Unit-2.pmd

26 CHEMISTRY. The rich diversity of chemical behaviour of different elements can be traced to the differences in the internal structure of atoms of these elements..

27 STRUCTURE OF ATOM. ••••• to explain the formation of different kinds of molecules by the combination of different atoms and,.

28 CHEMISTRY. (iii) In the absence of electrical or magnetic field, these rays travel in straight lines (Fig. 2.2)..

29 STRUCTURE OF ATOM. 2.1.3 Charge on the Electron.

30 CHEMISTRY Table 2.1 Properties of Fundamental Particles atomic models were proposed to explain the distributions of these charged particles in an atom. Although some of these models were not able to explain the stability of atoms, two of these models, proposed by J. J. Thomson and Ernest Rutherford are discussed below. 2.2.1 Thomson Model of Atom J. J. Thomson, in 1898, proposed that an atom possesses a spherical shape (radius approximately 10–10 m) in which the positive charge is uniformly distributed. The electrons are embedded into it in such a manner as to give the most stable electrostatic arrangement (Fig. 2.4). Many different names are given to this model, for example, plum pudding, raisin pudding or watermelon. This model In the later half of the nineteenth century different kinds of rays were discovered, besides those mentioned earlier. Wilhalm Röentgen (1845-1923) in 1895 showed that when electrons strike a material in the cathode ray tubes, produce rays which can cause fluorescence in the fluorescent materials placed outside the cathode ray tubes. Since Röentgen did not know the nature of the radiation, he named them X-rays and the name is still carried on. It was noticed that X-rays are produced effectively when electrons strike the dense metal anode, called targets. These are not deflected by the electric and magnetic fields and have a very high penetrating power through the matter and that is the reason that these rays are used to study the interior of the objects. These rays are of very short wavelengths (∼0.1 nm) and possess electro-magnetic character (Section 2.3.1). Henri Becqueral (1852-1908) observed that there are certain elements which emit radiation on their own and named this phenomenon as radioactivity and the elements known as radioactive elements. This field was developed by Marie Curie, Piere Curie, Rutherford and Fredrick Soddy. It was observed that three kinds of rays i.e., α, β- and γ-rays are emitted. Rutherford found that α-rays consists of high energy particles carrying two units of positive charge and four unit of atomic mass. He Fig.2.4 Thomson model of atom can be visualised as a pudding or watermelon of positive charge with plums or seeds (electrons) embedded into it. An important feature of this model is that the mass of the atom is assumed to be uniformly distributed over the atom. Although this model was able to explain the overall neutrality of the atom, but was not consistent with the results of later experiments. Thomson was awarded Nobel Prize for physics in 1906, for his theoretical and experimental investigations on the conduction of electricity by gases. 2015-16.

31 STRUCTURE OF ATOM. concluded that α- particles are helium nuclei as when α- particles combined with two electrons yielded helium gas. β-rays are negatively charged particles similar to electrons. The γ-rays are high energy radiations like X-rays, are neutral in nature and do not consist of particles. As regards penetrating power, α-particles are the least, followed by β-rays (100 times that of α–particles) and γ-rays (1000 times of that α-particles)..

32 CHEMISTRY. this difference in size by realising that if a cricket ball represents a nucleus, then the radius of atom would be about 5 km..

