Modern ABC Chemistry XI

2/90 MODERN’S abc + OF CHEMISTRY–XI 9. What is the difference in angular momentum of an 7. The limiting line of any spectral series in the hydrogen electron present in 2p and 3p orbital? spectrum is the line when n2 in the Rydberg’s formula is ................. 10. How many electrons are possible in all sub-shells with n + l = 4? 8. Lyman series of hydrogen spectrum lies in ................. region. III. Readthe followingpassageandanswerquestions 11–15 that follow: 9. The quantum number which tells about the orientation of different orbitals of an atom is called ................. . 10. The quantum number which describes the subshells present in any main shell is called ................. . ©The shapes of orbitals may be represented by boundary Modern Publishers. All rights reserved. surface diagrams. These boundary surface diagrams give the most probable regions. s-orbitals are non-directional while p-, d-and f-orbitals have different orientations hgaivveensphbyerimcall values. These Assertion Reason Questions boundary surfaces also nodes or radial Note : In the following questions a statement of assertion nodes and nodal planes or angular nodes which depend followed by a statement of reason is given. Choose the correct answer out of the following choices. upon the .values of n and l. (a) Assertion and reason both are correct statements and 11. How many orbitals are possible for l = 2 subshell? reason is correct explanation for assertion. 12. How many spherical nodes are present in 3p-orbital? (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. 13. How do 3s and 4s orbitals differ in terms of nodes present in these? (c) Assertion is correct statement but reason is wrong statement. 14. Does dz2 orbital has zero electron density in xy plane? 15. How many angular nodes are present in 3dyz orbital? (d) Assertion is wrong statement but reason is correct statement. True or False Questions Predict which of the following statements are true or false. 1. Assertion : Hydrogen has one electron in its orbit but 1. 2s orbital has one node. it produces several spectral lines. 2. Phosphorus (Z = 15) has three unpaired electrons. Reason : There are many excited energy levels 2. Fe3+ ion has four unpaired electrons. available. 4. Number of radial nodes in 5f orbital is four. 2. Assertion : The 19th electron in potassium atom enters 5. It takes less energy to ionize (or remove) an electron into 4s-orbital and not in the 3d-orbital. from first excited state than the ground state of Reason : (n + l) rule is followed for determining the H-atom. orbital of lowest energy state. 6. Paschen, Brackett and Pfund series fall in infra red (A.I.I.M.S. 1999) region. 3. Assertion : The energy of an electron is largely 7. The splitting of spectral lines in electrical field is determined by its principal quantum called Zeeman effect. number. 8. Electronic energy is negative because electron has Reason : The principal quantum number (n) is a negative charge. measure of the probable distance of finding 9. ψ2 represents probability density. the electron around the nucleus. (A.I.I.M.S. 1998) 10. Heisenberg principle is applicable to microscopic 4. Assertion : For the outermost electron in Na atom, the objects. orbital angular momentum is zero. Reason : For 3s electron , l = 0 and orbital angular 11. 3s and 4p orbitals have same number of radial momentum is zero. nodes. 5. Assertion : The configuration of Cexccalnunsiootnbep1risn2c2isp2le2pa2xn. Reason : According to Pauli 12. Angular momentum for 2p and 3p orbital is same. orbital can have maximum of two electrons. 13. Copper (I) is diamagnetic. 6. Assertion : 2s-orbital has one node. 14. Spin quantum number is not originated from Reason : Number of nodes in an orbital is equal to (n – l – 1) value. Schrodinger wave equation. 7. Assertion : All microscopic bodies in motion have wave 15. The de-Broglie wavelength associated with hydro- gen atom is larger than that of deuterium atom if both are travelling at the same speed. Fill in the blanks Questions character. Reason : Microscopic bodies have very large mass. 1. Out of 5d, 2f, 3d, 5s, the subshell which does not 8. Assertion : Cl– ions and K+ ions are isoelectronic. make sense is ................. Reason : Isoelectronic ions have same charge. 9. Assertion : It is impossible to determine the exact 2. For n = 3, the total number of orbitals is ................. position and exact momentum of an electron simultaneously. 3. There are ................. orbitals corresponding to l = 2. Reason : The path of an electron in an atom is clearly 4. Dual character of electrons was verified by ................. defined. 10. Assertion : Photoelectric effect is easily given by and ................. . cesium metal. 5. The series in hydrogen spectrum falling in visible Reason : Photoelectric effect is easily given by the metals having high ionization enthalpy. region is ................. series. 6. The number of unpaired electrons in Fe2+ is ............ .

STRUCTURE OF ATOM 2/91 Multiple Choice Questions 2. How does the existence of line spectra support for  Refer TOPICWISE MULTIPLE CHOICE QUESTIONS, the Bohr model of the atom ? COMPETITION FILE Page 2/100. 3. Why was Bohr model abandoned ? 4. In terms of Bohr theory of the structure of the atom, why is it that electrons do not spiral into the nucleus ? © Modern Publishers. All rights reserved.One Word/Very Short Sentence Answer5. Name the three quantum numbers which are necessary to describe an orbital. What are the 1. What is the difference between a quantum and a permitted values for these ? photon? 6. What is the maximum number of electrons in : 2. Can an electron have the quantum number values as n = 2, l = 2 and m = + 2 ? (i) a principal quantum number 3. How many sub-levels are there in M shell ? What (ii) an orbital are their designations ? (iii) p-subshell 4. What quantum numbers n and l are assigned to a (iv) s, p and d-subshells in an atom ? 3p-orbital ? 7. Give the number of orbitals in : 5. Write the electronic configuration of chromium (i) a p-subshell (Z = 24). (ii) a d-subshell 6. An atom of an element has 19 electrons. What is (iii) second shell. the total number of p-electrons ? 8. Describe the orbital with the following quantum 7. What is the sequence of energies of 3s, 3p and numbers : 3d-orbitals in (i) n = 1, l = 0 (ii) n = 2, l = 1, m = 0 (i) a hydrogen atom, and (iii) n = 3, l = 2 (iv) n = 4, l = 1 (ii) a multielectron atom ? (v) n = 3, l = 0, m = 0 (vi) n = 3, l = 1 8. According to which principle an atom cannot have more than two electrons ? 9. Give the electronic configurations of : 9. What designations are given to the following (i) Scandium (Z = 21) subshells having: (ii) Chromium (Z = 24). (i) n = 4, l = 3 10. How many orbitals and how many electrons are there in each of the first two principal quantum (ii) n = 3, l = 2 numbers ? 10. If n = 3, what are values of quantum number l? 11. Give the electronic configurations of the following 11. What is the relation between the shapes of 3dxy ions : and 3dx2 − y2 orbitals? (i) H– (ii) Na+ 12. How many electrons are present in 3d orbitals in (iii) F– (iv) Mg2+. chromium (Z = 24)? 12. Fill in the blanks : 13. If the quantum number ‘l ’ has a value of 2, what are (i) The size of an orbital is dependent on the value the permitted values of quantum number m ? of ...... 14. How many electrons s and p-subshell can (ii) The orbitals having the same energies are accommodate ? called ...... 15. Which energy levels do not have p-orbital ? (iii) The number of unpaired electrons in carbon is ...... and in nitrogen is ...... 16. What do you mean by saying that energy of electron is quantized ? (iv) The shape of 1s-orbital is .............. . 17. Out of electron and proton which one will have, a (v) ...... filled orbitals have extra stability. higher velocity to produce matter waves of the same wavelength ? Explain it. 13. Define Aufbau principle. Which of the following orbitals are possible ? 18. What is the total number of orbitals associated with principal quantum number, n = 3? 1p, 2s, 2p and 3f 19. Arrange 3s, 3p and 3d orbitals in the increasing 14. With the help of Pauli’s exclusion principle and the order of energy for hydrogen atom. concept of atomic numbers for orbitals, show that an M shell cannot accommodate more than 18 20. The atomic number of an element is 20. How many electrons. electrons have l = 0 in this ? 15. What is an orbital ? How will you differentiate 21. Define Hund’s rule of maximum multiplicity. between an orbit and an orbital ? 22. Define Pauli exclusion principle. 16. Compare the shapes of 1s- and 2s-orbitals. 17. What is the difference between the symbols l 1. List some important postulates of Bohr’s model of an atom. and L ? 18. How many electrons are there in the valence quantum level of copper (atomic number = 29) atom ? Give reasons.

2/92© Write the electronic configurations of the following MODERN’S abc + OF CHEMISTRY–XI 19.Modern Publishers. All rights reserved.elements : Carbon (Z = 6), neon (Z = 10), magnesium 2. What are quantum numbers ? What permitted 20. (Z = 12), chlorine (Z = 17), calcium (Z = 20), chromium values can these have ? Give the significance of each (Z = 24), iron (Z = 26) and rubidium quantum number. 21. (Z = 37). 22. Identify the atoms that have the following ground 3. What is photoelectric effect ? State the result of 23. state electronic configurations : photoelectric effect experiment that could not be (i) 1s2 2s2 2p6 3s2 explained on the basis of laws of classical physics. 24. (ii) 1s2 2s2 2p5 Explain this effect on the basis of quantum theory 25. (iii) 1s2 2s2 2p6 3s2 3p6 4s1 3d10. of electromagnetic radiations. 26. 27. In building up of the atoms, the filling of 4s-orbital 4. Why was a change in the Bohr Model of atom 28. takes place before the 3d-orbital. Explain. required? Due to which important development (s), 29. The expected electronic configuration of copper is concept of movement of an electron in an orbit was [Ar] 3d9 4s2 though actually it is [Ar] 3d10 4s1. Give replaced by, the concept of probability of finding 30. reasons. electron in an orbital ? What is the name given to 31. An atom of an element has 19 electrons. Find out : the changed model of atom? (a) its atomic number 32. (b) total number of s-electrons 5. (a) What is radial probability distribution curve? 33. (c) total number of p-electrons. Draw radial probability distribution curves for 34. Give the shapes of s and p-orbitals. 1s and 2s orbitals. What is Hund’s rule of maximum multiplicity ? 35. Illustrate this by taking the example of carbon. (b) Discuss the similarities and differences between Explain why half-filled and completely-filled 1s and 2s orbitals. orbitals have extra stability. What is Aufbau principle ? Write the electronic (c) How many nodes are present in 1s and 2s configurations of the elements of atomic numbers orbitals? 16, 20, 24 and 35. 6. (a) How many orbitals are possible for a d-subshell? Draw the shape of an orbital which has l = 0. (b) Draw the shapes of dxy and dx2 − y2 orbitals? State Aufbau principle. Write electronic configurations of the elements with atomic numbers What is common between these and what is 17 and 24. difference between these orbitals? What is the angle between the lobes of these orbitals? Why in the building of the atom, the filling of 4s-orbital takes place before 3d-orbital ? (c) Name a 3d orbital which has electron density along all the three axes. State and explain Pauli’s exclusion principle. Write the electronic configuration of the element with 1. What is the wavelength of light emitted when the atomic number 24. electron in a hydrogen atom undergoes transition from the energy level with n = 4 to the energy level What is Hund’s rule of maximum multiplicity ? with n = 1 ? In which region of the electromagnetic Explain by taking an example of nitrogen. spectrum does this radiation fall ? Discuss the significance of magnetic quantum number. 2. Calculate the wave number of the radiation emitted when an electron in a hydrogen atom undergoes a What is the experimental evidence in support of transition from 4th energy level to the 2nd energy the idea that electronic energies in an atom are level. quantized ? 3. In the Balmer series of atomic spectra of hydrogen The effect of uncertainty principle is significant only atom, a line corresponding to wavelength for motion of microscopic particles and is negligible 656·4 nm was obtained. Calculate the number of for the macroscopic particles. Justify the statement higher orbit from which the electron drops to with the help of a suitable example. produce this line. 1. Give the essential postulates of Bohr’s model of an 4. Calculate the wavelength (in nanometers) emitted atom. How did it explain by a photon during a transition from n = 6 to n = 4 state in the H-atom. (i) the stability of an atom 5. Calculate the wavelength associated with a H-atom (ii) origin of spectral lines in hydrogen atom ? (mass = 1.676 × 10–27 kg) moving with velocity of 8.0 × 102 cm s–1. 6. Calculate the value of Δx · Δv for an electron (mass of electron = 9.1 × 10–31 kg).

STRUCTURE OF ATOM 2/93 ©Passage Based Questions 13. True Modern Publishers. All rights reserved. 14. True 1. En = − 13.6 eV 15. True. n2 Energy of first excited state (n = 2), Fill in the Blanks Questions E2 = − 13.6 eV = −3.4 eV 1. 2f 2. Nine 3. Five 22 7. 6 2. Balmer series. 4. Davisson and Germer, Thomson 0.529 n2 3. rn = Z A° 5. Balmer 6. 4 0.529 × 16 8. uv 1 r4 (H) = 9. magnetic quantum number (Li2+ ) 0.529 × 9 (∵ Z = 3) 10. azimuthal quantum number. 3 r3 = r4 (H) 16 Assertion Reason Questions r3 (Li2+) 3 = 1. (a) 2. (a) 3. (a) 4. (a) 5. (b) 6. (a) 7. (c) 8. (c) 9. (c) 10. (c) 4. Between 4th orbit to 1st orbit. 5. Line spectra 6. Possible Very Short Answer Questions 7. 5 8. Nine 1. The smallest packet of energy of any radiation is 9. No difference because angular momentum depends called a quantum while that of light is called photon. only upon value of l. 2. No Angular momentum = h l(l + 1) 3. Three, 3s, 3p, 3d 2π 4. n = 3, l = 1 5. 1s2 2s2 2p6 3s2 3p6 4s1 3d5 The value of l (1) is same for 2p and 3p orbital. 6. 12 10. Subshells with n + l = 4 are 4s and 3p. Hence 7. (i) 3s = 3p = 3d electrons present in these subshells are 2 + 6 = 8. (ii) 3s < 3p < 3d 11. Five 8. Pauli’s exclusion principle 12. Nodes = n – l – 1 9. (i) 4f (ii) 3d 10. For n = 3, l = 0, 1, 2 =3–1–1=1 11. Rotation of 45° around X and Y axis of dxy orbital 13. 3s orbital has two nodes while 4s has three nodes. 14. No, dz2 orbital has electron density in xy-plane also gives dx2 − y2 orbital and vice versa. as shown by ring. 12. Cr (Z = 24): 1s22s22p63s23p63d5 4s1 15. Angular nodes = 2 No. of electrons in 3d orbitals = 5. True or False Questions 1. True 2. True 3. False. Fe3+ ion has five unpaired electrons. 13. m = – 2, – 1, 0, +1, +2 4. False. (n – l – 1) i.e. (5 – 3 – 1) = 1 node. 14. 2, 6 5. True 6. True 15. First 7. False. Zeeman effect is the splitting of the spectral 16. This means that the electrons in an atom have only lines in magnetic field. definite values of energies. 8. False 9. True 17. Electron (λ = h or v = h or v ∝ 1 for same λ) mv mλ m 10. False. Heisenberg principle is applicable to all objects but has significance only for microscopic bodies. 18. Nine 11. True. Both have 2 radial nodes. 19. 3s = 3p = 3d 12. True. Because angular momentum depends only upon 20. Electronic configuration : 1s22s22p63s23p64s2 h No. of electrons having l = 0 (i.e. s-orbitals) = 8. 2π the value of l as l(l + 1) .

