will have the lower energy. Pauli’s exclusion principle, the Hund’s rule The Table 2.5 illustrates the (n + l ) rule and of maximum multiplicity and the relative Fig. 2.16 depicts the energy levels of multi- energies of the orbitals. electrons atoms. It may be noted that different subshells of a particular shell have different Aufbau Principle energies in case of multi-electrons atoms. The word ‘aufbau’ in German means ‘building However, in hydrogen atom, these have the up’. The building up of orbitals means the Reprint 2025-26 62 chemistry filling up of orbitals with electrons. The the top, the direction of the arrows gives the principle states : In the ground state of the order of filling of orbitals, that is starting atoms, the orbitals are filled in order of from right top to bottom left. With respect to their increasing energies. In other words, placement of outermost valence electrons, electrons first occupy the lowest energy orbital it is remarkably accurate for all atoms. for available to them and enter into higher energy example, valence electron in potassium orbitals only after the lower energy orbitals must choose between 3d and 4s orbitals and are filled. As you have learnt above, energy as predicted by this sequence, it is found of a given orbital depends upon effective in 4s orbital. The above order should be nuclear charge and different type of orbitals assumed to be a rough guide to the filling are affected to different extent. Thus, there of energy levels. In many cases, the orbitals is no single ordering of energies of orbitals are similar in energy and small changes in which will be universally correct for all atoms. atomic structure may bring about a change in the order of filling. Even then, the above However, following order of energies of series is a useful guide to the building of thethe orbitals is extremely useful: electronic structure of an atom provided that1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, it is remembered that exceptions may occur.4f, 5d, 6p, 7s... Pauli Exclusion Principle The order may be remembered by using the method given in Fig. 2.17. Starting from The number of electrons to be filled in various orbitals is restricted by the exclusion principle, given by the Austrian scientist Wolfgang Pauli (1926). According to this principle : No two electrons in an atom can have the same set of four quantum numbers. Pauli exclusion principle can also be stated as : “Only two electrons may exist in the same orbital and these electrons must have opposite spin.” This means that the two electrons can have the same value of three quantum numbers n, l and ml, but must have the opposite spin quantum number. The restriction imposed by Pauli’s exclusion principle on the number of electrons in an orbital helps in calculating the capacity of electrons to be present in any subshell. For example, subshell 1s comprises one orbital and thus the maximum number of electrons present in 1s subshell can be two, in p and d subshells, the maximum number of electrons can be 6 and 10 and so on. This can be summed up as : the maximum number of electrons in the shell with principal quantum number n is equal to 2n2. Hund’s Rule of Maximum Multiplicity This rule deals with the filling of electrons into the orbitals belonging to the same subshell (that is, orbitals of equal energy, called Fig.2.17 Order of filling of orbitals degenerate orbitals). It states : pairing of Reprint 2025-26 structure of atom 63 electrons in the orbitals belonging to the 1s orbital. Its configuration is, therefore, 1s2. same subshell (p, d or f) does not take As mentioned above, the two electrons differ place until each orbital belonging to that from each other with opposite spin, as can be subshell has got one electron each i.e., it seen from the orbital diagram. is singly occupied. Since there are three p, five d and seven f orbitals, therefore, the pairing of electrons will The third electron of lithium (Li) is not start in the p, d and f orbitals with the entry allowed in the 1s orbital because of Pauli of 4th, 6th and 8th electron, respectively. It exclusion principle. It, therefore, takes the has been observed that half filled and fully next available choice, namely the 2s orbital. filled degenerate set of orbitals acquire extra The electronic configuration of Li is 1s22s1. stability due to their symmetry (see Section, The 2s orbital can accommodate one more 2.6.7). electron. The configuration of beryllium (Be) 2.6.5 Electronic Configuration of Atoms atom is, therefore, 1s2 2s2 (see Table 2.6, page 66 for the electronic configurations ofThe distribution of electrons into orbitals of an elements).atom is called its electronic configuration. In the next six elements—boronIf one keeps in mind the basic rules which (B, 1s22s22p1), carbon (C, 1s22s22p2), nitrogengovern the filling of different atomic orbitals, the electronic configurations of different (N, 1s22s22p3), oxygen (O, 1s22s22p4), fluorine atoms can be written very easily. (F, 1s22s22p5) and neon (Ne, 1s22s22p6), the 2p orbitals get progressively filled. This process The electronic configuration of different is completed with the neon atom. The orbital atoms can be represented in two ways. For picture of these elements can be represented example : as follows : (i) sa pbdc ...... notation (ii) Orbital diagram s p d In the first notation, the subshell is represented by the respective letter symbol and the number of electrons present in the subshell is depicted, as the super script, like a, b, c, ... etc. The similar subshell represented for different shells is differentiated by writing the principal quantum number before the respective subshell. In the second notation each orbital of the subshell is represented by a box and the electron is represented by an arrow (↑) a positive spin or an arrow (↓) The electronic configuration of the a negative spin. The advantage of second elements sodium (Na,1s22s22p63s1) to argon notation over the first is that it represents all (Ar,1s22s22p63s23p6), follow exactly the same the four quantum numbers. pattern as the elements from lithium to neon with the difference that the 3s and 3p orbitals The hydrogen atom has only one electron are getting filled now. This process can bewhich goes in the orbital with the lowest energy, namely 1s. The electronic configuration of the simplified if we represent the total number hydrogen atom is 1s1 meaning that it has of electrons in the first two shells by the one electron in the 1s orbital. The second name of element neon (Ne). The electronic electron in helium (He) can also occupy the configuration of the elements from sodium to Reprint 2025-26 64 chemistry argon can be written as (Na, [Ne]3s1) to (Ar, After this, filling of 6p, then 7s and finally 5f [Ne] 3s23p6). The electrons in the completely and 6d orbitals takes place. The elements filled shells are known as core electrons and after uranium (U) are all short-lived and all of the electrons that are added to the electronic them are produced artificially. The electronic shell with the highest principal quantum configurations of the known elements (as number are called valence electrons. For determined by spectroscopic methods) are example, the electrons in Ne are the core tabulated in Table 2.6 (page 66). electrons and the electrons from Na to Ar are One may ask what is the utility of knowing the valence electrons. In potassium (K) and the electron configuration? The modern calcium (Ca), the 4s orbital, being lower in approach to the chemistry, infact, depends energy than the 3d orbitals, is occupied by almost entirely on electronic distribution to one and two electrons respectively. understand and explain chemical behaviour. A new pattern is followed beginning with For example, questions like why two or more scandium (Sc). The 3d orbital, being