33 STRUCTURE OF ATOM. Problem 2.1. Calculate the number of protons, neutrons and electrons in 80 35Br ..

34 CHEMISTRY Fig.2.6 The electric and magnetic field components of an electromagnetic wave. These components have the same wavelength, frequency, speed and amplitude, but they vibrate in two mutually perpendicular planes. the dense nucleus and the electrons would pull the electrons toward the nucleus to form a miniature version of Thomson’s model of atom. Another serious drawback of the Rutherford model is that it says nothing about the electronic structure of atoms i.e., how the electrons are distributed around the nucleus and what are the energies of these electrons. 2.3 DEVELOPMENTS LEADING TO THE BOHR’S MODEL OF ATOM Historically, results observed from the studies of interactions of radiations with matter have provided immense information regarding the structure of atoms and molecules. Neils Bohr utilised these results to improve upon the model proposed by Rutherford. Two developments played a major role in the formulation of Bohr’s model of atom. These were: (i) Dual character of the electromagnetic radiation which means that radiations possess both wave like and particle like properties, and (ii) Experimental results regarding atomic spectra which can be explained only by assuming quantized (Section 2.4) electronic energy levels in atoms. 2.3.1 Wave Nature of Electromagnetic Radiation James Maxwell (1870) was the first to give a comprehensive explanation about the interaction between the charged bodies and the behaviour of electrical and magnetic fields on macroscopic level. He suggested that when electrically charged particle moves under accelaration, alternating electrical and magnetic fields are produced and transmitted. These fields are transmitted in the forms of waves called electromagnetic waves or electromagnetic radiation. Light is the form of radiation known from early days and speculation about its nature dates back to remote ancient times. In earlier days (Newton) light was supposed to be made of particles (corpuscules). It was only in the 19th century when wave nature of light was established. Maxwell was again the first to reveal that light waves are associated with oscillating electric and magnetic character (Fig. 2.6). Although electromagnetic wave motion is complex in nature, we will consider here only a few simple properties. (i) The oscillating electric and magnetic fields produced by oscillating charged particles are perpendicular to each other and both are perpendicular to the direction of propagation of the wave. Simplified picture of electromagnetic wave is shown in Fig. 2.6. (ii) Unlike sound waves or water waves, electromagnetic waves do not require medium and can move in vacuum. (iii) It is now well established that there are many types of electromagnetic radiations, which differ from one another in wavelength (or frequency). These constitute what is called electromagnetic spectrum (Fig. 2.7). Different regions of the spectrum are identified by different names. Some examples are: radio frequency region around 106 Hz, used for broadcasting; microwave region around 1010 Hz used for radar; infrared region around 1013 Hz used for heating; ultraviolet region 2015-16.

35 STRUCTURE OF ATOM. around 1016Hz a component of sun’s radiation. The small portion around 1015.

36 CHEMISTRY frequency. Substituting the given values, we have c = λ ν 8 –1 8 –1 3 –1 3.00 10 m s = 1368 kHz 3.00 10 m s = 1368 10 s = 219.3m × × × This is a characteristic radiowave wavelength. Problem 2.4 The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wavelengths in frequencies (Hz). (1nm = 10–9 m) Solution Using equation 2.5, frequency of violet light 8 –1 – 9 3.00 10 m s c = = 400 10 m × λ × ν = 7.50 × 1014 Hz Frequency of red light 8 –1 –9 3.00 10 ms c = = 750 10 m × λ × ν = 4.00 × 1014 Hz The range of visible spectrum is from 4.0 × 1014 to 7.5 × 1014 Hz in terms of frequency units. Problem 2.5 Calculate (a) wavenumber and (b) frequency of yellow radiation having wavelength 5800 Å. Solution (a) Calculation of wavenumber (ν ) –8 –10 =5800Å =5800 × 10 cm = 5800 ×10 m λ * 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. –10 6 –1 4 –1 1 1 = = 5800×10 m =1.724×10 m =1.724×10 cm λ ν (b) Calculation of the frequency (ν ) 8 –1 14 –1 –10 3×10 m s c = = =5.172×10 s 5800×10 m λ ν 2.3.2 Particle Nature of Electromagnetic Radiation: Planck’s Quantum Theory Some of the experimental phenomenon such as diffraction* and interference** can be explained by the wave nature of the electromagnetic radiation. However, following are some of the observations which could not be explained with the help of even the electromagentic theory of 19th century physics (known as classical physics): (i) the nature of emission of radiation from hot bodies (black -body radiation) (ii) ejection of electrons from metal surface when radiation strikes it (photoelectric effect) (iii) variation of heat capacity of solids as a function of temperature (iv) line spectra of atoms with special reference to hydrogen. It is noteworthy that the first concrete explanation for the phenomenon of the black body radiation was given by Max Planck in 1900. This phenomenon is given below: When solids are heated they emit radiation over a wide range of wavelengths. For example, when an iron rod is heated in a furnace, it first turns to dull red and then progressively becomes more and more red as the temperature increases. As this is heated further, the radiation emitted becomes white and then becomes blue as the temperature becomes very high. In terms of 2015-16.