2/94 MODERN’S abc + OF CHEMISTRY–XI Short Answer Questions = 109678 × 12 cm−1 = 8.2 × 104 cm–1 16 11. Hydrogen, about 1837 times 15. (i) 2n2 (ii) 2 (iii) 6 (iv) 2, 6, 10 3. 1 = 109678 ⎡1 − 1 ⎤ cm–1 16. (i) 3 (ii) 5 (iii) 4 λ ⎢ n22 ⎥ 17. (i) 1s (ii) 2pz (iii) 3d (iv) 4p (v) 3s (vi) 3p ⎣⎢ n12 ⎥⎦ 18. (i) 1s2 2s2 2p6 3s2 3p6 3d1 4s2© Modern Publishers. All rights reserved. For Balmer series n1 = 2, λ = 656.4 × 10–9 m (ii) 1s2 2s2 2s6 3s2 3p6 3d5 4s1 = 656.4 × 10–7 cm 20. (i) 1s2 (ii) 1s2 2s2 2p6 1 = 109678 ⎡1 − 1 ⎤ cm−1 (iii) 1s2 2s2 2p6 656.4×10−7cm ⎢ n22 ⎥ 21. (i) principal quantum number ⎣⎢ 22 ⎥⎦ (ii) degenerate or 1 − 1 =1 (iii) two, three 4 n22 109678×656.4×10−7 (iv) spherical (v) half-filled and completely. or 1 − 1 =1 22. 2s, 2p 4 n22 7.20 27. [Ar] 3d10 4s1 ; one. 28. 1s2 2s2 2p2, 1s2 2s2 2p6, 1s2 2s2 2p6 3s2, 1s2 2s2 2p6 3s2 or – 1 = 1 – 1 3p5,1s2 2s2 2p6 3s2 3p6 4s2, n22 7.20 4 1s2 2s2 2p6 3s2 3p6 3d5 4s1, 1s2 2s2 2p6 3s2 3p6 3d6 4s2, 1s2 2s2 2p6 3p6 3d10 4s2 4p6 5s1. = 4 − 7.20 = − 3.20 29. (i) Magnesium 28.80 28.80 (ii) fluorine (iii) copper. or 1 = 3.20 or n22 = 28.80 =9 32. (a) 19 (b) 7 (c) 12. n22 28.80 3.20 36. 1s2 2s2 2p6 3s2 3p4, 1s2 2s2 2p6 3s2 3p6 4s2, 1s2 2s2 2p6 4s1 3d5, 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5. n2 = 9 = 3. Third energy level. 40. 1s2 2s2 2p6 3s2 3p5, 1s2 2s2 2p6 3s2 3p6 3d5 4s1 4. For transition from ni = 6 to nf = 4, 41. 1s2 2s2 2p6 3s2 3p6 3d5 4s1. ΔE = 2.18 × 10–18 J ⎛1 − 1⎞ ⎝⎜ 62 42 ⎠⎟ = 2.18 × 10−18 ⎛ 16 − 36 ⎞ J ⎜⎝ 16 × 36 ⎟⎠ Numerical Problems = –7.57 × 10–20 J 1. According to Rydberg equation, It is emission energy. The frequency of the photon (taking magnitude) of energy, ⎛1 ⎞ 1 = 109678 ⎜⎜⎝ n12 − 1 ⎟⎠⎟ cm−1 7.57 × 10−20 λ n22 6.626 × 10−34 ν= ∆E = = 1.142 × 1014 s−1 Here n1 = 1, n2 = 4 h ∴ 1 = 109678 ⎛1 − 1 ⎞ cm−1 or λ= c = 3.0 × 108 m s−1 = 2.63 × 10–6 m λ ⎜⎝ 12 42 ⎠⎟ ν 1.142 × 1014 = 109678 × 15 cm−1 or = 2.63 × 103 nm 16 5. λ = h or λ= 16 = 9.72 × 10–6 cm mν 109678×15cm−1 = 6.626 × 10−34 Js = 97.2 × 10–9 m (1.676 × 10−27 kg) × (8 m s−1) = 97.2 nm = 4.94 × 10–8 m = 49.4 nm 2. ν = 1 = 109678 ⎛ 1 − 1 ⎞ cm–1 h= 6.626 × 10−34 λ ⎜⎜⎝ n12 n22 ⎟⎟⎠ 4πm 4 × 3.14 × 9.1 × 10−31 6. Δx · Δv = Here n1 = 2, n2 = 4 = 5.797 × 10–5 m2 s–2 ∴ ν = 1 = 109678 ⎛ 1 − 1 ⎞ cm–1 λ ⎜⎝ 22 42 ⎠⎟

STRUCTURE OF ATOM 2/95 Q.1.How did wave mechanical model of an atom Ans. By de-Broglie equation : © overruled the circular orbits proposed by Bohr ? Modern Publishers. All rights reserved. λ= h Ans. The wave mechanical model of an atom was proposed p by Schrodinger in terms of mathematical equation known as Schrodinger wave equation. The solution of this equation But, λ = Δx (given) gives wave function ψ and the square of the wave function, i.e., ψ2 measures the probability density of finding the electron. ∴ Δx = h So, according to this model, it is possible to locate the regions p around the nucleus where the probability of finding the electron is maximum. These regions where the probability of finding According to Heisenberg uncertainty principle, the electron is maximum are called orbitals. Thus, wave mechanical model leads to the concept of orbital. Δx . Δp ≥ h 4π This concept overruled the idea of circular orbits proposed by Bohr. According to Bohr’s model of an atom, the electrons h .Δp ≥ h revolve around the nucleus in certain well defined circular p 4π paths called orbits. This means that electron remains only in the circular orbits at a definite distance from the nucleus. or h . mΔv ≥ h However, according to wave mechanical model, there are mv 4π regions where the probability of finding the electrons is maximum. The electron may be even outside this regions. or Δv 1 v ≥ 4π Q.2. Which of the following sets of orbitals are degenerate and why ? or Δv ≥ v 4π (i) 1s, 2s, and 3s in Mg atom (ii) 2px, 2py and 2pz in C atom Thus, uncertainty in velocity is so large that its velocity (iii) 3s, 3px and 3d orbitals in H atom . is uncertain. Ans. (i) 1s, 2s and 3s orbitals in Mg atom are not Q.5. Show that ground state energy of an electron degenerate because these have different values in H-atom is equal to the first excited state energy of of n. electron in He+ ion (assuming their Rydberg’s constants (ii) 2px, 2py and 2pz orbitals in C atom are degenerate to be equal). because these belong to same subshell. (iii) 3s, 3px and 3d orbitals in H atom are degenerate Ans. Energy of electron in nth shell is because for H atom, the subshells having same value of n have same energies. En = −RH Z2 n2 Q.3. An electron in a hydrogen atom is excited from the ground state to the n = 4 state. Predict which of the = −13.60 Z2 eV following statements are true or false: n2 (i) n = 4 is the first excited state. For H-atom in the ground state, n = 1 and Z = 1 (ii) It takes more energy to ionize (remove) the ∴ E1 = –13.60 eV electron from n = 4 than in the ground state. For He+ ion in the first excited state, Z = 2, n = 2 (iii) The wavelength of light emitted when the ∴ E2 = −13.60 × 22 = −13.60 eV electron drops from n = 4 to n = 2 is longer than that 22 from n = 4 to n = 1. Q.6. What is the difference in the orbital angular (iv) The wavelength which the atom absorbs in momentum of 2p and 3p electron ? Explain. going from n = 1 to n = 4 is the same as emitted when it goes from n = 4 to n = 1. Ans. Orbital angular momentum is given as : (v) The electron is farther from the nucleus (on Orbital angular momentum average) in n = 4 than in the ground state. = h l(l + 1) Ans. (i) False (ii) False (iii) True (iv) True (v) True. 2π Q.4. If uncertainty in position of a moving electron Since it is independent of the principal quantum number is equal to its de-Broglie wavelength, show that its ‘n’ and depends only on the azimuthal quantum number ‘l’, it velocity is completely uncertain. will be same for 2p and 3p electron. Hence, there will be no difference. Q.7. Predict the number of nodes and nodal planes in (i) 3px (ii) 4s and (iii) 3dx2 – y2 orbital. Ans. No. of radial nodes = n – l –1 and No. of nodal planes = l

2/96 MODERN’S abc + OF CHEMISTRY–XI ∴ Nodes Nodal planes (i) 3px 1 1 (ii) 4s 3 0 0 2 (iii) 3dx2–y2 Q.8. What will become to the wavelength associated with a moving particle if its velocity is doubled ? Ans. Wavelength becomes half of the original value © Modern Publishers. All rights reserved.(λ =h). mv Q.9. Cu2+ is more stable than Cu+ in aqueous Q.12. For a multielectron atom, the maximum of solution. Explain. 2p-orbital in radial probability distribution graph is nearer the nucleus than that of 2s-orbital. Therefore, Ans. Cu+ has outermost electronic configuration as 3d10 2p-orbital should be closer to the nucleus and lower in and therefore, should be most common and most stable state energy than 2s-orbital. But 2s-orbital has lower energy because of extra stability associated with completely filled than 2p-orbital. Explain. d-subshell (d10). However, this is not true and Cu2+ with outer electronic configuration 3d9 is more stable than Cu+. This is Ans. The radial probability distribution graphs for 2s- due to high hydration energy in aqueous solution and and 2p- orbitals are shown below : high lattice energy in solid state of Cu2+ as compared to Cu+. The equilibrium : 2 Cu+ Cu2+ + Cu K = ⎣⎡Cu2+ ⎤⎦ = 1.6 ×106 ⎣⎡Cu+ ⎦⎤2 The constant for disproportionation of Cu+ in aqueous solution shows that Cu2+ is very stable in aqueous solution as compared to Cu+. It is clear from the figure that in case of 2s-orbital, there Q.10. How do dx2 −y2 and dxy orbitals differ in their is a small additional peak or hump. This indicates that a orientation in space ? 2s-electron spends some of its time near the nucleus. In other words, 2s-electron penetrates the 1s2-core (or K shell, shown shaded in the figure). Due to penetration, a 2s-electron gets Ans. The four lobes of dx2 −y2 lie on the X and Y axis while less shielding from other electrons and therefore, feels more nuclear charge. As a result, a 2s-electron is attracted more the lobes of dxy lie in between X and Y axis (i.e. at an angle of strongly by the nucleus than a 2p-electron. Thus, 2s has lower 45°). So, dx2 −y2 orbital is exactly like dxy except that it is energy than a 2p-electron. rotated through 45° around Z-axis. 13. For which hydrogen like ion the wavelength difference between the first lines of the Lyman and Q.11. For H atom the Bohr radius for first orbit is Balmer series is equal to 59.3 nm ? 0.529Å and the radius of maximum probability for H-atom according to wave mechanical model is also Ans. For a spectral line, 0.529Å. How do the two approaches differ ? L O1 1 1 Ans. Bohr predicted that the electron will always be NMM QPPλ found moving around the nucleus in a circular path of radius = RZ2 n 2 − n 2 0.529 Å. According to his model, electron cannot be found at 1 2 distance less than or more than 0.529 Å. However according to wave mechanical model, electron is most likely to be found For Lyman line, at this distance but there is definite probability of finding the electron at distances both shorter and larger than 0.529 Å. In L O1 1 1 = 3RZ2 other words, according to wave mechanical model, the electron NM QPλLyman 12 22 keeps on moving towards or away from the nucleus and the = RZ2 − 4 maximum probability of locating it is at a radius of 0.529 Å from the nucleus. Fig. (a) shows the electron cloud picture of or λLyman = 4 1s-orbital having radius of maximum probability and Fig. (b) 3 RZ2 gives the Bohr picture of orbit in which electron is found only at this distance. For Balmer line L O1 = RZ2 1 − 1 = 5RZ2 NM QPλBalmer 22 32 36 or λBalmer = 36 5 RZ2

STRUCTURE OF ATOM 2/97 λBalmer – λLyman = 36 – 4 FHG KJI e j e j4 ×2 ⇒ 59.3 × 10–7 cm = 5 RZ2 3 RZ2 22 9.1 × 10−28 × 4.8 × 10−10 4 × L O1 36 − 4 7 NM QPRZ2 5 3 = 33 × (6.626 × 10−24)3 = 2.42 × 1014 . © Modern Publishers. All rights reserved.⇒ 59.3 × 10–7 =1 ⎡108 − 20 ⎤16. Calculate the energy emitted when electrons 109677.8 Z2 ⎢⎣ 15 ⎦⎥ of 1.0 g atom of hydrogen undergo transition giving the spectral line of largest energy in the visible region of = 59.3 × 10–7 its atomic spectrum. (RH = 1.1 × 107 m–1, c = 3 × 108 m s–1, h = 6.62 × 10–34 J s) or Z2 = 1 × 88 =9 109677.8 15 × 59.3 × 10−7 Ans. The spectral line in visible region corresponds to Balmer series i.e. n1 = 2 and n2 = 3 for lowest energy. or Z = 3. This corresponds to Li2+ ion. Q. 14. Calculate the wavelength of radiation HFG KJINow, emitted producing a line in Lyman series when an electron falls from fourth energy level in hydrogen = RH 1 − 1 ν n12 n22 atom. (RH = 1.1 × 107 m–1) FHG KJI= 1.1 × 107 Ans. According to Rydberg equation, 1 1 F I1 22 − 32 m–1 HG KJλ 1 1 = RH n12 − n22 For Lyman series, n1 = 1 and n2 = 4 GHF JIK= 1.1 × 107 1 − 1 4 9 ∴ 1 = 1.1 × 107 ⎛1 − 1 ⎞ m−1 = 1.1 × 107 × 5 λ ⎜⎝ 12 42 ⎠⎟ 36 1 = 1.1 × 107 × 15 ∴ λ = 36 λ 16 1.1 × 107 × 5 or λ = 1 × 16 = 9.70 × 10–8 m = 6.55 × 10–7 m 1.1 × 107 × 15 or λ = 97.0 nm. Now, E = hν = hc λ Q.15. Calculate the velocity (cm s–1) of an electron placed in third orbit of the hydrogen atom. Also calcu- = 6.62 × 10−34 × 3.0 × 108 late the number of revolutions per second that this 6.55 × 10−7 electron makes around the nucleus. Ans. Velocity of electron is given as : v = 2πe2 = 3.03 × 10–19 J nh 1 gram atom of hydrogen = 6.02 × 1023 atoms Now, e = 4.8 × 10–10 e.s.u., n = 3, h = 6.63 × 10–27 erg sec ∴ Energy corresponding to 1 gram atom of 2 × 22 × (4.8 × 10−10)2 H = 3.03 × 10–19 × 6.02 × 1023 7 × 3 × 6.63 × 10−27 ∴ v = = 18.25 × 104 J = 7.27 × 107 cm s–1 = 182.5 kJ No. of revolutions per second = Velocity Q.17. What is the distance of separation between Circumference second and third orbits of H-atom ? v Ans. Radius of nth orbit of H-atom is given as : 2πr = rn = 0.529 × n2 Å but r = 4π2me2 r3 = 0.529 × 32 = 0.529 × 9 n2h2 r2 = 0.529 × 22 = 0.529 × 4 ∴ No. of revolutions = 2πe2 × 4π2me2 nh 2π n2h2 ∴ r3 – r2 = 0.529 (9 – 4)Å = 2.645 Å = 4π2me4 Q.18. Calculate the product of uncertainties in n3h3 displacement and velocity for an electron of mass 9.1×10–31 kg according to Heisenberg’s uncertainty principle. (h = 6.6 × 10–34 kg m2 s–1).

2/98 MODERN’S abc + OF CHEMISTRY–XI K.E. of electrons emitted by using λ = 3 × 103Å or 3000Å Ans. According to Heisenberg’s uncertainty principle, Δx × Δp ≥ h (K.E.)1 = hc ⎛ 1 − 1⎞ (... λ0 = 6000 Å) 4π ⎜⎝ 3000 6000⎟⎠ or Δx × m Δ v ≥ h (... m is constant) K.E. of electrons by using wavelength λ (to be calculated) 4π © orModern Publishers. All rights reserved.Δx × Δv ≥h (K.E.)2 = hc ⎛ 1 − 1⎞ 4mπ ⎝⎜ λ 6000⎟⎠ m = 9.1 × 10–31 kg, h = 6.6 × 10–34kg m2 s–1, π = 3.14 Now, (K.E.)2 = 2 (K.E.)1 Δx × Δv ≥ 6.6 × 10−34 kg m2s−1 hc ⎛ 1 − 1⎞ = 2hc ⎛ 1 − 1⎞ 4 × 9.1 × 10−31 kg × 3.14 ⎜⎝ λ 6000 ⎟⎠ ⎝⎜ 3000 6000⎟⎠ ( )∴ or 1 = 2 − 1 + 1 = 1 or Δx × Δv ≥ 6.6 × 10−3 m2s−1 λ 3000 3000 6000 2000 4 × 9.1 × 3.14 ∴ λ = 2000Å or Δx × Δv ≥ 5.77 × 10–5 m2 s–1 Q.21. The angular momentum of an electron in Thus, product of uncertainties in position and velocity Bohr’s orbit of H-atom is 3.02 × 10–34 kg m2 s–1. Calculate of an electron is equal to or greater than 5.77 × 10–5 m2 s–1. the wavelength of the spectral line emitted when the electron jumps from this level to the next lower level. Q.19. Which state of the triple ionized beryllium (Be3+) has the same orbit radius as that of the ground Ans. Angular momentum (mvr) state of hydrogen atom ? = n h = 3.02 × 10−34 kg m2 s−1 Ans. For H-atom, radius of ground state is 2π r1 = h2 ....(i) n = 3.02 × 10−34 × 2π 4π2me2 h For hydrogen like atom, 3.02 × 10−34 × 2 × 3.14 6.3 × 10−34 r′n = n2h2 ....(ii) = =3 4π2mZe2 When the electron jumps from n = 3 to n = 2, the Dividing Eq. (ii) by Eq. (i) wavelength of spectral line, rn′ = n2 1 = ⎛ 1 − 1 ⎞ r1 Z λ 109677 ⎜⎜⎝ n22 n12 ⎠⎟⎟ For Be3+ ion, Z = 4 ∴ rn′ = n2 = 109677 ⎛ 1 − 1 ⎞ r1 4 ⎝⎜ 22 32 ⎠⎟ Now, r′n = ri or n = 2 1 = 109677 × 5 ∴ n2 = 4 λ 36 Thus, the second orbit of Be3+ has the same radius as or λ = 36 the Bohr’s radius of hydrogen atom. 5 × 109677 Q.20. The threshold wavelength for emitting = 6.56 × 10–5 cm or 656 nm. photons from a metal is 6.0 × 103Å. What would be the Q.22. An electron in certain Bohr orbit has velocity wavelength of radiation to produce photoelectrons 1/275 of the velocity of light. Calculate the orbit in which the electron is revolving. having twice the kinetic energy of those produced by radiation of wavelength 3 × 103 Å ? Ans. Velocity of electron Ans. K.E. of photoelectrons emission is : = 1 × 3.0 × 1010 cm s−1 275 K.E. = hν − hν0 = hc − hc λ λ0 = 1.09 × 108 cm s–1 Velocity of electron in nth orbit, or K.E. = ⎛ 1 1 ⎞ hc ⎜ λ − λ0 ⎟ ⎠ 2πZe2 ⎝ vn = nh