lower atoms combine to form molecules, why some in energy than the 4p orbital, is filled first. elements are metals while others are non- Consequently, in the next ten elements, metals, why elements like helium and argon scandium (Sc), titanium (Ti), vanadium (V), are not reactive but elements like the halogens chromium (Cr), manganese (Mn), iron (Fe), are reactive, find simple explanation from the cobalt (Co), nickel (Ni), copper (Cu) and zinc electronic configuration. These questions have (Zn), the five 3d orbitals are progressively no answer in the Daltonian model of atom. occupied. We may be puzzled by the fact A detailed understanding of the electronic that chromium and copper have five and ten structure of atom is, therefore, very essential electrons in 3d orbitals rather than four and for getting an insight into the various aspects nine as their position would have indicated of modern chemical knowledge.with two-electrons in the 4s orbital. The reason is that fully filled orbitals and half- 2.6.6 Stability of Completely Filled and filled orbitals have extra stability (that is, Half Filled Subshells lower energy). Thus p3, p6, d5, d10,f 7, f14 etc. The ground state electronic configuration of configurations, which are either half-filled the atom of an element always corresponds to or fully filled, are more stable. Chromium the state of the lowest total electronic energy. and copper therefore adopt the d5 and The electronic configurations of most of the d10 configuration (Section 2.6.7)[caution: atoms follow the basic rules given in Section exceptions do exist] 2.6.5. However, in certain elements such as With the saturation of the 3d orbitals, Cu, or Cr, where the two subshells (4s and the filling of the 4p orbital starts at gallium 3d) differ slightly in their energies, an electron (Ga) and is complete at krypton (Kr). In the shifts from a subshell of lower energy (4s) to a next eighteen elements from rubidium (Rb) subshell of higher energy (3d), provided such to xenon (Xe), the pattern of filling the 5s, a shift results in all orbitals of the subshell 4d and 5p orbitals are similar to that of 4s, of higher energy getting either completely 3d and 4p orbitals as discussed above. Then filled or half filled. The valence electronic comes the turn of the 6s orbital. In caesium configurations of Cr and Cu, therefore, are (Cs) and the barium (Ba), this orbital contains 3d5 4s1 and 3d10 4s1 respectively and not 3d4 one and two electrons, respectively. Then from 4s2 and 3d9 4s2. It has been found that there is lanthanum (La) to mercury (Hg), the filling up extra stability associated with these electronic of electrons takes place in 4f and 5d orbitals. configurations. Reprint 2025-26 structure of atom 65 Causes of Stability of Completely Filled and Half-filled Subshells The completely filled and completely half-filled subshells are stable due to the following reasons: 1. Symmetrical distribution of electrons: It is well known that symmetry leads to stability. The completely filled or half filled subshells have symmetrical distribution of electrons in them and are therefore more stable. Electrons in the same subshell (here 3d) have equal energy but different spatial distribution. Consequently, their shielding of one- another is relatively small and the electrons are more strongly attracted by the nucleus. 2. Exchange Energy : The stabilizing effect arises whenever two or more electrons with the same spin are present in the degenerate orbitals of a subshell. These electrons tend to exchange their positions and the energy released due to this exchange is called exchange energy. The number of exchanges that can take place is maximum when the subshell is either half filled or completely filled (Fig. 2.18). As a result the exchange energy is maximum and so is the stability. You may note that the exchange energy is at the basis of Hund’s rule that electrons which enter orbitals of equal energy have parallel spins as far as possible. In other words, the extra stability of half-filled and completely filled subshell is due to: (i) relatively small shielding, (ii) smaller coulombic repulsion energy, and (iii) larger exchange energy. Details about the exchange energy will be Fig. 2.18 Possible exchange for a d5 configuration dealt with in higher classes. Reprint 2025-26 66 chemistry Table 2.6 Electronic Configurations of the Elements * Elements with exceptional electronic configurations Reprint 2025-26 structure of atom 67 ** Elements with atomic number 112 and above have been reported but not yet fully authenticated and named. Reprint 2025-26 68 chemistry Summary Atoms are the building blocks of elements. They are the smallest parts of an element that chemically react. The first atomic theory, proposed by John Dalton in 1808, regarded atom as the ultimate indivisible particle of matter. Towards the end of the nineteenth century, it was proved experimentally that atoms are divisible and consist of three fundamental particles: electrons, protons and neutrons. The discovery of sub-atomic particles led to the proposal of various atomic models to explain the structure of atom. Thomson in 1898 proposed that an atom consists of uniform sphere of positive electricity with electrons embedded into it. This model in which mass of the atom is considered to be evenly spread over the atom was proved wrong by Rutherford’s famous alpha-particle scattering experiment in 1909. Rutherford concluded that atom is made of a tiny positively charged nucleus, at its centre with electrons revolving around it in circular orbits. Rutherford model, which resembles the solar system, was no doubt an improvement over Thomson model but it could not account for the stability of the atom i.e., why the electron does not fall into the nucleus. Further, it was also silent about the electronic structure of atoms i.e., about the distribution and relative energies of electrons around the nucleus. The difficulties of the Rutherford model were overcome by Niels Bohr in 1913 in his model of the hydrogen atom. Bohr postulated that electron moves around the nucleus in circular orbits. Only certain orbits can exist and each orbit corresponds to a specific energy. Bohr calculated the energy of electron in various orbits and for each orbit predicted the distance between the electron and nucleus. Bohr model, though offering a satisfactory model for explaining the spectra of the hydrogen atom, could not explain the spectra of multi-electron atoms. The reason for this was soon discovered. In Bohr model, an electron is regarded as a charged particle moving in a well defined circular orbit about the nucleus. The wave character of the electron is ignored in Bohr’s theory. An orbit is a clearly defined path and this path can completely be defined only if both the exact position and the exact velocity of the electron at the same time are known. This is not possible according to the Heisenberg u principle. Bohr model of the hydrogen atom, therefore, not only ignores the dual behaviour of electron but also contradicts Heisenberg u Erwin Schrödinger, in 1926, proposed an equation called Schrödinger equation to describe the electron distributions in space and the allowed energy levels in atoms. This equation incorporates de Broglie’s concept of wave-particle duality and is consistent with Heisenberg u hydrogen atom, the solution gives the possible energy states the electron can occupy [and the corresponding wave function(s) (ψ) (which in fact are the mathematical functions) of the electron associated with each energy state]. These quantized energy states and corresponding wave functions which are characterized by a set of three quantum numbers (principal quantum number n, azimuthal quantum number l and magnetic quantum number ml) arise as a natural consequence in the solution of the