37 STRUCTURE OF ATOM. frequency, it means that the frequency of emitted radiation goes from a lower frequency to a higher frequency as the temperature increases. The red colour lies in the lower frequency region while blue colour belongs to the higher frequency region of the electromagnetic spectrum. The ideal body, which emits and absorbs radiations of all frequencies, is called a black body and the radiation emitted by such a body is called black body radiation. The exact frequency distribution of the emitted radiation (i.e., intensity versus frequency curve of the radiation) from a black body depends only on its temperature. At a given temperature, intensity of radiation emitted increases with decrease of wavelength, reaches a maximum value at a given wavelength and then starts decreasing with further decrease of wavelength, as shown in Fig. 2.8..

38 CHEMISTRY. Photoelectric effect. The results observed in this experiment were: (i) The electrons are ejected from the metal surface as soon as the beam of light strikes the surface, i.e., there is no time lag between the striking of light beam and the ejection of electrons from the metal surface..

39 STRUCTURE OF ATOM could explain the black body radiation and photoelectric effect satisfactorily but on the other hand, it was not consistent with the known wave behaviour of light which could account for the phenomena of interference and diffraction. The only way to resolve the dilemma was to accept the idea that light possesses both particle and wave-like properties, i.e., light has dual behaviour. Depending on the experiment, we find that light behaves either as a wave or as a stream of particles. Whenever radiation interacts with matter, it displays particle like properties in contrast to the wavelike properties (interference and diffraction), which it exhibits when it propagates. This concept was totally alien to the way the scientists thought about matter and radiation and it took them a long time to become convinced of its validity. It turns out, as you shall see later, that some microscopic particles like electrons also exhibit this wave-particle duality. Problem 2.6 Calculate energy of one mole of photons of radiation whose frequency is 5 ×10 14 Hz. Solution Energy (E) of one photon is given by the expression E = hν h = 6.626 ×10 –34 J s ν = 5×10 14 s –1 (given) E = (6.626 ×10 –34 J s) × (5 ×10 14 s –1) = 3.313 ×10 –19 J Energy of one mole of photons = (3.313 ×10 –19 J) × (6.022 × 10 23 mol –1) = 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 by the bulb. Solution Power of the bulb = 100 watt = 100 J s –1 Energy of one photon E = hν = hc/λ 34 8 1 9 6.626 10 J s 3 10 m s = 400 10 m − − − × × × × 19 = 4.969 10 J − × Number of photons emitted 1 20 1 19 100 J s 2.012 10 s 4.969 10 J − − − = × × Problem 2.8 When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68 ×10 5 J mol –1. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted ? Solution The energy (E) of a 300 nm photon is given by 34 8 –1 9 = c/ 6.626 10 J s 3.0 10 m s = 300 10 m − − λ × × × × h h ν = 6.626 × 10 -19 J The energy of one mole of photons = 6.626 ×10 –19 J × 6.022 ×10 23 mol –1 = 3.99 × 10 5 J mol –1 The minimum energy needed to remove one mole of electrons from sodium = (3.99 –1.68) 10 5 J mol –1 = 2.31 × 10 5 J mol –1 The minimum energy for one electron Metal Li Na K Mg Cu Ag W0 /eV 2.42 2.3 2.25 3.7 4.8 4.3 Table 2.2 Values of Work Function (W0) for a Few Metals C:\Chemistry XI�nit-2�nit-2(2)-Lay-3(reprint).pmd 27.7.6, 16.10.6 (Reprint) 2015-16.

40 CHEMISTRY. 5 –1. 23 –1. 19. 2.31 10 J mol =. 6.022 10 electrons mol.

41 STRUCTURE OF ATOM. red to violet, rather they emit light only at specific wavelengths with dark spaces between them. Such spectra are called line spectra or atomic spectra because the emitted radiation is identified by the appearance of bright lines in the spectra (Fig, 2.10) Line emission spectra are of great interest in the study of electronic structure. Each element has a unique line emission spectrum. The characteristic lines in atomic spectra can be used in chemical analysis to identify unknown atoms in the same way as finger prints are used to identify people. The exact matching of lines of the emission spectrum of the atoms of a known element with the lines from an unknown sample quickly establishes the identity of the latter, German chemist, Robert Bunsen (1811-1899) was one of the first investigators to use line spectra to identify elements..

