Modern ABC Chemistry XI

2/40 The velocity associated with a proton moving in a MODERN’S abc + OF CHEMISTRY–XI 38. potential difference of 1000 V is m s–1. If the hockey ball of mass 0.1 kg is moving with the velocity, be less than h/4π. The sign of equality refers to calculate the wavelength associated with this. minimum uncertainty and is equal to h/4π. The constancy of the product of uncertainties means that : (i) If Δx is small i.e., the position of the particle is measured accurately, Δp would be large, i.e., there would be large uncertainty in its momentum. (ii) On the other hand, if Δp is small, the momentum of the particle is measured more accurately, Δx would be large i.e., there would be large uncertainty with regard to the position of the particle. In other words, if the position of a particle is measured accurately, there will be more error in the measurement of momentum. Conversely, if momentum is measured more accurately, the position will not be accurately known. Since momentum p = mv, therefore, Δp = mΔv because mass is constant. The above relation may also be written as : ©39. What accelerating potential is needed to produce Modern Publishers. All rights reserved.an electron beam with wavelength of 9 pm ? 40. Calculate the de-Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1kV. 41. The kinetic energy of a subatomic particle is 5.85 × 10–25 J. Calculate the frequency of the particle wave. 42. What must be the velocity of a beam of electrons if they are to display a de-Broglie wavelength of 100 Å? 43. Calculate the wavelength associated with a moving electron having kinetic energy of 1.375 × 10–25 J. (mass of e = 9.1 × 10–31 kg, h = 6.63 × 10–34 kg m2 s–1). 34. 6.626 × 10–34m Δx × m(Δv) ≥ h 35. 6.139 × 10–33 kg 4π 36. 6.63 × 10–24 kg m s–1 37. 2.30 × 10–19 J or Δx × Δv ≥ h 38. 1.52 × 10–38 m 4πm 39. 1.86 × 104 V 40. 3.88 × 10–11 m This also means that the position and velocity 41. 1.76 × 109 s–1 of an object cannot be simultaneously known 42. 7.28 × 104 m s–1 with accuracy. 43. 1.32 × 10–6 m Physical concept of uncertainty principle. Hints & Solutions on page 68 In order to understand the physical concept of uncertainty principle, let us study as to how the position of a material object is determined. To determine the position of an object, we must be able to see the object. This can be done with the help of light of suitable wavelength. When a beam of light falls on an object, the photons of this incident light are scattered at the object and the reflected light enters our eye. Now, if 2. Heisenberg’s Uncertainty principle the object is large, its position and velocity will not So far, we have been believing that anything can change by the impact of the striking photons. Thus, it be specified to any desired degree of accuracy. will be possible to determine both the position and However, Heisenberg in 1927, put forward a principle velocity of the object simultaneously. known as Heisenberg’s uncertainty principle. It states that However, in case of microscopic objects, such as electrons, the interaction of the striking photon of light it is not possible to measure simultaneously with the particle will cause appreciable displacement both the position and momentum (or velocity) of the particle from its normal path. As a result of this, of a microscopic particle, with absolute the particle undergoes a considerable change in its accuracy. path and velocity (or momentum) due to the impact of Mathematically, this law may be expressed as : Δx × Δp ≥ h a single photon used to observe it. Thus, the very act 4π of measuring the position of the microscopic particle where Δx = uncertainty in position causes a change in its momentum. Fig. 28 shows as to Δp = uncertainty in momentum how the position of an electron changes by the impact of photon. The collision of photon with the electron The sign > means that the product of Δx and Δp sends it in an unpredictable direction. can be either greater than or equal to h/4π. It can never

STRUCTURE OF ATOM 2/41 Fig. 28. The change in the path of electron by proposed by Bohr. Since for a subatomic particle like the impact of striking photon. an electron, it is not possible to simultaneously determine the position and velocity at any moment Thus, it is not possible to determine simultaneously with good degree of precision, therefore, it is not the exact position and momentum of the electron or possible to talk about the trajectory of an electron or of any other microscopic object. well defined circular orbits. © Modern Publishers. All rights reserved. Let us consider a thought experiment for It should be borne in mind that the uncertainty determining the position of an electron by using is not due to lack of sufficiently refined techniques but photons of light. According to principles of optics, if it is due to the fact that we cannot observe microscopic we use light of wavelength λ, then the position of things without disturbing them. No instrument can electron cannot be located more accurately than +λ. observe the position of an electron without affecting The shorter the wavelength, the greater is the its motion. In other words, uncertainty principle is accuracy. Therefore, to observe the position of the the fundamental limitation of nature. Thus, we electron accurately, light of appropriately small cannot design an experiment to obtain an accurate wavelength should be used. But the photons of value of both the position and momentum for radiations of smaller wavelength will have higher microscopic objects. momentum (p = h/λ). When even a single photon of this light used to observe the position of electron However, in our daily life, this principle has no strikes against it, a large amount of momentum will significance. This is because we come across only large be transferred to the electron at the time of collision. objects, i.e., the objects which we can observe with naked This will change the velocity of the electron and eye without altering their motion. The position and consequently will result into greater uncertainty in velocity of these objects can be determined accurately velocity or momentum. because in these cases, during the interaction between the object and the measuring device, the changes in On the other hand, in order to minimise the position and velocity are negligible. change in momentum we have to use light having photons with small values of p. This will require This may also be illustrated as follows : radiations of longer wavelengths (λ = h/p). Now, if we use photons of light having larger wavelengths If uncertainty principle is applied to a microscopic (low momentum), the velocity or momentum will not object like an electron, Δx × Δv comes out to be of the change appreciably but we will not be able to measure order of 10–5 m2 s–1 as : position accurately with larger wavelength. Therefore, uncertainty in position will increase. Thus, we cannot Δx × Δv = h simultaneously measure the position and momentum 4πm of a small moving object like electron accurately. = 6.626 × 10−34 kg m2 s−1 Significance of Uncertainty Principle in Our 4 × 3.1416 × 9.11 × 10−31 kg Daily Life = 5.78 × 10–5 m2 s–1 One of the important implications of the Heisenberg's uncertainty principle is that it rules out It also means that if one tries to find the exact the well defined circular paths (orbits) or trajectories position of the electron, say to an uncertainty of about 1Å (1 × 10–8 m), then uncertainty in velocity, Δv would be Δv = 5.78 × 10−5 = 5.78 × 103 m s–1 10−8 This is very large. On the other hand, if uncertainty principle is applied to an object of mass say about a milligram (10–6 kg), then Δx × Δv = h 4πm = 6.626 × 10−34 kg m2 s−1 4 × 3.1416 × 10−6 kg = 5.27 × 10–29 m2 s–1 The value of Δx. Δv for milligram sized or heavier objects is extremely small. Therefore, uncertainty of such small values have no significance. Example 40. Calculate the uncertainty in the position of an electron if the uncertainty in its velocity is

2/42 MODERN’S abc + OF CHEMISTRY–XI 5.7 × 105 m s–1. (h = 6.6 × 10–34 J s and mass of Example 43. electron = 9.1 × 10–31 kg) Calculate the uncertainty in the position of an Solution : According to Heisenberg's uncertainty electron if uncertainty in its velocity is (i) 0.001% principle, (ii) zero. (The mass of electron = 9.1 × 10–31 kg, velocity of Δx × Δp≥ h or Δx × mΔv ≥ h electron = 300 m s–1). 4π 4π Solution : Mass of electron = 9.1 × 10–31 kg, h = 6.6 × 10–34 J s (i) When Δv = 0.001% of velocity of electron ©∴ Δx ≥ h Modern Publishers. All rights reserved.4πmΔv Δv = 5.7 × 105 m s–1, m = 9.1 × 10–31 kg, h = 6.6 × 10–34 J s = 0.001 × 300 = 3.0 × 10–3 m s–1 100 6.6 ×10−34kg m2 s−1 ∴ Δx = 4 × 3.142 × (9.1 ×10−31 kg) × (5.7 ×105 m s−1) Now, Δx × m Δv = h 4π = 1.012 × 10–10 m. ∴ Δx = h 4π m Δv Example 41. Calculate the uncertainty in the velocity of a wagon 6.6 ×10–34 kg m2 s–1 of mass 2000 kg whose position is known to an = 4 × 3.142 × (9.1 ×10–31kg) × (3.0 ×10–3ms–1) accuracy of ± 10 m. = 1.92 × 10–2 m. Solution : Mass of wagon = 2000 kg (ii) When Δv = 0 Uncertainty in position, Δx = ± 10 m According to Heisenberg uncertainty principle Δx = h 4π m Δv Δx × Δp = h or Δx × Δv = h or Δv = 4π 4πm If Δv is zero, then denominator in the above expression becomes zero and, therefore, uncertainty in position is h infinity. 4πmΔx Example 44. An electron has a speed of 500 m s–1 with = 6.626 × 10−34 kg m2 s−1 uncertainty of 0.02%. What is the uncertainty in 4 × 3.1416 × (2000 kg) × (10 m) locating its position ? Solution : Velocity of electron = 500 m s–1 = 2.636 × 10–39 m s–1. Uncertainty in velocity, Δv = 500 × 0.02% Example 42. On the basis of Heisenberg uncertainty principle, Mass of electron, = 500 × 0.02 show that electron cannot exist within the atomic 100 nucleus of radius 10–15m. Solution : The radius of the atomic nucleus is 10–15 m. = 0.1 m s–1 Now, if the electron were to exist within the nucleus, then the maximum uncertainty in its position would m = 9.1 × 10–31 kg have been 10–15 m. or Δx = 10–15 m h = 6.6 × 10–34 kg m2 s–1 Now according to uncertainty principle, According to uncertainty principle, Δx × Δp ≥ h 4π or Δx × m Δv ≥ h 4π Δx × Δp > h 4π or Δx ≥ h 4πmΔv or Δx × mΔv > h 6.6 ×10–34 kg m2s–1 4π ≥ 4 × 22 × (9.1 ×10–31 kg) × (0.1m s–1) or Δv > h 7 4π mΔx ≥ 5.77 × 10–4 m Mass of electron, m = 9.1 × 10–31 kg, ∴Uncertainty in position = 5.77 × 10–4 m. Δx = 1 × 10–15 m Example 45. 6 .6 × 1 0 −34 A golf ball has a mass of 40 g and a speed of 45 m/s. 4 × 3 .1 4 × 9 .1 × 1 0 −31 × 1 × 1 0 −15 If the speed can be measured with accuracy of 2%, ∴ Δv = = 5.77 × 1010 m s–1. calculate the uncertainty in position. The value of uncertainty in velocity, Δv is much higher than the velocity of light (3.0 × 108 m s–1) and therefore, Solution : The uncertainty in speed, it is not possible. Hence an electron cannot be found Δv = 45 × 2 = 0.9 ms–1 within the atomic nucleus. 100 m = 40 × 10–3 kg

STRUCTURE OF ATOM 2/43 Δx = h 44. 5.79 × 106 m s–1. 4πmΔv 45. 3.6 m s–1, 1.46 × 10–33 m. 46. 1.59 × 104 m s–1. = 6.626 × 10−34 kg m2s−1 47. 0.10 kg. 4 × 3.14 × 40 × 10−3 kg × 0.9 m s−1 48. 1.06 × 10–13 m. 49. 2.11 × 10–13 m. = 1.46 × 10–33m Example 46. Hints & Solutions on page 68 If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron. If suppose the momentum of the electron is h 4π × 0.05 nm is there any problem in defining this value? © QUANTUM MECHANICAL MODEL OF ATOM Modern Publishers. All rights reserved.Solution : According to uncertainty principle,AND CONCEPT OF ATOMIC ORBITAL Δx × Δp = h The wave character of an electron and uncertainty 4π in its position and momentum gave a serious blow to Δx = ± 0.002 nm = ± 0.002 × 10–9 m Bohr’s model of an atom. According to Bohr, the electrons revolve around the nucleus in certain well Δp = h = 6.626 ×10−34 kg m2s−1 defined circular orbits. But the idea of uncertainty 4πΔx 4 × 22 × 0.002 ×10−9 ms−1 in position and velocity overruled the Bohr’s 7 picture of fixed orbits. Thus, the classical mechanics = 2.63 × 10–23 kg m s–1. could not describe the behaviour of electrons in atoms correctly. Therefore, the scientists started looking for Momentum of electron = h a model which could incorporate the dual character of matter and uncertainty principle. This resulted in a 4π × 0.05 nm new approach called quantum mechanics or wave mechanics. 6.626 × 10–34 kg m2s–1 Quantum mechanics was developed independently = 4 × 22 × 0.05 ×10–9 m in 1926 by Werner Heisenberg and Erwin Schrodinger. 7 On the basis of quantum mechanics, Schrodinger proposed a model of an atom which considered wave- = 1.05 × 10–24 kg m s–1. particle duality of matter and Heisenberg uncertainty Since the actual momentum of electron is smaller than principle. This model was known as quantum the uncertainty, it cannot be defined. mechanical model of atom. In 1927, Erwin Schrodinger described the wave motion of the electron in three 44. A microscope using suitable photons is employed dimensional space around the nucleus by a mathematical to locate an electron in atom within a distance of equation known as Schrodinger wave equation. 0.1 Å. What is the uncertainty involved in the measurement of its velocity ? (mass of electron = The Schrodinger wave equation is : 9.11 × 10–31 kg, h = 6.626 × 10–34 Js) ∂2ψ + ∂2ψ + ∂2ψ + 8π2m (E − V)ψ = 0 ∂x2 ∂y2 ∂z2 h2 45. Table-tennis ball has a mass 10 g and speed of 90 m s–1. If speed can be measured within an accuracy where ψ is the amplitude of the wave, x, y and z are of 4%, what will be the uncertainty in speed and the coordinates of the position of electron, E is the position. total energy of the electron, V is the potential energy, m is the mass of the electron and h is the Planck’s 46. Calculate the minimum uncertainty in velocity of a constant. ∂2ψ/∂x2 represents second differential particle of mass 1.1 × 10–27 kg if uncertainty in its derivative of ψ w.r.t. x, ∂2ψ/∂y2 represents second position is 3 × 10–10 cm (h = 6.6 × 10–34 kg m2 s–1). differential derivative of ψ w.r.t. y and ∂2ψ/∂z2 represents second differential derivative of Ψ w.r.t. z. 47. The uncertainties in the position and velocity of a In short, the Schrodinger wave equation may be written particle are : 10–10 m and 5.27 × 10–24 m s–1. as : Calculate the mass of particle (h = 6.626 × 10–34 J s). Hˆ Ψ = EΨ where Hˆ is a mathematical operator called 48. A proton is accelerated to one-tenth of the velocity Hamiltonian operator. The solutions of Schrodinger of light. If its velocity can be measured with a wave equations give E and Ψ. precision of ± 1%, what must be its uncertainty in position ? (h = 6.6 × 10–34 J s, mass of proton = 1.66 × 10–27 kg). 49. A proton is accelerated to a velocity of 3 × 107 m s–1. If the velocity can be measured with a precision of ±0.5%, calculate the uncertainty in position of proton. (h = 6.6 × 10–34 Js, mass of proton = 1.66 × 10–27kg)

2/44 MODERN’S abc + OF CHEMISTRY–XI Although the Schrodinger wave equation was motion, the square of the wave function, ψ2 may be difficult to accept at first, it has now gained wide taken as intensity of electron at any point. In other acceptance. This is because the results obtained from words, ψ2 determines the probability of finding the Schrodinger wave equation are in excellent agreement moving electron in a given region i.e. it gives the with the experimental findings. One of the major probability density. Thus, ψ2 has been called the achievement of Schrodinger wave equation is that it probability density and ψ the probability successfully interpreted the experimental information amplitude. Hence, the solutions of Schrodinger wave about atoms and molecules. equation replace the discrete energy levels or orbits proposed by Bohr and led to the concept of most KEY NOTE probable regions in space in terms of ψ2. A large value of ψ2 means a high probability of finding the electron Operators at that place and a small value of ψ2 means low probability. If ψ2 is almost zero at a particular point, it While studying the state of a system, we make means that the probability of finding the electron at various measurements of its properties such as mass, that point is negligible. volume, momentum, position and energy. Each individual property is called observable. In order to determine the Energy of electron in an atom value of the observable property, we have to perform certain mathematical operations. This operation is represented When the Schrodinger wave equation was solved by an operator. Therefore, operator is a mathematical for hydrogen atom, it gave the following expression command or instruction which acts on a for the energy of electron : mathematical function. For example, in the equation 4 × 5 = 20, the operation is multiplication and the operator is '×'. We can also express the multiplication operation with some symbols, designated as M^ . For example, M^ (4 × 5) = 20. In this case, M^ is multiplication operator, where the ^ signifies an operator. © En = – 2π2 me e4 Modern Publishers. All rights reserved. n2 h2 The above expression is same as Bohr’s equation for the energy of electron in a hydrogen atom. Schrodinger Wave Equation for Hydrogen Substituting the values of me and h, the above Atom expression becomes :  When Schrodinger wave equation was solved En = – 2.18 × 10−18 J per atom n2 for hydrogen atom, the solution gives the possible energy states (or energy levels) that an electron can or = – 1.312 × 106 J per mol n2 occupy. These values of energy are called eigen values. Important Features of the Quantum Mechanical The corresponding values of wave functions (Ψ) of the Model of Atom electron associated with these energy values are called eigen functions.  The wave functions and the corresponding The important features of the quantum mechanical model of an atom are summed up below : energy states are characterized by a set of three 1. The energy of electrons in atoms is quantized quantum numbers (principal quantum number n, i.e., they can have only certain specific values. azimuthal quantum number l and magnetic quantum 2. The existence of quantized electronic energy levels is a direct result of the wave like properties of ncounmsebqeurenmcel).inTthheesesolnuutimonbeorf s ar ise as a na tural electrons and are the allowed solutions of the Schrodinger wave Schrodinger equation. equation. This quantum mechanical model of the 3. This gives the most probable regions in an atom where the probability of finding the electrons hydrogen atom successfully predicts all aspects of is maximum. Therefore, this model is in accordance with the Heisenberg uncertainty principle and does hydrogen atom spectrum and other phenomena which not specify the exact position and momentum of the electron. But it talks about the most probable regions could not be explained by Bohr model. called orbitals.  The Schrodinger wave equation cannot be 4. An atomic orbital is described with the help of solved exactly for multielectron atoms. However, wave function ψ for an electron in an atom. Whenever, approximate methods can be used to get solutions of an electron is described by a wave function, we say reasonable accuracy. that the electron occupies that orbital. Therefore, the Physical significance of wave function. In the physical sense, ψ gives the amplitude of the wave associated with the electron. We know that in the case of light waves, the square of the amplitude of the wave at a point is proportional to the intensity of light. Extending the same concept to electron wave

STRUCTURE OF ATOM 2/45 wave functions for an electron in an atom are also Differences between Orbit and Orbital called orbital wave functions or simply atomic The main differences between orbit and orbital orbitals. Since many wave functions are possible for are summed up below : an electron, there are many atomic orbitals in an atom. In each orbital, an electron has definite energy. Orbit Orbital Thus, all the information about the electron in an © atom is provided by its orbital wave function ψ. 1. Orbit is a well-defined 1. It represents the region in Modern Publishers. All rights reserved. circular path around space around the nucleus 5. The probability of finding the electron at a the nucleus in which an in which the probability of point within an atom is proportional to the square of electron revolves. finding the electron is the wave function i.e., |ψ|2 at that point. Therefore, maximum. |ψ|2 is known as probability density and is always 2. It represents the positive. planar motion of an 2. It represents the three- electron. dimensional motion of an 6. The probability of finding the electron in a electron around the region having volume dV can be obtained as ψ2. dV, 3. Orbit gives a definite nucleus. where dV is the infinitesimally small volume path of an electron and (dx . dy . dz). this concept is not in 3. Orbital does not specify accordance with the definite path and accor- 7. The quantum mechanical model gave three uncertainty principle. ding to this concept, constants, known as quantum numbers which are electron may be any-where required to specify the position and energy of the in this region. This concept electron in an atom. These are discussed later. is in accordance with the Probability Picture of an Electron : Concept of uncertainty principle. atomic orbital. The solution of Schrodinger wave equation led to the 4. All orbits are circular. 4. Orbitals have different concept of most probable regions in place of well- shapes. For example, defined circular paths proposed by Bohr. According to s-orbital is spherical, this approach, we cannot say simply that the electron while p-orbital is dumb exists at a particular point, but we talk about certain bell shaped. regions in space around the nucleus where the probability (chances) of finding the electron is 5. Orbits do not have 5. Except s-orbitals, all other maximum (90–95%). These most probable regions in directional characte- orbitals have space are called orbitals. Thus, ristics. directional characte- ristics. an orbital may be defined as a region in space around the nucleus where the probability of 6. An orbit can 6. An orbital cannot finding the electron is maximum. accommodate electrons accommodate more than If a boundary is drawn which encloses a region equal to 2n2 where n two electrons. where there is high probability (90–95%) of finding the electron, the figure obtained, gives the general picture represents the of orbital. For the sake of simplicity, it may be drawn principal quantum as shown in Fig. 29 (a). The orbital, here is shown by number. dotted figure representing electron cloud. The intensity of dots gives the relative probability of finding QUANTUM NUMBERS the electron in that particular region. It may be noted As discussed earlier, orbitals represent regions that there are some chances of finding the electron even outside the figure. Fig. 29 provides a comparison in space around the nucleus where the probability of of an orbital with the circular orbit proposed by Bohr. finding the electron is maximum. A large number of Fig. 29. (a) Electron cloud representation of an electron orbitals are possible in atom. These can be orbital (b) Bohr’s orbit. distinguished by their size, shape and orientation. To describe each electron in an atom in different orbitals, we need a set of three numbers known as quantum numbers. These are designated as n, l and mqula. nItnumadnduitmiobnertois these, three numbers another also needed which specifies the spin of the electron. These four numbers are called quantum numbers. These are discussed below : 1. Principal quantum number (n). This quantum number determines the main energy shell or level in which the electron is present. It is denoted by n. It can have whole number values starting from 1 such as n = 1, 2, 3, 4 .... This quantum number also identifies as shell. The shell with n = 1 is called the first shell. The shell with n = 2 is called the second shell and so on. The various shells are also called K, L, M, N as :