Schrödinger equation. The restrictions on the values of these three quantum numbers also come naturally from this solution. The quantum mechanical model of the hydrogen atom successfully predicts all aspects of the hydrogen atom spectrum including some phenomena that could not be explained by the Bohr model. According to the quantum mechanical model of the atom, the electron distribution of an atom containing a number of electrons is divided into shells. The shells, in turn, are thought to consist of one or more subshells and subshells are assumed to be composed of one or more orbitals, which the electrons occupy. While for hydrogen and hydrogen like systems (such as He+, Li2+ etc.) all the orbitals within a given shell have same energy, the energy of the orbitals in a multi-electron atom depends upon the values of n and l: The lower the value of (n + l ) for an orbital, the lower is its energy. If two orbitals have the same (n + l ) value, the orbital with lower value of n has the lower energy. In an atom many such orbitals are Reprint 2025-26 structure of atom 69 possible and electrons are filled in those orbitals in order of increasing energy in accordance with Pauli exclusion principle (no two electrons in an atom can have the same set of four quantum numbers) and Hund’s rule of maximum multiplicity (pairing of electrons in the orbitals belonging to the same subshell does not take place until each orbital belonging to that subshell has got one electron each, i.e., is singly occupied). This forms the basis of the electronic structure of atoms. EXERCISES 2.1 (i) Calculate the number of electrons which will together weigh one gram. (ii) Calculate the mass and charge of one mole of electrons. 2.2 (i) Calculate the total number of electrons present in one mole of methane. (ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron = 1.675 × 10–27 kg). (iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP. Will the answer change if the temperature and pressure are changed ? 2.3 How many neutrons and protons are there in the following nuclei ? 13C6 , 168 O, 2412 Mg, 5626 Fe, 8838 Sr 2.4 Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A) (i) Z = 17, A = 35. (ii) Z = 92, A = 233. (iii) Z = 4, A = 9. 2.5 Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wavenumber ( ) of the yellow light. 2.6 Find energy of each of the photons which (i) correspond to light of frequency 3×1015 Hz. (ii) have wavelength of 0.50 Å. 2.7 Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0 × 10–10 s. 2.8 What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy? 2.9 A photon of wavelength 4 × 10–7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV= 1.6020 × 10–19 J). 2.10 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol–1. 2.11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm. Calculate the rate of emission of quanta per second. 2.12 Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0 ) and work function (W0 ) of the metal. 2.13 What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2? Reprint 2025-26 70 chemistry 2.14 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit). 2.15 What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state? 2.16 (i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18 J atom–1. What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom. 2.17 Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen. 2.18 What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10–11 ergs. 2.19 The electron energy in hydrogen atom is given by En = (–2.18 × 10–18 )/n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition? 2.20 Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 m s–1. 2.21 The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength. 2.22 Which of the following are isoelectronic species i.e., those having the same number of electrons? Na+, K+, Mg2+, Ca2+, S2–, Ar. 2.23 (i) Write the electronic configurations of the following ions: (a) H– (b) Na+ (c) O2– (d) F– (ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5 ? (iii) Which atoms are indicated by the following configurations ? (a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1. 2.24 What is the lowest value of n that allows g orbitals to exist? 2.25 An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron. 2.26 An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element. 2.27 Give the number of electrons in the species 2.28 (i) An atomic orbital has n = 3. What are the possible values of l and ml ? (ii) List the quantum numbers (ml and l) of electrons for 3d orbital. (iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f 2.29 Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n=1, l=0; (b) n = 3; l=1 (c) n = 4; l =2; (d) n=4; l=3. 2.30 Explain, giving reasons, which of the following sets of quantum numbers are not possible. (a) n = 0, l = 0, ml = 0, ms = + ½ (b) n = 1, l = 0, ml = 0, ms = – ½ (c) n = 1, l = 1, ml = 0, ms = + ½ (d) n = 2, l = 1, ml = 0, ms = – ½ Reprint 2025-26 structure of atom 71 (e) n = 3, l = 3, ml = –3, ms = + ½ (f) n = 3, l = 1, ml = 0, ms = + ½ 2.31 How many electrons in an atom may have the following quantum numbers? (a) n = 4, ms = – ½ (b) n = 3, l = 0 2.32 Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit. 2.33 What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum ? 2.34 Calculate the energy required for the process He+ (g)  He2+ (g) + e– The ionization energy for the H atom in the ground state is 2.18 × 10–18 J atom–1 2.35 If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long. 2.36 2 ×108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm. 2.37 The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise. 2.38 A certain particle carries 2.5 × 10–16C of static electric charge. Calculate the number of electrons present in it. 2.39 In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10–18C, calculate the number of electrons present on it. 2.40 In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ? 2.41 Symbols 79Br35 and 79Br can be written, whereas symbols 7935Br and 35Br are not acceptable. Answer briefly. 2.42 An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol. 2.43 An ion with mass number 37 possesses one unit of negative charge. If the ion conatins 11.1% more neutrons than the electrons, find the symbol of the ion. 2.44 An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion. 2.45 Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays. 2.46 Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 1024, calculate the power of this laser. 2.47 Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy. Reprint 2025-26 72 chemistry 2.48 In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10–18 J from the radiations of 600 nm, calculate the number of photons received by the detector. 2.49 Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source. 2.50 The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calcualte the frequency of each transition and energy difference between two excited states. 2.51 The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron. 2.52 Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant. λ (nm) 500 450 400 v × 10–5 (cm s–1) 2.55 4.35 5.35 2.53 The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal. 2.54 If the photon of the wavelength 150 pm strikes an atom and one of tis inner bound electrons is ejected out with a velocity of 1.5 × 107 m s–1, calculate the energy with which it is bound to the nucleus. 2.55 Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represeted as v = 3.29 × 1015 (Hz) [1/32 – 1/n2] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum. 2.56 Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum. 2.57 Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 106 ms–1, calculate de Broglie wavelength associated with this electron. 