STRUCTURE OF ATOM 2/99 Z = 1, e = 4.80 × 10–10 esu, h = 6.63 × 10–27 erg sec ∴ 2.30 × 108 = 1.09 × 108 cm s–1 n 2 × 3.14 × 1 × 4.80 × 10−10 2 n × 6.3 × 10−27 ( ( ) )vn = ∴ n= 2.30 × 108 =2 1.09 × 108 = 2.30 × 108 n © ∴ n=2 Modern Publishers. All rights reserved. BOHR MODEL: RADIUS OF ORBIT, ENERGY AND rn = n2h2 ...(v) VELOCITY OF AN ELECTRON 4π2me2 According to Bohr model, an electron with charge Energy of Electron –e revolves around the nucleus in an orbit of radius r. Let the charge on the nucleus be +Ze. For an electron to remain The total energy, E of electron revolving in the in its orbit, the electrostatic attraction between the electron nth orbit is equal to the sum of the kinetic and the nucleus which tends to pull the electron towards the nucleus must be equal to the centrifugal force which energy ⎛ 1 mv2 ⎞ and the potential energy tends to throw the electron out of its orbit. ⎝⎜ 2 ⎟⎠ Centrifugal force = mv2 ⎛ −Ze2 ⎞ . Thus, Coulombic force of attraction bretween nucleus of charge ⎝⎜⎜ rn ⎠⎟⎟ +Ze and electron of charge –e is E = 1 mv2 – Ze2 ...(vi) 2 rn Coulombic force of attraction = Ze × e = Ze2 r2 r2 From eq. (i) When the two forces balance, mv2 = Ze2 rn mv2 = Ze2 r r2 Subtracting the value of mv2 in eq. (vi) Hence, v2 = Ze2 ...(i) E = Ze2 – Ze2 = − Ze2 ...(vii) mr 2rn rn 2rn Radius of Orbit, get Substituting the value of rn from eq. (v) in eq. (vii), we According to Bohr’s model of an atom, angular E = – Ze2 × 4π2Zme2 momentum is integral multiple of h/2π, 2 n2h2 mvr = nh Thus, the energy of electron in nth orbit, En may be 2π given as or v = nh ...(ii) En = – 2π2me4Z2 ...(viii) 2πmr ...(iii) n2h2 or v2 = n2h2 This is the Bohr equation of energy of electron in the From eq. (i) and (iii) 4π2m2r2 nth orbit for hydrogen and hydrogen like ions (such as Ze2 = n2h2 He+, Li2+, etc.) mr 4π2m2r2 Velocity of Electron in an Orbit or Ze2 = n2h2 Substituting the value of r from eq. (iv) in eq. (ii), 4π2mr we get or r = n2h2 ...(iv) v = nh × 4π2me2Z 4π2me2Z 2πm n2h2 Since for hydrogen atom, Z = 1, hence the radius of nth or v = 2πe2Z ...(ix) orbit of H-atom is nh

2/100 MODERN’S abc + OF CHEMISTRY–XI © TopicwiseMULTIPLE CHOICE QUESTIONS A10. The ratio of the difference in energy between the first Modern Publishers. All rights reserved. and the second Bohr orbit to that between second and Select the Correct Answers : third Bohr orbit is (a) 1 (b) 1 2 3 Bohr’s Model and its Limitations (c) 27 (d) 4 5 9 A1. In the Bohr’s orbit, the ratio of total kinetic energy and A11. Which of the following transitions will have minimum the total energy of the electron is wavelength ? (a) –2 (b) –1 (a) n2 ⎯→ n1 (b) n3 ⎯→ n1 (c) +2 (d) 0 (c) n4 ⎯→ n2 (d) n4 ⎯→ n1 A2. The spectral line in hydrogen spectrum obtained when A12. In hydrogen atom, energy of the first excited state is the electron jumps from n = 5 to n = 2 energy level belongs to : –3.04 eV. Then find out the K.E. of the same orbit of (a) Lyman series (b) Balmer series H-atom. (c) Paschen series (d) Pfund series. (a) + 3.4 eV (b) + 6.8 eV A3. The energy of the electron in the nth orbit of hydrogen (c) – 13.6 eV (d) + 13.6 eV atom is given as : A13. The ratio of radii of the first three Bohr orbits of H-atom 1311.8 is n2 En = – kJ mol–1 (a) 1 : 2 : 3 (b) 1 : 4 : 9 What is the energy emitted per atom when an electron (c) 1 : 3 : 27 (d) 1 : 2 : 3 jumps from third energy level to second energy level ? A14. The ratio of the frequency corresponding to the third line (a) 329·7 kJ (b) 3·03 × 10–19 J in Lyman series of hydrogen atomic spectrum to that of the first line in Balmer series of Li2+ spectrum is (c) 182·2 kJ (d) 145·7 kJ. 4 (b) 5 A4. The transition in He+ spectrum from n = 4 to n = 2 (a) 5 4 corresponds to which transition in the H atom spectrum ? 4 (d) 3 (a) n = 2 to n = 1 (b) n = 3 to n = 1 (c) 3 4 (c) n = 3 to n = 2 (d) n = 4 to n = 2. A15. TprhordeueceenderbgyytlreavneslistEio1n,sEa2raensdhEo3wannbdetlhoeww: ave lengths A5. The ionization energy of H atom is 13.6 eV. The ioniza- E3 tion energy of Li2+ ion will be : E2 E1 (a) 54.4 eV (b) 40.8 eV λ1 λ2 (c) 27.2 eV (d) 122.4 eV. λ3 A6. The radius of first orbit of hydrogen is 0.53 Å. The Which one of the following relationship is correct? radius of second orbit would be : (a) λ3 = λ1 + λ2 (b) λ1 + λ2 + λ3 = 0 (a) 1.06 Å (b) 0.26 Å (c) 0.53 Å (d) 2.12 Å. (c) λ3 = λ1λ2 (d) λ3 = λ1 + λ2 λ1 + λ2 λ1λ2 A7. How many spectral lines are produced in the spectrum of hydrogen atom from 5th energy level ? Wave Mechanical Model of Atom and Quantum Numbers (a) 5 (b) 10 (c) 15 (d) 4 A8. In Bohr’s theory, the radius, r of orbit is proportional to (a) n (b) n2 A16. The de-Broglie wavelength of an electron is 600 nm. (c) n–1 (d) n–2 The velocity of the electron is : A9. What transition in He+ ion shall have the same wave (h = 6.6 × 10–34J sec, m = 9.0 × 10–31 kg) number as the first line in Balmer series of H atom ? (a) 1.8 × 103 m s–1 (b) 1.2 × 105 m s–1 (a) 7 ⎯⎯→ 5 (b) 5 ⎯⎯→ 3 (c) 5.4 × 103 m s–1 (d) 1.2 × 103 m s–1 (c) 6 ⎯⎯→ 4 (d) 4 ⎯⎯→ 2 A17. If travelling at same speeds, which of the following matter waves have the shortest wavelength ? (a) Electron (b) Alpha particle (He2+) (c) Neutron (d) Proton A1. (b) A2. (b) A3. (b) A4. (a) A5. (d) A6. (d) A7. (b) A8. (b) A9. (c) A10. (c) A11. (d) A12. (a) A13. (b) A14. (d) A15. (c) A16. (d) A17. (b)

STRUCTURE OF ATOM 2/101 A18. If uncertainty in the position of an electron is zero, the A29. The orbital angular momentum for an electron revolving uncertainty in its momentum would be in an orbit for an s-electron is ©(a) zero (b) ≥ h (a) zero (b) h Modern Publishers. All rights reserved. 2π 4π (c) 2 h (d) 1 . h (c) < h (d) infinite 2π 2 2π 4π A19. Which of the following orbitals does not make sense ? (a) 6s (b) 3p A30. The following quantum numbers are possible for how many orbitals ? (c) 2d (d) 4f. n = 3, l = 2, ml = +2 A20. Two electrons occupying the same orbital are (a) 1 (b) 2 distinguished by : (a) spin quantum number (c) 3 (d) 4 (b) azimuthal quantum number (c) magnetic quantum number Shapes of Orbitals and Electronic Configurations (d) principal quantum number. A31. The correct ground state electronic configuration of chromium atom (Z = 24) is : A21. The designation of an orbital with n = 4 and l = 3 is : (a) 4s (b) 4p (a) [Ar] 3d5 4s1 (b) [Ar] 3d4 4s2 (c) 4d (d) 4f. (c) [Ar] 3d6 4s0 (d) [Ar] 4s1 4p5. A22. For each value of l, the number of ms values are : A32. In manganese atom, Mn (Z = 25), the total number of (a) 2l (b) nl orbitals populated by one or more electrons (in ground state) is : (c) 2l + 1 (d) n – l. A23. The maximum number of 4d-electrons having spin (a) 15 (b) 14 quantum number s = + 1 are : (c) 12 (d) 10. 2 (b) 7 (a) 10 A33. Which of the following has maximum number of unpaired electrons ? (c) 1 (d) 5. A24. The maximum number of electrons in a subshell is (a) Mg2+ (b) Ti3+ given by the expression : (a) 4l – 2 (b) 4l + 2 (c) V3+ (d) Fe3+. (c) 2l + 1 (d) 2n2. A34. Azimuthal quantum number for the last electron in Na atom is : A25. A subshell with n = 6, l = 2 can accommodate a maximum of : (a) 1 (b) 0 (a) 10 electrons (b) 12 electrons (c) 2 (d) 3. (c) 36 electrons (d) 72 electrons. A35. Orbital angular momentum depends on A26. For which one of the following sets of four quantum (a) l (b) n and l numbers an electron will have the highest energy : (c) n and m (d) m and s nl ms A36. How many electrons in Argon have m = 0 ? (a) 3 2 1 1/2 (a) 12 (b) 10 (b) 4 1 0 –1/2 (c) 8 (d) 6. (c) 4 2 –1 1/2 A37. Magnetic quantum number for the valence electron of potassium is : (d) 5 0 0 –1/2 A27. An orbital with the quantum numbers n = 4, l = 3, (a) 0 (b) 1 ml = 0 and ms = – 1 is (c) 2 (d) 7. (a) 3s 2 (b) 3p A38. Consider the following ions : (c) 4d (d) 4f 1. Ni2+ 2. Co2+ 3. Cr2+ 4. Fe3+ A28. Which of the following orbital designations is not (Atomic number : Cr = 24, Fe = 26, Co = 27 and correct corresponding to quantum numbers ? Ni = 28) (a) n = 5, l = 2 ⎯⎯→ 5d The correct sequence of increasing number of unpaired electrons in these ions is (b) n = 2, l = 0 ⎯⎯→ 2s (c) n = 4, l = 3 ⎯⎯→ 4f (a) 1, 2, 3, 4 (b) 4, 2, 3, 1. (d) n = 7, l = 2 ⎯⎯→ 7p (c) 1, 3, 2, 4 (d) 3, 4, 2, 1 A18. (d) A19. (c) A20. (a) A21. (d) A22. (c) A23. (d) A24. (b) A25. (a) A26. (c) A27. (d) A28. (d) A29. (a) A30. (a) A31. (a) A32. (a) A33. (d) A34. (b) A35. (a) A36. (b) A37. (a) A38. (a)

2/102 MODERN’S abc + OF CHEMISTRY–XI A39. If the value of (n + l) is more than 3 and less than 6, A43. The number of radial and angular nodes in 3p orbital what will be the possible number of orbitals are © (a) 6 (b) 9 (a) 1, 0 (b) 2, 1 Modern Publishers. All rights reserved. (c) 10 (d) 13 (c) 1, 1 (d) 2, 0 A40. In an atom, the signs of lobes indicate the A44. Consider the ground state of Cr atom (Z = 24). The number of electrons with azimuthal quantum numbers (a) sign of charges l = 1 and 2 are respectively (b) sign of probability distribution (a) 16 and 4 (b) 12 and 5 (c) sign of wave function (c) 12 and 4 (d) 16 and 5 (d) presence or absence of electron A45. Which of the following 3d-orbital has electron density in all the three axes? A41. The number of radial nodes possible for 3d orbital is (a) 3 (b) 1 (a) 3dxy (b) 3dx2 − y2 (c) 2 (d) 0 (c) 3dz2 (d) 3dyz A42. The radial part of wave function depends on the A46. How many spherical nodes are present in 4s orbital in quantum numbers a hydrogen atom? (a) n, l (b) n only (a) 0 (b) 2 (c) l, ml (d) l only (c) 3 (d) 4 A39. (d) A40. (c) A41. (d) A42. (a) A43. (c) A44. (b) A45. (c) A46. (c) B MULTIPLE CHOICE QUESTIONS (a) 1 h (b) h 2m π 2π from competitive examinations (c) 1 h (d) h AIPMT, NEET & Other State Boards’ mπ π Medical Entrance (C.B.S.E. P.M.T. 2008) B1. Consider the following sets of quantum numbers : B4. The measurement of the electron position is associated nl m s with an uncertainty in momentum which is equal to (A) 3 0 0 +1/2 1 × 10–18 g cm s–1. The uncertainty in velocity of electron (B) 2 2 1 +1/2 (mass of electron is 9 × 10–28g) is (C) 4 3 –2 –1/2 (a) 1 × 109 cm s–1 (b) 1 × 106 cm s–1 (D) 1 0 –1 –1/2 (c) 1 × 105 cm s–1 (d) 1 × 1011 cm s–1 (C.B.S.E. P.M.T. 2008) (E) 3 2 3 +1/2 B5. The velocity of particle A is 0.1 m/s and that of particle Which of the following sets of quantum numbers is not possible ? B is 0.05 m/s. If the mass of particle B is five times that (a) A, B, C and D (b) B, D and E of particle A, then the ratio of de-Broglie wavelength (c) A and C (d) B, C and D associated with particles A and B is (C.B.S.E. Med. 2007) (a) 2 : 5 (b) 3 : 4 B2. The maximum kinetic energy of photoelectrons ejected (c) 6 : 4 (d) 4 : 3 from a metal, when it is irradiated with radiation of frequency 2 × 1014 s–1 is 6.63 × 10–20 J. The threshold (e) 5 : 2 (Kerala PMT 2009) frequency of the metal is B6. Maximum number of electrons in a subshell of an atom is determined by the following : (a) 2 × 1014 s–1 (b) 3 × 1014s–1 (a) 2l + 1 (b) 4l – 2 (c) 2 × 10–14 s–1 (d) 1 × 10–14 s–1 (c) 2n2 (d) 4l + 2 (C.B.S.E. PMT 2009) (e) 1 × 1014 s–1 (Kerala P.M.T. 2008) B7. Which of the following is not permissible arrangement of electrons in an atom ? B3. If uncertainty in position and momentum are equal, then uncertainty in velocity is (a) n = 5 l = 3 m = 0 s = +1/2 (b) n = 3 l = 2 m = –3 s = –1/2 (c) n = 3 l = 2 m = –2 s = –1/2 (d) n = 4 l = 0 m = 0 s = –1/2 (C.B.S.E. PMT 2009) B1. (b) B2. (e) B3. (a) B4. (a) B5. (e) B6. (d) B7. (b)

STRUCTURE OF ATOM 2/103 B8. Which one of the following ion has electronic configura- (d) Larger the value of n, the larger is the orbit radius. tion [Ar] 3d6 ? (NEET 2013) (a) Fe3+ (b) Co3+ © (c) Ni3+ (d) Mn3+ B16. The value of Planck's constant is 6.63 × 10–34 J s. The Modern Publishers. All rights reserved.(C.B.S.E. Med. 2010) speed of light is 3 × 1017 nm s–1. Which value is closest to the wavelength in nanometer of a quantum of light B9. The total number of atomic orbitals in fourth energy with frequency of 6 × 1015 s–1 ? level of an atom is (a) 50 (b) 75 (a) 8 (b) 16 (c) 10 (d) 25 (NEET 2013) (c) 32 (d) 4 (AIPMT 2011) B10. The energies E1 and E2 of two radiations are 25 eV and B17. The ratio of de Broglie wavelengths of a deuterium 50 eV respectively. The relation between their atom to that of an α-particle, when the velocity of the former is five times greater than that of the latter, is wavelengths i.e., λ1 and λ2 will be (a) λ1 = λ2 (b) λ1 = 2λ2 (a) 4 (b) 0.2 (c) λ1 = 4λ2 (d) λ1 = 1 (AIPMT 2011) (c) 2 (d) 0.4 2 λ2 (e) 5 (Kerala PMT 2014) B11. If n = 6, the correct sequence for filling of electrons will be B18. The uncertainty in the velocity of a particle of mass 6.626 × 10–31 kg is 1 × 106 m s–1. What is the uncertainty (a) ns → (n – 2)f → (n – 1)d → np in its position (in nm)? (h = 6.626 × 10–34 J s) (b) ns → (n – 1)d → (n – 2)f → np ⎛ 1⎞ ⎛ 2.5⎞ ⎝⎜ 2π⎠⎟ ⎜⎝ π ⎠⎟ (c) ns → (n – 2)f → np → (n – 1)d (a) (b) (d) ns → np(n – 1)d → (n – 2)f (AIPMT 2011) ⎛ 4⎞ ⎛ 1⎞ ⎝⎜ π⎠⎟ ⎝⎜ 4π⎠⎟ B12. Maximum number of electrons in a subshell with (c) (d) = 3 and n = 4 is (a) 14 (b) 16 (c) 10 (d) 12 (e) ⎛ 5⎞ (Kerala PMT 2014) ⎝⎜ π⎠⎟ (A.I.P.M.T. 2012) B19. What is the maximum number of orbitals that can be B13. The correct set of four quantum numbers for the identified with the following quantum numbers? valence electron of rubidium atom (Z = 37) is n = 3, l = 1, ml = 0 (a) 5, 1, 1 + 1/2 (b) 6, 0, 0, + 1/2 (a) 1 (b) 2 (c) 5, 0, 0, + 1/2 (d) 5, 1, 0, + 1/2 (c) 3 (d) 4 (AIPMT 2014) (A.I.P.M.T. 2012) B20. Calculate the energy in joule corresponding to light of B14. What is the maximum number of electrons that can wavelength 45 nm. be associated with the following set of quantum numbers ? (Planck’s constant, h = 6.63 × 10–34 J s, speed of light, c = 3 × 108 m s–1) n = 3, l = 1 and m = – 1 (a) 6.67 × 1015 (b) 6.67 × 1011 (a) 4 (b) 2 (c) 4.42 × 10–15 (d) 4.42 × 10–18 (c) 10 (d) 6 (NEET 2013) (AIPMT 2014) B15. Based on equation E = – 2.178 × 10–18 J ⎛ Z2 ⎞ , certain B21. Which one of the following sets of quantum numbers conclusions are written. Which of them i⎜⎜⎝s nn2ot⎠⎟⎟ correct ? is possible? (a) Equation can be used to calculate the change in (a) n = 3, l = 3, ml = −3, ms = +1/2 energy when the electron changes orbit. (b) n = 2, l = 1, ml = 2, ms = −1/2 (c) n = 2, l = 0, ml = 0, ms = +1/2 (b) For n = 1, the electron has a more negative energy (d) n = 1, l = 0, ml = 0, ms = 0 than it does for n = 6 which means that the electron (e) n = 3, l = 2, ml = 3, ms = −1/2 is more loosely bound in the smallest allowed orbit. (Kerala PMT 2015) (c) The negative sign in equation simply means that B22. The angular momentum of electron in 'd' orbital is the energy of electron bound to the nucleus is lower equal to ? than it would be if the electrons were at the infinite distance from the nucleus. (a) 2 3 (b) 0 (c) 6 (d) 2 (AIPMT 2015) B8. (b) B9. (b) B10. (b) B11. (a) B12. (a) B13. (c) B14. (b) B15. (b) B16. (a) B17. (d) B18. (d) B19. (a) B20. (d) B21. (c) B22. (c)