2/46 MODERN’S abc + OF CHEMISTRY–XI n 1 2 3 4 ... The various subshells or values of l are also shell K L M N .... designated by letters s, p, d, f ... as The principal quantum number gives the following Value of l 0 1 2 3 4 5 ... informations: Designation s p d f g h ... (i) It gives the average distance of the electron from the nucleus. If n = 1 (first shell), it is closest to the nucleus and has lowest energy. As the value of n increases, the distance of the electron from the nucleus increases. (ii) It determines the energy of the electron, according to the formula, © The letters s, p, d and f originate from the Modern Publishers. All rights reserved.En =– 2π2me4 terms sharp, principal, diffuse and fundamental, n2h2 which were used in the atomic emission spectra. For l = 4 and higher values, the letters follow = – 1311.8 kJ mol–1 alphabetical order after f, i.e., l = 4 is designated n2 as g; l = 5 is designated as h and so on. For example, the energy of electron in K-shell Thus, for n = 1, l = 0 (n = 1), is This means that the first principal energy level consists of only one sublevel termed as s. E1 = – 1311.8 = –1311.8 kJ mol–1 Similarly, for n = 2, l = 0, l i.e., the second 12 principal energy level consists of two sublevels, s (l = 0) and p (l = 1). Similarly, the energy of the electron in L shell For n = 3, l = 0, 1, 2 i.e., third principal energy (n = 2), E2, M-shell (n = 3), E3....... are respectively, level consists of three sublevels s (l = 0), p (l = 1) and E2 = –327.9 kJ mol–1 E3 = –145.7 kJ mol–1 d (l = 2). Thus, principal quantum number gives the The different subshells or sublevels are average distance of the electron from the nucleus (size represented by first writing the value of n (1, 2, 3...) of electron cloud or shell) and energy associated with it. and then the letter designation for the value of l (s, p, d, f...). For example, an orbital with n = 1 and l = 0 is denoted as ls, an orbital with n = 3, l = 2 is denoted as 3d. The designations of subshells for n = 1 to n = 4 are given below : 2. Azimuthal quantum number (l). This Subshell No. of subshells quantum number determines the angular momentum of the electron. This quantum number is also known n l designation in a shell as orbital angular momentum or subsidiary quantum number. 10 ls One 20 Two This is denoted by l. The value of l gives the {2s Three subshell or sublevel in a given principal energy shell 1 to which an electron belongs. It can have positive 30 2p Four integer values ranging from zero to (n –1) where n is the principal quantum number. That is, 1 {3s 2 l = 0, 1, 2, 3 ... (n –1) 40 3p 1 3d The azimuthal quantum number gives the 2 following informations: {4s 3 4p 4d 4f (i) It gives the number of subshells present in a (ii) It gives the relative energies of various shells. principal shell. The value of ‘l’ depends upon the value of n. Except for hydrogen, the subshells within a given shell differ slightly in energy. The energy of a For example, for n = 1, l has only one value: subshell increases with increasing value of l. This l = 0 i.e., one subshell means that within a given shell, the s-subshell (l = 0) has lowest energy, p-subshell (l = 1) has next For n = 2, l has two values : l = 0, 1 i.e., two to lowest, followed by d, then f and so on. For subshells. example, in fourth energy shell, the energies of subshell increases as : For n = 3, l has three values : l = 0, 1, 2 i.e., three subshells. 4s < 4p < 4d < 4f Thus, for each value of n, there are n possible values of l. In other words, the number of subshells in a principal shell is equal to the value of n. Increasing energy

STRUCTURE OF ATOM 2/47 (iii) This quantum number gives the energy of the If l = 2, ml may be –2, –1, 0, +1, +2. electron due to the angular momentum of the electron. i.e., d-subshell contains five orbitals called The angular momentum of the electron is related d-orbitals. to l as : Similarly, if l = 3, ml may be –3, –2, –1, 0, +1, +2, +3. Angular momentum = h l (l + 1) i.e., f-subshell contains seven orbitals called f-orbitals. 2π The number of orbitals in a given subshell are given below : KEY NOTE © Modern Publishers. All rights reserved.• Some books express angular momentum as : Subshell s pd f g Angular momentum = l (l + 1) No. of orbitals 1 3 5 7 9 The use of is a shorthand way of writing h/2π and is It can be generalised that there are (2l + 1) extensively used in Quantum mechanics. orbitals (or m values) for each value of l (or subshell). • It may be noted that angular momentum depends only on the value of l and not on the value of n. This means that 1s, By working out different combinations of these 2s, 3s, etc., electrons will have same angular momentum. quantum numbers, it can be easily calculated that there Similarly, all p-electrons or d-electrons or f-electrons will is one orbital for n = 1 (1s), four for n = 2 (one 2s and have same angular momentum. three 2p), nine for n = 3 (one 3s, three 3p, five 3d). These are given above in Table 4. Thus, azimuthal quantum number determines the subshell in a given principal shell, angular momentum of the electron present in the subshell and relative energies of various subshells. 3. Magnetic quantum number (omf el)l.ecTtrhoins Thus, magnetic quantum number determines the quantum number describes the behaviour number of orbitals present in a given subshell. in a magnetic field. We know that the movement of Table 4. Permitted combinations of n, l and ml. electrical charge is always associated with magnetic field. Since the revolving electron possesses angular momentum, it will give rise to a very small magnetic Value Value Value Subshell No. of field which will interact with the external magnetic of n of l of ml orbitals field of the earth. Under the influence of external 10 0 1s 1 magnetic field, the electrons in a given subshell orient themselves in certain preferred regions of space around 20 0 2s {1 the nucleus. These are called orbitals. Thus, this 2p quantum number gives the number of orbitals in a 1 –1, 0, +1 3 =4 mogfivmceanlndsehupabevsnehdevslal.uluIpteoissn–dtehlsetihgvnraoaluutegedhobf0yl.tmoFl+o.rTl.ahTeghiavaleltoniwsveadluvealoufels, 30 0 3s {1 1 –1, 0, +1 3p 3 =9 5 ml = – l .....0..... + l 2 –2, –1, 0, +1, +2 3d In other words, there are (2l + 1) values of m for each value of l. that 4. Spin quantum nautommbeisr(nmost).onItlyisroebvsoelrvvinedg the electron in an This quantum number gives the number of around the nucleus but is also spinning around its own orbitals in a subshell. For example, axis. In 1925, George Uhlenbeck and Samuel Goudsmit If l = 0, m has only one value, i.e., ml = 0 proposed fourth quantum number known as electron i.e., s subshell has only one orbital called spin quantum number. This quantum number s-orbital. describes the spin orientation of the electron. It is tdwesoigwnaaytse—d bclyocmksw. iSsienocre the electron can spin in only If l = 1, ml may be – 1, 0, + 1. anti-clockwise and, therefore, i.e., p-subshell contains three orbitals called the spin quantum number can take only two values : p-orbitals. These are indicated by numerical subscripts + ½ or – ½. This quantum number has a value (ospurb+b1is,tcaprl0isp,,tpds–e1s()pigxo,nrapttyheeadsnaedsap2rpze)x.d, T2eshpiyugasn,natdthe2deprbze.yaarelpthharbeeeti2cpa-l independent of the values of the other three quantum numbers. Instead of giving number for mdessi=gn+at½edobry–ar½rowthseptowinotionrgieunptaatniodndsowarne usually NOTE up) or ↓ (spin down) respectively. : ↑ (spin Some books use different values of m for px, py and pz orbitals. However, it may be noted that for a pz-orbital, the Thus, an electron has also spin angular value of m has been conventionally taken to be equal to momentum commonly called spin. The magnitude of zero, i.e., m = 0. But a px or a py-orbital does not designate spin angular momentum of an electron is given as : only one value of m but a linear combination of orbitals with m = +1 and m = –1. Therefore, it is not correct to Spin angular momentum = s(s + 1) h designate single value of m for a px orbital as +1 or –1. 2π For example, for spin quantum number, s = 1 2

2/48 MODERN’S abc + OF CHEMISTRY–XI Spin angular momentum = 1 ⎛ 1 + 1⎠⎞⎟ h = 3h number of orbitals. For example, one s-orbital 2 ⎜⎝ 2 2π 4π (l = 0), three p-orbitals (l = 1), five d-orbitals (l = 2) per subshell. elect4r.omn.s refers to the orientation of the spin of the © Thus, the four quantum numbers describe the Modern Publishers. All rights reserved. position of an electron in an atom by specifying its main shell (n), subshell (l), the orientation of the owrobridtsa, lth(mesle) and direction of istesrvsepains a(nmas)d. dInreossthfeorr Fig. 30. Two possible orientations of the spin of an an electron. quantum numbers electron ms = +½ and ms = –½ The various permitted values of the quantum To sum up the four quantum numbers provide numbers are summarized below. the following informations about orbitals : Name Symbol Information Permitted provided values 1. n describes the shell, determines the size of the orbital and also to the large extent the energy of the orbital. Principal n shell 1, 2, 3, 4... 2. l determines the subshell and shape of the Azimuthal subshell 0, 1, 2, 3 ... (n – 1) orbital. There are n subshells in nth shell. To some Magnetic l orbital – l ... 0 ... + l extent, l also determines the energy of the orbital in Spin spin + ½, – ½ ml a multi-electron atom. ms For 3a.gmivlednesviaglnuaeteosf the orientation of the orbital. The number of subshells and orbitals in first three l, ml can have (2l + 1) values or shells (K, L and M) are shown in Fig. 31. Fig. 31. Number of subshells and orbitals in first three shells (K, L and M). PAULI’S EXCLUSION PRINCIPLE electrons provided their spin quantum numbers are different. For example, consider K shell where n = 1. After setting the rules for the possible values of It will have one value of l (as l = 0) and one value of m quantum numbers, we can calculate the maximum (as m = 0) but it can have either of the two values of s number of electrons that can be added in each shell and subshell in an atom. For this, there is a guiding as + 1 or – 1 i.e., principle, known as Pauli’s exclusion principle. It 22 states that n=1 l=0 m=0 s=+ 1 no two electrons in an atom can have same 2 values for all the four quantum numbers. n=1 l=0 m=0 s=– 1 This principle can be easily understood. We have 2 learnt that an orbital is specified by three quantum numbers n, l and m. Since an electron in an orbital This leads to the conclusion that an orbital can must have the same values of n, l and m, it follows have maximum of two electrons. Moreover, if an orbital that it can have two values of s as either + 1 or – 1 . has two electrons, these must be of opposite spins. In other words, an orbital can have at the2most tw2o Thus, Pauli’s exclusion principle also states that an orbital can contain maximum of two electrons.

STRUCTURE OF ATOM 2/49 From the above discussion, it can be concluded (b) An atomic orbital has n = 3, what are the that possible values of l ? (c) An atomic orbital has l = 3, what are the possible s-subshell (containing only one orbital) can values of ml ? have a maximum of 2 electrons Solution : (a) If l = 2, the permitted values of m are : p-subshell (containing three orbitals) can have maximum of 6 electrons d-subshell (containing five orbitals) can have maximum of 10 electrons f-subshell (containing seven orbitals) can have maximum of 14 electrons © ml = –2, –1, 0, + 1, + 2 Modern Publishers. All rights reserved. (b) For n = 3, l may have the value l = 0, 1, 2 (c) For l = 3, ml may have the values ml = –3, –2, –1, 0, + 1, + 2, + 3. Example 48. Thus, it may be concluded that List all the values of l and m for n = 2. No. of subshells in nth shell = n No. of orbitals in a subshell = (2l + 1) Solution : When n = 2, l can have values 0 and 1 No. of electrons in a subshell = 2(2l + 1) = 4l + 2 For l = 0, ml = 0 For l = 1, ml = – 1, 0, + 1. It may be noted that the number of electrons in a Example 49. subshell do not depend upon the value of principal Using the s, p, d notations, describe the orbital quantum number. From this we can calculate the with the following quantum numbers: number of electrons in various shells as given below : (a) n = 1, l = 0 (b) n = 3, l = 2 (c) n = 3, l = 1 (d) n = 2, l = 1 (n = 1) (l = 0) (m = 0) (e) n = 4, l = 3 (f) n = 4, l = 2 K-shell 1s-subshell one orbital (n = 2) 2 electrons L-shell (l = 0) 2s-subshell (m = 0) 2 electrons Solution : (n = 3) one orbital 6 electrons M-shell (l = 1) (m = – 1, 0, +1) 8 electrons (a) n = 1, l = 0 : 1s-orbital 2p-subshell three orbitals (b) n = 3, l = 2 : 3d-orbital (l = 0) 4 orbitals 3s-subshell (c) n = 3, l = 1 : 3p-orbital (l = 1) (d) n = 2, l = 1 : 2p-orbital 3p-subshell (m = 0) 2 electrons (e) n = 4, l = 3 : 4f-orbital (l = 2) one orbital 6 electrons 3d-subshell (m = –1, 0, +1) 10 electrons (f ) n = 4, l = 2 : 4d-orbital three orbitals (m = –2, –1, 0, +1, +2) Example 50. five orbitals (a) What subshells are possible in n = 3 energy level ? (b) How many orbitals (of all kinds) are possible in this level ? 9 orbitals 18 electrons Solution : (a) Subshells in n = 3 energy level (n = 4) (l = 0) (m = 0) 2 electrons We know that the subshells are given by different N-shell 4s-subshell one orbital values of l. (l = 1) (m = –1, 0, +1) 4p-subshell 6 electrons For n = 3, the possible values of l are 0, 1 and 2. (l = 2) three orbitals The corresponding subshells are : 4d-subshell (m = –2, –1, 0, +1, +2) 10 electrons l = 0, s-subshell; l = 1, p-subshell; l = 2, d-subshell. (l = 3) five orbitals (m = –3, –2, –1, 0, +1, 14 electrons (b) Number of orbitals 4f-subshell +2, +3) For n = 3, there are one s, three p and five d-orbitals. seven orbitals This makes total of nine orbitals in n = 3 level. 16 orbitals 32 electrons Example 51. Therefore, it may be concluded that the maximum Explain giving reasons, which of the following sets number of orbitals in each shell is n2 and maximum of quantum numbers are not possible number of electrons is 2n2 as shown below : (a) n = 0, l = 0, mmmmmlllll=====–000–03,,,,, mmmmmsssss = + 11111/////22222 (b) n = 1, l = 0, ml = 0, = – Shell symbol K L MN (c) n = 1, l = 1, = + Shell number (n) 1234 (d) n = 2, l = 1, = – No. of orbitals (n2) 1 4 9 16 (e) n = 3, l = 3, = + (f) n = 3, l = 2, No. of electrons (2n2) 2 8 18 32 ms = +1/2 Example 47. Solution : (a) Not possible because n cannot have zero (a) If the quantum number ‘l’ has value of 2, what value. are the permitted values of the quantum number (b) Possible ml ? (c) Not possible because for n = 1, l = 1 is not possible. l can have values 0, 1 ... (n – 1) only.

2/50 MODERN’S abc + OF CHEMISTRY–XI (d) Possible (e) Not possible because for n = 3, l cannot have 3 However, for 2s-orbital [Fig. 32 (b)] , the wave function value. decreases in the beginning with increase in r, becomes (f) Possible zero at a particular distance and then becomes negative. The wave function for 2p-orbital [Fig. 32(c)] rises to ©50. Give all possible values of l, ml and ms for electrons maximum and then decreases. It is very interesting to Modern Publishers. All rights reserved.when n = 3. note that the 2s-wave function can be positive or negative 51. How many electrons in a given atom can have the depending upon the distance. following quantum numbers ? NOTE (a) n = 3, l = 1 (b) n = 3, l = 2, ml = 0 (c) n = 3, l = 2, ml 1 It may be noted that the positive and negative = + 2, ms = + 2 (d) n = 3. signs refer to the sign of the wave function and have no connection to the positive and negative charges. 52. If the quantum number ‘n’ has a value of 3, what are the permitted values of the quantum number ‘l ’ ? As we here seen for 2s orbtial, at a point, the wave function also becomes zero. This point at which 53. (i) An atomic orbital has n = 3, What are the possible radial wave function becomes zero is called radial values of l and ml? nodal surface or simply node. At the node, the value (ii) List the quantum numbers (ml and l) of electrons of wave function changes from positive to negative. In for 3d-orbital. general, it has been found that ns orbitals have (n – 1) nodes. Similarly, np orbitals have (n – 2) nodes and so (iii) Which of the following orbital are possible ? on. For example, the radial wave function for 2p-orbital has no node as shown in Fig. 32. 1p, 2s, 2p and 3f. In all cases, Ψ approaches zero as r approaches 50. l = 0, ml = 0; l = 1, ml = –1, 0, + 1; infinity. l = 2, ml = –2, –1, 0, +1, +2 and ms =+ 1 and It may be noted that like Ψ, the plots of Ψ also do 2 not have physical meaning. However, the importance of these plots lies in the fact that they give information – 1 for each value of ml. about how the radial wave function changes with 2 distance r and about the presence of nodes where the sign of Ψ changes. These have particular importance in understanding the chemical bonding. 51. (a) 6 (b) 2 (c) 1 (d) 18 52. 0, 1, 2. 53. (i) l = 0, ml = 0 ; l = 1, m = – 1, 0, + 1 ; l = 2, ml = – 2 – 1, 0, + 1, + 2 (ii) l = 2, ml = –2, –1, 0 + 1, + 2, (iii) 2s, 2p Hints & Solutions on page 68 SHAPES OF ATOMIC ORBITALS OR Fig. 32. Plots of wave function (Ψ) with distance from BOUNDARY SURFACE DIAGRAMS the nucleus for It has been learnt that the probability of finding (a) 1s-orbital (b) 2s-orbital (c) 2p-orbital the electron does not become zero even at large B. Probability Density (Ψ2) Graphs distances from the nucleus. Therefore, it is not possible According to the German physicst, Max Born, the to draw any sort of geometrical figure that will enclose square of the wave function, Ψ2 at a point gives the a region of 100% probability. However, for the sake of probability density of finding the electron at that point. pictorial clarity, we draw figures in which the These variations of Ψ2 as a function of r are probability of finding the electron is maximum (about 90—95%). These most probable regions are called obtained by plotting ψ2 against r. boundary surface diagrams or orbitals. The plots for 1s-, 2s- and 2p orbitals are shown in A. Radial Wave Functions Fig. 33 ahead. These diagrams look similar to Fig. 32 The shapes of orbitals are obtained from the except that these become positive throughout (square variation of wave function Ψ as a function of r (distance of negative quantity is always positive). These graphs from the nucleus). This is also called radial dependence are called probability density graphs. or radial wave function. This can be shown in a simple method by plotting a graph between wave function (ψ) and distance (r) from the nucleus. These graphs are shown for 1s, 2s and 2p orbitals in Fig. 32. It is clear from Fig. 32 (a) that wave function for 1s-orbital continuously decreases with increase in r.

STRUCTURE OF ATOM 2/51 © Fig. 33. Variation of ψ2 with distance from the nucleus for (a) 1s-orbital (b) 2s-orbital and (c) 2p-orbital. Modern Publishers. All rights reserved. It is clear from Fig. 32 that for s-orbitals (1s and 2s), the maximum electron density is at the nucleus and for p-orbitals (2p), it has zero electron density at the nucleus. It may be noted that all orbitals except s-orbitals have zero electron density at the nucleus. Radial Probability Functions (4πr2Ψ2) The Ψ2 versus r plots give the probability density for the electron around the nucleus. However, in order to determine the total probability in an infinitesimally small region, we have to multiply probability density (Ψ2) by the volume of region, i.e. Probability = Ψ2 × dV where dV is the volume of the region. Since the atoms have spherical symmetry, it is more useful to discuss the probability of finding the electron in a spherical shell between the spheres of radius (r + dr) and r. To understand this, consider the space around the nucleus to be divided into infinite number of concentric shells. The volume of such a shell of extremely small thickness, dr, is 4πr2 dr* so that Probability = Ψ2 × 4πr2 dr = 4πr2 dr Ψ2 This gives the total probability of finding the electron at a particular distance (r). This is called radial probability. The plots of probability (4πr2 Ψ2) as a function of distance from the nucleus (r) are called radial probability distribution function (r.d.f.) graphs. It is clear that the radial probability distribution graphs depend not only upon the probability density but also on the volume of the shell. The probability density (Ψ2) for 1s is maximum near the nucleus and it goes on decreasing with distance (Fig. 33). However, the volume of the shell goes on increasing with increase in distance as shown in Fig. 34 (a). The product of probability density and volume of shell gives the radial probability (4πr2Ψ2) and is plotted against the distance from the nucleus [Fig. 34 (b)]. Fig. 34. Radial probability distribution function curve for 1s-orbital. * The volume of the shell is 4πr2dr and it should not be confused with the volume of sphere which is 4 πr3. It can be calculated as : 3 Volume of shell = ⎡Volume of sphere ⎤ − ⎡Volume of sphere⎤ = 4 π(r + dr)3 – 4 πr3 ⎢⎣with radius (r + dr)⎥⎦ ⎣⎢with radius r ⎥⎦ 3 3 = 4 π[r3 + 3r2dr + 3rdr2 + dr3] – th34eπhr3ig=he34rπp[or3w+er3sr2odfrd]r–su34cπhra3 s 3 (As dr represents an extremely small thickness, dr2 and dr3 may be neglected) 4 Volume of shell = 3 π × 3r2dr = 4πr2dr

2/52 MODERN’S abc + OF CHEMISTRY–XI The graph for 1s-orbital shows that the probability of finding the electron is zero at nucleus, it keeps on increasing and becomes maximum at a particular distance from the nucleus and then gradually decreases. The peak of the curve gives the distance from the nucleus where the probability of finding the electron is maximum. This is called the radius of maximum probability. For hydrogen atom, this distance has been found to 52.9 pm (or 0.529 Å). This value agrees well with the value calculated by Bohr for the radius of first orbit. © Modern Publishers. All rights reserved.It is important to note that Bohr predicted that the electron will always be at r = 52.9 pm for H atom. However, according to wave mechanics, the electron is most likely to be found at this distance but there is probability of finding the electron at distances shorter and larger than 52.9 pm. The radial probability distribution function for 2s orbital is shown in Fig. 35(b). The curve shows that there are two regions of high probability (maxima). In between the regions of maximum probabilities, there is a region where the probability of finding the electron is zero. It is called node. The radial probability function for 2p orbital is also shown in Fig. 35(c). The radial probability functions versus distance r from the nucleus for 1s, 2s and 2p orbitals are shown in Fig. 35. 1s-orbital Fig. 35. Radial probability functions as a function of distance, r from the nucleus for (a) 1s (b) 2s and (c) 2p orbital. NOTE It may be noted that the distance of maximum probability for a 2p orbital is slightly less than that for a 2s electron. However, in contrast to a 2p-curve, there is a small additional maxima in the 2s curve. This indicates that the electron in 2s orbital spends some of its time near the nucleus. In other words, the 2s electron penetrates a little closer to the nucleus than the 2p-electron. As a result, 2s electron is attracted more strongly by the nucleus than a 2p-electron. That is why, 2s electron is more stable and, hence, has lower energy than a 2p-electron. Boundary Surface Diagrams The shapes of orbitals may be represented in terms of boundary surfaces diagrams. In this representation, a boundary surface is drawn for an orbital representation which encloses maximum probability (about 90%) of finding the electron. In other words, boundary surface diagrams give the most probable regions. These boundary surface diagrams of some orbitals are discussed below : 1. Shapes of s-orbitals s-orbitals are non-directional and spherically symmetrical. This means that the probability of finding the electron is same in all directions at a particular distance from the nucleus. The 1s-orbital is shown in Fig. 36 ahead. It is observed that density of charge cloud is maximum at the nucleus and decreases with increase in distance from the nucleus. The 2s-orbital is also non-directional and spherically symmetrical. In this case, the probability density is maximum at the nucleus and becomes small at large distances. However, the effective volume or size of 2s is larger than 1s-orbital. An important feature of 2s-orbital is that there is a spherical shell within 2s-orbital (region without dots) where the probability of finding the electron is practically zero. This is called a node or a nodal surface. Thus, a 2s-orbital differs from 1s-orbital in being larger in size and having a nodal surface. For the sake of simplicity, the boundary surface diagram for 1s orbital is shown in Fig 36 (c) instead of giving its charge cloud.