2.58 Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron. 2.59 If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms–1, calculate the de Broglie wavelength associated with it. 2.60 The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms–1. If the hockey ball of mass 0.1 kg is moving with this velocity, calcualte the wavelength associated with this velocity. 2.61 If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the u electron is h/4πm × 0.05 nm, is there any problem in defining this value. 2.62 The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists: 1. n = 4, l = 2, ml = –2 , ms = –1/2 2. n = 3, l = 2, ml = 1 , ms = +1/2 Reprint 2025-26 structure of atom 73 3. n = 4, l = 1, ml = 0 , ms = +1/2 4. n = 3, l = 2, ml = –2 , ms = –1/2 5. n = 3, l = 1, ml = –1 , ms = +1/2 6. n = 4, l = 1, ml = 0 , ms = +1/2 2.63 The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge ? 2.64 Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p. 2.65 The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ? 2.66 Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr. 2.67 (a) How many subshells are associated with n = 4 ? (b) How many electrons will be present in the subshells having ms value of –1/2 for n = 4 ? Reprint 2025-26 Unit 3 Classification of Elements and Periodicity in Properties The Periodic Table is arguably the most important concept in chemistry, both in principle and in practice. It is the everyday support for students, it suggests new avenues of research to professionals, and it provides a succinct After studying this Unit, you will be organization of the whole of chemistry. It is a remarkable able to demonstration of the fact that the chemical elements are not a random cluster of entities but instead display trends• appreciate how the concept of and lie together in families. An awareness of the Periodic grouping elements in accordance Table is essential to anyone who wishes to disentangle to their properties led to the the world and see how it is built up from the fundamental development of Periodic Table. building blocks of the chemistry, the chemical elements. • understand the Periodic Law; • understand the significance of Glenn T. Seaborg atomic number and electronic configuration as the basis for periodic classification; In this Unit, we will study the historical development of the• n a m e t h e e l e m e n t s w i t h Periodic Table as it stands today and the Modern Periodic Z >100 according to IUPAC nomenclature; Law. We will also learn how the periodic classification follows as a logical consequence of the electronic• classify elements into s, p, d, configuration of atoms. Finally, we shall examine some of f blocks and learn their main characteristics; the periodic trends in the physical and chemical properties of the elements.• recognise the periodic trends in physical and chemical properties 3.1 WHY DO WE NEED TO CLASSIFY ELEMENTS ? of elements; We know by now that the elements are the basic units of• compare the reactivity of elements and correlate it with their all types of matter. In 1800, only 31 elements were known. occurrence in nature; By 1865, the number of identified elements had more than • explain the relationship between doubled to 63. At present 114 elements are known. Of ionization enthalpy and metallic them, the recently discovered elements are man-made. character; Efforts to synthesise new elements are continuing. With • use scientific vocabulary such a large number of elements it is very difficult to appropriately to communicate study individually the chemistry of all these elements and ideas related to certain important their innumerable compounds individually. To ease out p r o p e r t i e s o f a t o m s e . g . , this problem, scientists searched for a systematic way to atomic/ionic radii, ionization organise their knowledge by classifying the elements. Not enthalpy, electron gain enthalpy, only that it would rationalize known chemical facts about electronegativity, valence of elements. elements, but even predict new ones for undertaking further study. Classification of Elements and Periodicity in Properties 75

4.5 Valence Bond Theory of the valence bond theory is based on the knowledge of atomic orbitals, electronicAs we know that Lewis approach helps in configurations of elements (Units 2), thewriting the structure of molecules but it overlap criteria of atomic orbitals, thefails to explain the formation of chemical hybridization of atomic orbitals and thebond. It also does not give any reason for the principles of variation and superposition. Adifference in bond dissociation enthalpies and rigorous treatment of the VB theory in termsbond lengths in molecules like H2 (435.8 kJ of these aspects is beyond the scope of this mol-1, 74 pm) and F2 (155 kJ mol-1, 144 pm), book. Therefore, for the sake of convenience,although in both the cases a single covalent valence bond theory has been discussed in bond is formed by the sharing of an electron terms of qualitative and non-mathematical pair between the respective atoms. It also treatment only. To start with, let us consider gives no idea about the shapes of polyatomic the formation of hydrogen molecule which is molecules. the simplest of all molecules. Similarly the VSEPR theory gives the Consider two hydrogen atoms A and B geometry of simple molecules but theoretically, approaching each other having nuclei NA it does not explain them and also it has limited and NB and electrons present in them are applications. To overcome these limitations represented by eA and eB. When the two atoms the two important theories based on quantum are at large distance from each other, there mechanical principles are introduced. These is no interaction between them. As these two are valence bond (VB) theory and molecular atoms approach each other, new attractive orbital (MO) theory. and repulsive forces begin to operate. Valence bond theory was introduced Attractive forces arise between: by Heitler and London (1927) and developed (i) nucleus of one atom and its own electron further by Pauling and others. A discussion that is NA – eA and NB– eB. Reprint 2025-26 118 chemistry (ii) nucleus of one atom and electron of together to form a stable molecule having the other atom i.e., NA– eB, NB– eA. bond length of 74 pm. Similarly repulsive forces arise between Since the energy gets released when the bond is formed between two hydrogen atoms,(i) electrons of two atoms like eA – eB, the hydrogen molecule is more stable than (ii) nuclei of two atoms NA – NB. that of isolated hydrogen atoms. The energy Attractive forces tend to bring the two so released is called as bond enthalpy, which atoms close to each other whereas repulsive is corresponding to minimum in the curve forces tend to push them apart (Fig. 4.7). depicted in Fig. 4.8. Conversely, 435.8 kJ of energy is required to dissociate one mole of H2 molecule. H2(g) + 435.8 kJ mol–1 → H(g) + H(g) Fig. 4.8 The potential energy curve for the formation of H2 molecule as a function of internuclear distance of the H atoms. The minimum in the curve corresponds to the most stable state of H2. 4.5.1 Orbital Overlap Concept In the formation of hydrogen molecule, there is a minimum energy state when two hydrogen atoms are so near that their atomic orbitals undergo partial interpenetration. ThisFig. 