2/104 MODERN’S abc + OF CHEMISTRY–XI B23. Which is the correct order of increasing energy of the B30. 4d, 5p, 5f and 6p orbitals are arranged in the order of listed orbitals in the atom of titanium ? decreasing energy. The correct option is (At. no. = 22) © (a) 5f > 6p > 4d > 5p (b) 5f > 6p > 5p > 4d Modern Publishers. All rights reserved.(a) 3s 3p 3d 4s(b) 3s 3p 4s 3d (c) 6p > 5f > 5p > 4d (d) 6p > 5f > 4d > 5p (c) 3s 4s 3p 3d (d) 4s 3s 3p 3d (NEET 2019) (AIPMT 2015) B31. Which of the following series of transitions in the B24. The number of d–electrons in Fe2+ (Z = 26) is not spectrum of hydrogen atom falls in visible region? equal to the number of electrons in which one of the following? (a) Brackett series (b) Lyman series (c) Balmer series (d) Paschen series (a) d-electrons in Fe (Z = 26) (NEET 2019) (b) p-electrons in Ne (Z = 10) (c) s-electrons in Mg (Z = 12) JEE (Main) & Other State Boards’ Engineering Entrance (d) p-electrons in Cl (Z = 17) (AIPMT 2015) B25. The electrons occupying the same orbital are B32. Which of the following sets of quantum numbers distinguished by represents the highest energy of an atom ? (a) azimuthal quantum number (b) spin quantum number (a) n = 3, l = 2 , m = 1, s = +1/2 (b) n = 4, l = 0 , m = 0, s = +1/2 (c) principal quantum number (c) n = 3, l = 0 , m = 0, s = +1/2 (d) n = 3, l = 1 , m = 1, s = +1/2 (d) magnetic quantum number (NEET 2016) B26. How many electrons can fit in the orbital for which (A.I.E.E.E. 2007) n = 3 and l = 1? B33. The ionization enthalpy of hydrogen atom is 1.312 × 106 (a) 2 (b) 6 J mol–1. The energy required to excite the electron in the atom from n = 1 to n = 2 is (c) 10 (d) 14 (NEET 2016) B27. Which of the following pairs of d-orbitals will have (a) 9.84 × 105 J mol–1 (b) 8.51 × 105 J mol–1 electron density along the axes? (c) 6.56 × 105 J mol–1 (d) 7.56 × 105 J mol–1 (a) dz2 , dxz (b) dxz, dyz (A.I.E.E.E. 2008) (c) dz2 , dx2 − y2 (d) dzy, dx2 − y2 (NEET 2016) B34. In an atom, an electron is moving with a speed of 600 B28. Which one is the wrong statement? m/s with an accuracy of 0.005%. Certainty with which (a) The uncertainty principle is Δx × Δv ≥ h the position of an electron can be located is (h = 6.6 × 4π 10–34 kg m2 s–1; mass of electron, em = 9.1 × 10–31 kg) (b) Half-filled and fully-filled orbitals have greater (a) 1.52 × 10–4 m (b) 5.10 × 10–3 m stability due to greater exchange energy, greater symmetry and more balanced arrangement. (c) 1.92 × 10–3 m (d) 3.84 × 10–3 m (c) The energy of 2s-orbital is less than the energy of (A.I.E.E.E. 2009) 2p-orbital in case of hydrogen like atoms. B35. Calculate the wavelength associated with a proton (d) de-Broglie's wavelength is given by λ = h , where moving at 1.0 × 103 m/s. (Mass of proton = 1.67 × 10–27 kg mv and h = 6.63 × 10–34 Js) m = mass of the particle, v = velocity of the particle (a) 0.032 nm (b) 0.40 nm (c) 2.5 nm (d) 14.0 nm (NEET 2017) (A.I.E.E.E. 2009) B29. Which one is a wrong statement? B36. What is the maximum number of emission lines (a) Total orbital angular momentum of electron in obtained when the excited electrons of a hydrogen atom s-orbital is equal to zero. in n = 5 drop to ground state ? (b) An orbital is designated by three quantum numbers (a) 10 (b) 5 while an electron in an atom is designated by four (c) 12 (d) 15 (J.K. CET 2010) quantum numbers. B37. The wave number of the spectral line in the emission (c) The electronic configuration of N atom is 1s2 2s2 2p1x 2p1y 2p1z spectrum of hydrogen will be equal to 8 times the Rydberg’s constant if the electron jumps f9rom ↑↑ ↓ (a) n = 3 to n = 1 (b) n = 10 to n = 1 (d) The value of m for dz2 is zero. (NEET 2018) (c) n = 9 to n = 1 (d) n = 2 to n = 1 (Karnataka CET 2010) B23. (a) B24. (d) B25. (b) B26. (a) B27. (c) B28. (c) B29. (c) B30. (b) B31. (c) B32. (a) B33. (a) B34. (c) B35. (b) B36. (a) B37. (a)

STRUCTURE OF ATOM 2/105 B38. Ionisation energy of He+ is 19.6 × 10–18 J atom–1. The B44. The electronic configuration of Cu2+ ion is energy of first stationary state (n =1) of Li2+ is (a) [Ar] 3d84s1 (b) [Ar] 3d9 4s0 (a) – 2.2. × 10–15 J atom–1 © (c) [Ar] 3d74s2 (d) [Ar] 3d8 4s0 Modern Publishers. All rights reserved.(b) 8.82 × 10–17 J atom–1 (Karnataka CET 2013) (c) 4.41 × 10–16 J atom–1 (d) – 4.41 × 10–17 J atom–1 (A.I.E.E.E. 2010) B45. The ionization enthalpy of He+ ion is 19.60 × 10–18 J atom–1. The ionization enthalpy of Li2+ ion will be B39. Which transition in the hydrogen atomic spectrum will have the same wavelength as the transition, n = 4 to (a) 84.2 × 10–18 J atom–1 n = 2 of He+ spectrum ? (b) 44.10 × 10–18 J atom–1 (a) n = 4 to n = 3 (b) n = 3 to n = 2 (c) n = 4 to n = 2 (d) n = 3 to n = 1 (c) 63.20 × 10–18 J atom–1 (e) n = 2 to n = 1 (Kerala PET 2011) (d) 21.20 × 10–18 J atom–1 B40. For Balmer series in the spectrum of atomic hydrogen, (e) 2.17 × 10–19 J atom–1 (Kerala P.E.T. 2013) the wave number of each lines is given by ⎛ 1 1 ⎞ B46. Energy of an electron is given by ⎜⎜⎝ n12 n22 ⎠⎟⎟ ν = RH − where RH is a constant and n1 and 1e0le–1c8tJro⎜⎛⎝⎜nZni22n⎞⎟⎟⎠ (na2raer)ecoirnrteecgte?rs. Which of the following statement(s) is E = – 2.178 × . Wavelength of light required to excite an a hydrogen atom from level 1. As wavelength decreases, the lines in the series n = 1 to n = 2 will be (h = 6.62 × 10–34 J s and c = 3.0 × 108 converge. m s–1) 2. The integer n1 is equal to 2 (a) 8.500 × 10–7 m (b) 1.214 × 10–7 m 3. The ionization energy of hydrogen can be calculated from the wave number of these lines (c) 2.816 × 10–7 m (d) 6.500 × 10–7 m 4. The line of longest wavelength corresponds to (JEE Main 2013) n2 = 3. (a) 1, 2 and 3 (b) 2, 3 and 4 B47. As per de Broglie’s formula, a macroscopic particle of mass 100 g and moving at a velocity of 100 cm s–1 will (c) 1, 2 and 4 (d) 2 and 4 only have a wavelength of (e) 2 only (Kerala PET 2011) (a) 6.6 × 10–29 cm (b) 6.6 × 10–30 cm B41. The frequency of light emitted for the transition n = 4 (c) 6.6 × 10–31 cm (d) 6.6 × 10–32 cm to n = 2 of He+ is equal to the transition in H-atom (WB JEE 2014) corresponding to which of following ? (a) n = 2 to n = 1 (b) n = 3 to n = 2 B48. The shortest wavelength of the line in hydrogen atomic (c) n = 4 to n = 3 (d) n = 3 to n = 1 spectrum of Lyman series when RH = 109678 cm–1 is (AIEEE 2011) (a) 1002.7 Å (b) 1215.67 Å B42. The ratio of the frequency corresponding to the third (c) 1127.30 Å (d) 911.7 Å line in Lyman series of hydrogen atomic spectrum to that of the first line in Balmer series of Li2+ spectrum is (e) 1234.7 Å (Kerala PET 2014) (a) 4 (b) 5 B49. The correct set of four quantum numbers for the valence 5 4 electrons of rubidium atom (Z = 37) is (c) 4 (d) 3 (a) 5, 0, 1, + 1 (b) 5, 0, 0, + 1 3 4 2 2 (e) 3 (Kerala P.E.T. 2012) (c) 5, 1, 0, + 1 (d) 5, 1, 1, + 1 8 2 2 B43. The electrons identified by quantum numbers n and : (JEE Main 2014) (1) n = 4, = 1 (2) n = 4, = 0 B50. Which of the following is the energy of a possible excited state of hydrogen? (3) n = 3, = 2 (4) n = 3, = 1 can be placed in order of increasing energy as (a) −3.4 eV (b) + 6.8 eV (a) (4) < (2) < (3) < (1) (b) (2) < (4) < (1) < (3) (c) + 13.6 eV (d) −6.8 eV (c) (1) < (3) < (2) < (4) (d) (3) < (4) < (2) < (1) (JEE Main 2015) (A.I. E.E.E. 2012) B38. (d) B39. (e) B40. (c) B41. (a) B42. (d) B43. (a) B44. (b) B45. (b) B46. (b) B47. (c) B48. (d) B49. (b) B50. (a)

2/106 MODERN’S abc + OF CHEMISTRY–XI B51. Consider the following sets of quantum numbers. (Planck’s constant, h = 6.6262 × 10–34Js; mass of Which of the below setting is not permissible electron = 9.1091 × 10–31 kg; charge of electron arrangement of electrons in an atom? e = 1.60210 × 10–19 C; permittivity of vaccum nlm s © ∈0 = 8.854185 × 10–12 kg–1 m–3 A2) Modern Publishers. All rights reserved. (a) 4 0 0 −1 (a) 1.65 Å (b) 4.76 Å 2 (c) 0.529 Å (d) 2.12 Å (b) 5 3 0 +1 (JEE Main 2017) 2 B58. Which of the following set of quantum numbers is not possible? (c) 3 2 –2 −1 2 (a) n = 3, l = 0, m = 0 (b) n = 3, l = 1, m = –1 (d) 3 2 –3 +1 (Kerala PET 2016) (c) n = 2, l = 0, m = –1 (d) n = 2, l = 1, m = 0 2 (WB JEE 2018) B59. The number of unpaired electrons in Ni (atomic B52. A stream of electrons from a heated filament was number = 28) are passed between two charged plates kept at a potential (a) 0 (b) 2 difference V e.s.u. If e and m are charge and mass of an (c) 4 (d) 8 (WB JEE 2018) electron respectively, then the value of h/λ (where λ is B60. With respect to atomic spectrum, each line in the wavelength associated with electron wave) is given by Lyman series is due to electrons returning (a) meV (b) 2meV (a) from a particular higher energy level to n = 3 (c) meV (d) 2meV (b) from a particular higher energy level to n = 2 (JEE Main 2016) (c) from a particular higher energy level to n = 1 B53. If the given four electronic configurations (d) from a particular higher energy level to n = 4 (a) n = 4, l = 1 (b) n = 4, l = 0 (J.K. CET 2018) (c) n = 3, l = 2 (d) n = 3, l = 1 B61. The orbital nearest to the nucleus is are arranged in order of increasing energy, then the order will be (a) 4f (b) 5d (c) 4s (d) 7p (a) (iv) < (ii) < (iii) < (i) (Karnataka CET 2018) (b) (ii) < (iv) < (i) < (iii) B62. What is the work function of the metal if the light of wavelength 4000 Å generates photoelectrons of (c) (i) < (iii) < (ii) < (iv) velocity 6 × 105 ms–1 from it? (d) (iii) < (i) < (iv) < (ii) (WB JEE 2017) B54. Which of the following set of quantum numbers (Mass of electron = 9 × 10–31 kg represents the 19th electron of Cr (Z = 24)? Velocity of light = 3 × 108 ms–1 Planck’s constant = 6.626 × 10–34 Js (a) 1 (b) 1 Charge of electron = 1.6 × 10–19 JeV–1) (4, 1, –1, + 2 ) (4, 0, 0, + 2 ) (a) 4.0 eV (b) 2.1 eV 1 1 (c) 3.1 eV (d) 0.9 eV (c) (3, 2, 0, – 2 ) (d) (3, 2, – 2, + 2 ) (WB JEE 2017) (JEE Main 2018) B55. The correct set of quantum numbers for the unpaired B63. Two particles A and B are in motion. If the wavelength electron of chlorine atom is associated with ‘A’ is 33.33 nm, the wavelength associated with ‘B’ whose momentum is 1/3rd of ‘A’ is 1 1 (a) 1.0 × 10–8 m (b) 2.5 × 10–8 m (a) 2, 0, 0, + 2 (b) 3, 0, 0, + 2 (c) 1.25 × 10–7 m (d) 1.0 × 10–7 m (c) 1 (d) 1 (Karnataka CET 2019) 2, 1, –1, + 2 3, 1, 1, + 2 B64. Maximum number of photons emitted by a bulb (Karnataka CET 2017) capable of producing monochromatic light of B56. The energy of an electron in the 3s orbital (excited state) of H-atom is wavelength 550 nm is _______, if 100 V and 1A is supplied for one hour. (a) – 1.5 eV (b) – 13.6 eV (a) 1 × 1024 (b) 5 × 1024 (c) – 3.4 eV (d) – 4.53 eV (c) 1 × 1023 (d) 5 × 1023 (e) 4.53 eV (Kerala PET 2017) (e) 5 × 1022 (Kerala PET 2019) B57. The radius of the second Bohr orbit for hydrogen atom is : B51. (d) B52. (d) B53. (a) B54. (b) B55. (d) B56. (a) B57. (d) B58. (c) B59. (b) B60. (c) B61. (c) B62. (b) B63. (d) B64. (a)

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2/108 MODERN’S abc + OF CHEMISTRY–XI B76. The number of radial nodes of 3s and 2p-orbitals are thickness dr, at a distance r from the nucleus. The volume of this shell is 4πr2dr. The qualitative sketch respectively : of the dependence of P on r is? (a) 2, 0 (b) 0, 2 (a) (b) (c) (d) (JEE Advance 2016) © (c) 1, 2 (d) 2, 1 Modern Publishers. All rights reserved. (I.I.T. Screening 2005) B77. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom [a0 is Bohr radius] (a) h2 h2 4π2ma02 (b) 16π2ma02 h2 h2 (c) 32π2ma02 (d) 64π2ma02 (I.I.T. J.E.E. 2012) B78. P is the probability of finding the 1s electron of hydrogen atom in a spherical shell of infinitesimal B76. (a) B77. (c) B78. (c) C MULTIPLE CHOICE QUESTIONS (c) Manganese (Z = 25) (d) Chromium (Z = 24) with more than one correct answer C6. In which of the following the first orbital has higher energy than the second in H-atom ? C1. Which of the following statements are correct ? (a) n = 4, l = 3 and n = 5, l = 0 (a) Radial wave function gives the shape of the orbital (b) n = 3, l = 2 and n = 3, l = 1 (b) Radial wave function depends upon n and l. (c) n = 3, l = 1 and n = 3, l = 2 (c) 3p orbital has two spherical nodes (d) n = 3, l = 2 and n = 2, l = 1 (J.K. C.E.T. 2006) (d) 3d orbital has no spherical node C7. Identify the correct statement(s). C2. Which of the following are correct notations for the The findings from the Bohr model for H-atom are orbitals ? (a) n = 3, l = 2 ⎯→ 3d (a) angular momentum of the electron is expressed as (b) n = 4, l = 1 ⎯⎯→ 4p integral multiples of h 2π (c) n = 4, l = 3 ⎯→ 4d (b) the first Bohr radius is 0.529 Å (d) n = 6, l = 1 ⎯⎯→ 6f (c) the energy of the nth level, En is proportional to 1 n2 C3. The orbitals having the same number of spherical nodes are : (d) the spacing between adjacent levels increases with (a) 4p, 3d (b) 3p, 4d increase in 'n' (WB JEE 2017) (c) 4s, 5p (d) 2s, 3d C8. The ground state energy of hydrogen atom is –13.6 eV. Consider an electronic state Ψ of He+ whose energy, C4. Which of the following set of ions have the same number azimuthal quantum number and magnetic quantum number are –3.4 eV, 2 and 0, respectively. Which of the of electrons ? following statement(s) is (are) true for the state Ψ? (a) Cr3+, Fe2+ (b) Cu+, Zn2+ (a) It is a 4d state. (c) Mn2+, Fe3+ (d) Sc3+, V3+ (b) The nuclear charge experienced by the electron in this state is less than 2e, where e is the magnitude C5. Which of the following atoms have same number of of the electronic charge. unpaired electrons ? (c) It has 3 radial nodes. (a) Copper (Z = 29) (d) It has 2 angular nodes. (JEE Advanced 2019) (b) Scandium (Z = 21) C1. (b, d) C2. (a, b) C3. (b, c) C4. (b, c) C5. (a, b) C6. (a, d) C7. (a, b, c) C8. (a, d)