STRUCTURE OF ATOM 2/53 ©Fig. 36. Shapes of 1s and 2s orbitals. Fig. 37. Shape of 3s orbital. It has two nodes. Modern Publishers. All rights reserved. The higher s-orbitals have also spherical shapes and number of nodal surfaces in s-orbital for any given energy level is n – 1, where n represents energy level. For example, the shape of 3s orbital is shown in Fig. 37. It has two nodes. It is also observed that the size of s-orbital increases with increase in value of n i.e., 4s > 3s > 2s > 1s and the electron is located further away from the nucleus as the principal quantum number increases. 2. Shapes of p-orbitals ml2opbeyeaasnnFssdyotmrh2papmtz-.oetTrhtbrheiierctesaaellasatrh(eblrote=uhetr1eoa)er,pbtpaih-troaetrrlisbeciuatalraraelersetaiqhnxurieasea.lecDihpneopepsn-esseniubrdblgiesynhogberuuliletp.ndoTtnihaffteteihsoreeninasorrtceihoedrenreitresasiotgpironoiennantdoetifadnttghaiosetnoplsox.mb,Eepl asy=,caht–nhod1ers,pbe0zi;t,aafr+olerc1doeenxvsaasiimlgsuntpesasleto.e,fTd2thwpaixoss, 2olorpbbxe,its2apslyyaamrnemdsey2tmprzimcaacelctaorirrcdoauilnnagdraoxsu-anthxdeiszy-aaanxrdeiss2(yFpmyigom.r3be8ittr)a.iclTahhlaeasbsthowuaotpexlo,obyfeatshnsedyozmr-bamixteeatslririsceasclpaaellcretodiuvdneudlym.yT-bahxbaisteliwsl ,hs2hilpaexptoehr.beFiltooarbltehhsaeossfta2wkpeoz of simplicity a 2p orbital (e.g., 2px) is shown by simple boundary surface as given in Fig. 38 (d). Fig. 38. Shapes of 2p orbitals. It may be noted that the two lobes of a p-orbital are separated by a plane having zero electron density. This is called nodal plane. The nodal planes for 2px, 2py and 2pz orbitals are shown in Fig. 39 ahead. It should be noted that the probability of finding the electron in a particular p-orbital is equal in both lobes. The p-orbitals of higher energy levels (n = 3, 4, 5....etc.) have similar shapes although their sizes are bigger. The size of p-orbital increases with increase in value of n as 4p > 3p > 2p. Like s-orbitals, p-orbital also passes through zero probability region and number of nodes in p-orbitals are given by n – 2 i.e, for 3p-orbital number of radial nodes is one, two for 4p orbital and so on.

2/54 MODERN’S abc + OF CHEMISTRY–XI © Fig. 39. Nodal planes in 2p orbitals. Modern Publishers. All rights reserved. 3. Shapes of d-orbitals For d-orbital (l = 2), there are five possible orientations corresponding taoremdl e=si–gn2a,t–ed1,as0,3+dx1y,, + 2. This means that there are five orbitals in each d-subshell. For 3d-subshell, these 3dyz, 3dzx, 3dx2−y2 and 3dz2. These five orbitals are equal in energy but differ in their orientations. The shapes of these orbitals are described below : (i) The three orbitals zdxxyp,ladnyzesanredspdezxctaivreelys.imThileasrealnobdeesalcieh consists of four lobes of high electron density lying in xy, yz and in between the principal axes. For example, in cToarhbseietdaolxf2i−sdyxe2yxoarobcrtibtlyaitllaihlk,aestdhaxelysofoofruoburitrlaoolbfeehxsicgelhpietetlihencatxtryoitnpisdlaernnotesaititynedablteohntrwgouetghehne the x and y-axes. (ii) principal axes x and y. It may be noted that this 45° around the z-axis. (iii) The dz2 orbital consists of two lobes along the z-axis with a ring of high electron density in the xy plane. The shapes of 3d-orbitals are shown in Fig. 40. Fig. 40. Shapes of five 3d-orbitals. Nodes and Nodal Planes passing through the pnluacnleeus,sp. aFsosrinegxtahmropuleg,h3tdhxey orbital has two nodal A spherial surface within an orbital on which origin and bisecting the xy plane containing z-axis. the probability of finding the electron is zero is called The number of angular nodes or nodal planes spherical node or radial node. These are given by ‘l’ i.e, one angular node for p orbitals correspond to positions at which radial wave (l = 1); two angular nodes for d-orbitals (l = 2) and function passes through zero. so on. The number of spherical or radial nodes in In general, in an orbital : an orbital = (n –l–1) Total number of nodes = n – 1 For example, 1s orbital (n = 1, l = 0) has no Angular nodes = l node, 2s orbital; (n = 2, l = 0) has one node, 2p orbital (n = 2, l = 1) has no node, 3p orbital Radial nodes = n – l – 1. (n = 3, l = 1) has one node and so on.... For example, Beside radial nodes, the probability density Radial nodes functions for np and nd orbitals are zero at the Orbital No. of Orbital No. of planes passing through the nucleus. A plane nodes nodes passing through the nucleus on which the 1s 0 3d 0 probability of finding the electron is zero is called a 2s 1 4d 1 nodal plane. For example as shown in Fig. 39, in 3s 2 5d 2 case of 2caplxleodrbaitnalg,uylzaprlannoedies.aSnimodialalrplyla, nfoeu. Tr h3ids is also 2p 0 4f 0 orbitals (3dxy, 3dyz, 3dzx and 3dx2–y2) have two 3p 1 5f 1 perpendicular nodal planes that intersect in a line 4p 2 6f 2

STRUCTURE OF ATOM 2/55 Angular nodes No. of angular nodes ENERGY LEVEL DIAGRAM FOR ELECTRONS Orbital 0 IN AN ATOM s 1 plane for each orbital p 2 planes for each orbital* The relative energies of various orbitals can be d shown by an arrangement known as energy level diagram. The energy level diagram for hydrogen atom is given in Fig. 41. It is clear from the figure that the © *It may be noted that dz2 orbital has one conical surfaceModern Publishers. All rights reserved. which is counted as two angular nodes. NOTE Fig. 41. Energies of different orbitals in a hydrogen atom It may be noted that the shapes of orbitals energies of various orbitals in a hydrogen atom depend represent plots of probability density (ψ2) and are only upon the value of principal quantum number (n) therefore, always positive. But sometimes sketches and independent of the value of l. In other words, all of wave function (ψ) are also plotted. These have +ve the subshells in a given principal shell i.e., s, p, d and and –ve lobes. The signs in the lobes represent f have same energies. For example, 2s- and 2p-orbitals the sign of the wave function in different have same energies. Similarly, all the orbitals of third directions. The wave functions may have +ve or –ve shell (n = 3) i.e., 3s, 3p and 3d have same energies and signs. These graphs of wave functions are very all the orbitals of fourth shell (n = 4), i.e., 4s, 4p, 4d important in understanding the bonding between and 4f have same energies. atoms (discussed in unit 4). These +ve and –ve signs have nothing to do with positive or negative charge. 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f It may be remembered that The orbitals having the same energy are called  s-orbitals are always positive. degenerate orbitals. The electron in 1s orbital in  The opposite lobes of p-orbitals have opposite hydrogen atom corresponds to most stable state and is called ground state. An electron in the 2s, 2p or higher signs. orbital in a hydrogen atom is called excited state.  The opposite lobes of d-orbitals have same sign (two However, in the case of multielectron atom, the energies of orbitals having the same value of n but opposite lobes have +ve sign while the other two different values of l are different. This means that opposite lobes have –ve sign). These are shown below: energies of the orbitals depend upon the value of n as well as l. In other words, the different subshells of the same energy shell have different values. For example, 2s and 2p-orbitals have different energies. The three subshells of n = 3 i.e., 3s, 3p and 3d have different energies. Similarly, the subshells for n = 4, i.e. 4s, 4p, Fig. 42. Energies of different subshells and orbitals in a multielectron atom.

2/56 MODERN’S abc + OF CHEMISTRY–XI 4d and 4f-subshell have different energies as shown in energy shell. For example, the energy of Fig. 42. The main reason for having different energies 4s-orbital is less than that of a 3d-orbital. of the subshells is the mutual repulsion among Similarly, 5s-orbital is lower in energy than electrons in a multielectron atom. In case of hydrogen 4d-orbital. atom, the only interaction present is the attraction (iv) The energy of various orbitals increases in between the negatively charged electron and the the order : ©positively charged nucleus. However, in multielectron 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, Modern Publishers. All rights reserved.atoms, besides the presence of attraction between the 4f, 5d, 6p, 7s. electron and the nucleus, there are repulsion terms ELECTRONIC CONFIGURATION OF ATOMS between every electron and other electrons present The filling of orbitals in an atom is a hypothetical in the atom. process in which the atom is built up by feeding Cause of different energy patterns of electrons in orbitals, one at a time and by placing hydrogen atom and multielectron atoms : The each new electron in the lowest available energy energy on an electron in an orbital depends upon the orbital. The distribution of electrons in different following factors : orbitals is known as electronic configuration of (i) nuclear charge the atom. This characterises each electron in an (ii) principle energy level, atom. For the sake of presentation, the following (iii) the presence of electrons in the lower energy symbols are commonly used : levels. In a multielectron atom, the electrons occupying A box for an orbital (square or circular); an arrow the inner energy levels tend to reduce the effect of for an electron, the direction of the arrow giving the nuclear charge on the electrons present in the orientation of its spin. Two arrows are shown for two outermost energy level. In other words, the electrons electrons with opposite orientations of spin. in the inner shells screen or shield the outermost electrons form the nucleus. This effect is known as Alternatively, electronic configuration is screening or shielding effect. This is different in expressed by indicating the principal quantum number different types of orbitals. The net positive charge and its respective orbital along with the number of experienced by the electron from the nuclear is called electrons present in it. For example, the notation 3px1 effective nuclear charge s(Zphefef)r.ical indicates that there is one electron in px orbital of For example, being in shape, s-orbital third principal shell. spends more time closer to the nucleus in comparison to p-orbital and p-orbital spends more time near the The filling of orbitals is governed by the following nucleus in comparison to d-orbital. In other words, principles : s-orbital will be screened less than a p-orbital and a p-orbital will be less screened than a d-orbital for a 1. Aufbau principle. The Aufbau principle given shell. In other words, effective nuclear charge states that o(Zfefaf)zeimxpuetrhieanl cqeduabnyttuhme orbital decreases with increase number (l). This means the in the ground state of the atoms, the orbitals s-orbitals will be more tightly bound to the nucleus are filled in order of their increasing than p-orbitals while p-orbitals will be more tightly energies. In other words, electrons first occupy bound to the nucleus than d-orbitals and so on. Thus, the lowest-energy orbital available to them and the energy of different orbitals will increase as enter into higher energy orbitals only after the lower energy orbitals are filled. s < p < d ..... However, in the case of hydrogen atom, there is only one electron and therefore, no shielding effect. As a result, the different orbitals (s, p, d and f) of the same energy level have same energies. From Fig. 42, the following points are clear : (i) For the same value of n, the higher the value of l, the higher is the energy. Thus, 2p-orbital (n = 2, l = 1) has more energy than 2s-orbital (n = 2, l = 0). Similarly, the energies of subshells for n = 3 are in the order: 3s (n = 3, l = 0) < 3p (n = 3, l = 1) < 3d (n = 3, l = 2) (ii) The difference of energies of s and p-orbitals is small for the same value of n but the difference of p- and d-orbitals is large and so on. (iii) In some cases, the sub-level of higher energy shell falls below the sub-level of a lower

STRUCTURE OF ATOM 2/57 The word aufbau in German means building up. as (n + l) rule or Bohr Bury’s rule. According to The building up of the orbitals means the filling up of this : orbitals with electrons. From an energy level diagram (Fig. 42) for multi-electron atoms, the following (i) Orbitals fill in the order of increasing sequence is observed for orbitals in the increasing value of n + l. For example, 3s-orbital order of energy: (n + l = 3 + 0 = 3) will be filled before 4s (n + l = 4 + 0 = 4) orbital. Similarly, out of 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 3d and 4s, the 4s (n + l = 4 + 0 = 4) orbital will 6p, 7s be filled before 3d (n + l = 3 + 2 = 5) orbital. According to Aufbau principle, the orbital should (ii) If the two orbitals have same value of (n + l), be filled in the above sequence. then the orbital with lower value of n will be filled first. For example, 2p-orbital (n + l = 2 It is very important to remember that the + 1 = 3) and 3s-orbital (n + l = 3 + 0 = 3) have sequence of energy levels pertains up to 3p and the same (n + l) value but 2p-orbital has lower then 4s-orbital comes first instead of 3d. In fact, the value of n and therefore, it will be filled first. energy of an orbital is determined by the quantum This is shown below in Table 5. numbers n and l with the help of important rule known © Modern Publishers. All rights reserved.Table 5. Arrangement of orbitals with increasing energy on the basis of (n + l) rule Orbital Value Value of Value of of n l (n + l) 1s 2s 1 0 1+0=1 2p 2 0 2+0=2 3s 2 1 2+1=3 VUW 2p (n = 2) has lower energy 3p 3 0 3+0=3 WVU than 3s (n = 3) 4s 3 1 3+1=4 WVU 3p (n = 3) has lower energy 3d 4 0 4+0=4 than 4s (n = 4) 4p 3 2 3d (n = 3) has lower energy 4 1 3+2=5 than 4p (n = 4) 4+1=5 This rule also helps to account for the fact that Fig. 43. Schematic diagram to remember sequence of certain orbitals with higher value of n but lower value filling atomic orbitals. of l have less energy than orbitals with lower value of n and higher value of l. For example, let us apply this electron pairing will not take place in rule to 4s- and 3d-orbitals. For 3d-orbital (n + l) value orbitals of same energy (same subshell) until = 3 + 2 = 5 while for 4s-orbital (n + l) value = 4 + 0 = 4. each orbital is singly filled. Thus, 4s-orbital has lesser energy than 3d-orbital and, therefore, is filled first. This suggests that it is difficult for an electron to enter an orbital which already has an electron than to The sequence of energy levels can be easily enter an unoccupied orbital of same energy. This remembered by the systematic diagram as shown in principle is very important in guiding the filling of p, d Fig. 43. Starting from the top, the direction of the arrows gives the order of filling of orbitals in starting from right top to bottom left. 2. Pauli’s exclusion principle. According to this principle, an orbital can accommodate maximum of two electrons and these must have opposite spins. This principle thus limits the accommodation of electrons in an orbital. This means that an orbital can have 0, 1 or 2 electrons. Moreover, if an orbital has two electrons, they must be of opposite spins. 3. Hund’s rule of maximum multiplicity. According to this rule,

2/58 MODERN’S abc + OF CHEMISTRY–XI and f orbitals, which have more than one kind of orbitals. For example, we know that there are three p-orbitals (px, py and pz) of the p-subshell in a principal energy level. According to Hund’s rule, each of the three p-orbitals must get one electron of parallel spin before any one of them receives the second electron of opposite spin. Electronic Configurations of Atoms Based upon the above rules and the sequence of energy levels, let us write the electronic configurations of some atoms. © Modern Publishers. All rights reserved.Hydrogen (At. No. = 1). Since hydrogen has only one electron, it must go to 1s-orbital which has lowest energy. 1s H : (Z = 1) 1s1 or ↑ Helium (At. No. = 2). In helium atom, the second electron can also go into 1s-orbital. The two electrons must have opposite spins (Pauli’s exclusion principle). 1s He : (Z = 2) 1s2 or ↑↓ Lithium (At. No. = 3). Since 1s-orbital is filled with two electrons, it cannot have any more. Therefore, the third electron goes to the next lowest energy orbital, namely 2s-orbital. 1s 2s Li : (Z = 3) 1s22s1 or ↑↓ ↑ Beryllium (At. No. = 4). The fourth electron in beryllium fills 2s-orbital. 1s 2s Be : (Z = 4) 1s2 2s2 or ↑↓ ↑↓ Boron (At. No. = 5). In this case, the fifth electron goes into any one of the 2p orbitals (say 2px) : 1s 2s 2px2py2pz B : (Z = 5) 1s22s22px1 or ↑↓ ↑↓ ↑ It may be remembered here that the fifth electron can enter any of the tohrrbeitea2lspxi,n2tphyeoorr2dpezr orbitals because all are of the same energy. It is only a matter of convention that we fill the of 2px, 2py and 2pz. Carbon (At. No. = 6). In carbon atom, the sixth electron is also to go into the 2p-orbitals because it can accommodate six electrons. Here Hund’s rule applies, i.e., the electrons enter the orbitals of same energy with parallel spin until all are singly filled. Therefore, the sixth electron cannot enter the 2px orbital, rather it can go into either 2py or 2pz in accordance with Hund’s rule : 1s 2s 2px 2py 2pz C : (Z = 6) 1s2 2s2 2px1 2py1 or ↑↓ ↑↓ ↑↑ Nitrogen (At. No. = 7). Applying Hund’s rule, nitrogen atom has three unpaired electrons in 2p-orbitals as : 1s 2s 2px 2py 2pz N : (Z = 7) 1s2 2s2 2px1 2py1 2pz1 or ↑↓ ↑↓ ↑ ↑↑ Oxygen (Z = 8), fluorine (Z = 9) and neon (Z = 10). Beginning with oxygen, the 2p-orbitals start getting filled by second electron till each of these is completely filled. O : (Z = 8) 1s2 2s2 2px2 2py1 2pz1 or 1s 2s 2px 2py 2pz F : (Z = 9) 1s2 2s2 2px2 2py2 2pz1 or Ne : (Z = 10) 1s2 2s2 2px2 2py2 2pz2 or ↑↓ ↑↓ ↑↓ ↑ ↑ 1s 2s 2px 2py 2pz ↑↓ ↑↓ ↑↓ ↑↓ ↑ 1s 2s 2px 2py 2pz ↑↓ ↑↓ ↑↓ ↑↓ ↑↓

STRUCTURE OF ATOM 2/59 Sodium (Z = 11) to Argon (Z = 18). The electronic configurations of these atoms are written exactly in the same manner as discussed above. Here, 3s- and 3p-orbitals are filled as did 2s-and 2p-orbitals in Li to Ne. Furthermore, for simplicity a common convention is used. In this, the detailed electronic configuration of the noble gas core preceding the valence shell is represented by the symbol of the noble gas in square brackets. Then, the configuration of the valence shell is written after the symbol. © Modern Publishers. All rights reserved.For example, the electronic configuration of sodium may be written as Na (Z = 11) : [Ne]3s1. The electronic configurations of these elements are given below : Sodium, Na (Z = 11) [Ne] 3s1 Magnesium, Mg (Z = 12) [Ne] 3s2 Aluminium, Al (Z = 13) [Ne] 3s2 3px1 Silicon, Si (Z = 14) [Ne] 3s2 3px1 3py1 Phosphorus, P (Z = 15) [Ne] 3s2 3px1 3py1 3pz1 Sulphur, S (Z = 16) [Ne] 3s2 3px2 3py1 3pz1 Chlorine, Cl (Z = 17) [Ne] 3s2 3px2 3py2 3pz1 Argon, Ar (Z = 18) [Ne] 3s2 3px2 3py2 3pz2 Potassium (Z = 19) and calcium (Z = 20). In these atoms, electrons enter the 4s-orbital in accordance with the fact that 4s-orbital is slightly lower in energy than 3d-orbital. 4s 3d Potassium, K (Z = 19) : [Ar] 4s1 ↑ 4s 3d Calcium, Ca (Z = 20) : [Ar] 4s2 ↑↓ Scandium (Z = 21) to zinc (Z = 30). With scandium (Z = 21), 3d-subshell comes into use and from scandium to zinc, this subshell is gradually filled up. The electronic configurations of the elements from scandium (Z = 21) to zinc (Z = 30) are given below : 3d 4s Scandium, Sc (Z = 21) : [Ar] 3d1 4s2 or ↑ ↑↓ Titanium, Ti (Z = 22) : [Ar] 3d2 4s2 or ↑↑ ↑↓ Vanadium, V (Z = 23) : [Ar] 3d3 4s2 or ↑↑↑ ↑↓ Chromium,Cr (Z = 24) : [Ar] 3d5 4s1 or ↑↑↑↑↑ ↑ Manganese, Mn (Z = 25) : [Ar] 3d5 4s2 or ↑↑↑↑↑ ↑↓ Iron, Fe (Z = 26) : [Ar] 3d6 4s2 or ↑↓ ↑ ↑ ↑ ↑ ↑↓ Cobalt, Co (Z = 27) : [Ar] 3d7 4s2 or ↑↓ ↑↓ ↑ ↑ ↑ ↑↓ Nickel, Ni (Z = 28) : [Ar] 3d8 4s2 or ↑↓ ↑↓ ↑↓ ↑ ↑ ↑↓ Copper, Cu (Z = 29) : [Ar] 3d10 4s1 or ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ Zinc, Zn (Z = 30) : [Ar] 3d10 4s2 or ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ Exceptional Configurations of Chromium and Copper From the above configurations, it may be noted that there are two irregularities in the general trend. The electronic configurations of Cr and Cu are different than what we expected. The electronic configurations for these atoms are expected to be Cr : [Ar] 3d4 4s2 and Cu : [Ar] 3d9 4s2 However the actual configurations are Cr : [Ar] 3d5 4s1 and Cu : [Ar] 3d10 4s1