4.7 Forces of attraction and repulsion partial merging of atomic orbitals is called during the formation of H2 molecule overlapping of atomic orbitals which results in Experimentally it has been found that the pairing of electrons. The extent of overlap the magnitude of new attractive force is decides the strength of a covalent bond. Inmore than the new repulsive forces. As a general, greater the overlap the stronger is theresult, two atoms approach each other and bond formed between two atoms. Therefore,potential energy decreases. Ultimately a stage is reached where the net force of attraction according to orbital overlap concept, the balances the force of repulsion and system formation of a covalent bond between two acquires minimum energy. At this stage atoms results by pairing of electrons present two hydrogen atoms are said to be bonded in the valence shell having opposite spins. Reprint 2025-26 Chemical Bonding And Molecular Structure 119 4.5.2 Directional Properties of Bonds As we have already seen, the covalent bond is formed by overlapping of atomic orbitals. The molecule of hydrogen is formed due to the overlap of 1s-orbitals of two H atoms. In case of polyatomic molecules like CH4, NH3 and H2O, the geometry of the molecules is also important in addition to the bond formation. For example why is it so that CH4 molecule has tetrahedral shape and HCH bond angles are 109.5°? Why is the shape of NH3 molecule pyramidal ? The valence bond theory explains the shape, the formation and directional properties of bonds in polyatomic molecules like CH4, NH3 and H2O, etc. in terms of overlap and hybridisation of atomic orbitals. 4.5.3 Overlapping of Atomic Orbitals When orbitals of two atoms come close to form bond, their overlap may be positive, negative or zero depending upon the sign (phase) and direction of orientation of amplitude of orbital wave function in space (Fig. 4.9). Positive and negative sign on boundary surface diagrams in the Fig. 4.9 show the sign (phase) of orbital wave function and are not related to charge. Fig.4.9 Positive, negative and zero overlaps ofOrbitals forming bond should have same sign s and p atomic orbitals(phase) and orientation in space. This is called positive overlap. Various overlaps of s and p hydrogen. The four atomic orbitals of carbon, orbitals are depicted in Fig. 4.9. each with an unpaired electron can overlap with the 1s orbitals of the four H atoms which The criterion of overlap, as the main factor are also singly occupied. This will result in the for the formation of covalent bonds applies formation of four C-H bonds. It will, however, uniformly to the homonuclear/heteronuclear be observed that while the three p orbitals of diatomic molecules and polyatomic molecules. carbon are at 90° to one another, the HCH We know that the shapes of CH4, NH3, and angle for these will also be 90°. That is three H2O molecules are tetrahedral, pyramidal C-H bonds will be oriented at 90° to one and bent respectively. It would be therefore another. The 2s orbital of carbon and the 1s interesting to use VB theory to find out if these orbital of H are spherically symmetrical and geometrical shapes can be explained in terms they can overlap in any direction. Therefore of the orbital overlaps. the direction of the fourth C-H bond cannot Let us first consider the CH4 (methane) be ascertained. This description does not fit molecule. The electronic configuration of in with the tetrahedral HCH angles of 109.5°. carbon in its ground state is [He]2s2 2p2 which Clearly, it follows that simple atomic orbital in the excited state becomes [He] 2s1 2px1 2py1 overlap does not account for the directional 2pz1. The energy required for this excitation is characteristics of bonds in CH4. Using similar compensated by the release of energy due to procedure and arguments, it can be seen that in overlap between the orbitals of carbon and the the case of NH3 and H2O molecules, the HNH Reprint 2025-26 120 chemistry and HOH angles should be 90°. This is in above and below the plane of the disagreement with the actual bond angles of participating atoms. 107° and 104.5° in the NH3 and H2O molecules respectively. 4.5.4 Types of Overlapping and Nature of Covalent Bonds The covalent bond may be classified into two types depending upon the types of overlapping: (i) Sigma(σ) bond, and (ii) pi(π) bond (i) Sigma(σ) bond : This type of covalent 4.5.5 Strength of Sigma and pi Bonds bond is formed by the end to end (head- Basically the strength of a bond depends on) overlap of bonding orbitals along the upon the extent of overlapping. In case of internuclear axis. This is called as head sigma bond, the overlapping of orbitals takes on overlap or axial overlap. This can be place to a larger extent. Hence, it is stronger formed by any one of the following types as compared to the pi bond where the extent of combinations of atomic orbitals. of overlapping occurs to a smaller extent. • s-s overlapping : In this case, there is Further, it is important to note that in the overlap of two half filled s-orbitals along formation of multiple bonds between two the internuclear axis as shown below : atoms of a molecule, pi bond(s) is formed in addition to a sigma bond. 4.6 Hybridisation In order to explain the characteristic geometrical shapes of polyatomic molecules • s-p overlapping: This type of overlap like CH4, NH3 and H2O etc., Pauling introduced occurs between half filled s-orbitals of one the concept of hybridisation. According to him atom and half filled p-orbitals of another the atomic orbitals combine to form new set of atom. equivalent orbitals known as hybrid orbitals. Unlike pure orbitals, the hybrid orbitals are used in bond formation. The phenomenon is known as hybridisation which can be defined as the process of intermixing of the orbitals of • p–p overlapping : This type of overlap slightly different energies so as to redistribute takes place between half filled p-orbitals their energies, resulting in the formation of of the two approaching atoms. new set of orbitals of equivalent energies and shape. For example when one 2s and three 2p-orbitals of carbon hybridise, there is the formation of four new sp3 hybrid orbitals. Salient features of hybridisation: The main (ii) pi( ) bond : In the formation of π bond features of hybridisation are as under : the atomic orbitals overlap in such a 1. The number of hybrid orbitals is equal to way that their axes remain parallel to the number of the atomic orbitals that get each other and perpendicular to the internuclear axis. The orbitals formed hybridised. due to sidewise overlapping consists 2. The hybridised orbitals are always of two saucer type charged clouds equivalent in energy and shape. Reprint 2025-26 Chemical Bonding And Molecular Structure 121 3. The hybrid orbitals are more effective in vacant 2p orbital to account for its bivalency. forming stable bonds than the pure atomic One 2s and one 2p-orbital gets hybridised to orbitals. form two sp hybridised orbitals. These two sp hybrid orbitals are oriented in opposite4. These hybrid orbitals are directed in direction forming an angle of 180°. Each of space in some preferred direction to have the sp hybridised orbital overlaps with the minimum repulsion between electron 2p-orbital of chlorine axially and form two Be- pairs and thus a stable arrangement. Cl sigma bonds. This is shown in Fig. 4.10. Therefore, the type of hybridisation indicates the geometry of the molecules. Important conditions for hybridisation (i) The orbitals present in the valence shell of the atom are hybridised. (ii) The orbitals undergoing hybridisation Be should have almost equal energy. (iii) Promotion of electron is not essential condition prior to hybridisation. (iv) It is not necessary that only half filled orbitals participate in hybridisation. In some cases, even filled orbitals of valence shell take part in hybridisation. Fig.4.10 (a) Formation of sp hybrids from s and 4.6.1 Types of Hybridisation p orbitals; (b) Formation of the linear There are various types of hybridisation BeCl2 molecule involving s, p and d orbitals. The different (II) sp2 hybridisation : In this hybridisationtypes of hybridisation are as under: there is involvement of one s and two (I) sp hybridisation: This type of hybridisation p-orbitals in order to form three equivalent involves the mixing of one s and one p orbital sp2 