42 CHEMISTRY. The series of lines described by this formula are called the Balmer series. The Balmer series of lines are the only lines in the hydrogen spectrum which appear in the visible region of the electromagnetic spectrum. The Swedish spectroscopist, Johannes Rydberg, noted that all series of lines in the hydrogen spectrum could be described by the following expression :.

43 STRUCTURE OF ATOM. 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..

44 CHEMISTRY. stationary states or energy levels of hydrogen atom. This representation is called an energy level diagram..

45 STRUCTURE OF ATOM 15 2 2 i f 1 1 3.29 10 Hz n n � � = × − � � � � (2.19) and in terms of wavenumbers (ν ) ν ν � � = = − � � � � H 2 2 i f R 1 1 c c h n n (2.20) 15 1 8 s 2 2 i f 3.29 10 s 1 1 = 3 10 m s n n − − � � × − � � × � � 7 1 2 2 i f 1 1 = 1.09677 10 m n n − � � × − � � � � (2.21) In case of absorption spectrum, nf > ni and the term in the parenthesis is positive and energy is absorbed. On the other hand in case of emission spectrum ni > nf , ∆ E is negative and energy is released. The expression (2.17) is similar to that used by Rydberg (2.9) derived empirically using the experimental data available at that time. Further, each spectral line, whether in absorption or emission spectrum, can be associated to the particular transition in hydrogen atom. In case of large number of hydrogen atoms, different possible transitions can be observed and thus leading to large number of spectral lines. The brightness or intensity of spectral lines depends upon the number of photons of same wavelength or frequency absorbed or emitted. Problem 2.10 What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom? Solution Since ni = 5 and nf = 2, this transition gives rise to a spectral line in the visible region of the Balmer series. Fr om equation (2.17) 18 2 2 19 1 1 = 2.18 10 J 5 2 = 4.58 10 J E − − � � ∆ × − � � � � − × It is an emission energy The frequency of the photon (taking energy in terms of magnitude) is given by = E h ∆ ν –19 –34 4.58×10 J = 6.626×10 Js = 6.91×1014 Hz 8 1 14 3.0 10 m s c = = = 434 nm 6.91 10 Hz − × λ × ν Problem 2.11 Calculate the energy associated with the first orbit of He+ . What is the radius of this orbit? Solution 18 2 n 2 (2.18 10 J)Z E n − × = − atom–1 For He +, n = 1, Z = 2 18 2 18 1 2 (2.18 10 J)(2 ) 8.72 10 J 1 E − − × = − = − × The radius of the orbit is given by equation (2.15) 2 (0.0529 nm) rn n Z = Since n = 1, and Z = 2 2 (0.0529 nm)1 r 0.02645 nm 2 n = = 2.4.2 Limitations of Bohr’s Model Bohr’s model of the hydrogen atom was no doubt an improvement over Rutherford’s nuclear model, as it could account for the stability and line spectra of hydrogen atom and hydrogen like ions (for example, He+, Li2+, Be3+, and so on). However, Bohr’s model was too simple to account for the following points. i) It fails to account for the finer details (doublet, that is two closely spaced lines) of the hydrogen atom spectrum observed 2015-16.

46 CHEMISTRY. Louis de Broglie (1892 – 1987). Louis de Broglie, a French physicist, studied history as an undergraduate in the early 1910’s. His interest turned to science as a result of his assignment to radio communications in World War I. He received his Dr. Sc. from the University of Paris in 1924. He was professor of theoretical physics at the University of Paris from 1932 untill his retirement in 1962. He was awarded the Nobel Prize in Physics in 1929..