2/60 MODERN’S abc + OF CHEMISTRY–XI This is attributed to the fact that half-filled (containing one electron per orbital) and completely filled (containing two electrons per orbital) electronic configurations have lower energy and, therefore, extra stability. Consequently, the configurations, d5, d10, f 7 and f 14 which are either half-filled or completely-filled, are more stable. Thus, to acquire increased stability, one of the 4s-electrons goes into the nearby 3d-orbitals so that 3d- orbitals get half-filled in Cr and completely-filled in Cu. The extra stability of half-filled and completely-filled electronic configurations is due to their (i) symmetrical arrangement and (ii) large exchange energy as explained below : (i) Stability on the basis of symmetrical distribution of electrons. The symmetry leads to stability. Therefore, the electronic configurations in which all the orbitals of the same subshell are either completely- filled or exactly half-filled are more stable because of symmetrical distribution of electrons. For example, the expected configuration of Cr (Z = 24) is : © Modern Publishers. All rights reserved. 3d4 4s2 ↑↑↑↑ ↑↓ But if one electron is shifted from 4s- to 3d-subshell, the distribution of electrons become more symmetrical and more stable. 3d5 4s1 ↑↑↑↑↑ ↑ Similarly, in case of copper, when an electron is shifted from 4s- to 3d-subshell, the 3d-subshell becomes more stable due to symmetrical arrangement. 3d10 4s1 ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ (ii) Stability due to exchange energy. The half-filled and completely-filled electronic configurations are also stable due to large exchange of energy of stabilization. The exchange means the shifting of electrons from one orbital to another in the same subshell. This can be easily illustrated by considering the example of chromium. If the electronic configuration of Cr is 3d4 4s2, then the electron in d-subshell can exchange in the following ways : 3d 4s Cr (Z = 24) ↑↑↑↑ ↑↓ 1234 (i) The first electron (number 1) can exchange its positions with electrons numbered 2, 3 and 4 i.e., in 3 ways. (ii) The second electron (number 2) can exchange its positions in two ways with electrons 3 and 4 only because the exchange of position between 1 and 2 has already been considered in step (i). (iii) The electron (number 3) can exchange only in one way i.e. with electron 4. ↑↑↑↑ ↑ ↑↑↑ ↑↑↑↑ 1 2 3 Thus, in 3d4 arrangement, electrons can exchange in 3 + 2 + 1 = 6 ways or there are six possible arrangements with parallel spins in 3d4 configuration. On the other hand, in 3d5 configuration, the electron can exchange in 10 ways as shown below : ↑ ↑ ↑ ↑↑ ↑↑↑↑↑ 1 2 ↑ ↑ ↑ ↑↑ ↑↑↑↑↑ 3 4

STRUCTURE OF ATOM 2/61 It is clear that the electron 1 can exchange in four ways, the electron 2 in three ways, electron 3 in two ways and electron 4 in one way. Thus, total number of ways of exchanging electron in 3d5 configuration are 4 + 3 + 2 + 1 = 10. Since the number of exchanges in 3d5 4s1 configuration is more than that in 3d4 4s2 configuration, therefore, the greater exchange is responsible for the extra stability of this configuration. Similarly, it can be seen that the exchange energy for d10 electronic configuration is more than for d9 configuration. Thus, the greater exchange for half-filled and completely-filled configurations gives them extra stability. © Modern Publishers. All rights reserved.Gallium (Z = 31) to Krypton (Z = 36). With gallium onwards, 4p-orbitals get filled up as : Gallium Ga (Z = 31) : [Ar] 3d10 4s2 4px1 Germanium Ge (Z = 32) : [Ar] 3d10 4s2 4px1 4py1 Arsenic As (Z = 33) : [Ar] 3d10 4s2 4px1 4py1 4pz1 Selenium Se (Z = 34) : [Ar] 3d10 4s2 4px2 4py1 4pz1 Bromine Br (Z = 35) : [Ar] 3d10 4s2 4px2 4py2 4pz1 Krypton Kr (Z = 36) : [Ar] 3d10 4s2 4p6 Note. To write the electronic configuration of an atom, it is better if we remember the atomic number of noble gases and the orbitals which follow the noble gas. Then the next orbitals to be filled can be seen from the general sequence. The noble gases with their atomic numbers and orbitals which follow these noble gas cores are given below : [He]2 2s; [Ne]10 3s; [Ar]18 4s; [Kr]36 5s; [Xe]54 6s; [Rn]86 7s. For example, let us try to write the electronic configuration of I(Z = 53). The noble gas having atomic number less than 53 is [Kr]36 and orbital which follows this is 5s. After 5s the orbitals with increasing energies are 4d, 5p ..... so that I (Z = 53) : [Kr] 4d10 5s2 5p5 In the next eighteen elements from rubidium (Rb) to Xenon (Xe), the orbitals which are being gradually filled are 5s, 4d and 4p similar to that of 4s, 3d and 4p orbitals as discussed above. Then 6s orbitals are filled for Cs and Ba with one and two electrons respectively. Then from lanthanum (La) to mercury (Hg), the filling up of electrons takes place in 4f and 5d orbitals. After this, the filling of 6p, then 7d and finally 5f and 6d orbitals takes place. However, it may be noted that there are many anomalous configurations due to stability of half filled and completely filled orbitals. The electronic configurations of all elements are given in Table 6. It may be noted that it is very important to know the electronic configuration of atoms. The chemical behaviour of an atom depends on the distribution of electrons in various shells, subshells or orbitals. The tendency of atoms to take part in chemical combination also depends on the electrons in the outermost shell (called valence shell). Therefore, a detailed understanding of the electronic structure of atoms is very essential for getting an insight into the various aspects of modern chemical knowledge. Table 6. Electronic configuration of different elements. Atomic Element Electronic Atomic Element Electronic No. configuration No. configuration 1H 1s1 16 S – 3s2 3p4 2 He 1s2 17 Cl – 3s2 3p5 3 Li [He]2 2s1 18 Ar – 3s2 3p6 4 Be – 2s2 5B – 2s2 2p1 19 K – [Ar]18 4s1 6C – 2s2 2p2 20 Ca – 4s2 7N – 2s2 2p3 21 Sc – 3d1 4s2 8O – 2s2 2p4 22 Ti – 3d2 4s2 9F – 2s2 2p5 23 V – 3d3 4s2 10 Ne – 2s2 2p6 24 Cr* – 3d5 4s1 11 Na [Ne]10 3s1 25 Mn – 3d5 4s2 12 Mg – 3s2 26 Fe – 3d6 4s2 13 Al – 3s2 3p1 27 Co – 3d7 4s2 14 Si – 3s2 3p2 28 Ni – 3d8 4s2 15 P – 3s2 3p3 29 Cu* – 3d10 4s1 30 Zn – 3d10 4s2

2/62 MODERN’S abc + OF CHEMISTRY–XI Atomic Element Electronic Atomic Element Electronic No. configuration No. configuration ©31 Ga – 3d10 4s2 4p1 72 Hf – 4f 14 5d2 6s2 Modern Publishers. All rights reserved.32 Ge– 3d10 4s2 4p2 73 Ta – 4f 14 5d3 6s2 33 As – 3d10 4s2 4p3 74 W – 4f 14 5d4 6s2 34 Se – 3d10 4s2 4p4 75 Re – 4f 14 5d5 6s2 35 Br – 3d10 4s2 4p5 76 Os – 4f 14 5d6 6s2 36 Kr – 3d10 4s2 4p6 77 Ir – 4f 14 5d7 6s2 37 Rb – [Kr]36 5s1 78 Pt* – 4f 14 5d9 6s1 38 Sr – 5s2 79 Au* – 4f 14 5d10 6s1 39 Y – 4d1 5s2 80 Hg – 4f 14 5d10 6s2 40 Zr – 4d2 5s2 81 Tl – 4f 14 5d10 6s2 6p1 41 Nb* – 4d4 5s1 82 Pb – 4f14 5d10 6s2 6p2 42 Mo* – 4d5 5s1 83 Bi – 4f14 5d10 6s2 6p3 43 Tc – 4d5 5s2 84 Po – 4f14 5d10 6s2 6p4 44 Ru* – 4d7 5s1 85 At – 4f14 5d10 6s2 6p5 45 Rh* – 4d8 5s1 86 Rn – 4f14 5d10 6s2 6p6 46 Pd* – 4d10 87 Fr – [Rn]86 7s1 47 Ag* – 4d10 5s1 88 Ra – 7s2 48 Cd – 4d10 5s2 89 Ac – 6d1 7s2 49 In – 4d10 5s2p1 90 Th – 6d2 7s2 50 Sn – 4d10 5s2p2 91 Pa – 5f26d17s2 51 Sb – 4d10 5s2p3 92 U – 5f36d17s2 52 Te – 4d10 5s2p4 93 Np – 5f46d17s2 53 I – 4d10 5s2p5 94 Pu – 5f67s2 54 Xe – 4d10 5s2p6 95 Am – 5f77s2 55 Cs – [Xe]54 6s1 96 Cm – 5f76d17s2 56 Ba – 6s2 97 Bk – 5f97s2 57 La* – 5d1 6s2 98 Cf – 5f107s2 58 Ce* – 4f 2 6s2 99 Es – 5f117s2 59 Pr – 4f 3 6s2 100 Fm – 5f127s2 60 Nd – 4f 4 6s2 101 Md – 5f137s2 61 Pm – 4f 5 6s2 102 No – 5f147s2 62 Sm – 4f 6 6s2 103 Lr – 5f146d17s2 63 Eu – 4f 7 6s2 104 Rf – 5f146d27s2 64 Gd* – 4f 7 5d1 6s2 105 Ha – 5f146d37s2 65 Tb – 4f 9 6s2 106 Sg – 5f146d57s1 66 Dy – 4f 10 6s2 107 Bh – 5f146d57s2 67 Ho – 4f 11 6s2 108 Hs – 5f146d67s2 68 Er – 4f 12 6s2 109 Mt – 5f146d77s2 69 Tm – 4f 13 6s2 110 Ds – 5f146d87s2 70 Yb – 4f 14 6s2 71 Lu – 4f 14 5d1 6s2 111 Rg – 5f146d107s1 112 Cn – 5f146d107s2 * Elements with exceptional electronic configurations. Electronic Configuration of Ions neutral atom. However, it may be remembered that The electronic configuration of the ions may be electrons with the highest value of n are removed first.If more than one subshell of the valence shell are written almost similar to those of the atoms. In case occupied, the one with the highest value of l loses the of negatively charged ions, the extra electrons equal electron first. For example, to the charge are added to the appropriate orbitals. Mg (Z = 12) : 1s2 2s2 2p6 3s2 For example : Mg2+ (Z = 12) : 1s2 2s2 2p6 Fe (Z = 26) : 1s2 2s2 2p6 3s2 3p6 3d6 4s2 Cl (Z = 17) : 1s2 2s2 2p6 3s2 3px2 3p 2 3p 1 Fe2+ (Z = 26) : 1s2 2s2 2p6 3s2 3p6 3d6 Cl– (Z = 17) : y z S (Z = 16) : It is very important to remember here that in case 1s2 2s2 2p6 3s2 3p 2 3p 2 3p 2 of Fe2+ ion, the electrons are removed from 4s-orbital x y z first (highest value of n) and not from 3d-orbitals. 1s2 2s2 2p6 3s2 3p 3p 1 3p1z 2 y x S2– (Z = o1f6)po:si1tisv2e2lys2c2hpa6r3ges2d3ipon2x s3, pth2ye3pel2zectrons In case equal to the charge on the ion are removed from the

STRUCTURE OF ATOM 2/63 Ni (Z = 28) : 1s2 2s2 2p6 3s2 3p6 3d8 4s2 Example 55. Ni2+ (Z = 28) : 1s2 2s2 2p6 3s2 3p6 3d8 What atoms are indicated by the following Co (Z = 27) : 1s2 2s2 2p6 3s2 3p6 3d7 4s2 configuration ? Are they in the ground state or Co3+ (Z = 27) : 1s2 2s2 2p6 3s2 3p6 3d6 excited state ? REMEMBER (a) 1s2 2s2 2p2x 2p2y 2p1z While writing the electronic configuration of cations, the electronic configuration of the atom should be written first. The atomic number Z, which is equal to number of protons, remains the same. Only the number of electrons decreases. Therefore, the number of electrons equal to the units of positive charge should be removed from the outermost shell. © (b) 1s2 2s1 2p1x 2p1y 2p1z Modern Publishers. All rights reserved. (c) 1s2 2s2 2p6 3s2 3p1x 3p1y (d) 1s2 2s2 2p6 3s1 3p1x 3p1y 3p1z 3d1 (e) [Ar] 3d5 4s2 Solution : (a) 1s2 2s2 2p2 2p2y 2p1z . Atomic number is x 9 and it is configuration of fluorine in ground state. Example 52. (b) 1s2 2s1 2p1 2p1y 2p1 . Atomic number is 6 and it is Which of the following are isoelectronic species : x z Na+, K+, Mg2+, Ca2+, S2–, Ar ? configuration of carbon in excited state. Solution : Isoelectronic species are those which (c) 1s2 2s2 2p6 3s2 3p1x 3p1y Atomic number is 14 and it contain same number of electrons is configuration of silicon in ground state. Na+, Mg2+ are isoelectronic (contain 10 electrons) K+, Ca2+ S2– and Ar are isolectronic (contain 18 (d) 1s2 2s2 2p6 3s1 3p 1 3p1y 3p1z 3d1. Atomic number electrons). x Example 53. Write the electronic configuration of elements of is 15 and it is configuration of phosphorus in atomic numbers 10, 17, 25, 29 and 37. excited state. Solution : Electronic configuration of elements : (e) [Ar] 3d5 4s2. Atomic number is 25 and it is Atomic No. 10 : 1s2 2s2 2p6 configuration of manganese in ground state. Atomic No. 17 : 1s2 2s2 2p6 3s2 3p2x 3p2y 3p1z Example 56. Give the symbol of the atom whose ground state Atomic No. 25 : [Ar] 3d5 4s2 corresponds to each of the following Atomic No. 29 : [Ar] 3d10 4s1 configurations: Atomic No. 37 : [Kr] 5s1 (i) 1s2 2s2 2p6 Example 54. (ii) 1s2 2s2 2p6 3s2 3p6 How many unpaired electrons are present in the Give two examples of negative and two positive ground state of (i) P (Z = 15) (ii) Fe2+ (Z = 26) (iii) ions corresponding to the above configurations. Cl– (Z = 17) ? Solution : The number of unpaired electrons can be Solution : (i) 1s2 2s2 2p6. Atomic number is 10 and it predicted by writing the electronic configurations : corresponds to Ne (neon). The ions corresponding to this configuration are : (i) P (Z = 15) : 1s2 2s2 2p6 3s2 3p 1 3p1y 3p1z O2–, F–, Na+, Mg2+. x (ii) 1s2 2s2 2p6 3s2 3p6. Atomic number is 18 and it ∴ No. of unpaired electrons = 3 corresponds to Ar (argon). The ions corresponding to this configuration are : (ii) Fe2+ (Z = 26). The electronic configuration of Fe is 1s2 2s2 2p6 3s2 3p6 3d6 4s2. For Fe2+, two electrons S2–, Cl–, K+, Ca2+. are removed from 4s-orbtial so that the configuration of Fe2+ is : Example 57. Write the electronic configuration of the following 1s2 2s2 2p6 3s2 3p6 3d6 ions : All orbitals are completely filled except 3d- (i) H– (ii) Na+ (iii) O2– (iv) F– orbitals. The arrangement of five 3d-orbitals is : Solution : ↑↓ ↑ ↑ ↑ ↑ (i) 1s2 ∴ No. of unpaired electrons = 4 (ii) 1s2 2s2 2p6 (iii) 1s2 2s2 2p6 (iii) Cl– (Z = 17). The electronic configuration of Cl atom (iv) 1s2 2s2 2p6 is 1s2 2s2 2p6 3s2 3p5. Cl– ion has one extra electron and its configuration is : Cl– : 1s2 2s2 2p6 3s2 3p6 Since all orbitals are filled, there is no unpaired electron.

2/64 MODERN’S abc + OF CHEMISTRY–XI Example 58. Example 61. The quantum numbers of six elements are given below. Arrange them in order of increasing An atom has 2 electrons in the first (K) shell, 8 energies. If any of these combinations has/have electrons in the second (L) shell and the same energies, list them. 2 electrons in the third (M) shell. Give its electronic configuration and find out the following : (a) Atomic number (b) Total number of principal quantum numbers (c) Total number of sublevels (d) Total number of s-orbitals (e) Total number of p-electrons. Solution : The electronic configuration of the atom is : 1s2 2s2 2p6 3s2 (a) Atomic number = 2 + 2 + 6 + 2 = 12 (b) Number of principal quantum numbers = 3 (c) Number of sub-levels = 4 (1s, 2s, 2p, 3s) (d) Number of s-orbitals = 3 (1s, 2s, 3s) (e) Total number of p-electrons = 6. ©(1) n = 4, l = 2, ml = –2, ms = −1 Modern Publishers. All rights reserved. 2 (2) n = 3, l = 2, ml = 1, ms = +1 2 (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 Solution : 5(3p) < 2 (3d) = 4 (3d) < 6 (4p) Example 62. = 3 (4p) < 1 (4d) What is the maximum number of unpaired Example 59. electrons in Cu (Z = 29), Br– (Z = 35) and K+ (Z = 19) ? Indicate the number of unpaired electrons in (i) P (ii) Si (iii) Cr (iv) Fe (v) Kr Solution : Cu (Z = 29) Solution : : [Ne] 3s2 3p3 : 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (i) P(Z = 15) unpaired electrons = 3 Br (Z = 35) Unpaired electrons = 1 ∴ Br– (36e) (ii) Si (Z = 16) : [Ne] 3s23p2 : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5 unpaired electrons = 2 K (Z = 19) : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 (iii) Cr (Z = 24) : K+ (18 e) (iv) Fe (Z = 26) : [Ar] 3d5 4s1 Unpaired electrons = 0 unpaired electrons = 6 : 1s2 2s2 2p6 3s2 3p6 4s1 : 1s2 2s2 2p6 3s2 3p6 [Ar] 3d64s2 unpaired electrons = 4 Unpaired electrons = 0 (v) Kr (Z = 36) : [Ar] 3d104s2 4p6 unpaired electrons = 0. Example 60. 54. How many unpaired electrons are present in the ground state of phosphorus (Z = 15) ? Which of the following do and which do not make 55. Name the elements which correspond to each of sense ? the following configurations : Solution : 7p, 2d, 3s3, 3py3, 4f. (a) 1s2 2s2 2px2 2py1 2pz1 (b) 1s2 2s2 2p6 3s2 3px1 7p : makes sense (c) 1s2 2s2 2p6 3s2 3p6 4s1 2d : does not make sense. (There are no (d) 1s2 2s2 2p6 3s2 3p6 3d5 4s1 d-orbitals for n = 2) 56. List the values of n and l for the following orbitals: 3s3 : does not make sense (3s-orbital cannot have more than two electrons) (a) 3s (b) 4p (c) 4f (d) 3d 3py3 : does not make sense (3py orbital cannot 57. Write down all the four quantum numbers for have more than two electrons) outermost electron of sodium atom (Z = 11). 4f : makes sense

STRUCTURE OF ATOM 2/65 58. What is the maximum number of electrons that may 54. Three be present in all the atomic orbitals with principal quantum number 3 and azimuthal quantum 55. (a) Oxygen (b) Aluminium number 2 ? (c) Potassium (d) Chromium. 59. Write down the values of all the quantum numbers for 19th electron of Cr (Z = 24). 60. How many electrons in zinc (Z = 30) have n + l value equal to 4 ? 61. Write the electronic configuration of chromium (Z = 24) and predict in it (i) number of subshells (ii) number of electrons in subshell having l = 0 (iii) number of electrons having n + l value equal to 3 (iv) number of electrons in highest value of n. 62. How many electrons in p-subshell of argon (Z = 18) have same spin ? © 56. (a) n = 3, l = 0 (b) n = 4, l = 1 Modern Publishers. All rights reserved. (c) n = 4, l = 3 (d) n = 3, l = 2 57. n = 3, l = 0, m = 0, s = + ½ 58. Ten. 59. n = 4, l = 0, m = 0, s = + ½ 60. 8 (electrons in 4s and 3p-subshells) 61. (i) 7 (ii) 7 (iii) 8 (iv) 1 62. 6 Hints & Solutions on page 68 2 Q. 1. What will happen to the wavelength associated with a moving particle if its velocity is reduced to half? Ans. Q. 2. Wavelength becomes double of the original value because λ = h or λ ∝1. Ans. mv v Q. 3. Can we apply Heisenberg’s uncertainty principle to a stationary state ? Ans. Q. 4. No, because velocity is zero and position can be measured accurately. An electron beam after hitting a nickel crystal produces a diffraction pattern. What do you conclude ? Electron has wave nature. A molecule of O2 and O3 travel with the same velocity. What is the rates of their wavelength ? Ans. λ = h or λ ∝ 1 . mv m Mass of O2 molecule is 32 u and mass of O3 molecule is 48 u. ∴ λ(O2) = 48 = 1.5 ∴ λ(O2) = 1.5 times λ (O3). λ(O3) 32 Q. 5. What is the sequence of energies of 4s, 4p, 4d and 4f-orbitals in (i) a hydrogen atom and (ii) a zinc atom? Ans. Q. 6. (i) 4s = 4p = 4d = 4f (ii) 4s < 4p < 4d < 4f. Ans. Q. 7. Which shell would be the first to have a g-subshell ? Ans. 5th energy shell. Q.8. The 4f subshell of an atom contains 12 electrons. What is the maximum number of electrons Ans. having same spin in it ? Q.9. Ans. Seven. Which of the following sets of quantum numbers for orbitals in hydrogen atom has larger energy ? n = 3, l = 2, ml = +1, n = 3, l = 2, ml = – 1 Both are orbitals of the same subshell and, therefore, have same energy. How many nodes are present in 3p-orbital ? No. of nodes in an orbital = (n – l – 1) ∴ No. of nodes in 3p-orbital = 3 – 1 – 1 = 1.