STRUCTURE OF ATOM 2/109 )602 D5. The wavelength of particles constituting a beam of helium atoms moving with a velocity of 2.0 × 104 m s–1 is © Modern Publishers. All rights reserved.Passage I. (a) 4.99 pm (b) 49.9 pm In 1924, de-Broglie proposed that every particle (c) 499 nm (d) 499 pm possesses wave properties with a wavelength, λ given by Passage II. λ = h where m is the mass of the particle, v is its mv The position and energy of an electron is specified with the help of four quantum numbers namely, principal velocity and h is Planck’s constant. The de-Broglie quantum number (n), azimuthal quantum number (l), prediction was confirmed experimentally when it was magnetic qu antum number v(amlul)esaonfdthsepsien quantum found that an electron beam undergoes diffraction, a number (ms). The permissible are : phenomenon characteristic of waves. The de-Broglie wavelength can be estimated by measuring kinetic energy n = 1, 2 of an electron accelerating by a potential V as : l = 0, 1, ......(n – 1) 1 mv2 = eV where 1 eV = 1.6 × 10–19C, h = 6.6 × 10–34 J. 2 ml = – l....... 0, ..... + l ms = + 1 and – 1 for each value of ml. 2 2 Answer the following questions : The angular momentum of electron is given as D1. The mass of a photon moving with velocity of light l(l + 1). h having wavelength same as that of an α-particle (mass = 6.6 × 10–27 kg) moving with velocity of 2π 2.5 × 102 m s–1 is While spin angular momentum is given as s(s + 1) . h (a) 7.92 × 10–21 kg where s = 1 2π (b) 5.5 × 10–33 kg 2 (c) 5.65 × 10–31 kg The electrons having the same value of n, l and Pmalualri’es said to belong to the same orbital. According to (d) 7.92 × 10–28 kg exclusion principle, an orbital can have maximum of two electrons and these two must have opposite spin. D2. The wavelength of matter wave associated with an electron passing through an electric potential of Answer the following questions : 100 million volts is D6. For an electron having n = 3 and l = 0, the orbital (a) 104 (2 me)1/2h (b) h×10−4 angular momentum is (2 me)1/ 2 (a) 3 h (b) 6 h 10−3 h h × 104 π 2π (2 me)1/ 2 (2 me)−1/ 2 (c) (d) (c) zero (d) 2 3 h π D3. The proton and He2+ are accelerated by the same D7. The maximum number of electrons having n + l = 5 in potential, then their de-Broglie wavelengths λHe2+ an atom is and λp are in the ratio of (mHe2+ = 4mp) : (a) 32 (b) 18 (a) 1 (b) 1 2 22 (c) 10 (d) 8 (c) 2 2 (d) 1 D8. Which of the following statements is not correct ? 2 (a) For sodium, the outermost electron has n = 3, D4. If λ is the wavelength associated with the electron in l = 0, ml = 0, s = + 1/2 the 4th circular orbit of hydrogen atom, then (b) The orbitals having n = s3a,ml e=e2n,emrglie=s.+ 2 and radius of the orbit is n = 3, l = 2, ml = –2 have (a) λ (b) λ (c) For 4f electron, n = 4, l = 3, ml = 0, s = + 1/2 is not 2π π possible. (c) 2 (d) 2π (d) The orbitals 2d, 3f and 4g are not possible. πλ λ Passage I. D1. (b) D2. (b) D3. (d) D4. (c) D5. (a) Passage II. D6. (c) D7. (b) D8. (c)

2/110 MODERN’S abc + OF CHEMISTRY–XI Passage III. D10. Energy sotfattheeesnteartgeySi1s in units of the hydrogen atom ground The hydrogen like species Li2+ is in a spherically symmetrical state S1 with one radial node. Upon absorbing light the ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state of the hydrogen atom. © (a) 0.75 Modern Publishers. Ψn,All rl,mlights reserved. (b) 1.50 (c) 2.25 (d) 4.50 Answer the following questions : D11. The orbital angular momentum quantum number of the state S2 is D9. The state S1 is (a) 0 (b) 1 (a) 1s (b) 2s (c) 2p (d) 3s (c) 2 (d) 3 (I.I.T. 2010) Passage IV. The wave φifenuliesfnoccarttrzmioiomanntu,iaotnψnhdn.w, Icln,hhmiatclhrhiaesicsmtaeaarvmtizahaeieldtmahbbealymteicitanahtleitcfhauqrluneacfeutnicnotounclutmsimognninvusewmnohfbiotneshretsehvtneaa,tbllaulebea,lneadd,neZpmsewisln. eadHrtsoetmruhepeicorfnnoisulslmdpoihbwseetirarninacgcanelqdfuparoeo0lsmaitsrinoBcunoochsol:rreduriasn,daiθtueisss. (r, θ, φ,) of the colatitude and Using this Column 1 Column 2 Column 3 3 ⎛ Z ⎞ 2 e−(Zr / a0 ) (r ) ⎝⎜ a0 ⎠⎟ (I) 1s orbital (i) Ψ n,l,ml ∝ (P) (II) 2s orbital (III) 2pz orbital 0 r/a0 (ii) One radial node (Q) Probability density at nucleus ∝ 1 a03 5 ⎛ Zr ⎞ ⎜ ⎟ ⎛ ⎞2 − (iii) ⎜ Z ⎟ r e ⎝ 2a0 ⎠ cos θ (R) Probability density is maximum at nucleus. ψn, l, ml ∝ ⎝ a0 ⎠ (IV) 3dz2 orbital (iv) xy-plane is a nodal plane (S) Energy needed to excite electron from n = 2 state to n = 4 state is 27 times the 32 energy needed to excite electron from n = 2 state to n = 6 state D12. For He+ ion, the only incorrect combination is List-I List-II (a) (I) (i) (R) (b) (II) (ii) (Q) (I) Radius of the nth orbit (P) ∝ n–2 (II) Angular momentum of (Q) ∝ n–1 (c) (I) (i) (S) (d) (I) (iii) (R) the electron in the nth orbit D13. For the given orbital in Column 1, the only correct combination for any hydrogen like species is (III) Kinetic energy of the (R) ∝ n0 (a) (I) (ii) (S) (b) (IV) (iv) (R) electron in the nth orbit (c) (III) (iii) (P) (d) (II) (ii) (P) (IV) Potential energy of the (S) ∝ n1 electron in the nth orbit D14. For hydrogen atom, the only correct combination is (a) (II) (i) (Q) (b) (I) (iv) (R) (T) ∝ n2 (U) ∝ n1/2 (c) (I) (i) (P) (d) (I) (i) (S) (JEE Advance 2017) D15. Which of the following options has the correct Passage V. combination considering List-I and List-II? Consider the Bohr’s model of a one-electron atom (a) (III), (S) (b) (IV), (Q) where the electron moves around the nucleus. In the (c) (III), (P) (d) (IV), (U) following, List-I contains some quantities for the nth orbit (JEE Advance 2019) D16. Which of the following options has the correct of the atom and List-II contains options showing how they combination considering List-I and List-II? depend on n. (a) (II), (R) (b) (II), (Q) (c) (I), (P) (d) (I), (T) (JEE Advance 2019) Passage III. D9. (b) D10. (c) D11. (b) Passage IV. D12. (d) D13. (d) D14. (d) D15. (c) D16. (d)

STRUCTURE OF ATOM 2/111 2. Match the entries of List I with its discoverer given in List II and select the correct answer using the code given below the lists: ©1. Match the orbital given in List-I with its description Modern Publishers. All rights reserved.given in List-II and select the correct answer using theList IList II code given below the lists: (P) discovery of proton 1. Chadwick List-I List-II 2. Goldstein (Q) discovery of neutron 3. Moseley (P) 3pz 1. total number of nodes = 2 (Q) 4s 2. has 3 radial nodes and 0 angular (R) determination of atomic 4. J.J. Thomson number node (S) determination of charge/ (R) 3dx2–y2 3. corresponds to n = 3, l = 0, ml = 0 mass of electron (S) 4p 4. has no radial node Code Code PQR S PQR S (a) 4 2 1 3 (a) 2 4 3 1 (b) 4 1 2 3 (b) 4 2 3 1 (c) 2 1 4 3 (c) 4 1 3 2 (d) 2 1 3 4 (d) 3 2 4 1 1. (a) 2. (d) Matrix Match Type Questions Each question contains statements given in two columns, If the correct matches are A-q, A-r, which have to be matched. Statements in Column I are B-p, B-s, C-r, C-s and D-q, then the labelled as A, B, C and D whereas statements in Column II correctly bubbled matrix will look are labelled as p, q, r and s. Match the entries of Column I like as shown: with appropriate entries of Column II. Each entry in Column I may have one or more than one correct option from Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. 1. Match the spectral series in Column I with its 2. Match the quantum number given in Column I with characteristics in Column II. its description given in Column II Column I Column II Column I Column II (A) n (p) designates the orientation of (A) Balmer series (p) falls in infra red region (B) l orbital (B) Pfund series (q) involve jumps from higher (q) orientation of the spin of the orbits to n = 3 (C) ml (D) ms electron (C) Brackett series (r) falls in visible region (r) has values corresponding to ‘n’ (s) defines the shell (D) Paschen series (s) involve jumps from higher orbits to n = 5 (1) : (A) – (r) (B) – (p), (s) (C) – (p) (D) – (p), (q) (2) : (A) – (s) (B) – (r) (C) – (p) (D) – (q)

2/112 MODERN’S abc + OF CHEMISTRY–XI 3. Match the entries in Column I with the correctly related quantum number(s) in ColumnII. © Column I Column II Modern Publishers. All rights reserved. (p) Principal quantum number (A) Orbital angular momentum of the electron in a hydrogen-like atomic orbital. (q) Azimuthal quantum number (B) A hydrogen-like one-electron wave function (r) Magnetic quantum number obeying Pauli principle (s) Electron spin quantum number (C) Shape, size and orientation of hydrogen like atomic orbitals. (D) Probability density of electron at the nucleus in hydrogen-like atom. (3) : (A)– (q) (B)– (p), (q), (r), (s) (C) – (p), (q) (r) (D) – (p), (q) Integer Type and Numerical Value Type Questions 7. Number of times radius of the 3rd shell of the H-atom as compared to that of radius of first shell is Integer Type: The answers to each of the following question 8. The atomic masses of He and Ne are 4 and 20 a.m.u., is a single digit integer ranging from 0 to 9. respectively. The value of the de Broglie wavelength of He gas at –73°C is M times that of the de Broglie 1. The maximum number of electrons that can have wavelength of Ne at 727°C. M is principal quantum number, n = 3 and spin quantum number, ms = +1/2 is (I.I.T. 2011) (JEE Advanced 2013) 2. How many times is the ionization energy of He+ ion as compared to that of H-atom? 9. In an atom, the total number of electrons having quantum numbers, n = 4, |ml| = 1 and ms = –1/2 is 3. Total number of orbitals in a f-sub-shell is (JEE Advance 2014) 4. Number of unpaired electrons in Fe2+ ion is 5. The maximum number of electrons in an atom with 10. Not considering the electronic spin, the degeneracy n = 4 and l = 1 is of the second excited state (n = 3) of H atom is 9, while the degeneracy of the second excited state of 6. The value of angular momentum for an electron in H– is (JEE Advance 2015) 3d orbital is x h . The value of x is 2π 1. (9) 2. (4) 3. (7) 4. (4) 5. (6) 6. (6) 7. (9) 8. (5). 9. (3). 10. (3). Hints & Explanations for Objective Type Questions Difficult ∴ Total energy = 1 mv2 + (– mv2) = – 1 mv2 2 2 A1. (b) K.E. = 1 mv2 , P.E. = − Ze2 K.E. 1 mv2 2 r =2 = –1 ∴ Total energy − 1 mv2 and electrostatic force = centrifugal force 2 Ze2 = mv2 A3. (b) Energy emitted = 1311.8 ⎛ 1311.8⎞ mol−1 r2 r 32 ⎝⎜ 22 ⎠⎟ − − − kJ or Ze2 = mv2 r or = 1311.8 ⎛ 1 − 1 ⎞ kJ mol–1 ⎝⎜ 22 32 ⎠⎟ or PE = –mv2

STRUCTURE OF ATOM 2/113 = 1311.8 ⎛ 1 − 1 ⎞ Ratio, ΔE1 = 3 × 36 = 27 ⎝⎜ 4 9 ⎠⎟ ΔE2 4 5 5 © Modern Publishers. All rights reserved.= 1311.8 × 5 kJ mol–1 A11. (d) ΔE = hν = hc = or λ = hc 36 λ ΔE ∴ Energy emitted per atom = 1311.8 × 103 × 5 J i.e., λ ∝ 1 6.02 × 1023 × 36 ΔE = 3.03 × 10–19 J ΔE = E4 – E1 will be maximum and hence λ will be A4. (a) For transition in He+ (Z = 2) minimum. 1 ⎛1 − 1 ⎞ A12. (a) Energy in the excited state is same as the K.E. λ = RZ2 ⎜⎝⎜ n12 n22 ⎟⎟⎠ A13. (b) rn = 0.059 n2 =R×4 ⎛ 1 − 1 ⎞ = 3R ⎝⎜ 4 16 ⎟⎠ 4 r1 = 0.059, r2 = 0.059 × 4, r3 = 0.059 × 9 For H atom (Z = 1) ∴ r1 : r2 : r3 = 1 : 4 : 9 1 = R ⎛ 1 − 1 ⎞ λ ⎜⎜⎝ n12 n22 ⎟⎟⎠ A14. (d) Frequency of a line in spectral series can be calculated as =R ⎛1 − 1 ⎞ = 3R ν= c = cR (Z) 2 ⎛ 1 − 1⎞ ⎜⎜⎝ n12 n22 ⎟⎟⎠ 4 λ ⎝⎜ n12 n22 ⎠⎟ H 1 − 1 = 3 For third line in Lyman series, n12 n22 4 n1 = 1, n2 = 4 ∴ n2 = 2, n1 = 1, νH = c.RH (1)2 ⎡1 1 ⎤ 15 A5. (d) I.E.Li2+ = Z2 × I.E. (H) ⎣⎢12 − 42 ⎥⎦ = 16 RHC = 9 × 13.6 eV = 122.4 eV For first line in Balmer series for Li2+ A6. (d) rn = 0.53 × n2 ∴ r2 = 0.53 × 4 = 2.12 Å n1 = 2, n2 = 3 A7. (b) No. of spectral lines = (n2 − n1) (n2 − n1 + 1) νLi2+ = cR (Z )2 ⎡ 1 − 1⎤ ⎢ n12 ⎥ 2 H ⎢⎣ n22 ⎥⎦ = (5 − 1) (5 − 1 + 1) = 4 × 5 = 10 = cRH (3)2 ⎡ 1 − 1⎤ 22 ⎢⎣ 22 32 ⎥⎦ A8. (b) rn = 0.529 n2Å = c × RH ×9× 5 = 5 cR H A9. (c) 36 4 For H atom, first Balmer line in the series is E3 – E2 = E1 (H) – E1 (H) = 5E1 (H) νH = 15 × 4 = 3 9 4 νLi2+ 16 5 4 36 For He+ ion (Z = 2) A15. (c) Energies are additive so that −E1 (H) ×22 E1 (H) ×22 E3 = E1 + E2 E6 – E4 = 62 – 42 hc = hc + hc λ3 λ1 λ2 ⎛ 16 − 36 ⎞ = – E1(H) × 22 ⎜⎝ 16 × 36 ⎠⎟ 1 = 1+ 1 = λ1 + λ2 λ3 λ1 λ2 λ1λ2 5E1 (H) or = 4 × 20 E1(H) = 36 λ1λ2 16 × 36 λ1 + λ2 λ3 = A10. (c) En = – 1.316 × 106 A16. (d) λ = h or v = h n2 mv λm ∴ Difference between first and second orbit 6.6 ×10−34 9.0 × 10−31 × 600 × 10−9 ΔE1 = –1.316 × 106 ⎛ 1 − 1 ⎞ = – 1.316 × 106 × 3 ∴v= = 1.22 × 103 m s–1 ⎜⎝ 1 22 ⎟⎠ 4 Difference between second and third orbit A18. (d) Δx × Δp = h 4π ΔE2 = –1.316 × 106 ⎛1 − 1 ⎞ = – 1.316 × 106 × 5 ⎝⎜ 22 32 ⎠⎟ 36 or Δp = h = h = ∞ 4πΔx 4π × 0