47 STRUCTURE OF ATOM. Solution. Since K. E. = ½ mv2.

48 CHEMISTRY simultaneously to determine the position and velocity at any given instant to an arbitrary degree of precision, it is not possible to talk of the trajectory of an electron. The effect of Heisenberg Uncertainty Principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects. This can be seen from the following examples. If uncertainty principle is applied to an object of mass, say about a milligram (10–6 kg), then –34 –28 2 –1 –6 v. x = 4 . 6.626×10 Js = 10 m s 4×3.1416×10 kg h m ∆ ∆ π ≈ 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 uncertainties are hardly of any real consequence. In the case of a microscopic object like an electron on the other hand. ∆v.∆x obtained is much larger and such uncertainties are of real consequence. For example, for an electron whose mass is 9.11×10–31 kg., according to Heisenberg uncertainty principle v. x = 4 h m ∆ ∆ π –34 –31 –4 2 –1 6.626×10 Js = 4× 3.1416×9.11×10 kg =10 m s It, therefore, means that if one tries to find the exact location of the electron, say to an uncertainty of only 10–8 m, then the uncertainty ∆v in velocity would be ≈ –4 2 –1 4 -1 –8 10 m s 10 m s 10 m which is so large that the classical picture of electrons moving in Bohr’s orbits (fixed) cannot hold good. It, therefore, means that the precise statements of the position and momentum of electrons have to be replaced by the statements of probability, that the electron has at a given position and momentum. This is what happens in the quantum mechanical model of atom. Problem 2.15 A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 Å. What is the uncertainty involved in the measurement of its velocity? Solution x = or x v = 4 4 h h p m ∆ ∆ ∆ ∆ π π v = 4 x h m ∆ π∆ –34 –10 –31 6.626×10 Js v 4×3.14×0.1×10 m 9.11 10 kg ∆ = × × = 0.579×107 m s–1 (1J = 1 kg m2 s–2) = 5.79×106 m s –1 Problem 2.16 A golf ball has a mass of 40g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position. 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. 2015-16.

49 STRUCTURE OF ATOM. Erwin Schrödinger (1887-1961).

50 CHEMISTRY. mechanics was developed by Schrödinger and it won him the Nobel Prize in Physics in 1933. This equation which incorporates wave- particle duality of matter as proposed by de Broglie is quite complex and knowledge of higher mathematics is needed to solve it. You will learn its solutions for different systems in higher classes..

51 STRUCTURE OF ATOM. is why, as you shall see later on, one talks of only probability of finding the electron at different points in an atom..

52 CHEMISTRY. Table 2.4 Subshell Notations. Value of l 0 1 2 3 4 5.

53 STRUCTURE OF ATOM. quantum number known as the electron spin quantum number (m s). An electron spins around its own axis, much in a similar way as earth spins around its own axis while revolving around the sun. In other words, an electron has, besides charge and mass, intrinsic spin angular quantum number. Spin angular momentum of the electron — a vector quantity, can have two orientations relative to the chosen axis. These two orientations are distinguished by the spin quantum numbers ms which can take the values of +½ or –½. These are called the two spin states of the electron and are normally represented by two arrows, ↑ (spin up) and ↓ (spin down). Two electrons that have different ms values (one +½ and the other –½) are said to have opposite spins. An orbital cannot hold more than two electrons and these two electrons should have opposite spins..

54 CHEMISTRY Fig. 2.13 (a) Probability density plots of 1s and 2s atomic orbitals. The density of the dots represents the probability density of finding the electron in that region. (b) Boundary surface diagram for 1s and 2s orbitals. Fig. 2.12 The plots of (a) the orbital wave function ψ(r ); (b) the variation of probability density ψ2(r) as a function of distance r of the electron from the nucleus for 1s and 2s orbitals. away from it. On the other hand, for 2s orbital the probability density first decreases sharply to zero and again starts increasing. After reaching a small maxima it decreases again and approaches zero as the value of r increases further. The region where this probability density function reduces to zero is called nodal surfaces or simply nodes. In general, it has been found that ns-orbital has (n – 1) nodes, that is, number of nodes increases with increase of principal quantum number n. In other words, number of nodes for 2s orbital is one, two for 3s and so on. These probability density variation can be visualised in terms of charge cloud diagrams [Fig. 2.13(a)]. In these diagrams, the density 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 which the probability of finding the electron is very high, say, 90%. The boundary surface diagram for 1s and 2s orbitals are given in Fig. 2.13(b). One may ask a question : Why do we not draw a boundary surface diagram, 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%. * 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. 2015-16.