2/66 MODERN’S abc + OF CHEMISTRY–XI Q.10. What is the difference in the angular momentum of an electron present in 2p and that present in 3p-orbital ? Ans. Angular momentum is same because it depends upon the quantum number l and not on n as : Angular momentum = h l(l + 1) . 2π © Modern Publishers. All rights reserved.Q.11. What will be the maximum number of electrons present in an atom having n + l = 4 ? Ans. The subshells which can have n + l = 4 are 4 s (4 + 0) and 3p ( 3 + 1). Therefore, these will accommodate maximum of 2 + 6 = 8 electrons. Q.12. How many spherical nodes do you expect in a 4s-orbital ? Ans. Three nodes. Q.13. What is the maximum number of electrons which can be accommodated in an atom in which the highest principal quantum number value is 4 ? Ans. 36 corresponding to Kr (4s2 4p6). Q.14. How many quantum numbers are required to specify an orbital ? Ans. Three (n, l and ml) Q.15. What are the quantum numbers of the valence electrons in potassium atom (Z = 19) in ground state ? Ans. For 3p-orbital, n = 4, l = 0, ml = 0, ms = + 1 . 2 Q.16. In which atom, the outermost electron can have the following set of quantum numbers ? n = 3, l = 0, ml = 0 ms = – 1 2 Ans. Sodium. Q.17. What is the value of orbital angular momentum for an electron in 2s-orbital ? Ans. Angular momentum is given as h l (l + 1) 2π Since for 2s orbital, l = 0 ∴ Angular momentum = 0 Q.18. How many unpaired electrons are present in P(Z = 15) ? Ans. Electronic configuration of P is : 1s2 2s2 2p6 3s2 3p3 No. of unpaired electrons = 3. Q.19. In what ways do the spatial distribution of the orbitals in 1s and 2s differ ? Ans. 2s orbital is larger while 1s orbital is smaller. 2s orbital has a spherical node while 1s orbital does not have. Q. 20. Explain the meaning of the symbol 4d6 . Ans. It means that 4d subshell has 6 electrons. 4 represents fourth energy shell and d is a subshell and 6 electrons are present in d orbitals of subshell. Q. 21. How many nodes are present in 4d orbital ? Ans. No. of nodes = (n – l – 1) = 4 – 2 – 1 = 1 (one node). Q. 22. What is the lowest shell which has an f-subshell ? Ans. Fourth Q. 23. Which quantum number does not follow from the solutions of Schrodinger wave equation ? Ans. Spin quantum number Q. 24. What is value of angular momentum for an electron in 5th orbit according to Bohr’s theory? Ans. 5h . 2π Q. 25. How are dxy and dx2 – y2 orbitals related ? Ans. The dxy orbital is exactly like dx2 – y2 orbital except that its lobes are at an angle of 45° to the lobes of dx2 – y2 orbital. Q. 26. How many electrons are possible in all shells with n + l = 5 ? Ans. Subshell with n + l = 5 are 5s, 4p and 3d. Hence electrons present in these subshells are 2 + 6 + 10 = 18. Q. 27. Write n, l and ml values for (i) 3s (ii) 2py (iii) 4pz. Ans. (i) n = 3, l = 0, ml = 0 (ii) n = 2, l = 1, ml = + 1 or – 1 (iii) n = 4, l = 1, ml = 0 Q. 28. Does d 2 orbital has zero electron density in xy plane ? z Ans. No. d 2 orbital has electron density in xy plane also as shown by ring. z

STRUCTURE OF ATOM 2/67 Q. 29. Which oefntehregyfooufraqnuenleactturmonniunmabheyrdsr(ong,eln, matl,omms)adnedtmerumltiineelectron (a) the atoms (b) the size of an orbital (c) the shape of an orbital (d) the orientation of an orbital in space ? ©Ans. (a) n (b) n (c) l (d) ml. Modern Publishers. All rights reserved. Q. 30. What physical meaning is attributed to the square of the absolute value of wave function |Ψ2| ? Ans. It measures the electron probability density at a point in an atom. Q. 31. Discuss the similarities and differences between a 1s and a 2s orbital. Ans. Similarities. (i) Both the orbitals have spherical shape. (ii) Both have same angular momentum (equal to zero). Differences. (i) Size of 2s orbital is larger than that of 1s orbital (ii) 1s has no node while 2s has one node. (iii) Energy of 2s orbital is more than that of 1s. Q. 32. What is the difference between the notation l and L ? Ans. The notation l represents azimuthal quantum number, which can have values 0, 1, 2, ... etc., whereas L represents second Bohr orbit for which n = 2. Q. 33. How many electrons in an atom may have the following quantum number ? (i) n = 4, ms = +½ (ii) n = 3 , l = 0 Ans. (i) n = 4, ms = +½ = 16 electrons (ii) n = 3 , l = 0 = 2 electrons Q. 34. For each of the following pair of hydrogen orbitals, indicate which is higher in energy : (i) 1s, 2s (ii) 2p, 3p (iii) 3dxy, 3dyz (iv) 3s, 3d (v) 4f, 5s. Ans. (i) 2s (ii) 3p (iii) both have same energy (iii) both have same energy (v) 5s. Q. 35. Which orbital in each of the following pairs is lower in energy in a many electron atom ? (i) 2s, 2p (ii) 3p, 3d (iii) 3s, 4s (iv) 4d, 5f Ans. (a) 2s (b) 3p (c) 3s (d) 4d. Q. 36. Nickel atom can lose two electrons to form Ni2+ ion. The atomic number of nickel is 28. From which orbital will nickel lose two electrons ? Ans. The electronic configuration of nickel is Ni : [Ar] 3d84s2 It will lose two electrons from 4s orbital. Q.37. Which of the following orbitals are degenerate ? 3dxy, 4dxy, 3dz2, 3dyz , 4dyz, 4dz2 Ans. The degenerate orbitals are : 3dxy, 3dz2, 3dyz, and 4dxy, 4dyz, 4dz2 Q.38. Calculate the total number of angular nodes and radial nodes present in 3p orbital. Ans. For 3p orbital, n = 3 and l = 1 No. of radial nodes = n – l – 1 = 3 – 1 – 1 = 1 No. of angular nodes = l = 1 Q.39. (a) Based on Bohr Bury’s rules arrange the following orbitals in the increasing order of energy. (i) 1s, 2s, 3s, 2p (ii) 4s, 3s, 3p, 4d (iii) 5p, 4d, 5d, 4f, 6s (iv) 5f, 6d, 7s, 7p (b) Based on these rules solve the questions given below : (i) Which of the following orbitals has the lowest energy ? 4d, 4f, 5s, 5p (ii) Which of the following orbitals has the highest energy ? 5p, 5d, 5f, 6s, 6p Ans. (a) (i) 1s < 2s < 2p < 3s (ii) 3s < 3p < 4s < 4d (iii) 4d < 5p < 6s < 4f < 5d (iv) 7s < 5f < 6d < 7p (b) (i) 5s has lowest energy (ii) 5f has highest energy.

2/68 MODERN’S abc + OF CHEMISTRY–XI Accelerate Your Potential (for JEE Advance) Problem 9. When would the wavelength associated = 2π r λ with an electron become equal to the wavelength asso- ciated with proton ? (Mass of electron = 9.1095 × 10–31kg, But λ = h mass of proton = 1.6725 × 10–27 kg). mv No. of waves = 2πr = 2πr . mv h / mv h or = 2π (mvr) h The angular momentum of Bohr's 3rd orbit is ©Solution According to de-Broglie equation, Modern Publishers. All rights reserved. λ = h mv For electron, λe = h meve mvr = 3h For proton, λp = h 2π But mpvp λe = λp No. of waves = 2π × 3h = 3 ∴ h 2π ∴ h= h No. of waves in Bohr's 3rd orbit = 3. meve mpvp Problem 12. An electron in a H-atom in its ground or meve = mpvp state absorbs 1.50 times as much energy as the mpvp minimum energy required for it to escape from the or ve = me atom. What is the wavelength of the emitted electron? = 1.6725 × 10–27 × vp = 1836 vp Solution The energy required to remove an electron 9.1095 × 10−31 Velocity of electron = 1836 velocity of proton. from H atom = Ionisation energy of H atom = 2.18 × 10−18 J atom−1 Problem 10. Calculate the ratio between the wavelength of an electron and a proton if the proton The amount of energy absorbed = 1.50 × 2.18 × 10−18 J is moving with half the velocity of electron. = 3.27 × 10−18 J (mass of electron = 9.11 × 10–31 kg, mass of proton = 1.67 × 10–27 kg). Amount of energy converted to kinetic energy = 3.27 × 10−18 − 2.18 × 10−18 J Solution λe = h , λp = h = 1.09 × 10−18 J meve mpvp K.E = 1 mν2 λe = mpvp , vp = 1 ve or vp = 1 2 λ p meve 2 ve 2 2 × (1.09 × 10−18 J) λe = 1 mp = 1 × 1.67 × 10−27 = 916.6 . ν= 2 K.E = (9.1 × 10−31kg) λp 2 me 2 9.11 × 10−31 m Problem 11. Find out the number of waves made = 1.55 × 106 m s−1 by a Bohr electron in one complete revolution in its According to de-Broglie equation, 3rd orbit. λ = h mν Solution In general, the number of waves in any orbit is = 6.63 × 10−34 Js (9.1 × 10−31 kg) × (1.55 × 106 ms−1) Number of waves = Circumference of orbit Wavelength = 4.70 × 10−10m Solution File Hints & Solutions for Practice Problems 1. (i) p = Atomic no. = 6 (iii) p = atomic no. = 38 Mass number = p + n n + p = Mass no. 12 = 6 + n ∴ n = 6 (ii) p = Atomic no. = 26 n + 38 = 88 ∴ n = 50 n + p = Mass no. (iv) p = atomic no = 92 n + 26 = 56 ∴ n = 30 n + p = 238 ∴ n = 238 – 92 = 146.

STRUCTURE OF ATOM 2/69 8. Since it carries – 1 charge, it will have one electron ∴ E = 6.63×10−34 ×3.0×108 more than the number of protons. 0.50×10−10 Let number of electrons = x (ii) No. of protons = x – 1 16. = 3.98 × 10–15 J E = 6.63 × 10–34 Js × 3 × 1015 s–1 = 1.98 × 10–18 J λ = 25.0 μm = 25.0 × 10–6 m © No. of neutrons = x + x × 11.1 = 1.111 x Modern Publishers. All rights reserved. 100 Now, x – 1 + 1.111 x = 37 or 2.111x = 38 x = 18 3.0×108 No. of e = 18 ν = c = 25.0×10−6 = 1.2 × 1013 λ No. of p = 18 – 1 = 17 No. of n = 35 – 17 = 18 Energy of photon, E = hν = 6.626 × 10–34 × 1.2 × 1013 Symbol 35 Cl– = 7.96 × 10–21 J 17 17. Energy of photon, 9. ν = c λ E = hc λ = 3.0×108 = 6.52 × 1014 sec–1 460×10–9 ∴ E = (6.63×10−34 Js)×(3×108 ms−1) 400×10−9 m 10. Frequency, ν = 980 kHz = 980 × 103 s–1 = 4.97 × 10–19 J Wavelength λ = c ν No. of photons in 1J ∴ λ= 3.0 × 108 m s−1 = 306 m. = 1 = 2.01 × 1018 photons 980 ×103s−1 4.97×10−19 11. Wave number,ν = ν 18. Wavelength of light, c λ = 250 nm = 250 × 10–9 m ∴ ν = 4×1014 s−1 3.0×108m s−1 E = 6.63×10−34 ×3.0×108 = 1.33 × 106 m–1. 250×10−9 12. Frequency, ν = c = 7.956 × 10–19 J ∴ λ Energy required per molecule ν = 3.0×108m s−1 = 9.6 × 10–17 J 480 ×10−9 m No. of photons per molecule = 6.25 × 1014 s–1 Wave number, ν =1 = 9.6×10−17 = 121 photons λ 7.956×10−19 =1 19. K.E. = hν – hν0 = h (ν – ν0) 480×10−9 h = 6.63 × 10–34 J s, = 2.08 × 106 m–1 v = 1.0 × 1015 s–1, v0 = 7.0 × 1014 s–1 ∴K.E. = 6.63 × 10–34 (10.0 × 1014 – 7.0 × 1014) 13. (i) Wave number, ν = 1 5800 × 10−10 m = 6.63 × 10–34 × 3.0 × 1014 = 1.99 × 10–19 J = 1.724 × 106 m–1 = 3.44 × 10–22 J (ii) Frequency, = c = 3.0 ×108 m s−1 ν λ 5800 ×10−10 m 20. K.E. = hν – hν0 If the velocity is zero, K.E. = 0 = 5.172 × 1014 s–1 14. Wavelength, λ = 25.0 μm = 25.0 × 10–6 m ∴ 0 = hν – hν0 Frequency, ν = c = 3.0 × 108 m s−1 or hν = hν0 or ν = ν0 λ 25.0 × 10−6 m c 3.0 ×108 m s−1 = 1.2 × 1013 Hz ∴ ν= λ = 6800 ×10−10 m Wave number, ν = 1 = 25.0 1 m = 4.41 × 1014 s–1 λ × 10−6 = 4.0 × 104 m Threshold frequency = 4.41 × 1014 s–1 15. (i) E = hν = hc Work function = hν0 = 6.626 × 10–34 × 4.41 × 1014 λ = 2.92 × 10–19 J

2/70 MODERN’S abc + OF CHEMISTRY–XI 21. 1 = 109678 ⎛1 − 1 ⎞ cm–1 Now, E = 6.63×10−34 ×3.0×108 λ ⎝⎜⎜ n12 n22 ⎠⎟⎟ 9.11×10−8 © Modern Publishers. All rights reserved.For Balmer series, = 2.183 × 10–18 J 24. For Balmer series, the longest wavelength n1 = 2, corresponds to transition from n = 3 to n = 2 1 ⎛ 1 1 ⎞ cm−1 λ = 109678 ⎝⎜ 22 − n2 ⎟⎠ ⎛1 1⎞ ⎝⎜ 22 n2 ⎠⎟ For the first line, ν = 109678 − cm–1 n2 = 3 so that = 109678 ⎛ 1 − 1 ⎞ ⎝⎜ 22 32 ⎟⎠ 1 = 109678 ⎛1 − 1⎞ cm–1 λ ⎝⎜ 4 9 ⎟⎠ = 15233.0 cm–1 or = 1.523 × 106 m–1 = 109678 × 5 25. No. of lines produced when an electron in 6th shell 36 drops to ground state or λ = 36 cm = 6.56 × 10–5 cm (n2 − n1) (n2 − n1 + 1) = (6 − 1) × (6 − 1 + 1) 5× 109678 22 = 656 nm = 5 × 6 = 15 lines For the limiting line, 2 n2 = ∞ These lines correspond to ∴ 1 = 109678 ⎛ 1 − 1⎞ cm–1 6 → 5, 6 → 4, 6 → 3, 6 → 2, 6 → 1 (5 lines) λ ⎜⎝ 22 ∞2 ⎟⎠ 5 → 4, 5 → 3, 5 → 2, 5 → 1 (4 lines) 4 → 3, 4 → 2, 4 → 1 (3 lines) = 109678 × 1 3 → 2, 3 → 1 (2 lines) 4 2 → 1 (1 line) or λ = 4 cm = 3.647 × 10–5 cm 26. v = ΔE = 399.1 kJ mol−1 = 1 × 1015 Hz 109678 h 3.98 × 10−13 kJ mol−1 s = 364.7 nm. 1 ⎛1 − 1 ⎞ c 3.0×108 msec−1 22. λ = 109678 ⎝⎜⎜ n22 ⎟⎠⎟ cm–1 ν 1×1015 sec−1 n2 Now, λ = = = 3.0 × 10–7 m 1 ∴ 1 = 109678 ⎛1 − 1 ⎞ cm–1 = 109678 × 8 = 300 nm λ ⎝⎜ 12 32 ⎟⎠ 9 2.18 ×10−18 9 cm = 102.6 × 10–7 cm 27. En = – n2 J atom–1 109678×8 or λ= For ionization from 5th orbit, = 102.6 nm n1 = 5 and n2 = ∞ 23. According to Rydberg equation, = 2.18 × 10–18 ⎛1 1 ⎞ 1 ⎛ 1 1 ⎞ ⎜⎝⎜ n12 − n22 ⎟⎟⎠ λ ⎝⎜⎜ n12 n22 ⎟⎠⎟ = 109678 − Here n1 = 1, n2 = ∞ = 2.18 × 10–18 ⎛1 − 1⎞ = 8.72 × 10–20 J ⎜⎝ 52 ∞ ⎠⎟ 1 ⎛1 1 ⎞ ∴ λ = 109678 ⎜⎝ 12 − ∞2 ⎠⎟ 1 = 109678 For ionization of H atom, λ n1 = 1, n2 = ∞ ΔE = 2.18 × 10–18 ⎛1 − 1 ⎞ ⎝⎜ 12 ∞ ⎟⎠ or λ = 1 = 9.11 × 10–6 cm 109678 = 2.18 × 10–18 J or = 9.11 × 10–8 m

STRUCTURE OF ATOM 2/71 28. 1 = R ⎛1 − 1 ⎞ 31. ΔE = 2.18 × 10–18 ⎛1 − 1 ⎞ = 2.09 × 10–18J λ ⎝⎜⎜ n12 n22 ⎠⎟⎟ ⎜⎝ 12 52 ⎠⎟ © Modern Publishers. All rights reserved.For spectral line of electron jump from n = 5 to λ= hc = 6.626 × 10−34 × 3.0 × 108 n = 3, ΔE 2.09 × 10−18 1 = R ⎛ 1 − 1 ⎞ = R ⎛⎝⎜ 16 ⎞ ...(i) = 9.51 × 10–8 m = 951 Å 1281.8 ⎜⎝ 32 52 ⎠⎟ 225 ⎠⎟ For spectral line of electron jump from n = 3 to 32. ν = 109678 ⎛1 − 1⎞ n=1 ⎝⎜⎜ n12 n22 ⎟⎠⎟ 1 = R ⎛ 1 − 1 ⎞ = R ⎛ 8 ⎞ ...(ii) For longest wavelength, n2 = 3, n1 = 2 λ ⎜⎝ 12 32 ⎟⎠ ⎜⎝ 9 ⎟⎠ = 109678 ⎛ 1 1 ⎞ Dividing (i) by (ii) ν ⎝⎜ 22 − 32 ⎠⎟ λ = 16 × 9 = 109678 × 5 = 1.523 × 104 cm–1 1281.8 225 8 36 16×9 × 1281.8 = 102.5 nm = 1.523 × 106 m–1 225×8 or λ= 33. Ei = – 2.18 × 10−18 = 5.45 × 10–19 J 22 Ef = 0 29. ν = 1= RH Z2 ⎛1 − 1⎞ λ Z= 2, n⎜⎝⎜ 2n1=2 4, nn122=⎟⎠⎟2 For He+ ΔE = 0 – ( – 5.45 × 10–19) = 5.45 × 10–19 J spectrum, ν = 1 = RH × 4 ⎛ 1 − 1 ⎞ = 3RH λ= hc = 6.626 × 10−34 × 3.0 × 108 λ ⎜⎝ 22 42 ⎠⎟ 4 ΔE 5.45 × 10−19 For hydrogen spectrum, 1 = 3 R H = 3.647 × 10–7 or 3647 Å λ4 34. λ= 6.626 × 10−34kg m2 s−1 = 1 = RH × 1 ⎛1 1 ⎞ 0.1 kg × 10 m s−1 ∴ ν λ ⎝⎜⎜ n12 − n22 ⎟⎠⎟ = 6.626 × 10–34m RH ⎛1 − 1 ⎞ = 3RH 35. λ= h or m = h ⎝⎜⎜ n12 n22 ⎟⎠⎟ 4 mv λv 1 − 1 = 3 = 6.63×10−34 = 6.139 × 10–33 kg n12 n22 4 3.6×10−10 ×3.0 ×108 ∴ n1 = 1, n2 = 2 36. mv = h Thus, the transition is from n = 2 to n = 1. λ 30. Ionisation energy is the energy needed when an electron is removed from n1 = 1 to n2 = ∞. For H-atom, = 6.63 × 10−34 kg m2s−1 0.1 × 10−9 m I. E. = 2π2me 4 Since n = 1 n2h2 = 6.63 × 10–24 kg m s–1 37. Mass of α-particle = 4 = 6.64 × 10−24 g 6.022 × 1023 = 2π2me 4 ∴ I.E. energyho2f H = 13.6 eV or = 6.64 × 10–24 × 10–3 kg = 6.64 × 10–27 kg Ionisation like ion, h or v = h I. E = 2π2me 4Z2 λ= mv mλ h2 λ = 12 pm = 12 × 10–12 m I.E. for He+ ion (Z = 2) = I.E. for H – atom × (2)2 = 13.6 × 4 = 54.4 eV h = 6.63 × 10–34 Js Similarly, I.E. for Li2+ ion (Z = 3) = I.E. for v = 6.63 × 10−34 H-atom × (3)2 6.64 × 10−27 × 12 × 10−12 = 13.6 × 9 = 122.4 eV