hybridised orbitals. For example, in resulting in the formation of two equivalent BCl3 molecule, the ground state electronicsp hybrid orbitals. The suitable orbitals for configuration of central boron atom is sp hybridisation are s and pz, if the hybrid 1s22s22p1. In the excited state, one of the 2s orbitals are to lie along the z-axis. Each sp electrons is promoted to vacant 2p orbital as hybrid orbitals has 50% s-character and 50% p-character. Such a molecule in which the central atom is sp-hybridised and linked directly to two other central atoms possesses linear geometry. This type of hybridisation is also known as diagonal hybridisation. The two sp hybrids point in the opposite direction along the z-axis with projecting positive lobes and very small negative lobes, which provides more effective overlapping resulting in the formation of stronger bonds. Example of molecule having sp hybridisation BeCl 2: The ground state electronic configuration of Be is 1s22s2. In the exited Fig.4.11 Formation of sp2 hybrids and the BCl3 state one of the 2s-electrons is promoted to molecule Reprint 2025-26 122 chemistry a result boron has three unpaired electrons. ground state is 2S 2 2 p 1x 2 p 1y 2 p 1z having threeThese three orbitals (one 2s and two 2p) unpaired electrons in the sp3 hybrid orbitalshybridise to form three sp2 hybrid orbitals. and a lone pair of electrons is present in theThe three hybrid orbitals so formed are fourth one. These three hybrid orbitals overlaporiented in a trigonal planar arrangement with 1s orbitals of hydrogen atoms to formand overlap with 2p orbitals of chlorine to three N–H sigma bonds. We know that the form three B-Cl bonds. Therefore, in BCl3 force of repulsion between a lone pair and a(Fig. 4.11), the geometry is trigonal planar bond pair is more than the force of repulsionwith ClBCl bond angle of 120°. between two bond pairs of electrons. The (III) sp 3 hybridisation: This type of molecule thus gets distorted and the bond hybridisation can be explained by taking the angle is reduced to 107° from 109.5°. The example of CH4 molecule in which there is geometry of such a molecule will be pyramidal mixing of one s-orbital and three p-orbitals as shown in Fig. 4.13. of the valence shell to form four sp3 hybrid orbital of equivalent energies and shape. There is 25% s-character and 75% p-character in each sp3 hybrid orbital. The four sp3 hybrid orbitals so formed are directed towards the four corners of the tetrahedron. The angle between sp3 hybrid orbital is 109.5° as shown in Fig. 4.12. Fig.4.13 Formation of NH3 molecule In case of H2O molecule, the four oxygen orbitals (one 2s and three 2p) undergo sp3 hybridisation forming four sp3 hybrid orbitals out of which two contain one electron each and the other two contain a pair of electrons. These σ four sp3 hybrid orbitals acquire a tetrahedral geometry, with two corners occupied by σ σ hydrogen atoms while the other two by the lone pairs. The bond angle in this case is reduced to 104.5° from 109.5° (Fig. 4.14) σ and the molecule thus acquires a V-shape or angular geometry. Fig.4.12 Formation of sp3 hybrids by the combination of s, px , py and pz atomic orbitals of carbon and the formation of CH4 molecule The structure of NH3 and H2O molecules can also be explained with the help of sp3 hybridisation. In NH3, the valence shell (outer) electronic configuration of nitrogen in the Fig.4.14 Formation of H2O molecule Reprint 2025-26 Chemical Bonding And Molecular Structure 123 4.6.2 Other Examples of sp3, sp2 and sp used for making sp2–s sigma bond with two Hybridisation hydrogen atoms. The unhybridised orbital (2px sp3 Hybridisation in C2H6 molecule: In or 2py) of one carbon atom overlaps sidewise ethane molecule both the carbon atoms with the similar orbital of the other carbon assume sp3 hybrid state. One of the four atom to form weak π bond, which consists of sp3 hybrid orbitals of carbon atom overlaps two equal electron clouds distributed above axially with similar orbitals of other atom to and below the plane of carbon and hydrogen form sp3-sp3 sigma bond while the other three atoms. hybrid orbitals of each carbon atom are used Thus, in ethene molecule, the carbon-in forming sp3–s sigma bonds with hydrogen carbon bond consists of one sp2–sp2 sigmaatoms as discussed in section 4.6.1(iii). bond and one pi (π ) bond between p orbitalsTherefore in ethane C–C bond length is 154 which are not used in the hybridisation andpm and each C–H bond length is 109 pm. are perpendicular to the plane of molecule; thesp2 Hybridisation in C2H4: In the formation bond length 134 pm. The C–H bond is sp2–sof ethene molecule, one of the sp2 hybrid sigma with bond length 108 pm. The H–C–Horbitals of carbon atom overlaps axially with bond angle is 117.6° while the H–C–C anglesp2 hybridised orbital of another carbon atom is 121°. The formation of sigma and pi bondsto form C–C sigma bond. While the other two sp2 hybrid orbitals of each carbon atom are in ethene is shown in Fig. 4.15. Fig. 4.15 Formation of sigma and pi bonds in ethene Reprint 2025-26 124 chemistry sp Hybridisation in C2H2 : In the formation 4.6.3 Hybridisation of Elements of ethyne molecule, both the carbon atoms involving d Orbitals undergo sp-hybridisation having two The elements present in the third period unhybridised orbital i.e., 2py and 2px. contain d orbitals in addition to s and p orbitals. The energy of the 3d orbitals are One sp hybrid orbital of one carbon atom comparable to the energy of the 3s and 3poverlaps axially with sp hybrid orbital of the orbitals. The energy of 3d orbitals are also other carbon atom to form C–C sigma bond, comparable to those of 4s and 4p orbitals. while the other hybridised orbital of each As a consequence the hybridisation involving carbon atom overlaps axially with the half either 3s, 3p and 3d or 3d, 4s and 4p is filled s orbital of hydrogen atoms forming possible. However, since the difference in σ bonds. Each of the two unhybridised p energies of 3p and 4s orbitals is significant, no orbitals of both the carbon atoms overlaps hybridisation involving 3p, 3d and 4s orbitals sidewise to form two π bonds between the is possible. carbon atoms. So the triple bond between the The important hybridisation schemes two carbon atoms is made up of one sigma involving s, p and d orbitals are summarised and two pi bonds as shown in Fig. 4.16. below: Shape of Hybridisation Atomic molecules/ Examples type orbitals ions Square dsp2 d+s+p(2) [Ni(CN)4]2–, planar [Pt(Cl)4]2– Trigonal sp3d s+p(3)+d PF5, PCl5 bipyramidal Square sp3d2 s+p(3)+d(2) BrF5 pyramidal Octahedral sp3d2 s+p(3)+d(2) SF6, [CrF6]3– d2sp3 d(2)+s+p(3) [Co(NH3)6]3+ (i) Formation of PCl5 (sp3d hybridisation): The ground state and the excited state outer electronic configurations of phosphorus (Z=15) are represented below. Fig.4.16 Formation of sigma and pi bonds in sp3d hybrid orbitals filled by electron pairs ethyne donated by five Cl atoms. Reprint 2025-26 Chemical Bonding And Molecular Structure 125 Now the five orbitals (i.e., one s, three six sp3d2 hybrid orbitals overlap with singly p and one d orbitals) are available for occupied orbitals of fluorine atoms to form hybridisation to yield a set of five sp3d hybrid six S–F sigma bonds. Thus SF6 molecule has orbitals which are directed towards the five a regular octahedral geometry as shown in corners of a trigonal bipyramidal as depicted Fig. 4.18. in the Fig. 4.17. sp3d2 hybridisation Fig. 4.17 Trigonal bipyramidal geometry of PCl5 molecule It should be noted that all the bond angles in trigonal bipyramidal geometry are not equivalent. In PCl5 the five sp3d orbitals of phosphorus overlap with the singly occupied p orbitals of chlorine atoms to form five P–Cl sigma bonds. Three P–Cl bond lie in one plane and make an angle of 120° with each other; these bonds are