2/114 MODERN’S abc + OF CHEMISTRY–XI A24. (b) No. of orbitals in a shell = 2l + 1 B3. (a) Δx = Δp No. of electrons in a shell = 2(2l + 1) = 4l + 2 Δx . Δp = h A25. (a) It corresponds to 6d-orbitals and, therefore, can Δp . Δp = 4π accommodate maximum of 10 electrons.© h Modern Publishers. All rights reserved. 4π A26. (c) 4d has highest energy. A28. (d) n = 7, l = 2 corresponds to 7d and not 7p Δp2 = h A29. (a) Because l = 0 for s electrons. 4π Orbital angular momentum = l (l + 1). h = 0 Δp = h 2π 4π A30. (a) One set of quantum numbers represents only one mΔv = h orbital. 4π A32.(a) Electronic configuration of Mn (Z = 25) is Δv = 1 h 1s22s22p63s23p63d54s2 2m π Total orbitals populated B4. (a) Δp = mΔv = 1 × 10–18 g cm s–1 = 1 + 1 + 3 + 1 + 3 + 5 + 1 = 15 Δv = 1 ×10−18 g cm s−1 A33. (d) Fe3+ has 3d5 configuration and, therefore, has 5 9 ×10−28 g (maximum) number of electrons while Mg2+ has zero, Ti3+ has 1(3d1) and V3+ has 2(3d2). ∴ = 1 ×109 cm s–1 A34. (b) The last electron in Na atom is present in 3s B5. (e) λ = h subshell i.e., l = 0 mA = x mv mB = 5x vA = 0.1 mAvA = 0.1 x A36. (b) The electronic configuration of Ar is : 1s22s22p63s23p6 vB = 0.05 mBvB = 0.25 x No. of electrons having m = 0 are = 2(1s) + 2 (2s) + 2(2pz) + 2(3s) + 2(3pz) = 10 λ A = mBvB λB mAvA A38. (a) No. of unpaired electrons : 1. Ni2+ (Z = 28) : [Ar] 3d8 = 2 = 0.25x = 25 = 5 2. Co2+ (Z = 27) : [Ar] 3d7 = 3 0.1x 10 2 3. Cr2+ (Z = 24) : [Ar] 3d4 = 4 4. Fe3+ (Z = 26) : [Ar] 3d5 = 5 B6. (d) Maximum number of electrons in a subshell = 2 (2l + 1) or 4l + 2 ∴ Correct order = 1, 2, 3, 4 A39. (d) It corresponds to 3p (3), 3d (5), 4s (1), 4p (3), 5s (1) B7. (b) For l = 2, m = – 3 is not possible. = 13 B8. (b) Co : [Ar]3d74s2 A41. (d) No. of nodes = n – l – 1 = 3 – 2 – 1 = 0 Co3+ : [Ar] 3d6 B9. (b) No. of orbitals in an energy level = n2. A43. (c) Radial nodes = n – l – 1 = 3 – 1 – 1 = 1 ∴ No. of orbitals in 4th energy level = 42 = 16 Angular nodes = l = 1 B10. (b) E1 = hc , E2 = hc A44. (b) Electronic configuration of Cr (Z = 24): λ1 λ2 1s2 2s2 2p6 3s2 3p6 4d5 4s1 E1 = λ2 or λ2 = 25 = 1 E2 2λλ12 λ1 50 2 No. of electrons with l = 1 are 6(2p6) + 6(3p6) = 12 = B11. (a) ∴ n = λ61 No. of electrons with l = 2 are 5 (3d5) For 6s ⎯→ 4f ⎯→ 5d ⎯→ 6p A46. (c) No. of nodes = n – l – 1 = 4 – 0 – 1 = 3 B12. (a) This corresponds to 4f subshell and therefore, it can B1. (b) (B) Value of l cannot be equal to n have 14 electrons. (D) For l = 0, m cannot be –1 B13.(c) The electronic configuration of Rb is (E) For l = 2, m cannot be 3 [Ar]3d104s24p65s1. The outermost electron is 5s1. B2. (e) 1 mv2 = h (ν – ν0) Quantum numbers : 5, 0, 0, + 1 . 2 B14. (b) The orbital associated with2n = 3, l = 1 is 3p. 6.63 × 10–20 kg m2s–2 The number of electrons with m = –1 is 2. B15. (b) The electron is more tightly bound in the smallest = 6.63 × 10–34 kg m2 s–1 (2 × 1014 – ν0) allowed orbit. 6.63 ×10−20 = 2 × 1014 –ν0 B16. (a) ν = c or λ = c 6.63 ×10−34 λ ν 1 × 1014 – 2 × 1014 = – ν0 or or ν0 = 1 × 1014 s–1 λ= 3 × 1017 nm s−1 = 50 nm . 6 × 1015 s−1

STRUCTURE OF ATOM 2/115 B17. (d) Deuterium has 1 proton and 1 neutron B28. (c) The energy of 2s-orbital and 2p-orbital is equal in hydrogen like atoms. α-particle has 2 protons and 2 neutrons B29. (c) Option (c) is wrong according to Hund’s rule. ©∴ Mass of α-particle (m ) is twice that of deuterium Modern Publishers. All rights reserved.(mD)i.e.m=2mDα α 1.312 ×106 ⎛ 1.312 × 106 ⎞ Also, velocity of deuterium (vD) B33. (a) ΔE = E2 – E1 = – 22 – ⎜⎝⎜ − 12 ⎟⎟⎠ = 5 times velocity of α-particle (v ) or α i.e. vD = 5v v = vD/5 = 9.85 × 105 J mol–1. α α λD = h and B34. (c) Δv = 0.005 × 600 = 0.03 mDvD 100 λα = h =h ⎛⎜⎝ vD ⎞⎟⎠ Δx × mΔv = h mαvα 5 4π (2mD ) Δx = h or λD = h × (2mD) (vD) = 0.4 4πmΔv B18. (d) λα mDvD 5h m = 6.626 × 10–31 kg = 6.6 × 10−34 Δ v = 1 × 106 ms–1 4 × 3.14 × 9.1 × 10−31 × 0.03 Δx = ? = 1.92 × 10–3 m Δ x. mΔ v = h 4π h 6.63 × 10−34 mv 1.67 × 10−27 × 1 × 103 or Δx = h = 6.626 × 10−34 B35. (b) λ= = 4πmΔv 4 × π × 6.626 × 10−31 × 106 = 3.9 × 10–10 m = 1 × 10−9 m ≈ 4 × 10–10 m = 0.40 nm 4π B36. (a) Number of spectral lines emitted when electron jumps from n2 to n1 = ⎛ 1 ⎞ nm ⎝⎜ 4π ⎟⎠ = (n2 − n1)(n2 − n1 + 1) 2 B19. (a) It represents 3pZ orbital. = (5 − 1)(5 − 1 + 1) = 10 lines B20. (d) E = hc = 6.626 × 10−34 × 3 × 108 = 4.42 × 10–18 J 2 λ 45 × 10−9 B21. (c) n= 3, l = 0, ml = 0, ms = + 1 is possible and all other B37. (a) ν = RH ⎡1 − 1 ⎤ = 8 R are not possible. 2 ⎣⎢ n12 n22 ⎦⎥ 9 H B22. (c) Orbital angular momentum = l(l + 1) ∴ 1 − 1 = 8 n12 n22 9 For a d–orbital, l=2 ∴ Orbital angular momentum = 2(2 + 1) ∴ n1 = 1 n2 = 3 =6 B38. (d) IE of He+ ion = IE of H atom × Z2 B23. (a) After filling electrons in titanium, the configuration = I.E of H atom × (2)2 is 1s2 2s2 2p6 3s2 3p6 3d1 4s2 So, correct order of orbitals in titanium is 3s, 3p, 19.6 × 10–18 J atom–1 = 4 × I.E. of H atom 3d 4s ∴ I.E of H atom = 19.6 × 10−18 B24. (d) Fe2+ : [Ar]3d6 4 Number of d–electrons = 6 Fe : [Ar] 3d6 4s2 (No. of d-electrons = 6) = 4.9 × 10–18 J atom–1 Ne : 1s2 2s2 2p6 (No. of p-electrons = 6) I.E of Li2+ ion = I.E of H atom × (3)2 Mg : 1s2 2s2 2p6 3s2 (No. of s-electrons = 6) I.E of Li2+ ion = 4.9 × 10–18 × 9 Cl : 1s2 2s2 2p6 3s2 3p5 (No. of p-electrons = 11) = 4.41 × 10–17 J Energy = – 4.41 × 10–17 J B26. (a) For n = 3 and l = 1, the subshell is 3p and a particular 3p orbital can accommodate only 2 B39. (e) For He+, 1 = Z2R ⎛ 1 − 1 ⎞ electrons. λ ⎝⎜⎜ n12 n22 ⎟⎠⎟ B27. (c) dx2–y2 and dz2 orbitals have electron density along 1 = 22R ⎛ 1 − 1 ⎞ = 3R tdheensaitxyesb.etdwxye,ednyzthaendaxdeszx. orbitals have electron λ ⎜ 22 42 ⎟ 4 ⎝ ⎠

2/116 MODERN’S abc + OF CHEMISTRY–XI 1 ⎛ 1 1 ⎞ 3R For Li2+, Z = 3 and first energy state, n = 1 λ R ⎜⎜⎝ n12 n22 ⎠⎟⎟ 4 For H atom, = − = Z2 4.9 × 10−18 × 32 n2 12 © ∴ E1 = − k = − Modern Publishers. All rights reserved. or 1 − 1 = 3 = – 44.1 × 10–18 J atom–1 n12 n22 4 I.E. of Li2+ = 4E4∞.1– ×E11 0=–108 – (–44.1 × 10–18) = J atom–1. ∴ n1 = 1, n2 = 2 B40. (c) 1. Beyond a certain wavelength, line spectrum 12 appears as band spectrum. 12 B46. (b) E1 = –2.178 × 10–18 × = –2.178 × 10–18 J 2. For Balmer series n1 = 2. E2 = –2.178 × 10–18 × 12 = –5.445 × 10–19 J 4. For line of longest wavelength (lowest energy) 22 in this s=erνie(sH, en+2) = 3. B41. (a) ν (H) Energy required to excite electron from n = 1 to n = 2, RZ2 ⎛ 1 − 1⎞ = RZ2 ⎛ 1 − 1 ⎞ ⎜ n12 ⎟ ⎜⎝⎜ n12 n22 ⎟ ΔE = – 5.445 × 10–19 – (–2.178 × 10–18) ⎝ n22 ⎠ ⎠⎟He+ H = 16.335 × 10–19 J 1 − 1 = 4 ⎛ 1 − 1⎞ hc = 16.335 × 10–19 n12 n22 ⎝⎜ 4 16 ⎠⎟ λ 1 − 1 = 1− 1 = 3 λ= 6.63 × 10−34 × 3.0 ×108 n12 n22 44 16.335 × 10−19 ∴ n1 = 1 and n2 = 2 = 1.217 × 10–7 m B42. (d) For third line in Lyman series, B47. (c) λ = h mv n1 = 1, n2 = 4 v = 100 cm s–1 = 1 m s–1 m = 100 g = 0.1 kg c ⎛ 1 1⎞ νH = λ = c.RHZ2 ⎜⎝ n12 − n22 ⎠⎟ 6.6 × 10−34 0.1 × 1 λ= = 6.6 × 10–33 m = 6.6 × 10–31 cm = c.RH (1)2 ⎡1 − 1 ⎤ = 15 RHc ⎡1 1 ⎢⎣12 42 ⎦⎥ 16 n22 B48. (d) 1 = RH − ⎤ For first line in Balmer series for Li2+ λ ⎢ n12 ⎥ n1 = 2, n2 = 3 ⎣ ⎦ ⎡1 1 ⎤ Shortest wavelength line in hydrogen spectrum of n22 ⎥ Lyman series is from n2 = ∞ and n1 = 1 νLi2+ = cRH (Z)2 ⎢ n12 − ⎦ ⎣ 1 = 109678 ⎡1 − 1 ⎤ = 109678 cm–1 ⎡1 1 ⎤ 5 5 λ ⎣⎢ (1)2 ∞2 ⎥⎦ = cRH (3)2 ⎢⎣ 22 − 32 ⎥⎦ = c × RH × 9 × 36 = 4 cRH νH = 15 × 4 = 3 or λ = 1 cm = 9.11 × 10–6 cm νLi2+ 16 5 4 109678 = 911 × 10–10 m or 911 Å B43. (a) The orbitals are (1) 4p (2) 4s (3) 3d (4) 3p. B49. (b) The electronic configuration of Rb (Z = 37) is Increasing order of energies is : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s1 (4) 3p < (2) 4s < (3) 3d < (1) 4p. B44. (b) Cu : [Ar] 3d10 4s1 Valence electron is 5s1 Cu2+ : [Ar] 3d9 4s0 Quantum numbers ; B45. (b) En = 2π2me4 n = 5, l = 0, m = 0, s = + ½ n2h2 − (Z2 ) B50. (a) For first excited state of hydrogen (n = 2) or En ∝ E–∞nZ–22 E=1 − kZ2 Energy = −13.6 eV = −13.6 = −3.4 eV I.E. He+ = n2 n2 (2)2 of B51. (d) For l = 2, m cannot have a value of –3. The only =0 – ⎛ k(2)2 ⎞ = 4k permissible values of m for l = 2 are: ⎜⎝⎜ − 12 ⎠⎟⎟ –2, –1, 0, +1, +2 4k = 19.60 × 10–18 J atom–1 B52. (d) An electron of charge 'e' is passed through a potential difference of V, then its kinetic energy or k = 19.60 × 10−18 = 4.9 × 10–18 J atom–1 becomes eV. 4

STRUCTURE OF ATOM 2/117 1 mv2 = eV B63. (d) λA = 33.33 nm = 33.33 ×10−9 m 2 or v = 2eV m According to de-Broglie, © pB = 1 pA Modern Publishers. All rights reserved. 3 λA = h ...(i) pA ...(ii) λ= h = h mv m 2eV λB = h = h = 3h m pB pA /3 pA or λ= h From eq. (i) and (ii) 2meV λA = 1 ∴ h = 2meV λB 3 λ ∴ λB = 3 × λA = 3 × 33.33 × 10–9 m B53. (a) (i) 4p orbital (ii) 4s orbital (iii) 3d-orbital (iv) 3p-orbital = 1.0 × 10–7 m Order of increasing energy B64. (a) Energy of a photon = hc λ 3p (iv) < 4s (ii) < 3d (iii) < 4p (i) B54. (b) Electronic configuration of Cr (Z = 24) = 6.62 ×10−34 × 3.0 ×108 = 3.6 ×10−19 J Cr : 1s2 2s2 2p6 3s2 3p6 4s1 3d5 550 ×10−9 Energy absorbed per sec by bulb = V × I × t 19th electron is present in 4s-orbital = 100 × 1 × 3600 For 4s orbital; n = 4, l = 0, m = 0, s = + ½ B55. (d) Cl (Z = 17) : 1s2 2s2 2p6 3s2 3p5 n= 100 × 3600 = 1 × 1024 3.6 × 10−19 The unpaired electron of Cl is in 3p orbtial. Hence, quantum numbers for unpaired electron are: B65. (b) hν = 1 mv2 2 1 1 w0 + 2 2 n = 3, l = 1, m = + 1, 0, –1, s = + or – 1 mv2 2 – 13.6 w0 = hν − n2 B56. (a) En = eV 6.626 ×10−34 × 3.0 ×108 − 1 × 9 ×10−31 × (6 ×105)2 4000 ×10−10 2 For 3s-orbital, n = 3 = En = – 13.6 = –1.51eV = 4.96 × 10–19 –1.62 × 10–19 32 = 3.34 × 10–19 J B57. (d) r = 0.529n2 or = 3.34 ×10−19 Z 1.6 × 10−19 Z =1, n = 2 = 2.1 eV r = 0.529 × (2)2 B66. (d) For 71st electron : 1 Z = 70 [Xe]54 4f14 6s2 = 2.12Å Z = 71 [Xe] 4f14 5d1 6s2 B58. (c) For n = 2, l = 0, m cannot be –1 because m has values B67. (b) Ratio of the spectral lines, –l ....0 ...... + l B59. (b) Ni (28) : [Ar] 4s2 3d8 1 = R H ⎛ 1 − 1 ⎞ Z2 λ2 ⎜ n12 n22 ⎟ No. of unpaired electrons = 2. ⎝ ⎠ B60. (c) For Lyman series electrons return from n = 2, 3, 4, 5 1 = RH ⎛ 1 − 1 ⎞ Z 2 to n = 1. λ1 ⎜ m12 m22 ⎟ B62. (b) ⎝ ⎠ For shortest wavelengths, n2 = ∝ and m2 = ∝ hv = w0 + 1 mv2 ∴ λ1 = 9 = m12 2 λ2 1 n12 wo = hv − 1 mv2 Now, if m1 = 3 and n1 = 1, the ratio is justified. 2 Hence, these correspond to Lyman and Paschen = 6.626 × 10−34 × 3.0 × 108 − 1 × 9 × 10−31 × (6 × 105)2 series. 4000 × 10−10 2 B68. (a) In K, 2s feels maximum attraction due to more = 4.96 × 10−19 − 1.62 × 10−19 = 3.34 × 10−19 J Zeff.Therefore, it has lowest energy. B70. (c) I (4d), II (3d) III (4p) IV (3p) or = 3.34 × 10−19 = 2.1 eV 1.6 × 10−19 Order of energies is IV (3p) < II (3d) < III (4p) < I (4d)

2/118 MODERN’S abc + OF CHEMISTRY–XI B71. (d) For Lyman series B75. (d) Radius of an orbit, ⎛1 1 ⎞ r = 0.529n2 Å ⎜⎝ 12 ∝2 ⎟⎠ Z For H atom Z = 1 For Be3+, Z = 4 r1 = 0.529 n2 Å © νmax = RH − = RH Modern Publishers. All rights reserved. νmin = RH ⎛1 − 1 ⎞ = 3 R H ⎜⎝ 12 22 ⎠⎟ 4 ΔνLyman = RH − 3 R H = RH r Be3+ = 0.529n2 Å 4 4 4 For Balmer series Given that r = r ⎛ 1 1 ⎞ RH H Be3+ Be3+ ⎜⎝ 22 ∝2 ⎟⎠ 4 νmax = RH − = ∴ It is equal when n = 2 for νmin = RH ⎛ 1 − 1 ⎞ = 5 RH ∴ r2(Be3+) = r1(H) ⎜⎝ 22 32 ⎠⎟ 36 B77. (c) For Bohr orbit, angular momentum is ΔνBalmer = RH − 5RH mvrn = nh or v = nh 4 36 2π 2πmrn = 4RH Kinetic energy, KE = 1 mv2 36 2 = RH or K.E. = 1m× n2h2 n2h2 9 2 4π2m2rn2 8π2mrn2 ΔνLyman = RH /4 = 9 or 9 : 4 = ΔνBalmer RH /9 4 For n = 2, rn = a0 × n2 = 4a0 (a0 = Bohr radius) B72. (b) de-Broglie wavelength λ= h K.E. = 22 h2 = h2 2 K.E. × m 8π2m(4a0 )2 32 π 2 m a02 Now for photoelectron emission, C3. (b, c) : (b) 3p and 4d have 1 node (c) 4s and 5p have 3 nodes. hν = hν0 + KE KE = hν – hν0 C4. (b, c) : No. of unpaired electrons : Cr3+(3), Fe2+(4), Cu+(0), Zn2+(0), Mn2+(5), Fe3+(5), Sc3+(0), Ti3+(1). ∴ λ= h 2m (hν − hν0) C5. (a, b) : No. of unpaired electrons : Cu (1), Sc (1), Mn (5), Cr (6). or λ = h 2mh (ν − ν0) C6. (a, d) : (a) 4d > 5s (b) 3d = 3p (c) 3p = 3d (d) 3d > 2p. or λ∝ 1 (ν − ν0 ) 1 C7. (a, b, c) : (d) is not correct. The spacing between adjacent 2 levels decreases with increase in ‘n’. B73. (d) 1 = ν = RH Z2 ⎛ 1 − 1 ⎞ −13.6 × Z2 λ ⎜ n12 n22 ⎟ n2 ⎝ ⎠ C8. (a, d) : –3.4 = ν = RH ⎛ 1 − 1 ⎞ (Z = 1 for H) n2 = −13.6 × Z2 = 16 (∵ Z = 2) × ⎜⎝⎜ nf 2 82 ⎟⎠⎟ −3.4 ν = RH × 1 − RH n=4 nf 2 64 l=2 1 ∴ It corresponds to 4 d state. The plot of ν against n f 2 will be linear with slope No. of angular nodes = l = 2 = RH Radial nodes = n – l – 1 = 4 – 2 – 1 = 1 B74. (c) hν = hν0 + 1 mv2 2 1 mv2 (K.E.) = hν – hν0 D1. (b) Wavelength of α-particles 2 Plot of K.E. vs ν cannot have intercept equal to λ (α-particle) = h = h zero. mv 6.6 ×10−27 × 2.5 ×102