2/72 MODERN’S abc + OF CHEMISTRY–XI = 8.32 × 103 m s–1 = 6.626 × 10−34 kg m2s−1 (9.1 × 10−31 kg) × (1.876 × 107 ms−1) K.E. = 1 mv2 = 1 × 6.64 × 10−27 × (8.32 × 103)2 22 = 3.88 × 10–11 m = 2.30 × 10–19 J 38. λ = h mv © 41. λ = v or v = h ∴ ν = mv2 Modern Publishers. All rights reserved. ν ν mv h = 6.63 × 10−34 = 1.52 × 10–38m 1 mv2 = 5.85 × 10–25 0.1 × 4.37 × 10−5 2 39. The kinetic energy of the electron is provided by mv2 = 2 × 5.85 × 10–25 accelerating potential ∴ ν= 2 × 5.85 × 10−25 6.63 × 10−34 1 = 1.76 × 109 s–1 1 mv2 = eV or v = ⎛ 2 Ve ⎞2 42. λ = h 2 ⎝⎜ m ⎠⎟ mv λ= h = h = h ( )∴ mv 2Ve 1 or v = h m 2 1 mλ m (2Vem) 2 ∴ (2Vem)1/2 = h or 2 Vem = h2 = 6.63 × 10−34 kg m2s−1 λ λ2 (9.1 × 10−31 kg) × (100 × 10−10 m) = 7.28 × 104 m s–1 or V = h2 43. 1 mv2 = K.E. 2 λ2 ×2 × e × m λ = 9 pm = 9 × 10–12 m, or v2 = 2K.E. e = 1.6 × 10–19 C, m m = 9.1 × 10–31 kg h = 6.63 × 10–34 Js = 2 × 1.375 × 10−25 9.1 × 10−31 (6.63 × 10−34 Js)2 V= (9 × 10−12 m)2 × 2 × (1.6 × 10−19C) × 9.1 × 10−31 kg = 0.3022 × 106 m2s–1 = 1.86 × 104 kg m2 s–2/C = 1.86 × 104 V. ∴ v = 5.5 × 102 m s–1 40. Energy acquired = 1000 eV λ= 6.63 × 10−34 9.1 × 10−31 × 5.5 × 102 = 1000 × 1.602 × 10–19 = 1.602 × 10–16 J = 1.32 × 10–6 m 1 mv2 = 1.602 × 10–16 J 44. Δv = h Δx = 0.1 Å or = 0.1 × 10–10 m 2 4πmΔx 1 × 9.1 × 10−31v2 = 1.602 × 10–16 J Δv = 6.626 × 10−34 2 4 × 3.14 × 9.11 × 10−31 × 0.1 × 10−10 v2 = 2 × 1.602 × 10−16 = 3.521 × 1014 = 5.79 × 106 m s–1 9.1 × 10−31 45. Uncertainty in the speed of ball v = 1.876 × 107 m s–1 Now, λ= h = 90 × 4 = 3.6 m s–1 mv 100

STRUCTURE OF ATOM 2/73 Uncertainty in position, Δx =h 48. Δv = 1 × 3.0 × 108 = 3.0 × 105 m s–1 4πmΔv 49. 100 10 h Δx = 4π m. Δv = 6.626 ×10−34 Js 4 × 3.14 ×10 ×10−3 kg × 36 m s−1 = 1.46 × 10–33 m © = 6.626 × 10−34 Modern Publishers. All rights reserved. 4 × 3.14 × 1.66 × 10−27 × 3.0 × 105 46. Δv = h = 6.626 ×10−34 = 1.06 × 10–13 m 4πmΔx 4 × 3.14 ×1.1 ×10−27 × 3 ×10−12 Δv = 3 × 107 × 0.5 = 1.5 × 105 m s–1 100 = 1.59 × 104 m s–1. h 47. m= h Δx = = 4πmΔv 4πΔx . Δv 6.626 × 10−34 = 6.626 × 10−34 4 × 3.14 × 1.66 × 10−27 × 1.5 × 105 4 × 3.14 × 10−10 × 5.27 × 10−24 = 2.11 × 10–13 m = 0.10 kg Electromagnetic spectrum is the arrangement of different types of electromagnetic radiations in the increasing order of wavelengths (or decreasing frequencies). Planck’s quantum theory states that radiant energy is not emitted continuously but discontinuously in the form of small packets of energy called quantum or photon (for light), each having energy = hν. de Broglie principle. All microscopic material particles in motion have dual character i.e., particle and wave characteristics. λ = h or h mv p Dual nature of electrons was experimentally verified by Davisson and Germer (1927) and G. Thomson (1928). Heisenberg uncertainty principle states that it is impossible to measure simultaneously both the position and momentum (or velocity) with absolute accuracy for a microscopic particle. Mathematically, it may be expressed as: Δx× Δp ≥ h 4π Schrodinger wave equation is a mathematical equation to describe behaviour of electron in an atom. Schrodinger wave equation may be expressed as : Hˆ ψ = Eψ, where ψ gives the amplitude of electron wave and ψ2 is the probability density. Electronic configuration gives the distribution of electrons in various orbitals in an atom. Quantum numbers. A set of four numbers which give complete information about any electron in an atom. Aufbau rule states that the orbitals are filled in the order of increasing energy, starting with the orbital of lowest energy. Pauli’s exclusion principle. No two electrons in an atom can have same set of all the four quantum numbers. This means that an orbital can accommodate only two electrons having opposite spins. Hund’s rule. The pairing of electrons in orbitals of a same subshell cannot take place until all orbitals are singly occupied with parallel spin.

2/74 MODERN’S abc + OF CHEMISTRY–XI QUICK CHAPTER ROUND UP Energy of photon, E = hν = hc = hcν Bohr model λ Bohr model gave the concept of well defined circular orbits or energy levels in which energy remains Einstein equation for photoelectron emission Publishers. All rights reserved. constant. 1 mv2 = hν − hνo Angular momentum quantized, mvr = nh 2 ↓ 2π Energy of electron in an orbit work function = – 2π2mee4Z2 = – 2.18 × 10−18 Z2 J per atom n2h2 n2 ν0 = threshold frequency Rydberg equation = – 1.312 × 106 Z2 J per mol = – 13.6 Z2 eV atom–1 n2 n2 1 ⎛ 1 1⎞ λ = ν(cm−1) = R ⎜⎝ n12 − n22 ⎟⎠ Bohr radius of an orbit n2 > n1, R = 109677 rn = 0.059 n2 Å Spectral lines of H-atom Z Lyman n1 = 1, n2 = 2, 3, 4, 5, ⇒ ultraviolet e.g., For H atom, r1 = 0.059 Å, r2 = 2.12 Å Balmer n1 = 2, n2 = 3, 4, 5, 6, ⇒ visible For He+, r3 = 4.77 Å r2 = 1.06 Å Paschen n1 = 3, n2 = 4, 5, 6 ⎤ r1 = 0.0295 Å, Brackett n1 = 4, n2 = 5, 6, 7 ⎥⎦ ⇒ Infrared Pfund n1 = 5, n2 = 6, 7, 8 rn(H-like) = rn(H atom) Z de-Broglie equation, λ = h Velocity v = 2.19 × 106 Z = m s−1 mv n Relationship between K.E. and λ of moving particle No. of revolutions = 2πmvZe2 n2h2 h λ= Schrodinger wave equation 2 × K.E. × m ∂2ψ ∂2ψ ∂2ψ 8π2m Heisenberg uncertainly principle ∂x2 ∂y2 ∂z2 h2 ∆x × ∆p ≥ h ; ∆x × ∆v ≥ h + + + (E − V)ψ = 0 4π 4πm or Hˆ ψ = Eψ Quantum Numbers ψ = Radial wave function Principal quantum number (n) tells about the energy 1s orbital Y 2s orbital of the electron and size of the shell. (n = 1, l = 0) (n = 2, l = 0) Y Azimuthal quantum number (l) tells about the shape Node of the orbital (called subshell). r Magnetic quantum number (m) tells about the r orientation of the electron cloud in a subshell. ψ2 = Probability density Spin quantum number (s) tell about the spin of the electron. Modern Name Symbol Information Permitted provided values Y2 1s orbital 2s orbital Principal n shell 1, 2, 3, 4... Y2 Azimuthal l subshell 0, 1, 2, 3 ... (n – 1) Magnetic ml orbital Node Spin ms –l ... 0 ... + l spin +½, – ½ r r 2s orbital © For each orbital 4πr2ψ2 = Radial probability function No. of nodes = n – l – 1 No. of nodal planes = l 1s orbital 4pr2y2 3dz2 orbital has electron density in all the three axes. 4pr2y2 r 1 2 r3 4 5 6 7 8

STRUCTURE OF ATOM 2/75 Cr (3d54s1) and Cu (3d104s1) have anomalous Sequence of energy levels configurations because of extra stability associated with 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s. half filled and completely filled orbitals. © NCERT Textbook Exercises Modern Publishers. All rights reserved. Q. 1. (i) Calculate the number of electrons which = 6.022 × 1023 × 8 neutrons will together weigh one gram. = 4.8176 × 1024 neutrons (a) 14 g of 14C contain = 4.8176 × 1024 neutrons (ii) Calculate the mass and charge of one mole 7mg (7 × 10–3 g) of 14C contain of electrons. = 4.8176 × 1024 × 7 × 10–3 neutrons Ans. (i) Mass of one electron = 9.11 × 10–31 kg 14 ∴ 9.11 × 10–31 kg = 1 electron 1 g or 10–3 kg = 1 × 10–3 electrons = 2.4088 × 1021 neutrons 9.11 × 10−31 (b) Mass of 1 neutron = 1.675 × 10–27 kg = 1.098 × 1027 electrons ∴ Mass of 2.4088 × 1021 neutrons = 1.675 × 10–27 × 2.4088 × 1021kg (ii) Mass of one electron = 9.11 × 10–31 kg Mass of 1 mol of electrons = 9.11 × 10–31 × 6.022 × 1023 = 4.035 × 10–6 kg. = 5.486 × 10–7 kg. (iii) 1 mole of NH3 = 17g NH3 = 6.022 × 1023 molecules of NH3 Charge on one electron 1 molecule of NH3 contains protons = 7 + 3 = 10 = 1.602 × 10–19 C (a) 17g of NH3 contain molecules = 6.022 × 1023 34 mg of NH3 contain molecules Charge of 1 mol of electrons = 1.602 × 10–19 × 6.022 × 1023 Coulombs = 9.647 × 104 C = 6.022 ×1023 × 34 ×10−3 17 Q. 2. (i) Calculate the total number of electrons = 1.2044 × 1021 present in one mole of methane. No. of protons in 34 mg NH3 = 1.204 × 1021 × 10 (ii) Find (a) the total number and (b) the total = 1.204 × 1022 mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron = 1.675 × 10–27 kg). (b) Mass of proton = 1.6726 × 10–27 (iii) Find (a) the total number and (b) the total ∴ Mass of 1.204 × 1022 protons = 1.6726 × 10–27 × 1.204 × 1022 mass of protons in 34 mg of NH3 at STP. Will the answer change if the temperature = 2.0138 × 10–5 kg and pressure are changed ? There is no effect of temperature and pressure. How many neutrons and protons are there in the Ans. (i) 1 molecule of methane (CH4) contains electrons Q. 3. following nuclei ? = 6 + 4 = 10 Ans. 1 mole or 6.022 × 1023 molecules of CH4 will contain electrons = 6.022 × 1023 × 10 136C, 168O, 2142Mg, 5266Fe, 8388Sr = 6.022× 1024 Nucleus Z A p n (A–Z) (ii) In 14C 163C 6 13 6 13 – 6 = 7 168O 8 No. of neutrons + No. of Protons = 14 No. of protons = 6 16 8 16 – 8 = 8 ∴ No. of neutrons + 6 = 14 2142Mg 12 24 12 24 – 12 = 12 ∴ No. of neutrons = 14 – 6 = 8 5266Fe 26 56 26 56 – 26 = 30 Now 1 mol of 14C = 14 g C 8388Sr 38 88 38 88 – 38 = 50 or = 6.022 × 1023 atoms of 14C

2/76 MODERN’S abc + OF CHEMISTRY–XI Q. 4. Write the complete symbol for the atom with the emission, and (iii) the velocity of the photoelectron (1 eV = 1.602 × 10–19J) given atomic number (Z) and atomic mass (A) (i)Energy of photon (i) Z = 17, A = 35 Ans. (ii) Z = 92, A = 233 E = hν = hc λ (iii) Z = 4, A = 9 © Ans.Modern Publishers. All rights reserved.(i)35ligh(tiie) m29i32t3tUed(iii) 94Be λ = 4.0 × 10–7m, c = 3.0 × 108 m s–1 Q. 5. from a sodium 17Cl h = 6.626 × 10–34 J s, Yellow lamp has a wavelength (λ) of 580 nm. Calculate the ∴E= (6.626 ×10−34 Js) × (3.0 ×108 m) frequency (ν) and wavenumber (ν–) of the yellow light. (4.0 ×10−7 m) Ans. Refer Solved Example 6 (Page 13). = 4.97 × 10–19 J Q. 6. Find energy of each of the photons which : or = 4.97 ×10-19 = 3.10 eV. 1.6020 ×10-19 (i) correspond to light of frequency 3 × 1015 Hz. (ii) Kinetic energy of emission (ii) have wavelength of 0.50 Å K.E. = hν – hν0 hν = 3.10 eV and hν0 (work function) = 2.13 eV Ans. (i) E = hν ∴ K.E. = 3.10 – 2.13 = 0.97 eV = (6.626 × 10–34 Js) × (3 × 1015 s–1) (iii) Now, K.E. = 1 mv2 = 1.988 × 10–18 J 2 (ii) E = hν = hc or 1 mv2 = 0.97 eV = 0.97 × 1.602 × 10–19 J λ 2 Q. 7. ( ) ( )6.626 ×10−34 Js × 3 ×108 m s−1 or 1 × (9.11 × 10–31kg) v2 = 0.97 × 1.602 × 10–19 J 2 = v2 = 2 × 0.97 × 1.602 ×10−19 kg m2s−2 ( )0.50 ×10−10 m 9.11 × 10−31kg = 3.98 × 10–15 s = 34.12 × 1010 m2 s–2 Calculate the wavelength, frequency and wavenumber of a light wave whose period is ∴ v = 5.84 × 105 m s–1 2.0 × 10–10s. Ans. (i) Frequency = 1 = 1 = 5 × 109 s–1 Period 2.0 ×10−10s (ii) Wavelength, λ = c = 3 × 108 m s−1 = 6.0 × 10–2 m Q. 10. Electromagnetic radiation of wavelength 242 ν 5 × 109s−1 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in (iii) Wave number _ = 1 = 1 = 16.66 m–1 kJ mol–1. λ 6.0 × 10−2 ν Ans. Energy possessed by electromagnetic radiation with wavelength 242 nm will be equal to ionization energy, Q. 8. What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy ? hc λ Ans. Wavelength of light, λ = 4000 pm E = hν = = 4000 × 10–12 m = 4 × 10–9 m h = 6.626 × 10–34 J s Energy of 1 photon, c = 3.0 × 108 m, λ = 242 × 10–9 m E = hc ∴ E = (6.626 × 10−34 J s) × (3.0 × 108 m s−1) λ (242 × 10−9 m) h = 6.626 × 10–34 J s, c = 3.0 × 108 m s–1 = 8.21 × 10–19 J ∴ E = 6.626 × 10–34 × 3.0 × 108 Ionization energy per mol = 8.21 × 10–19 J × 6.023 × 1023 4 × 10–9 = 494 × 103 J = 494 kJ mol–1. = 4.97 × 10–17 J Q. 11. A 25 watt bulb emits monochromatic yellow No. of photons providing 1 J of energy light of wavelength of 0.57 μm. Calculate the rate of emission of quanta per second. = 1 = 2.01 × 1016 4.97 × 10–17 Ans. Energy of one photon, Q. 9. A photon of wavelength 4 × 10–7 m strikes on E = hν = hc metal surface, the work function of the metal λ being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the h = 6.626 × 10–34 J s c = 3.0 × 108 m s–1, λ = 0.57 μm

STRUCTURE OF ATOM 2/77 = 0.57 × 10–6 m s–1 Thus, the energy required to remove electron from Ist E = 6.63 × 10−34 × 3.0 × 108 orbit is 25 times than that required to remove electron 0.57 × 10−6 from 5th orbit. Q. 15. What is the maximum number of emission lines = 3.487 × 10–19 J when the excited electron of a H-atom in n = 6 Now, 25 Watt = 25 J/s drops to the ground state ? Ans. Refer Solution to Practice Problems 25 (Page 70). Q. 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 asociated with the fifth orbit ? (ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom. Ans. Refer Solved Example 22 (Page 28). Q. 17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen. © ∴ No. of photons emitted per second = 25 Modern Publishers. All rights reserved. 3.487×10−19 = 7.17 × 1019s–1. Q. 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. Ans. K.E. = hν –hν0 If the velocity is zero, K.E. = 0 ∴ 0 = hν – hν0 Ans. ν– = ⎛ 1 − 1 ⎞ or hν = hν0 or ν = ν0 R ⎜⎜⎝ n12 n22 ⎟⎟⎠ ∴ ν= c = 3.0 ×108 m s−1 FshooruBldalbmeemr isneirmieusmn1s=o 2, for longest wavelength, ν– λ 6800 ×10−10 m that n2= 3. = 4.41 × 1014 s–1 Threshold frequency = 4.41 × 1014 s–1 ν– = (1.097 × 107m–1) ⎛1 1 ⎞ ⎜⎝ 22 32 ⎟⎠ Work function = hν0 = 6.626 × 10–34 × 4.41 × 1014 − = 2.92 × 10–19 J. Q. 13. What is the wavelength of light emitted when = 1.097 × 107 × 5 the electron in a hydrogen atom undergoes 36 transition from an energy level with n = 4 to an energy level with n = 2 ? = 1.523 × 106 m–1. Q. 18. What is the energy in joules, required to shift the Ans. Refer Solved Example 17 (Page 21). electron of the hydrogen atom from the first Bohr Q. 14. How much energy is required to ionise a orbit to the fifth Bohr orbit and what is the H-atom if the electron occupies n = 5 orbit ? wavelength of the light emitted when the electron Compare your answer with the ionization returns to the ground state ? The ground state enthalpy of H-atom (energy required to remove electron energy is –2.18 × 10–11 ergs. the electron from n = 1 orbit). Ans. The ground state energy is –2.18 × 10–11 ergs., Ans. En = −21.8 ×10−19 J atom–1 ∴ En =– 2.18 ×10−11 ergs n2 n2 For ionization from 5th orbit, n1 = 5, n2 = ∞ ΔE = E5 – E1 = – 2.18 × 10–11 ⎛1 − 1 ⎞ ⎜⎝ 52 12 ⎠⎟ ⎛1 1 ⎞ ΔE = E2– E1 = – 21.8 × 10–19 ⎝⎜⎜ n22 − n12 ⎟⎠⎟ = 2.18 × 10–11 × 24 = 2.093 × 10–11 ergs 25 ⎛1 1 ⎞ = – 21.8 × 10–19 ⎝⎜ ∞ − 52 ⎠⎟ or = 2.093 × 10–18 J (1 erg = 10–7J) = – 21.8 × 10–19 × ⎛ − 1 ⎞ When electron returns to ground state (i.e., n = 1), ⎝⎜ 25 ⎠⎟ energy emitted = 2.093 × 10–18 J = 8.72 × 10–20 J For ionization from Ist orbit, n1 = 1, n2 = ∞ Now, ΔE = hc or λ = hc λ ΔE ΔE′ = – 21.8 × 10–19 ⎛1 − 1 ⎞ ⎝⎜ ∞ 12 ⎠⎟ ( ) ( )= 6.626 ×10−34 Js × 3.0 ×108 ms−1 = 21.8 × 10–19J ( )2.093 ×10−18 J ΔE' = 21.8 ×10−19 J = 9.50 × 10–8 m ΔE 8.72 × 10−20 or = 950 Å. = 25.

2/78 MODERN’S abc + OF CHEMISTRY–XI Q. 19. The electron energy in hydrogen atom is given Ans. (i) No. of protons = 29 by En = (–2.18 × 10–18)/n2J. Calculate the energy (ii) Atomic number = 29 required to remove an electron completely from Electronic configuration : [Ar]18 3d104s1. the n = 2 orbit. What is the longest wavelength of Q. 27. Give the number of electrons in the species H+2, light in cm that can be used to cause this H2 and O+2 . Ans. H+2 : 1 , H2 = 2, O+2 = 15. Q. 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 elec- trons for 3d orbital. (iii) Which of the following orbitals are possible ? 1p, 2s, 2p and 3f Ans. (i) For n = 3, l = 0, 1, 2 © transition ? Modern Publishers. All rights reserved. Ans. ΔE = E∞ – E2 = 0 – ⎛ − 2.18 × 10−18 J atom−1 ⎞ ⎜⎝ 22 ⎟⎠ = 5.45 × 10–19 J atom–1 ΔE = hc or λ = hc λ ΔE ( ) ( )∴ λ= 6.626 ×10−34 Js × 3.0 ×108 ms−1 When l = 0, ml = 0 5.45 ×10−19 J atom-1 When l = 1, ml = –1, 0, + 1 When l = 2, ml = –2, –1, 0, +1, + 2 = 3.647 × 10–7m (ii) n = 3, l = 2 and for l =2, ml = –2, –1, 0, +1, +2 (iii) 2s, 2p are possible. or = 3.647 × 10–5 cm. Q. 20. Calculate the wavelength of an electron moving Q. 29. Using s, p, d notations, describe the orbital with with a velocity of 2.05 × 107 ms–1 the following quantum numbers. Ans. Refer Solved Example 31 (Page 38). (a) n = 1, l = 0; (b) n = 3; l = 1 Q. 21. The mass of an electron is 9.1 × 10–31 kg. If its K.E. (c) n = 4; l = 2; (d) n = 4; l = 3 is 3.0 × 10–25 J, calculate its wavelength. Ans. Refer solved Example 49 (Page 49). Ans. Refer Solved Example 39. (Page 39). Q. 30. Explain, giving reasons, which of the following Q. 22. Which of the following are isoelectronic sets of quantum numbers are not possible. species i.e., those having the same number of electrons ? (a) n = 0, l = 0, ml = 0, ms = + 1/2 Na+, K+, Mg2+, Ca2+, S2–, Ar. (b) n = 1, l = 0, ml = 0, ms = – 1/2 (c) n = 2, l = 1, ml = 0, ms = + 1/2 Ans. Na+, Mg2+ (d) n = 2, l = 1, ml = 0, ms = – 1/2 K+, Ca2+, S2– and Ar are isoelectronic. (e) n = 3, l = 3, ml = –3, ms = + 1/2 (f) n = 3, l = 1, ml = 0, ms = + 1/2 Q. 23. (i) Write the electronic configurations of the Ans. Refer Solved Example 51 (Page 49). following ions : (a) H– (b) Na+ (c) O2– (d) F– Q. 31. How many electrons in an atom may have the (ii) What are the atomic numbers of elements following quantum numbers ? whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5 ? (a) n = 4, ms = –1/2 (b) n = 3, l = 0 Ans. (i) Total number of electrons in n = 4 are 2 × 42 = 32 (iii) Which atoms are indicated by the following Half of these i.e., 16 electrons have ms = –1/2 configurations ? (ii) It means 3s orbital. It can have 2 electrons. (a) [He] 2s1 (b) [Ne]3s2 3p3 (c) 4s2 3d1 Q. 32. Show that the circumference of the Bohr orbit Ans. (i) Refer Solved Example 57 (Page 63). for the hydrogen atom is an integral multiple of the de-Broglie wavelength associated with the (ii) (a) 11 (b) 7 (c) 17 electron revolving around the orbit. (iii) (a) Li (b) P (c) Sc Ans. According to Bohr’s postulation of angular momentum Q. 24. What is the lowest value of n that allows g mvr = nh or 2πr = n h ....(i) orbitals to exist ? 2π mv Ans. n = 5 According to de Broglie equation, λ = h mv Q. 25. An electron is in one of the 3d orbitals. Give the .....(ii) possible values of n, l and ml for this electron. From (i) and (ii) we get Ans. n = 3, l = 2 and for l = 2 , ml = –2, –1, 0, +1, +2. Q. 26. An atom of an element contains 29 electrons and 2πr = nλ 35 neutrons. Deduce (i) the number of protons Thus, the circumference (2πr) of Bohr orbit for H-atom is an integral multiple of de Broglie and (ii) the electronic configuration of the ele- wavelength. ment. For detail, Refer Text (Page 35–37).