55 STRUCTURE OF ATOM. 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 distance is equal in all the directions. It is also observed that the size of the s orbital increases with increase in n, that is, 4s > 3s > 2s > 1s and the electron is located further away from the nucleus as the principal quantum number increases. Boundary surface diagrams for three 2p orbitals (l = 1) are shown in Fig. 2.14. In these diagrams, the nucleus is at the origin. Here, unlike s-orbitals, the boundary surface diagrams are not spherical. Instead each p orbital consists of two sections called lobes that are on either side of the plane that passes through the nucleus. The probability density function is zero on the plane where the two lobes touch each other. The size, shape and energy of the three orbitals are identical. They differ however, in the way the lobes are oriented. Since the lobes may be considered to lie along the x, y or z axis, they are given the designations 2px, 2py, and 2pz. It should be understood, however, that there is no simple relation between the values of ml (–1, 0 and +1) and the x, y and z directions. For our purpose, it is sufficient to remember that,.

56 CHEMISTRY. Fig. 2.16 Energy level diagrams for the few electronic shells of (a) hydrogen atom and (b) multi-electronic atoms. Note that orbitals for the same value of principal quantum number, have the same energies even for different azimuthal quantum number for hydrogen atom. In case of multi-electron atoms, orbitals with same principal quantum number possess different energies for different azimuthal quantum numbers..

57 STRUCTURE OF ATOM. 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 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 shell electrons from the nucleus by the inner shell electrons, and the net positive charge experienced by the outer electrons is known as effective nuclear charge (Zeff e). Despite the shielding of the outer electrons from the nucleus by the inner shell electrons, the attractive force experienced by the outer shell electrons increases with increase of nuclear charge. In other words, the energy of interaction between, the nucleus and electron (that is orbital energy) decreases (that is more negative) with the increase of atomic number (Z )..

58 CHEMISTRY. Fig.2.17 Order of filling of orbitals.

59 STRUCTURE OF ATOM. 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..

60 CHEMISTRY. The electronic configuration of the elements sodium (Na, 1s 22s 22p63s1) to argon (Ar,1s22s22p63s23p6), follow exactly the same pattern as the elements from lithium to neon with the difference that the 3s and 3p orbitals are getting filled now. This process can be simplified if we represent the total number of electrons in the first two shells by the name of element neon (Ne). The electr onic configuration of the elements from sodium to argon can be written as (Na, [Ne]3s1) to (Ar, [Ne] 3s23p6). The electrons in the completely filled shells are known as core electrons and the electrons that are added to the electronic shell with the highest principal quantum number are called valence electrons. For example, the electrons in Ne are the core electrons and the electrons from Na to Ar are the valence electrons. In potassium (K) and calcium (Ca), the 4s orbital, being lower in energy than the 3d orbitals, is occupied by one and two electrons respectively. A new pattern is followed beginning with scandium (Sc). The 3d orbital, being lower in energy than the 4p orbital, is filled first. Consequently, in the next ten elements, scandium (Sc), titanium (Ti), vanadium (V), chromium (Cr), manganese (Mn), iron (Fe), cobalt (Co), nickel (Ni), copper (Cu) and zinc (Zn), the five 3d orbitals are progressively occupied. We may be puzzled by the fact that chromium and copper have five and ten electrons in 3d orbitals rather than four and nine as their position would have indicated with two-electrons in the 4s orbital. The reason is that fully filled orbitals and half- filled orbitals have extra stability (that is, lower energy). Thus p3, p6, d5, d10,f 7, f14 etc. configurations, which are either half-filled or fully filled, are more stable. Chromium and copper therefore adopt the d5 and d10.

61 STRUCTURE OF ATOM. The completely filled and completely half filled sub-shells 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..

62 CHEMISTRY. Table 2.6 Electronic Configurations of the Elements.

63 STRUCTURE OF ATOM. ** Elements with atomic number 112 and above have been reported but not yet fully authenticated and named..

64 CHEMISTRY. SUMMARY. Atoms are the building blocks of elements. They are the smallest parts of an element that.

65 STRUCTURE OF ATOM. 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..

66 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)..

67 STRUCTURE OF ATOM. (e) n = 3, l = 3, ml = –3, ms = + ½.

68 CHEMISTRY. 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..

69 STRUCTURE OF ATOM. 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.