STRUCTURE OF ATOM 2/119 λ (photon) = h D11. (b) S2 state is 3p state m × 3.0 × 108 Orbital angular momentum quantum number = 1 D12. (d) The 1s-orbital is non-directional and its wave function is independent of cos θ. Hence (d) is incorrect. D13. (d) For hydrogen like species, only (d) is correct. In (a) 1s orbital has no radial node. In (b) for 3dz2 , xy is not a nodal plane. In (c) 2pz has no radial node. D14. (d) Energy needed when electron is excited from n = 2 to n = 4 state. ©Since λ(α-particle) = λ (photon) Modern Publishers. All rights reserved. h = h 6.6 ×10−27 × 2.5 ×102 m × 3.0 ×108 m= 6.6 ×10−27 × 2.5 ×102 3.0 ×108 = 5.5 × 10–33 kg E4 – E2 = R ⎛ 1 − 1 ⎞ = 3R ⎝⎜ 22 42 ⎟⎠ 16 D2. (b) Wavelength of electron accelerated by potential of V volt is λ= h = h = h × 10−4 Energy needed when electron is excited from n = 2 to (2 Vem)1/ 2 (2 ×100 ×106em)1/ 2 (2 me)1/ 2 n = 6 state. D3. (d) λp = h , E6 – E2 = R ⎛1 − 1 ⎞ = 8 R (2π Ve mp)1/ 2 ⎝⎜ 22 62 ⎠⎟ 36 λ He2+ = h =h E4 – E2 3 × 36 = 27 (2π Ve mHe2+ )1 / 2 (2πVe 4mp)1/ 2 E6 – E2 16 8 32 = 1 ∴ λHe2+ = (2π Ve mp)2 = 1 Hence E4–E2 = 27 × (E6 – E2) λp 2 32 1 n2h2 (2π Ve 4mp)2 4π2Zme2 D4. (c) mvr = n h = 4h D15–D16. (I) Radius, rn = or Z ∝ n2 2π 2π (I) – (T) ∴ r = 2h = 2 (II) Angular momentum = nh or ∝ n πmv πλ 2π D5. (a) m = 4 = 6.64 × 10–27 kg (II) – (S) 6.022 ×1023 (III) Kinetic energy, K.E. = 1 mv2 = 1 Ze2 λ= 6.6 ×10−34 = 4.99 × 10–12m 2 2 rn 6.64 ×10−27 × 2.0 ×104 Since, rn ∝ n2 = 4.99 pm K.E. ∝ 1 n2 D6. (c) For 3s electron, l = 0 so that orbital angular momentum l(l + 1) h is zero. (III) – (P) 2π −Ze2 D7. (b) n + l = 5 is possible for 5s, 4p and 3d. (IV) Potential energy, P.E. = rn D8. (c) For 4f electron, n = 4, l = 3, ml = 0 and s = + 1 is 2 possible. P.E. ∝ 1 ∴ n2 D9. (b) S1 state is 2s because in 2s orbital, there is one node (IV) – (P) (n – l –1) = 2 – 0 – 1 = 1 D10. (c) Energy in S1 state D15. (c) (III) – (P) is correct combination. E = – 13.6 × 32 D16. (d) (I) – (T) is correct combination 22 = − 9 × 13.6 eV  1 (9) : No. of orbitals in a shell in n = 3 will be = n2 4 i.e., 32 = 9. Each orbital can have maximum of EH = – 13.6 eV (ground state) E = − 9 × EH = 2.25 EH two electrons, one with ms = + 1 and the second 4 2

2/120 MODERN’S abc + OF CHEMISTRY–XI with ms = – 1 . Hence number of electrons with λNe ∝ h 2 2mNeTNe © Modern Publishers. All rights reserved.ms = +1will be 9. λHe ∝ m Ne TNe 2 λNe mHeTHe  2 (4) : Ionization energy of He+ ion = Z2 × I.E. (H) = 22 TH = 273 – 73 = 200 K, × I.E. (H) = 4 times TNe = 727 + 273 = 1000  3 (7) : For f-subshell, l = 3 and hence it has (2 × 3 + 1) = 7 times  4 (4) : Fe2+ : [Ar] 3d6. Hence, it has four unpaired λHe = 20 ×1000 electrons. λNe 4 × 200  5 (6) : It corresponds to p-subshell and hence has maximum of 6 electrons. λHe = 5 λNe  6 (6) : Angular momentum is l(l + 1) h ∴ λHe = 5 λNe 2π  9 (3) : For n = 4 ; l = 0, 1, 2, 3. Since l = 2 for 3d orbital ET= hx1ce.erEpetfaolcrh=e,0ot,fhatrlhel eeostoehreobrrit‘bla’itlvsaalolsuf ecnsan=ca4ancchcaoanmvehmamovlde=amt1e.l Angular momentum = 2(2 + 1) h = 6 h . 2π 2π  7 (9) : rn = a0 n2 Z one electron with ms = −1. Thus, the total For H, Z = 1 and for n = 3 number of electrons is thr2ee. r3 = a0 × 32 = 9a0.  10 (3) : In 'H' atom, there is only 1 electron (1s1).  8 (5) : λ= h Therefore, there is no screening effect. Due to mv this, energy of all subshells in a given shell is same. Thus, order of energy of various sub- 1 mv2 = KE shells is 2 1s < 2s = 2p < 3s = 3 p = 3d < 4s = 4p = 4d = 4f mv2 = 2KE m2v2 = 2m KE Same energy ∴ mv = 2 m KE The second excited state of hydrogen corresponds to the third energy level. It has 9 ∴ λ= h orbitals of same energy (one 3s, three 3p, five Now 2m.KE 3d: Total = 1 + 3 + 5 = 9 orbitals). Thus, degeneracy is 9. K.E. ∝ T In H−, there are two electrons. Thus, screening effect is present. Therefore, order of energy of λ∝ h various sub-shells is λHe ∝ 2mT 1s < 2s < 2p < 3s < 3p ------- First excited state for H would be : 1s1 2s1 h Second excited state for H− would be 1s1, 2s0, 2p1 2mHeTHe Degeneracy of a p-subshell is 3.

STRUCTURE OF ATOM 2/121 ©Time Allowed: 1½ Hr. Maximum Marks: 25 Modern Publishers. All rights reserved. (1) 1. Maximum number of electrons in a subshell of an atom is determined by (a) 2 l + 1 (b) 2n2 (c) 4 l + 2 (d) 4l – 2 2. The orbital angular momentum of a p-electron is given as : (a) h (b) 2h 2π π (c) 3h (d) 0 (1) 2π 3. Which of the following d-orbital has electron density in all the three axes (a) 3 dxy (b) 3 dx2 − y2 (c) 3 dz2 (d) 3 dyz (1) Q. 4 and 5 are assertion reason type questions In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. 4. Assertion: 2s orbtial is spherically symmetrical. Reason: s-orbitals have no angular dependence. (1) 5. Assertion: Wavelength associated with a moving particle becomes double if its velocity is doubled. Reason: Wavelength associated with a moving object is inversely proportional to its velocity. (1) 6. Which has larger wavelength : an electron or a proton? (1) 7. What is the designation of an orbital having n = 4 and l = 2 ? (1) 8. Calculate the wavelength associated with an electron (mass = 9.1 × 10–31 kg) having kinetic energy 2.275 × 10–25 J. (2) 9. Write the electronic configurations and predict the number of unpaired electrons in the following: (ii) Cr (Z = 24) (ii) Fe (Z = 26) (2) 10. What is an orbital? Compare the shapes of 1s and 2s-orbital. (3) 11. An electron has a speed of 600 m s–1 with uncertainty of 0.025%. What is the uncertainty in locating (3) (3) its position ? 12. What are Balmer series? Calculate the wave number of the longest and shortest wavelength transition in the Balmer series of atomic spectrum.

2/122 MODERN’S abc + OF CHEMISTRY–XI 13. (a) What observations in scattering experiment led Rutherford to make the following conclusions ? (i) The most of the space in an atom is empty. (ii) The whole of the mass of the atom is present in the centre of the nucleus. (iii) Nucleus has positive charge. (b) What is the value of orbital angular momentum for an electron in 2s orbital ? (c) How many electrons in an atom may have n = 4 and ms = +1/2 ? (d) What physical meaning is attributed to the square of the absolute value of wave function Ψ2 ? © (5) Modern Publishers. All rights reserved. To check your performance, see HINTS and SOLUTIONS to some questions at the end of Part I of the book.

UNIT© Modern Publishers. All rights reserved. 3 CLASSIFICATION OF NonM-meetatallilclicCChhaararactceter r N ELEMENTS AND O PERIODICITY B IN PROPERTIES L E G A S E S Scientists like to find patterns. About 200 years ago, scientists were discovering lots of new elements. However, they were struggling hard to bring order and pattern to the vast amount of information they were discovering about the elements and their properties. The periodic table is an important triumph for the nineteenth century chemists and has become an important landmark in the history of chemistry. It is used as a sorting system for chemical elements. It has become the everyday support for students and teachers and provides a concise organization of the whole of chemistry. In the present unit, we will study the historical developments of the periodic table. We will also learn how the periodic classification of elements follows as a logical consequence of the electronic configuration of atoms. We shall also study some of the important periodic trends in the physical and chemical properties of elements. OBJECTIVES Building on..... Assessing..... Preparing for Competition..... Understanding Text 1 REVISION EXERCISES 53 Additional Useful Information 60 Conceptual Questions 16,38 Topicwise MCQs 60 Answers/Hints for Revision Competitive Examination Qs CHAPTER SUMMARY & QUICK Exercises 58 AIPMT, NEET & Other State Boards’ CHAPTER ROUND UP 41 HOTS & Advanced Level Medical Entrance 62 NCERT FILE Questions with Answers 58 JEE (Main) & Other State Boards’ Textbook Exercises with Engineering Entrance 63 Hints & Explanations for Difficult 68 Solutions 42 UNIT PRACTICE TEST 71 Objective Questions NCERT Exemplar Problems with Answers/Hints 46 3/1

3/2 MODERN'S abc + OF CHEMISTRY–XI NEED FOR CLASSIFICATION OF ELEMENTS However, this classification proved to be inadequate. Upto the end of seventeenth century, only 31 Some of the earlier important attempts to classify the elements are briefly summed up below: elements were known. Therefore, it was very easy to study and remember the properties of these elements. 1. Dobereiner’s triads However, during the later part of the eighteenth In 1817, a German scientist, Johann Dobereiner century, the pace of discovery of new elements classified the elements in groups of three elements quickened. Between 1800 to 1869, the number of called triads. The elements in a triad had similar identified elements had become nearly double to 63. properties and the atomic weight of the middle With such a large number of elements, it became member of each triad is very close to the arithmetic difficult to study individually the chemistry of all these mean (average) of the other two elements. The elements and their innumerable compounds. At this common triads of Dobereiner classification were : stage, it was realised that there should be some simple lithium, sodium and potassium; calcium, strontium way to study and remember the numerous properties and barium; chlorine, bromine and iodine; etc. of the elements and their compounds. This gave rise to For example, in the triad of lithium, sodium and necessity of classification of the elements into various potassium, groups having similar properties. This has been done Atomic weight of middle element (sodium) = 23 by arranging the elements in such a way that similar Mean of atomic mass of lithium and potassium elements are placed together while dissimilar elements© are separated from one another. This is known asModern Publishers. All rights reserved.= 7 + 39 = 23. classification of elements. Such a classification of The Dobereiner2’s relationship was also referred the elements has resulted in the formulation of the to as law of triads. However, it seemed to work only periodic table. Periodic table may be defined as for a few elements. It was dismissed as coincidence because all the known elements could not be arranged the arrangement of the known elements in triads. according to their properties in a tabular form. At present, about 118 elements are known. Of 2. De Chancourtois Classification these, the recently discovered elements are man The next reported attempt was made by a French made. Efforts to synthesize new elements are geologist, A.E.B.de Chancourtois in 1862. He arranged continuing. The periodic classification of the elements the known elements in order of increasing atomic has extremely simplified their study. Not only the weights and proposed a cylindrical table of elements periodic classification rationalizes the known chemical to display the periodic recurrence of properties. He facts about elements, but it also helps to predict new observed that the elements with similar properties fell ones for undertaking further study. in a vertical line from the centre of the spiral. However, this did not attract much attention. HISTORICAL DEVELOPMENT OF THE PERIODIC TABLE* 3. Newlands law of octaves Since the beginning of the nineteenth century, In 1865, an English chemist, John Newlands scientists have been trying to find a basis of grouping proposed a new system of grouping elements of similar elementshavingsimilarproperties.Lavoisier classified properties. According to him, when the elements are the elements simply as metals and non-metals. arranged in the increasing order of atomic weights, the properties of every eighth element Table 1. Dobereiner’s triads are similar to the first one. Newlands called this relation as the law of octaves due to similarity with Element Atomic weight Mean of 1st and 3rd the musical scale. Lithium 7 7 + 39 = 23 This relationship is just like every eighth note Sodium 23 2 resembles the first in octaves of music. The Newlands Potassium 39 law of octaves seemed to be true only for elements up to calcium as shown below : Calcium 40 Mean of 1st and 3rd 40 + 137 = 88.5 Table 2. Newlands’s law of octaves Strontium 88 Barium 137 2 Chlorine 35.5 Mean of 1st and 3rd Element Li Be B C N O F 35.5 + 127 = 81.25 Bromine 80 At. Wt. 7 9 11 12 14 16 19 Iodine 127 2 Element Na Mg Al Si P S Cl At. Wt. 23 24 27 29 31 32 35.5 Element K Ca *Not in the syllabus of CBSE (only for Recapitulation) At. Wt. 39 40

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 3/3 In the above table, sodium (Na), the eighth© elements with similar properties recur at regular element from lithium (Li) is similar to Li and similarlyModern Publishers. All rights reserved.intervals or periodically. As a result of this, the elements the next eighth element potassium is similar to Na. fall in certain groups or families. The elements in each The same is true for magnesium resembling beryllium, group were similar to each other in many properties. aluminium resembling boron, etc. The properties repeated periodically. Mendeleev’s periodic table published in 1905 is given in Table 3 4. Lother Meyer’s arrangement ahead. The horizontal rows in the periodic table are called periods and the vertical columns are called In 1869, a German chemist, Lother Meyer used groups. the physical properties such as atomic volume, melting point, boiling point, etc., to arrive at his table of Mendeleev’s system of classifying elements was elements. He showed that when the properties of the more elaborate than the earlier attempts. He realized elements such as atomic volume, melting point, boiling that some of the elements did not fit in very well with point, etc., are plotted as a function of their atomic his scheme of classification if the order of atomic weights, the elements with similar properties occupied weight was strictly followed. He showed courage to almost similar positions. On this basis, Lother Meyer ignore the order of atomic weights, thinking that the proposed that when the elements are arranged in atomic weight measurements might be incorrect. He the increasing order of their atomic weights, placed the elements with similar properties together. similarities in physical and chemical properties For example, iodine has lower atomic weight than that appear at regular intervals. of tellurium (of Group VI) but he placed iodine in Group VII alongwith fluorine, chlorine and bromine because All these attempts provided same clue that there of similarities in their properties. are certain regularities among the elements. The first most successful attempt in classification of elements The Mendeleev’s classification gave him so much was made by a Russian chemist Dmitri Mendeleev. confidence that he boldly left certain spaces or gaps for undiscovered elements. By considering the properties 5. Mendeleev’s Periodic Law of the adjacent elements in his table, he predicted the properties of the undiscovered elements. Later on, In 1869, Mendeleev made a remarkable when these elements were discovered, their properties contribution to the classification of elements. On the were found to be exactly similar to those predicted by basis of physical and chemical properties of the Mendeleev. For example, gallium and germanium elements, he gave a law known as the periodic law. were not discovered at that time, when Mendeleev The law states that formulated his periodic table and therefore, he left gaps for these elements. He not only predicted the existence the physical and chemical properties of elements are periodic function of their of the elements but he estimated their properties. He atomic weights. tentatively named these elements as eka-aluminium This means that when the elements are arranged in the order of their increasing atomic weights, the and eka-silicon (from the Sanskrit word eka meaning elements with similar properties recur at regular intervals. Such orderly recurring properties in a cyclic ‘next’) because he believed that these would be similar fashion are said to be occurring periodically. This is responsible for the name periodic law or periodic table. to aluminium and silicon respectively. When chemists Mendeleev’s Periodic Table discovered these elements, Mendeleev’s prediction of On the basis of his periodic law, Mendeleev their properties proved to be remarkably correct. arranged all the known elements in the form of a table known as periodic table. It was observed that the Some of the properties predicted by Mendeleev for these elements and those found experimentally are given in Table 4. Table 4. Comparison of the properties predicted for eka-aluminium and eka-silicon by Mendeleev with those observed for gallium and germanium. Property Eka-aluminium predicted Gallium Eka-silicon predicted Germanium by Mendeleev (found) by Mendeleev (found) Atomic mass 68 70 72 72.6 Density (g cm–3) Melting point (K) 5.9 5.94 5.5 5.36 Formula of oxide Formula of chloride Low 29.78 High 1231 E2O3 Ga2O3 EO2 GeO2 ECl3 GaCl3 ECl4 GeCl4