STRUCTURE OF ATOM 2/79 Q. 33. What transition in the hydrogen spectrum would Diameter of carbon atom = 2.4 have the same wavelength as the Balmer 2 ×108 transition n = 4 to n = 2 of He+ spectrum ? = 1.2 × 10–8 cm Ans. For He+ spectrum, = 1.2 × 10–10 m L O1 = RZ2 1 1 MMN PQPλ © n 2 − n 2 Radius of carbon atom = 1.2 ×10−10 Modern Publishers. All rights reserved. 1 2 2 Now, n1 = 2, n2 = 4 and Z = 2 = 0.6 × 10–10 m = 0.06 × 10–9 m 1 1 1 =3R = 0.06 nm 22 42 Q. 37. The diameter of zinc atom is 2.6 Å. Calculate LNM QPOλ 4 (a) radius of zinc atom in pm and (b) number of = R × (2)2 − ...(i) atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise. ML POFor H atom 1 = R 1 − 1 ...(ii) Ans. (i) Diameter of zinc atom = 2.6 Å = 2.6 × 10–10m NM PQλ n 2 n 2 Radius = 2.6 ×10−10 1 2 2 Since λ is the same, equating equations (i) and (ii) = 1.3 × 10–10m = 130 × 10–12m or = 130 pm 1 − 1 =3 (ii) No. of Zn atoms present on a 1.6 cm length = n22 4 n 2 1 Now, if n1 = 1 and n2 = 2. 1.6 = 6.154 × 107 2.6 ×10−10 ×102 Therefore, the transition from n = 2 to n = 1 in H atom will have the same wavelength as the transition from Q. 38. A certain particle carries 2.5 × 10–16C of static n = 4 to n = 2 in He+. electric charge. Calculate the number of electrons present in it. Q. 34. Calculate the energy required for the process He+(g) ⎯⎯→ He2+ (g) + e– Ans. Charge of one electron = 1.602 × 10–19 C The ionization energy for the H atom in the ground state is 2.18 × 10–18 J atom –1. Ans. For H like species, En = −2π 2m Z2e4 2.5 × 10−16 n2h2 1.602 × 10−19 For H atom, IE = En – E1 No. of electrons present = = 1.56 × 103 = 0 – ⎛ − 2π 2m e4 ⎞ Q. 39. In Millikan’s experiment, static electric charge ⎝⎜ 1 × h2 ⎠⎟ on the oil drops has been obtained by shining X- rays. If the static electric charge on the oil drop = 2π 2m e4 = 2.18 × 10–18 J atom–1 (given) is –1.282 × 10–18C, calculate the number of elec- h2 trons present in it. For the given process, −1.282 ×10−18C Energy required=En – E1 Ans. Number of electrons = −1.6022 ×10−19C =8 ⎛ 2π2me4 (2)2 ⎞ 2π 2m e4 Q. 40. In Rutherford’s experiment, generally the ( )=0– ⎜− ⎟ =4× h2 ⎜⎝ 12 × h2 ⎠⎟ thin foil of heavy atoms, like gold, platinum, etc., have been used to be bombarded by the = 4 × 2.18 × 10–18 J α-particles. If the thin foil of light atoms like aluminium, etc., is used, what difference would = 8.72 × 10–18 J. Q. 35. If the diameter of a carbon atom is 0.15 nm, be observed from the above results ? calculate the number of carbon atoms which can be placed side by side in a straight line across Ans. In Rutherford’s experiment, heavy atoms have heavy length of scale of length 20 cm long. nucleus carrying a large amount of positive charge. Therefore, some α-particles even got deflected back Ans. A carbon atom covers length = diameter of atom = after hitting the nucleus. Because of large positive 0.15 nm charge on the heavy nucleus, some α-particles are = 0.15 × 10–9 × 102 cm deflected through small angles which passed closer ∴ No. of carbon atoms which can be placed on 20 cm to the nucleus because of repulsion. However, if length lighter atoms are used, their nuclei will be light and they will have small positive charge on the nucleus. = 20 = 1.33 × 109 As a result, the number of particles deflected back 0.15 × 10−9 × 102 and those deflected through small angles will be very very small, almost negligible. Q. 36. 2 × 108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length Q. 41. Symbols 7395Br and 79Br can be written, whereas of this arrangement is 2.4 cm. sAynmswboelrsbr375i9eBfrly.and 35Br are not acceptable. Ans. An atom will cover length equal to its diameter

2/80 MODERN’S abc + OF CHEMISTRY–XI Ans. Atomic number of an element (say Br) is fixed but its Ans. Energy of 1 photon, mass number is not fixed depending upon the isotope. E = hc Hence, it is necessary to indicate mass number. λ Q. 42. An element with mass number 81 contains 31.7% h = 6.626 ×10–34Js, c = 3.0 ×108 ms–1, more neutrons as compared to protons. Assign λ = 600nm = 600 × 10–9m the atomic symbol. Ans. Refer Solved Example 4 (Page 10). © E = 6.626 × 10−34 × 3.0 × 108 Modern Publishers. All rights reserved.Q. 43. An ion with mass number 37 possesses one unit 600 × 10−9m of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the = 3.31 × 10–19 J symbol of the ion Total energy received = 3.15 × 10–18 J Ans. Solutions to Practice Problem 8 (Page 69). No. of photons = 3.15 × 10−18 = 9.52 10 3.31 × 10−19 Q. 44. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than Q. 49. Lifetimes of the molecules in the excited states electrons. Assign the symbol to this ion. are often measured by using pulsed radiation Ans. Refer Solved Example 5 (Page 10). source of duration nearly in the nano second Q. 45. Arrange the following type of radiations in range. If the radiation source has the duration of increasing order of frequency : (a) radiation from microwave oven (b) amber light from traffic 2 ns and the number of photons emitted during signal (c) radiation from FM radio (d) cosmic the pulse source is 2.5 × 1015, calculate the energy rays from outer space and (e) X-rays. of the source. Ans. Cosmic rays < X-rays < amber light < microwave Ans. Frequency = 1 = 0.5 × 109 s–1 < FM. 2 ×10−9 Q. 46. Nitrogen laser produces a radiation at a Energy = Nhν = (2.5 × 1015) × (6.62 × 10–34 Js) wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 1024, calculate the power of this × (0.5 × 109 s–1) = 8.275 × 10–10J. laser. Q. 50. The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and Ans. E = N hν = Nhc energy difference between two excited states. λ Ans. λ1 = 589 nm = 589 × 10–9 m ( ) ( ) ( )5.6 ×1024 × 6.62 ×10−34 Js × 3.0 ×108 ms−1 ν1 = c = 3.0 × 108 ms−1 = 5.093 × 1014 s–1 = λ1 589 × 10−9 m 337.1 × 10−9 m λ2 = 589.6 nm = 589.6 × 10–9 m = 3.3 × 106 J. ν2 = c = 3.0 ×108 ms−1 = 5.088 × 1014 s–1 Q. 47. Neon gas is generally used in the sign boards. If λ2 589.6 ×10−9 m it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance travelled by ΔE = h (ν2 –ν1) this radiation in 30 s (c) energy of quantum and = (6.626 × 10–34 Js) (5.093 – 5.088) × 1014 s–1 (d) number of quanta present if it produces 2 J of energy. = 3.44 × 10–22 J Ans. Wavelength of light, λ =616 nm = 616 ×10–9 m Q. 51. The work function for caesium atom is 1.9 eV. (a) Frequency = c = 3.0 × 108 m s−1 Calculate (a) the threshold wavelength and λ 616 × 10−9 m (b) the threshold frequency of the radiation. If = 4.87 × 1014 s–1 the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy (b) Distance traveled in 30 s and the velocity of the ejected photoelectron. = 30 × 616 × 10–9 × 4.87 × 1014 = 9.0 × 109 m Ans. Work function, (c) Energy of quanta, E = hν hν0 = 1.9 eV = 1.9 × 1.602 × 10–19 J = 6.626 × 10–34Js × 4.87 × 1014 s–1 = 3.04 × 10–19 J = 3.227 × 10–19 J ∴ Threshold frequency = 3.04 × 10−19 J 2 6.626 × 10−34 Js (d) No. of quanta in 2J of energy = 3.227 × 10−19 = 4.59 × 1014s–1 = 6.2 × 1018 photons. Threshold wavelength, λ0 = c ν0 Q. 48. In astronomical observations, signals observed from the distant stars are generally weak. = 3.0 × 108 ms−1 If the photon detector receives a total of 4.59 × 1014 s−1 3.15 × 10–18 J from the radiations of 600 nm, calculate the number of photons received by the = 6.54 × 10–7 m or 654 nm detector.

STRUCTURE OF ATOM 2/81 Now, energy of radiated light, E = hc h = 123.17 ×10−21 × 10−9 × 400 × 526 λ 3.0 ×108 ×126 = 6.626 × 10−34 × 3.0 × 108 = 3.98 × 10–19 J = 6.84 × 10–34 Js. 500 × 10−9 Q. 53. The ejection of the photoelectron from the silver ∴ K.E. of ejected electron = 3.98 × 10–19 – 3.04 × 10–19 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. Ans. Energy of incident radiation = Work function + K.E. of photoelectrons Energy of incident radiation, © = 9.4 × 10–20 J Modern Publishers. All rights reserved. K.E. = 1 mv2 = 9.4 × 10–20 J 2 v2 = 9.4 × 10−20 × 2 = 20.66 × 1010 9.1 × 10−31 ∴ v = (20.66 × 1010)½ = 4.54 × 105 m s–1 hc λ Q. 52. Following results are observed when sodium E = hν = metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant. ( ) ( )= 6.626 × 10−34 Js × 3.0 × 108 ms−1 ( )256.7 × 10−9 m λ(nm) 500 450 400 = 7.74 × 10–19 J v × 10–5 (cm s–1) 2.55 4.35 5.20 Ans. Let threshold wavelength be λ0 nm = λ × 10–9 m h(ν – ν0) = 1 mv2 or = 7.74 ×10−19 J = 4.83 eV 2 1.602 ×10−19 h ⎛c − c ⎞ = 1 mv2 The potential energy gives the kinetic energy to the ⎜ λ0 ⎟ 2 electrons. ⎝ λ ⎠ Hence K.E. of the electron = 0.35 eV Work function = 4.83 eV – 0.35 eV = 4.48 eV or hc ⎛ 1 − 1 ⎞ = 1 mv2 Q. 54. If the photon of the wavelength 150 pm strikes ⎜ λ λ0 ⎟ 2 an atom and one of its 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 hc ⎛ 500 1 − λ0 1 ⎞ = 1 mv2 the nucleus. ⎜⎜⎝ × 10−9 × 10−9 ⎠⎟⎟ 2 Ans. Energy of incident photon hc ⎛ 1 − 1 ⎞ = 1 m (2.55 × 105)2 ....(i) = hc 10−9 ⎜ 500 λ0 ⎟ 2 λ ⎝ ⎠ Similarly, ( ) ( )= 6.626 × 10−34 Js × 3.0 × 108 ms−1 ( )150 ×10−12 m hc ⎛ 1 − 1 ⎞ = 1 m (4.35 × 105)2 ....(ii) 10−9 ⎜ 450 λ0 ⎟ 2 = 1.325 × 10–15 J ⎝ ⎠ Energy of ejected electron hc ⎛ 1 − 1 ⎞ = 1 m (5.20 × 105)2 ....(iii) = 1 mv2 10−9 ⎜ 400 λ0 ⎟ 2 2 ⎝ ⎠ Dividing eq. (ii) by (i) = 1 × (9.11 × 10–31 kg) × (1.5 × 107ms–1)2 2 λ0 − 450 × 500λ0 = ⎛ 4.35 ⎞2 450λ0 λ0 − 500 ⎝⎜ 2.55 ⎟⎠ = 1.025 × 10–16J Energy with which the electron was bound to the λ0 − 450 = 450 × ⎛ 4.35 ⎞2 nucleus λ0 − 500 500 ⎜⎝ 2.55 ⎠⎟ = 13.25 × 10–16 – 1.025 × 10–16 λ0 − 450 = 2.91 = 12.225 × 10–16 J λ0 − 500 or = 12.225 ×10−16 = 7.63 × 103 eV. or λ0 – 450 = 2.91 λ0 – 1455 1.602 ×10−19 1.91 λ0 = 1005 Q. 55. Emission transitions in the Paschen series end at ∴ λ0 = 526.2 nm or = 526 nm orbit n = 3 and start from orbit n and can be Substituting the value of λ0 in eqn. (iii), we get represented as ν = 3.29 × 1015 (Hz) [1/32 – 1/n2] Calculate the value of n if the transition is ( )h × 3.0 ×108 ⎛ 1 − 1 ⎞ = 1 × 9.11 × 10–31 × (5.20 × 105)2 observed at 1285 nm. Find the region of the ⎝⎜ 400 526 ⎟⎠ 2 spectrum. 10−9 ( )h × 3.0 ×108 ⎛ 526 − 400 ⎞ = 123.17 × 10–21 Ans. ν = 3.29 × 10 15 ⎛ 1 − 1 ⎞ s–1 ⎜⎝ 400 × 526 ⎟⎠ ⎜ 32 n2 ⎟ 10−9 ⎝ ⎠

2/82 MODERN’S abc + OF CHEMISTRY–XI ∴ λ = 1285 nm 6.626 × 10−34 kg m2s−1 ( ) ( )Ans. λ = h = ν = c = 3.0 ×108ms−1 mv 9.11 ×10−31 kg × 1.6 ×106 ms−1 λ 1285 × 10−9 m = 4.55 × 10–10m = 2.33 × 1014 s–1 or = 455 pm. Q. 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. © 2.33 × 1014 = 3.29 × 1015 ⎛1 − 1 ⎞ Modern Publishers. All rights reserved. ⎝⎜ 9 n2 ⎠⎟ or 1 − 1 = 2.33×1014 = 0.0708 9 n2 3.29 ×1015 − 1 = 0.0708 – 1 = –0.0403 Ans. λ= h = or ν = h n2 9 mv mλ 1 = 0.0403 = 6.626 ×10−34 kg m2s−1 n2 1.675 ×10−27 kg × 800 ×10−12m or n2 = 24.82 = 4.94 × 102 ms–1 ∴ n= 5 Q. 59. If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms–1, calculate the de Broglie Transition is obtained in infrared region. wavelength associated with it. Q. 56. Calculate the wavelength for the emission Ans. Refer Solved Example 35 (Page 38). transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series Q. 60. The velocity associated with a proton moving in to which this transition belongs and the region a potential difference of 1000 V is 4.37 × 105 of the spectrum. ms–1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength Radius of nth orbit of H like species = 0.529 n2 Å associated with this velocity. Z Ans. Ans. λ= h mv or = 52.9n2 pm Z ( )= 6.626 × 10−34 kg m2s−1 52.9n12 r1 = 1.3225 nm = 1322.5 pm = Z (0.1 kg) × 4.37 × 105 ms−1 r2 = 211.6 pm = 52.9 n22 = 1.516 × 10–28 m Z Q. 61. If the position of the electron is measured within ∴ r1 = 1322.5 = n12 an accuracy of ± 0.002 nm, calculate the r2 211.6 n22 uncertainty in the momentum of the electron. Suppose the momentum of the electron is ∴ n12 = 6.25 or n1 = 2.5 h ,is there any problem in defining n22 n2 4π × 0.05 nm this value. sIfitnio2n= f2r,onm1 = 5, This transition corresponds to tran- 5th orbit to 2nd orbit. Hence, it belongs Ans. Refer Solved Example 46 (Page 43). to Balmer series. Q. 62. The quantum numbers of six electrons are given below. Arrange them in order of Now, ν– = 1.097 × 107 m–1 ⎛1 − 1 ⎞ increasing energies. If any of these ⎝⎜ 22 52 ⎠⎟ combination(s) has/have the same energy lists : = 1.097 × 107 × 21 m–1 (a) n = 4, l = 2, ml = –2, ms = – 1/2 100 (b) n = 3, l = 2, ml = 1, ms = + 1/2 (c) n = 2, l = 0, ml = 0, ms = + 1/2 ∴ λ = 1 = 1 (d) n = 2, l = 0, ml = 0, ms = – 1/2 ν 23.037 × 105 m−1 (e) n = 3, l = 3, ml = 0, ms = + 1/2 (f) n = 3, l = 1, ml = 0, ms = + 1/2 = 434 × 10–9m or 434 nm Refer Solved Example 58 (Page 64). It lies in the visible region. Q. 57. Dual behaviour of matter proposed by Ans. de Broglie led to the discovery of electron microscope often used for the highly magnified Q. 63. The bromine atom possesses 35 electrons. It images of biological molecules and other type of contains 6 electrons in 2p orbital, 6 electrons in material. If the velocity of the electron in this 3p orbital and 5 electron in 4p orbital. Which of microscope is 1.6 × 106 ms–1, calculate these electron experiences the lowest effective de Broglie wavelength associated with this nuclear charge ? electron.

STRUCTURE OF ATOM 2/83 Ans. 4p electrons are farthest from the nucleus and Ans. Unpaired electron in case of silicon will experience therefore, these electrons will experience the lowest more effective nuclear charge. effective nuclear charge. Q. 66. Indicate the number of unpaired electrons in: Q. 64. Among the following pairs of orbitals which (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr. orbital will experience the larger effective nuclear charge ? (i) 2s and 3s, (ii) 4d and 4f, Ans. Refer Solved Example 59 (Page 64). (iii) 3d and 3p. Q. 67. (a) How many sub-shells are associated with Ans. (i) 2s (ii) 4d (iii) 3p. Q. 65. The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ? © n = 4 ? (b) How many electrons will be present in Modern Publishers. All rights reserved. the sub-shell having ms value of –1/2 for n=4? Ans. Sub-shells in n = 4 are 4 i.e., 4s, 4p, 4d and 4f No. of electrons having ms = −1 2 for n = 4 will be 16. NCERT Exemplar Problems 1. Which one of the following conclusions could not be (a) They start from the cathode and move towards the derived from Rutherford’s α-paticle scattering anode. experiments ? (a) Most of the space in the atom is empty. (b) They travel in straight line in the absence of an (b) The radius of the atom is about 10–10 m while that external electrical or magnetic field. of nucleus is 10–15 m. (c) Electrons move in a circular path of fixed energy (c) Characteristics of cathode rays do not depend upon called orbits. the material of electrodes in cathode ray tube. (d) Electrons and the nucleus are held together by electrostatic forces of attraction. (d) Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube. 2. Which of the following options does not represent ground state electronic configuration of an atom ? 5. Which of the following statement about the electron is (a) 1s2 2s2 2p6 3s2 3p6 3d8 4s2 incorrect ? (b) 1s2 2s2 2p6 3s2 3p6 3d9 4s2 (a) It is a negatively charged particle (c) 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (b) The mass of electron is equal to the mass of neutron (d) 1s2 2s2 2p6 3s2 3p6 3d5 4s1 (c) It is a basic constituent of all atoms (d) It is a constituent of cathode rays 3. The probability density plots of 1s and 2s orbitals are given in Fig. 1 : 6. Which of the following properties of atom could be explained correctly by Thomson model of atom ? Fig. 1. (a) Overall neutrality of atom. The density of dots in a region represents the probability (b) Spectra of hydrogen atom. density of finding electrons in the region. (c) Position of electrons, protons and neutrons in atom. On the basis of above diagram which of the following (d) Stability of atom. statements is incorrect ? (a) 1s and 2s orbitals are spherical in shape. 7. Two atoms are said to be isobars if (b) The probability of finding the electron is maximum (a) they have same atomic number but different mass near the nucleus. number. (c) The probability of finding the electron at a given (b) they have same number of electrons but different distance is equal in all directions. number of neutrons. (d) The probability density of electrons for 2s orbital (c) they have same number of neutrons but different decreases uniformly as distance from the nucleus number of electrons. increases. 4. Which of the following statements is not correct about (d) sum of the number of protons and neutrons is same the characteristics of cathode rays ? but the number of protons is different. 8. The number of radial nodes for 3p orbital is (a) 3 (b) 4 (c) 2 (d) 1 9. Number of angular nodes for 4d orbital is (a) 4 (b) 3 (c) 2 (d) 1 10. Which of the following is responsible to rule out the existence of definite paths or trajectories of electrons ? (a) Pauli’s exclusion principle. (b) Heisenberg’s uncertainty principle (c) Hund’s rule of maximum multiplicity. (d) Aufbau principle.