termed as equatorial Fig. 4.18 Octahedral geometry of SF6 moleculebonds. The remaining two P–Cl bonds–one lying above and the other lying below the equatorial plane, make an angle of 90° with 4.7 Molecular Orbital Theory the plane. These bonds are called axial bonds. Molecular orbital (MO) theory was developed As the axial bond pairs suffer more repulsive by F. Hund and R.S. Mulliken in 1932. The interaction from the equatorial bond pairs, salient features of this theory are : therefore axial bonds have been found to (i) The electrons in a molecule are present be slightly longer and hence slightly weaker in the various molecular orbitals as the than the equatorial bonds; which makes PCl5 electrons of atoms are present in the molecule more reactive. various atomic orbitals. (ii) Formation of SF6 (sp3d2 hybridisation): (ii) The atomic orbitals of comparableIn SF6 the central sulphur atom has the energies and proper symmetry combineground state outer electronic configuration to form molecular orbitals.3s23p4. In the exited state the available six orbitals i.e., one s, three p and two d are (iii) While an electron in an atomic orbital singly occupied by electrons. These orbitals is influenced by one nucleus, in a hybridise to form six new sp3d2 hybrid molecular orbital it is influenced by orbitals, which are projected towards the six two or more nuclei depending upon the corners of a regular octahedron in SF6. These number of atoms in the molecule. Thus, Reprint 2025-26 126 chemistry an atomic orbital is monocentric while ψA and ψB. Mathematically, the formation of a molecular orbital is polycentric. molecular orbitals may be described by the linear combination of atomic orbitals that can(iv) The number of molecular orbital formed take place by addition and by subtraction of is equal to the number of combining wave functions of individual atomic orbitals atomic orbitals. When two atomic as shown below : orbitals combine, two molecular orbitals are formed. One is known as bonding ψMO = ψA + ψB molecular orbital while the other is Therefore, the two molecular orbitals called antibonding molecular orbital. σ and σ* are formed as : (v) The bonding molecular orbital has σ = ψA + ψB lower energy and hence greater stability σ* = ψA – ψB than the corresponding antibonding The molecular orbital σ formed by the molecular orbital. addition of atomic orbitals is called the bonding (vi) Just as the electron probability molecular orbital while the molecular orbital distribution around a nucleus in an σ* formed by the subtraction of atomic orbital atom is given by an atomic orbital, the is called antibonding molecular orbital as electron probability distribution around depicted in Fig. 4.19. a group of nuclei in a molecule is given by a molecular orbital. (vii) The molecular orbitals like atomic orbitals are filled in accordance with the aufbau principle obeying the Pauli’s exclusion principle and the Hund’s rule. 4.7.1 Formation of Molecular Orbitals Linear Combination of Atomic σ* = ψA – ψB Orbitals (LCAO) According to wave mechanics, the atomic ψA ψBorbitals can be expressed by wave functions (ψ ’s) which represent the amplitude of the σ = ψA + ψBelectron waves. These are obtained from the solution of Schrödinger wave equation. However, since it cannot be solved for any system containing more than one electron, molecular orbitals which are one electron wave functions for molecules are difficult Fig.4.19 Formation of bonding (σ) and antibonding (σ*) molecular orbitals by the linearto obtain directly from the solution of combination of atomic orbitals ψA andSchrödinger wave equation. To overcome this problem, an approximate method known ψB centered on two atoms A and B respectively.as linear combination of atomic orbitals (LCAO) has been adopted. Qualitatively, the formation of molecular Let us apply this method to the orbitals can be understood in terms of the homonuclear diatomic hydrogen molecule. constructive or destructive interference of the Consider the hydrogen molecule consisting electron waves of the combining atoms. In the of two atoms A and B. Each hydrogen atom formation of bonding molecular orbital, the in the ground state has one electron in 1s two electron waves of the bonding atoms orbital. The atomic orbitals of these atoms reinforce each other due to constructive may be represented by the wave functions interference while in the formation of Reprint 2025-26 Chemical Bonding And Molecular Structure 127 antibonding molecular orbital, the electron as the molecular axis. It is important to note waves cancel each other due to destructive that atomic orbitals having same or nearly interference. As a result, the electron density in the same energy will not combine if they do a bonding molecular orbital is located between not have the same symmetry. For example, the nuclei of the bonded atoms because of 2pz orbital of one atom can combine with 2pz which the repulsion between the nuclei is very orbital of the other atom but not with the less while in case of an antibonding molecular 2px or 2py orbitals because of their different orbital, most of the electron density is located symmetries. away from the space between the nuclei. 3. The combining atomic orbitals must Infact, there is a nodal plane (on which the overlap to the maximum extent. Greater electron density is zero) between the nuclei the extent of overlap, the greater will be the and hence the repulsion between the nuclei is electron-density between the nuclei of a high. Electrons placed in a bonding molecular molecular orbital. orbital tend to hold the nuclei together and 4.7.3 Types of Molecular Orbitalsstabilise a molecule. Therefore, a bonding molecular orbital always possesses lower Molecular orbitals of diatomic molecules are energy than either of the atomic orbitals that designated as σ (sigma), π (pi), δ(delta), etc. have combined to form it. In contrast, the In this nomenclature, the sigma ( ) electrons placed in the antibonding molecular molecular orbitals are symmetrical around orbital destabilise the molecule. This is the bond-axis while pi ( ) molecular orbitals because the mutual repulsion of the electrons are not symmetrical. For example, the linear in this orbital is more than the attraction combination of 1s orbitals centered on two between the electrons and the nuclei, which nuclei produces two molecular orbitals which causes a net increase in energy. are symmetrical around the bond-axis. Such It may be noted that the energy of the molecular orbitals are of the σ type and are antibonding orbital is raised above the designated as σ1s and σ*1s [Fig. 4.20(a), page energy of the parent atomic orbitals that 124]. If internuclear axis is taken to be in have combined and the energy of the bonding the z-direction, it can be seen that a linear orbital has been lowered than the parent combination of 2pz- orbitals of two atoms orbitals. The total energy of two molecular also produces two sigma molecular orbitals orbitals, however, remains the same as that designated as 2pz and *2pz. [Fig. 4.20(b)] of two original atomic orbitals. Molecular orbitals obtained from 2px and 4.7.2 Conditions for the Combination of 2py orbitals are not symmetrical around the Atomic Orbitals bond axis because of the presence of positive lobes above and negative lobes below theThe linear combination of atomic orbitals to molecular plane. Such molecular orbitals,form molecular orbitals takes place only if the are labelled as π and =π* [Fig. 4.20(c)]. Afollowing conditions are satisfied: π bonding MO has larger electron density1. The combining atomic orbitals must above and below the inter-nuclear axis. Thehave the same or nearly the same energy. π* antibonding MO has a node between theThis means that 1s orbital can combine with nuclei.another 1s orbital but not with 2s orbital because