2/84 MODERN’S abc + OF CHEMISTRY–XI 11. Total number of orbitals associated with third shell 15. For the electrons of oxygen atom, which of the f ollowing will be _______ statements is correct ? (a) 2 (b) 4 (a) Zfoerff for an electron in a 2s orbital is the same as Zeff an electron in a 2p orbital. (c) 9 (d) 3 (b) An electron in the 2s orbital has the same energy 12. Orbital angular momentum depends on _______ . as an electron in the 2p orbital. (a) l (b) n and l © (c) Zfoerff for an electron in 1s orbital is the same as Zeff (c) n and mModern Publishers. All rights reserved.(d) m and s an electron in a 2s orbital. 13. Chlorine exists in two isotopic forms, Cl-37 and Cl-35 (d) The two electrons present in the 2s orbital have but its atomic mass is 35.5. This indicates the ratio of spin quantum numbers ms but of opposite sign. Cl-37 and Cl-35 is approximately 16. If travelling at same speeds, which of the following (a) 1:2 (b) 1:1 matter waves have the shortest wavelength ? (c) 1:3 (d) 3:1 (a) Electron (b) Alpha particle (He2+) 14. The pair of ions having same electronic configuration is (c) Neutron (d) Proton (a) Cr3+, Fe3+ (b) Fe3+, Mn2+ (c) Fe3+, Co3+ (d) Sc3+, Cr3+ ANSWERS / HINTS MCQs Type-I 1. (c) The concept of circular paths of fixed energy was 9. (c) 10. (b) proposed by Bohr and not derived from Rutherford’s scattering experiment. 11. (c) No. of orbitals in 3rd shell (n = 3) = n2 = 32 = 9. 2. (b) Correct configuration in ground state should be 12. (a) 1s2 2s2 2p6 3s2 3p6 3d10 4s1. 13. (c) If Cl-37 and Cl-35 exist in the ratio of 1:3, then 3. (d) The probability density of electrons in 2s orbit first average atomic mass comes out to be 35.5. increases, then decreases and after that it begins to increase again. 14. (b) Cr3+ : [Ar]3d3, Fe3+ : [Ar]3d5 ; Mn2+ : [Ar]3d5 4. (d) 5. (b) Thus, Mn2+ and Fe3+ have same electronic configuration. 15. (d) 6. (a) Thomson model of atom could explain only the overall neutrality of the atom. 7. (d) 16. (b) λ ∝ 1 so that alpha particles having larger mass m 8. (d) No. of radial nodes in 3p = n – l – 1 has shortest wavelength. or 3 – 1 – 1 = 1 19. Which of the following sets of quantum numbers are correct ? nl ml 17. Identify the pairs which are not of isotopes ? (a) 1 1 2 (a) 12 X, 136 Y (b) 35 X, 3177Y (b) 2 1 + 1 6 17 (c) 14 X, 147 Y (d) 84X, 85Y (c) 3 2 – 2 6 (d) 3 4 – 2 18. Out of the following pairs of electrons, identify the pairs of electrons present in degenerate orbitals : 20. In which of the following pairs, the ions are iso- (a) (i) n = 3, l = 2, ml = – 2, ms = −1 electronic ? 2 (a) Na+ , Mg2+ (b) Al3+ , O– (ii) n = 3, l = 2, ml = – 1, ms = −1 (c) Na+ , O2– (d) N3– , Cl– 2 21. Which of the following statements concerning the quan- (b) (i) n = 3, l = 1, ml = 1, ms = +1 tum numbers are correct ? 2 (a) Angular quantum number determines the three dimensional shape of the orbital. (ii) n = 3, l = 2, ml = 1 ms = +1 2 (b) The principal quantum number determines the orientation and energy of the orbital. (c) (i) n = 4, l = 1, ml = 1, ms = +1 2 (c) Magnetic quantum number determines the size of the orbital. (ii) n = 3, l = 2, ml = 1, ms = +1 2 (d) Spin quantum number of an electron determines −1 the orientation of the spin of electron relative to the (d) (i) n = 3, l = 2, ml = + 2, ms = 2 chosen axis. (ii) n = 3, l = 2, ml = + 2, ms = +1 2

STRUCTURE OF ATOM 2/85 ANSWERS / HINTS Al3+ = 13 – 3 = 10 e–, O2– = 8 + 1 = 9 e– i.e. do not have same no. of electrons. MCQs Type-II Na+ = 11 – 1 = 10 e–, O2– = 8 + 2 = 10 e– i.e. have same 17. (c, d) are not isotopes because they have different atomic no. of electrons. numbers. N3– = 7 + 3 = 10 e–, Cl– = 17 + 1 = 18 e– i.e. do not have 18. (a, d) : Degenerate orbitals have same n and same same no. of electrons. l values. 21. (a, d) 19. (b, c) 20. (a, c) Na+ = 11 – 1 = 10 e–, Mg2+ = 12 – 2 = 10e–, i.e. have same no. of electrons. © Modern Publishers. All rights reserved.Q.22. Arrange s, p and d sub-shells of a shell in the(b) Which of the following orbitals has the increasing order of effective nuclear charge highest energy ? (Zeff) experienced by the electron present in them. 5p, 5d, 5f, 6s, 6p Ans. s-orbitals shield the electrons from the nucleus more Ans. Refer Conceptual Qs. 2 Q.39 (Page 67). effectively than p-orbitals which in turn shield more effectively than d-orbitals. Hence, the subshells in Q. 28. Which of the following will not show deflection the increasing order of effective nuclear charge is : from the path on passing through an electric field ? d<p<s Proton, cathode rays, electron, neutron. Q.23. Show the distribution of electrons in oxygen atom (atomic number 8) using orbital diagram. Ans. Neutron is a neutral particle. Hence, it will not show deflection from the path on passing through an electric field. Ans. O (Z = 8) : 1s 2s 2p Q. 29. An atom having atomic mass number 13 has 7neutrons.Whatistheatomicnumberoftheatom ? ↑↓ ↑↓ ↑↓ ↑ ↑ Ans. Atomic mass, A = 13, n = 7 Q.24. Nickel atom can lose two electrons to form Ni2+ ion. The atomic number of nickel is 28. From n + p = 13 ∴ p = 13 – 7 = 6 which orbital will nickel lose two electrons. Hence, atomic number, Z = p = 6 Ans. Refer Conceptual Qs. 2 Q.36 (Page 67). Q. 30. Wavelengths of different radiations are given below : Q.25. Which of the following orbitals are degenerate ? λ(A) = 300 nm λ(B) = 300 μm 3dxy, 4dxy , 3dz2 , 3dyz, 4dyz, 4dz2 λ(C) = 3 nm λ(D) = 30 Å Arrange these radiations in the increasing order Ans. Refer Conceptual Qs. 2 Q.37 (Page 67). of their energies. Q.26. Calculate the total number of angular nodes and Ans. E = hν or = hc or E ∝ 1 radial nodes present in 3p orbital. λλ Ans. Refer Conceptual Qs. 2 Q.38 (Page 67). λ(A) = 300 nm = 300 × 10–9 m or = 3 × 10–7 m ; Q.27. The arrangement of orbitals on the basis of λ(B) = 300 × 10–6 m = 3 × 10–4 m energy is based upon their (n + l) value. Lower λ(C) = 3 × 10–9 m, λ(D) = 30 × 10–10 m = 3 × 10–9 m the value of (n + l), lower is the energy. For ∴ Increasing order of energies is : orbitals having same values of (n + l), the orbital B<A<C=D with lower value of n will have lower energy. Q. 31. The electronic configuration of valence shell of Cu is 3d104s1 and not 3d94s2. How is this I. Based upon the above information, arrange configuration explained ? the following orbitals in the increasing order Ans. This is because completely filled and half filled orbitals have extra stability. In 3d104s1, d-orbitals of energy. are completely filled and hence this configuration is more stable. (a) 1s, 2s, 3s, 2p (b) 4s, 3s, 3p, 4d Q. 32. The Balmer series in the hydrogen spectrum (c) 5p, 4d, 5d, 4f, 6s (d) 5f, 6d, 7s, 7p corresponds to the transition from n1 = 2, to n2 = 3, 4, .........., This series lies in the visible region. II. Based upon the above information, solve the Calculate the wave number of line associated questions given below : (a) Which of the following orbtials has the lowest energy ? 4d, 4f, 5s, 5p

2/86 MODERN’S abc + OF CHEMISTRY–XI with the transition in Balmer series when the Ans. Wavelength is the distance between the successive electron moves to n = 4 orbit. crests. (RH = 109677 cm–1) ∴ Wavelength, λ = 4 × 2.16 = 8.64 pm Ans. ν = ⎛ 1 − 1⎞ cm−1 Q.37. Chlorophyll present in green leaves of plants 109677 ⎜ n12 ⎟ absorbs light at 4.620 × 1014 Hz. Calculate the ⎠ wavelength of radiation in nanometer. Which ⎝ part of the electromagnetic spectrum does it belong to ? © n22 Modern Publishers. All rights reserved. = 109677 ⎛ 1 − 1⎞ = 109677 ⎛ 1 − 1⎞ ⎜⎝ 22 42 ⎠⎟ ⎜⎝ 4 16 ⎟⎠ = 109677 × 3 = 20564.4 cm–1 Ans. λ = c 16 ν Q.33. According to de Broglie, matter should exhibit ∴ λ = 3.0 ×108 = 0.6494 × 10–6 dual behaviour, that is both particle and wave 4.620 × 1014 like properties. However, a cricket ball of mass 100 g does not move like a wave when it is thrown = 649.4 × 10–9 m = 649.4 nm by a bowler at a speed of 100 km/h. Calculate the wavelength of the ball and explain why it does It belongs to visible light. not show wave nature. Q.38. What is the difference between the terms ‘orbit Ans. m = 100 g = 0.1 kg and orbital’ ? v = 100 km/h = 100 × 1000 Ans. Refer Text (Page 45). 60 × 60 Q.39. Table-tennis ball has a mass 10 g and a speed of = 1000 ms−1 90 m/s. If speed can be measured within an 36 accuracy of 4% what will be the uncertainty in speed and position ? λ= h = 6.63 × 10−34 kg m2s−1 Ans. Uncertainty in speed of ball = 90 × 4 = 3.6 ms–1 mv (0.1 kg) (1000 / 36 ms−1) 100 = 2.387 × 10–34 m h 4πmΔv Since the wavelength is very small, the wave nature Uncertainty in position, Δx = cannot be detected. Q.34. What is the experimental evidence in support of = 6.626 × 10−34 Js idea that electronic energies in an atom are 4 × 3.14 × 10 × 10−3 kg × 3.6 ms−1 quantized ? = 1.46 × 10–33 m Ans. In the line spectrum of any element, lines of definite wavelength are obtained. These lines correspond to Q.40. The effect of uncertainty principle is electronic transitions between fixed energy levels. significant only for motion of microscopic Thus, the electrons in these energy levels have definite particles and is negligible for the macroscopic energy i.e, quantized values. particles. Justify the statement with the help of a suitable example. Q.35. Out of electron and proton which one will have, a higher velocity to produce matter waves of the Ans. Refer Text, significance of uncertainty principle same wavelength ? Explain it. (Page 41). Ans. λ = h . For the same value of λ, the electron will have Q.41. Hydrogen atom has only one electron, so mutual mv repulsion between electrons is absent. However, in multielectron atoms, mutual repulsion higher velocity because it is lighter particle. between the electrons is significant. How does this affect the energy of an electron in the orbitals Q.36. Ahypotheticalelectromagneticwaveisshownin of the same principal quantum number in Fig. 1. Find out the wavelength of the radiation. multielectron atoms ? Ans. In case of H-atom, the energies of electron in different orbitals depend only on the value of n. Hence, different orbitals of same shell have same energy. However, in case of multielectron atoms, the energies of the orbitals depend upon n + l values. Hence, for the same value of n but different values of l i.e., different subshells belonging to the same main shell have different energies. Fig. 1

STRUCTURE OF ATOM 2/87 Q.42. Match the following species with their corresponding ground state electronic configuration. Atom / Ion Electronic configuration (i) Cu (a) 1s2 2s2 2p6 3s2 3p6 3d10 (ii) Cu2+ (b) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 (iii) Zn2+ (c) 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (iv) Cr3+ (d) 1s2 2s2 2p6 3s2 3p6 3d9 (e) 1s2 2s2 2p6 3s2 3p6 3d3 © Modern Publishers. All rights reserved.Q.43. Match the quantum numbers with the information provided by these. Quantum number Information provided (i) Principal quantum number (a) orientation of the orbital (ii) Azimuthal quantum number (b) energy and size of orbital (iii) Magnetic quantum number (c) spin of electron (iv) Spin quantum number (d) shape of the orbital Q.44. Match the following rules with their statements : Column I Column II (i) Hund’s Rule (a) No two electrons in an atom can have the same set of four quantum numbers. (ii) Aufbau Principle (b) Half-filled and completely filled orbitals have extra stability. (iii) Pauli Exclusion Principle (c) Pairing of electrons in the orbitals belonging to the same subshell does not take place until each orbital is singly occupied. (iv) Heisenberg’s Uncertainty Principle (d) It is impossible to determine the exact position and exact momentum of a subatomic particle simultaneously. (e) In the ground state of atoms, orbitals are filled in the order of their increasing energies. Q.45. Match the following : Q.47. Match species given in Column I with the electronic configuration given in Column I Column II Column II. (i) X-rays (a) ν = 1 – 104 Hz Column I Column II (ii) UV (b) ν = 1010 Hz (iii) Long radio waves (c) ν = 1016 Hz (i) Cr (a) [Ar]3d84s0 (iv) Microwave (d) ν = 1018 Hz (ii) Fe2+ (b) [Ar]3d104s1 (iii) Ni2+ (c) [Ar]3d64s0 Q.46. Match the following : (iv) Cu (d) [Ar]3d54s1 (e) [Ar]3d64s2 Column I Column II (i) Photon (a) Value is 4 for N shell (ii) Electron (b) Probability density (iii) Ψ2 (c) Always positive value (iv) Principal (d) Exhibits both momentum quantum number n and wavelength ANSWERS / HINTS has also particle as well as wave nature. ψ2 represents probability density and is always positive. Principle Matching Type quantum number n = 4 for N shell and always has positive values. 42. : (i) – (c) (ii) – (d) (iii) – (a) (iv) – (e) 43. : (i) – (b) (ii) – (d) (iii) – (a) (iv) – (c) 47. : (i) – (d) (ii) – (c) (iii) – (a) (iv) – (b) 44. : (i) – (c) (ii) – (e) (iii) – (a) (iv) – (d) 45. : (i) – (d) (ii) – (c) (iii) – (a) (iv) – (b) Cr : [Ar] 3d5 4s1, Fe2+ : [Ar] 3d64s0; Ni2+ = [Ar] 3d84s0; Cu = [Ar] 3d104s1. Frequency decreases in the order : X-rays > UV-rays > microwaves > radiowaves. 46. : (i) – (d) (ii) – (d) (iii) – (b), (c) (iv) – (a), (c) Photon has particle as well as wave nature. Electron

2/88 MODERN’S abc + OF CHEMISTRY–XI In the following questions a statement of Assertion (A) Reason (R) : The chemical properties of an followed by a statement of Reason (R) is given. Choose atom are controlled by the num- the correct option out of the choices given below. ber of electrons in the atom. (a) Both A and R are true and R is the correct explanation of A. Q.49. Assertion (A) : Black body is an ideal body that (b) Both A and R are true but R is not the correct emits and absorbs radiations of all explanation of A. frequencies. (c) A is true but R is false. (d) Both A and R are false. Q.48. Assertion (A) : All isotopes of a given element show the same type of chemical behaviour. © Reason (R) : The frequency of radiation emitted Modern Publishers. All rights reserved. by a body goes from a lower frequency to higher frequency with an increase in temperature. Q.50. Assertion (A) : It is impossible to determine the exact position and exact momentum of an electron simultaneously. Reason (R) : The path of an electron in an atom is clearly defined. ANSWERS / HINTS Assertion Reason Type 48. (a) 49. (b) 50. (c) Q.51. What is photoelectric effect? State the result Q.53. When an electric discharge is passed through of photoelectric effect experiment that could hydrogen gas, the hydrogen molecules dissociate not be explained on the basis of laws of to produce excited hydrogen atoms. These classical physics. Explain this effect on the basis excited atoms emit electromagnetic radiation of quantum theory of electromagnetic of discrete frequencies which can be given by radiations. the general formula. Ans. Refer Text (Page 14–16). ν = 109677 ⎡ 1 − 1 ⎤ ⎢ ni2 n2f ⎥ ⎣⎢ ⎥⎦ Q.52. Threshold wfrheicqhueancpyh,otνo0nismtuhset minimum What points of Bohr’s model of an atom can be frequency possess to used to arrive at this formula? Based on these points derive the above formula giving eject an electron from a metal. It is different description of each step and each term. for different metals. When a photon of frequency 1.0 × 1015 s–1 was allowed to hit a metal surface, an electron having 1.988 × 10–19J Ans. The following two points of Bohr’s model can be used to derive the given formula : of kinetic energy was emitted. Calculate the (i) Electrons revolve around the nucleus in definite threshold frequency of this metal. Show that energy levels called orbits. an electron will not be emitted if a photon with (ii) Energy is emitted or absorbed only when electron jumps from one orbit to another. a wavelength equal to 600 nm hits the metal Derivation surface. (i) According to Bohr, energy of an electron in the Ans. hν = hν0 + 1 mu2 nth level is given as 2 or hν0 = hν – 1 mu2 En = − 2π2me4 2 n2h2 = 6.63 × 10–34 × 1.0 × 1015 – 1.988 × 10–19 where m = mass of electron, e = charge of electron, = 6.63 × 10–19 – 1.988 × 10–19 h = Planck’s constant. = 4.642 × 10–19 J (ii) When electron jumps from outer orbit (n2) to inner orbit (n1), then difference in energy (ΔE) is ∴ ν0 = 4.642 × 10−19 J = 6.988 × 1014 s–1 emitted as : 6.63 × 10−34 Js ΔE = E2 – E1 Now frequency of photon of wavelength 600 nm = 600 × 10–9 m 2π2me4 ⎛ 2π2me4 ⎞ ν= 3.0 × 108 ms−1 = 5.0 × 1014 s–1 = − n22h2 − ⎝⎜⎜ − n12h2 ⎟⎠⎟ 600 × 10−9 m Since frequency of photon is less than v0, electron will not be emitted. = 2π2me4 ⎛ 1 − 1 ⎞ h2 ⎜⎜⎝ n12 n22 ⎠⎟⎟

STRUCTURE OF ATOM 2/89 Now, ΔE = hν = hc = ΔE = hν λ hcν = hc = λ hcν or ν = ΔE = 2π2me4 ⎛ 1 − 1⎞ = (6.626 × 10–34 Js) (3.0 × 1010 cm s–1) hc ch3 ⎜ n12 ⎟ ⎝ n22 ⎠ × (15232.9 cm–1) © Modern Publishers. All rights reserved.Substituting the values of c, h, π, m, e in CGS units,= 3.028 × 10–19 J we get ν = cν ν = 109677 cm −1 ⎛ 1 − 1⎞ = (3.0 × 1010 cm s–1) × 15239.9 cm–1 ⎜ n12 ⎟ ⎠ = 4.57 × 1014s–1 ⎝ n22 Q.55. Why was a change in the Bohr model of atom required? Due to which important development Q.54. Calculate the energy and frequency of the (s), concept of movement of an electron in an radiation emitted when an electron jumps from orbit was replaced by the concept of probability n = 3 to n = 2 in a hydrogen atom. of finding electron in an orbital ? What is the name given to the changed model of atom ? Ans. ν = ⎛ 1 − 1⎞ cm−1 109677 ⎜ n12 ⎟ Ans. According to Bohr’s model, electrons move along fixed n22 ⎠ circular paths called orbits. Therefore, position and ⎝ velocity of electron can be well defined. This was contradicted by de Broglie concept of dual nature of For n1 = 2, n2 = 3 electron and Heisenberg’s uncertainty principle. Therefore, to incorporate the two principles, the well ν = 109677 ⎛ 1 − 1 ⎞ defined orbits were replaced by regions of maximum ⎜⎝ 22 32 ⎠⎟ probability called orbitals. The changed concept formed the basis of wave mechanical model of atom. = 109677 × 5 = 15232.9 cm–1 36 Passage Based Questions 2. Which series of hydrogen spectrum lies in the visible region? I. Read the following passage and answer questions 1–5 that follow: 3. What is the ratio of radius of 4th orbit of hydrogen and 3rd orbit of Li2+ ion? Bohr’s model enables us to derive the energy of an electron revolving in nth orbit. For H-atom and 4. Which transition between Bohr’s orbits corresponds to hydrogen like species: third line in Lyman series? En = − 2π2m e4Z2 5. What is the experimental evidence in support of the fact n2h2 that electronic energies in an atom are quantized? or = − 13.6 Z2 eV atom−1 = − 21.8 × 10−19 Z2 J atom−1 II. Read the following passage and answer questions n2 n2 6–10 that follow: This helps to calculate the radius of an orbit, The electrons are distributed around the nucleus in various energy levels called shells, subshells and rn = 0.529 n2 A° orbitals. A set of quantum numbers completely describe Z the position and total energy of electron in an atom. The various permitted values of quantum numbers are: Bohr’s model also explains the occurrence of different principal, n = 1, 2, 3, 4 ..... spectral lines. The wavelengths of different lines can be azimuthal, l = 0, 1, 2 ..... (n – 1) given as: magnetic, ml = –l .... 0..... + l 1 = ν (in cm−1) = ⎛ 1 − 1 ⎞ spin, ms = +1/2 and –1/2 λ R ⎜⎝⎜ n12 n22 ⎠⎟⎟ 6. Is the following set of quantum numbers possible or not? R = 109678 cm−1 and n2 > n1 n = 3, l = 2, m = 0, s = –1/2 1. What is the energy of first excited state of H atom? 7. What is the value of n that allows g subshell? 8. How many orbitals (of all kind) are possible for n = 3 energy level?