the energy of 2s orbital is appreciably 4.7.4 Energy Level Diagram for Molecular higher than that of 1s orbital. This is not true Orbitals if the atoms are very different. We have seen that 1s atomic orbitals on two 2. The combining atomic orbitals must atoms form two molecular orbitals designated have the same symmetry about the as σ1s and σ*1s. In the same manner, the 2s molecular axis. By convention z-axis is taken and 2p atomic orbitals (eight atomic orbitals Reprint 2025-26 128 chemistry Fig. 4.20 Contours and energies of bonding and antibonding molecular orbitals formed through combinations of (a) 1s atomic orbitals; (b) 2pz atomic orbitals and (c) 2px atomic orbitals. on two atoms) give rise to the following eight The energy levels of these molecular molecular orbitals: orbitals have been determined experimentally from spectroscopic data for homonuclearAntibonding MOs σ∗2s σ∗2pz π∗2px π∗2py diatomic molecules of second row elements Bonding MOs σ2s σ2pz π2px π2py of the periodic table. The increasing order of Reprint 2025-26 Chemical Bonding And Molecular Structure 129 energies of various molecular orbitals for O2 The rules discussed above regarding the and F2 is given below: stability of the molecule can be restated in terms of bond order as follows: A positive bond1s <  ∗1s <  2s <  ∗2s < 2pz < (π 2px=π 2py) order (i.e., Nb > Na) means a stable molecule < (π ∗2px= π∗ 2py) <  ∗2pz while a negative (i.e., Nb<Na) or zero (i.e., However, this sequence of energy levels Nb = Na) bond order means an unstable of molecular orbitals is not correct for the molecule. remaining molecules Li2, Be2, B2, C2, N2. For Nature of the bond instance, it has been observed experimentally Integral bond order values of 1, 2 or 3 that for molecules such as B2, C2, N2, etc. correspond to single, double or triple bonds the increasing order of energies of various respectively as studied in the classical molecular orbitals is concept. 1s <  ∗1s < 2s <  ∗2s < (π 2 px = π 2 py) Bond-length < 2pz < (π ∗2px =π∗2py) <  ∗2pz The bond order between two atoms in a The important characteristic feature molecule may be taken as an approximate of this order is that the energy of 2pz measure of the bond length. The bond length molecular orbital is higher than that decreases as bond order increases. of 2px and 2py molecular orbitals. Magnetic nature 4.7.5 Electronic Configuration and If all the molecular orbitals in a molecule are Molecular Behaviour doubly occupied, the substance is diamagnetic (repelled by magnetic field). However if one orThe distribution of electrons among various more molecular orbitals are singly occupied itmolecular orbitals is called the electronic is paramagnetic (attracted by magnetic field),configuration of the molecule. From the e.g., O2 molecule.electronic configuration of the molecule, it is possible to get important information about 4.8 BONDING IN SOME HOMONUCLEAR the molecule as discussed below. DIATOMIC MOLECULES Stability of Molecules: If Nb is the number In this section we shall discuss bonding in of electrons occupying bonding orbitals and some homonuclear diatomic molecules. Na the number occupying the antibonding 1. Hydrogen molecule (H2 ): It is formed byorbitals, then the combination of two hydrogen atoms. Each (i) the molecule is stable if Nb is greater hydrogen atom has one electron in 1s orbital. than Na, and Therefore, in all there are two electrons in (ii) the molecule is unstable if Nb is less hydrogen molecule which are present in σ1s than Na. molecular orbital. So electronic configuration of hydrogen molecule is In (i) more bonding orbitals are occupied and so the bonding influence is stronger and a H2 : (σ1s)2 stable molecule results. In (ii) the antibonding The bond order of H2 molecule can be influence is stronger and therefore the calculated as given below: molecule is unstable. N b  N a 2  0 Bond order Bond order =   1 2 2 Bond order (b.o.) is defined as one half the This means that the two hydrogen atoms difference between the number of electrons are bonded together by a single covalent bond. present in the bonding and the antibonding The bond dissociation energy of hydrogen orbitals i.e., molecule has been found to be 438 kJ mol–1 and bond length equal to 74 pm. Since no Bond order (b.o.) = ½ (Nb–Na) Reprint 2025-26 130 chemistry unpaired electron is present in hydrogen vapour phase. It is important to note that molecule, therefore, it is diamagnetic. double bond in C2 consists of both pi bonds 2. Helium molecule (He2 ): The electronic because of the presence of four electrons in configuration of helium atom is 1s2. Each two pi molecular orbitals. In most of the other helium atom contains 2 electrons, therefore, molecules a double bond is made up of a in He2 molecule there would be 4 electrons. sigma bond and a pi bond. In a similar fashion the bonding in N2 molecule can be discussed.These electrons will be accommodated in σ1s and σ*1s molecular orbitals leading to 5. Oxygen molecule (O2 ): The electronic electronic configuration: configuration of oxygen atom is 1s2 2s2 2p4. Each oxygen atom has 8 electrons, hence, He2 : (σ1s)2 (σ*1s)2 in O2 molecule there are 16 electrons. The electronic configuration of O2 molecule, Bond order of He2 is ½(2 – 2) = 0 therefore, is He2 molecule is therefore unstable and does not exist. O2 : (1s)2 ( ∗1s)2 ( 2s)2 ( ∗ 2s)2 (2pz)2 Similarly, it can be shown that Be2 molecule (π2px2 ≡ π2py2) (π∗2p1x ≡ π ∗2py1) (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 also does not exist. 3. Lithium molecule (Li2 ): The electronic O2 :configuration of lithium is 1s2, 2s1. There are six electrons in Li2. The electronic configuration of Li2 molecule, therefore, is From the electronic configuration of O2 molecule it is clear that ten electrons are Li2 : (σ1s)2 (σ*1s)2 (σ2s)2 present in bonding molecular orbitals and six The above configuration is also written electrons are present in antibonding molecular as KK(σ2s)2 where KK represents the closed orbitals. Its bond order, therefore, is K shell structure (σ1s)2 (σ*1s)2. From the electronic configuration of Li2 Bond order = [Nb – Na] = [10 – 6] =2 molecule it is clear that there are four electrons So in oxygen molecule, atoms are heldpresent in bonding molecular orbitals and two by a double bond. Moreover, it may be notedelectrons present in antibonding molecular that it contains two unpaired electrons inorbitals. Its bond order, therefore, is ½ (4 – π∗2px and π∗2py molecular orbitals, therefore,2) = 1. It means that Li2 molecule is stable and since it has no unpaired electrons it O2 molecule should be paramagnetic, a prediction that corresponds toshould be diamagnetic. Indeed diamagnetic experimental observation. In this way, theLi2 molecules are known to exist in the theory successfully explains the paramagneticvapour phase. nature of oxygen. 4. Carbon molecule (C2 ): The electronic Similarly, the electronic configurationsconfiguration of carbon is 1s2 2s2 2p2. There of other homonuclear diatomic molecules of [ ]are twelve electrons in C2. The electronic the second row of the periodic table can be configuration of C2 molecule, therefore, is written. In Fig. 4.21 are given the molecular C2 : (1s)2 ( ∗1s)2 ( ∗ 2s)2 (π2p2x = π2p2y) orbital occupancy and molecular properties for B2 through Ne2. The sequence of MOs and or KK (2s)2 ( ∗ 2s)2 (π2p2x = π2p2y) their electron population are shown. The bond energy, bond length, bond order, magnetic The bond order of C2 is ½ (8 – 4) = 2 properties and valence electron configurationand C2 should be diamagnetic. Diamagnetic appear below the orbital diagrams.C2 molecules have indeed been detected in Reprint 2025-26 Chemical Bonding And Molecular Structure 131 Fig. 4.21 MO occupancy and molecular properties for B2 through Ne2.

12.4 The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?