© MoTable 3. Mendeleev’s periodic table published in 1905 showing elements and their atomic weights. deSERIES 3/4 MODERN'S abc + OF CHEMISTRY–XI rn1 0 II III GROUPS OF ELEMENTS VII I IV V VI VIII — — PHe Hydrogen Beryllium Boron — H Be Br — — — — Helium 9.1 11.0 Carbon Nitrogen Oxygen Fluorine uNeon 1.008 Magnesium Aluminium C N O F 2 4.0 b19.9 Lithium 12.0 14.04 16.00 19.0 Li Mg Al Phosphorus lArgon 7.03 24.3 27.0 Silicon Sulphur Chlorine 3 Ne Si P S Cl is4 Sodium Calcium Scandium 31.0 Na Ca Sc 28.4 31.0 35.45 40.1 44.1 h— Ar 23.5 Titanium Vanadium Chromium Manganese Iron Cobalt Nickel (Cu) 38 Zinc Gallium Ti V Cr Mn Fe Co Ni 5 Potassium Zn Ga 48.1 51.4 52.1 55.0 55.9 59 59 K Germanium 39.1 65.4 70.0 Arsenic Selenium Bromine Ge As Se Br Copper Strontium Yttrium 72.3 75 79 erKrypton Cu Sr Y 79.95 s6 Kr 87.6 89.0 63.6 Zirconium Niobium Molybdenum — Ruthenium Rhodium Palladium 81.8 Cadmium Indium Zr Nb Mo Rubidium Cd In 90.6 94.0 96.0 Ru Rh Pd (Ag) . A7 — Rb 85.4 112.4 114.0 Tin Antimony Tellurium 101.7 103.0 106.5 Sn Sb Te Silver 119.0 Iodine Ag 120.0 127 I 107.9 127 l8 Xenon lXe Cesium Barium Lanthanum Cerium Tantalum Tungsten — Osmium Iridium Platinum (Au) Cs Ba La Ce Ta W 9 128 132.9 137.4 139 140 183 184 Os Ir Pt rig10 — — Ytterbium — Bismuth — 191 193 194.9 h11 Gold Mercury Yb Lead Bi ts12 — Au Hg 173 Pb Uranium — 197.2 200.0 206.9 208 U Thallium 239 — Radium Tl Thorium — Ra Th 224 204.1 232 — reFormulae sFormulae erved.of hydrides R R2O RO R2O3 GRROO2 UPS OF ERL2OE5MENTS RO3 R2O7 RO4 of oxides R RH RH2 RH3 RH4 RH3 RH2 RH

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 3/5 © The success of Mendeleev’s bold quantitative 3. Position of isotopes. Isotopes are the atoms Modern Publishers. All rights reserved.predictions convinced Chemists of the usefulness of of the same element having different atomic masses Mendeleev’s periodic table and led to its wide acceptance. but same atomic number. Therefore, according to Mendeleev’s classification, these should be placed at Important Contributions of Mendeleev’s Periodic different places depending upon their atomic masses. Table For example, isotopes of hydrogen with atomic masses 1, 2 and 3 should be placed at three places. However, Mendeleev’s periodic table was one of the greatest isotopes have not been given separate places in the achievements in the development of chemistry. Some of periodic table. the important contributions of his periodic table are : 4. Some similar elements are separated and 1. Systematic study of the elements. The dissimilar elements are grouped together. In the Mendeleev’s periodic table simplified the study of Mendeleev’s periodic table, some similar elements chemistry of elements. Knowing the properties of one were placed in different groups while some dissimilar element in a group, the properties of other elements in elements had been grouped together. For example, the group can be easily guessed. Thus, it became very copper and mercury resembled in their properties but useful in studying and remembering the properties of they had been placed in different groups. At the same a large number of elements. time, elements of group I A such as Li, Na and K were grouped with copper (Cu), silver (Ag) and gold (Au), 2. Correction of atomic masses. The though their properties are quite different. Mendeleev’s periodic table helped in correcting the atomic masses of some elements based on their positions 5. Cause of periodicity. Mendeleev did not in the table. For example, atomic mass of beryllium explain the cause of periodicity among the elements. was corrected from 13.5 to 9. Similarly, with the help of this table, atomic masses of indium, gold, platinum 6. Position of lanthanoids (or lanthanides) etc., were corrected. and actinoids (or actinides). The fourteen elements following lanthanum (known as lanthanoids, from 3. Prediction of new elements. At the time of atomic number 58–71) and the fourteen elements Mendeleev, only 56 elements were known. While following actinium (known as actinoids, from atomic arranging these elements, he left some gaps. These number 90—103) have not been given separate places gaps represented the undiscovered elements. Mendeleev in Mendeleev’s table. predicted the properties of these undiscovered elements on the basis of their positions. For example he predicted In order to cover more elements, Mendeleev the properties of gallium (eka-aluminium) and modified his periodic table. germanium (eka-silicon) which were discovered later. The observed properties of these elements were found ATOMIC NUMBER AND MODERN PERIODIC to be similar to those predicted by Mendeleev (Table 4). LAW Defects of Mendeleev’s Periodic Table In 1913, Moseley, a young English physicist discovered the relationship between X-ray spectra and In spite of many advantages, the Mendeleev’s the atomic number of the elements. When high energy periodic table had certain defects also. Some of these electrons were focused on a target made of the are given below : elements under study, X-rays were generated. He studied the frequencies of X-rays emitted from the 1. Position of hydrogen. Hydrogen is placed in elements and observed that the frequency (ν) of the group I. However, it resembles the elements of group prominent X-rays emitted by an element was I (alkali metals) as well as the elements of group VIIA proportional to the atomic number and not to the (halogens). Therefore, the position of hydrogen in the atomic weight. He gave a general equation : periodic table is not correctly defined. ν = a (Z – b) 2. Anomalous pairs. In certain pairs of elements, the increasing order of atomic masses was not obeyed. where ν is the frequency of the emitted X-ray and a In these cases, Mendeleev placed elements according and b are constants that are same for all the elements. to similarities in their properties and not in increasing A plot of ν versus atomic number (Z) gives a straight order of their atomic masses. For example, argon (Ar, line. This led Moseley to conclude that atomic atomic mass 39.9) is placed before potassium (K, atomic number and not atomic weight is the mass 39.1). Similarly, cobalt (Co, atomic mass 58.9) is fundamental property of the atoms. He, therefore, placed before nickel (Ni, atomic mas 58.6) and tellurium suggested that atomic number instead of atomic weight (Te, atomic mass 127.6) is placed before iodine (I, should be the basis of classification of elements. atomic mass 126.9). These positions were not justified.

© 3/6 MODERN'S abc + OF CHEMISTRY–XI Modern Publishers. All rights reserved. The acceptance of atomic number, as the International Union of Pure and Applied Chemistry important characteristic of an atom, led to the modern (IUPAC), the groups are numbered from 1 to 18 (Fig. 1). periodic law. Modern periodic law may be stated as These are discussed below : the physical and chemical properties of the Periods elements are periodic functions of their atomic numbers. A horizontal row in the periodic table is called period. This law implies that the physical and chemical In terms of electronic structure of the atom, a properties of the elements depend on atomic number period constitutes a series of elements whose atoms and this dependency shows periodicity. Consequently, have the same number of electron shells i.e. principal when the elements are arranged in the order of their quantum number (n). Thus, the period indicates the increasing atomic numbers, it is observed that the value of n for the outermost or valence shell of the elements of similar properties recur at regular atom of the element. Each successive period in the intervals or periodically. As a result of this, the periodic table is associated with the filling of the next elements fall in certain groups and lead to an higher principal energy level (n=1, n=2, n=3, etc.) There arrangement called the periodic table. are seven periods and each period starts with a different principal quantum number. With the modern periodic law as the guiding It can be seen that the number of elements in principle and the modern theory of atomic structure, each period is twice the number of atomic orbitals many new forms of the periodic table have been available in the energy level that is being filled. proposed from time to time. Some forms emphasise chemical reactions and valence, whereas other forms Electronic Configurations of Elements in Periods stress the electronic configurations of elements. However, the general plan of the table has remained Elements of First Period the same as proposed by Mendeleev. Furthermore, it is now recognized that the periodic law is essentially The first period corresponding to n = 1 is unique the consequence of the periodic variation in electronic because it contains only two elements. This is not configurations of atoms, which determine the physical surprising because first energy shell (K) has only one and chemical properties of the elements and their orbital (i.e. 1s) which can accommodate only two compounds. electrons. This means that there can be only two elements in which one and two electrons are present THE LONG FORM OF THE PERIODIC TABLE in first energy level. The first period contains hydrogen (1s1) and helium (1s2). The most widely used periodic table these days is ‘ The Long Form of the Periodic Table’. This is Elements of Second Period constructed on the basis of repeating electronic configuration of the atoms when the elements are The second period contains 8 elements because arranged in the order of increasing atomic numbers. for n = 2, there are four orbitals (one 2s and three 2p) Long form of the periodic table is given in Fig. 1 ahead. in second energy shell (L). In all, these four orbitals have a capacity of eight electrons and, therefore, second Structural Features of the Long Form of the period has eight elements in it. It starts with lithium Periodic Table (Z = 3) in which one electron enters the 2s-orbital. The period ends with neon (Z = 10) in which the second The long form of the periodic table consists of shell is complete (2s2 2p6). horizontal rows called periods and vertical columns called groups. There are altogether seven periods. Elements of Third Period These are numbered as 1, 2, 3, 4, 5, 6 and 7. The first period contains 2 elements and the subsequent periods In third period corresponding to n = 3, there consist of 8, 8, 18, 18, 32 and 32 elements respectively. are nine orbitals : one 3s, three 3p and five 3d. Fourteen elements of both sixth and seventh periods However, we know from energy level diagram for (called lanthanoids and actinoids respectively) are multielectron atoms (Unit 2) that 3d-orbitals are higher placed in separate columns at the bottom of the periodic in energy than 4s-orbitals. Consequently, 3d-orbitals table. are filled after filling 4s-orbital. Hence, this period involves the filling of only four orbitals (3s and 3p) and There are eighteen groups in the long form of contains eight elements from sodium the periodic table. Earlier, these groups were (Z = 11) to argon (Z =18). designated as IA, IB, IIA, IIB.... VIII or zero groups. According to new recommendations of the Elements of Fourth Period The fourth period corresponding to n = 4 involves the filling of one 4s-and three 4p-orbitals (4d and 4f orbitals are higher in energy than 5s-orbitals

© Mod⎯⎯→ ernGROUP = LONG FORM OF THE PERIODIC TABLE CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES IA IIA IIIB IVB VB VIB VIIIB VIIIB IB IIB IIIA IVA VA VIA VIIA VIIIA (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18) ←⎯⎯⎯⎯⎯⎯ p-block Elements ⎯⎯⎯⎯⎯⎯→ s–Block Elements ns2np1–6 Noble ns1–2 Gases Period – 1 1 H 2 He 1s2 P1s1 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ u2 3 Li 4 Be 2s1 2s2 b3 11 Na 12 Mg ←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ d-block Elements ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar l3s1 3s2 5 B 6 C 7 N 8 O 9 F 10 Ne 2s2 2p1 2s2 2p2 2s2 2p3 2s2 2p4 2s2 2p5 2s2 2p6 i4 19 K 20 Ca 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr s4s1 4s2 3d 1 4s2 3d 2 4s2 3d 3 4s2 3d 5 4s1 3d 5 4s2 3d 6 4s2 3d 7 4s2 3d 8 4s2 3d 10 4s1 3d10 4s2 4s2 4p1 4s2 4p2 4s2 4p3 4s2 4p4 4s2 4p5 4s2 4p6 (n–1)d1–10ns0–2 3s2 3p1 3s2 3p2 3s2 3p3 3s2 3p4 3s2 3p5 3s2 3p6 h5 37 R b 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 T c 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe 5s1 5s2 4d 1 5s2 4d 2 5s2 4d 4 5s1 4d 8 5s1 4d 5 5s2 4d 7 5s1 4d8 5s1 4d10 5s0 4d10 5s1 4d10 5s2 5s2 5p1 5s2 5p2 5s2 5p3 5s2 5p4 5s2 5p5 5s2 5p6 e6 55 Cs 56 Ba 57 La* 72 Hf 73 Ta 74 W 75 Re 76 O s 77 Ir 78 Pt 79 Au 80 Hg 81 TI 82 Pb 83 Bi 84 Po 85 At 86 Rn r6s1 6s2 5d1 6s2 5d 2 6s2 5d 3 6s2 5d 5 6s2 5d 5 6s2 5d 6 6s2 5d7 6s2 5d 9 6s1 5d10 6s2 5d10 6s2 6s2 6p1 6s2 6p2 6s2 6p3 6s2 6p4 6s2 6p5 6s2 6p6 s7 87 Fr 88 Ra 89 Ac** 104 Rf 105 Db 106 Sg 107 Bh 108 Hs 109 Mt 110 Ds 111Rg 112Cn 113 Nh 114 Fl 115 Mc 116 Lv 117 Ts 118 Og .7s1 7s2 6d1 7s2 6d 2 7s2 6d 3 7s2 6d 4 7s2 7s2 6d5 6d 6 7s2 6d7 7s2 6d8 7s2 6d10 7s1 6d10 7s2 7s2 7p1 7s2 7p2 7s2 7p3 7s2 7p4 7s2 7p5 7s2 7p6 Representative AElements Representative Elements ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ l⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ f-block Elements ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ ←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Transition Elements l(n–2)f 0–14( n–1) d 0–2ns 2 ←⎯⎯⎯⎯⎯⎯⎯→ ri58 Ce 59 Pr 60 Nd 61 Pm 62 Sm 63 Eu 64 Gd 65 Tb 66 Dy 67 Ho 68 Er 69 Tm 70 Yb 71 Lu gLanthanoids* 4f 2 h90 Th 91 Pa 92 U 93 Np 94 Pu 95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr 4f 3 4f 4 4f 5 4f 6 4f 7 4f 7 4f 9 4f 10 4f 11 4f 12 4f 13 4f 14 4f 14 4f1-145d0-16s2 5d0 6s2 5d0 6s2 5d 0 6s2 5d0 6s2 5d 0 6s2 5d 0 6s2 5d1 6s2 5d 0 6s2 5d 0 6s2 5d 0 6s2 5d 0 6s2 5d 0 6s2 5d 0 6s2 5d 1 6s2 tActinoids** s5f1–146d0-27s2 6d 2 7s2 6d1 7s2 6d1 7s2 6d1 7s2 6d 0 7s2 6d 0 7s2 6d1 7s2 6d 0 7s2 6d 0 7s2 6d 0 7s2 6d 0 7s2 6d 0 7s2 6d 0 7s2 6d 1 7s2 5f ° 5f 2 5f 3 5f 4 5f 6 5f 7 5f 7 5f 9 5f 10 5f 11 5f 12 5f 13 5f 14 5f 14 r←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Inner Transition Elements ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ ∗ Discovery of these elements have been claimed eserved.Fig 1. Long form of the periodic table. 3/7

3/8© MODERN'S abc + OF CHEMISTRY–XI Modern Publishers. All rights reserved. and are filled later). In between 4s-and 4p-orbitals, five This also provides a theoretical justification for 3d-orbitals are filled which have energies in between periodicity occurring at regular intervals of 2, 8, 8, these orbitals. Thus, in all nine orbitals are to be filled 18, 18 and 32. These numbers i.e., 2, 8, 18 and 32 are and, therefore, there are eighteen elements in fourth also called magic numbers. The number of elements period from potassium (Z = 19) to krypton (Z = 36). In and the corresponding orbitals being filled are given this period, we come across elements which involve in Table 5. the filling of 3d orbitals. These are known as transition series of elements. This starts from scandium (Z = The first three periods containing 2, 8 and 8 21) which has electronic configuration 3d1 4s2 and ends elements respectively are called short periods, the at zinc (Z = 30) with 3d orbitals completely filled having next three periods containing 18, 18 and 32 elements electronic configuration 3d10 4s2. respectively are called long periods. Groups Elements of Fifth Period A vertical column in the periodic table is The fifth period, like the fourth period also known as group. consists of 18 elements. It begins with rubidium In terms of electronic structure of the atom, a (Z = 37) with filling of 5s-orbital and ends at xenon group constitutes a series of elements whose atoms (Z = 54) with the filling up of the 5p-orbitals. This period have the same outermost electronic configurations. also contains ten elements of 4d transition series There are 18 groups in the long form of the periodic starting at yttrium (Z = 39). table. According to the new recommendations of International Union of Pure and Applied Chemistry Elements of Sixth Period (IUPAC), the groups are numbered from 1 to 18. Previously, these were numbered only from I to VIII The sixth period contains 32 elements (Z = 55 as A and B groups. Both the systems of numbering the to 86) and successive electrons enter into 6s, 4f, 5d groups is given in the periodic table. However, the old and 6p-orbitals, in that order. It starts with cesium convention is still used in many books. (Z = 55) and ends at radon (Z = 86). In addition to ten The first two groups on the extreme left and last elements of 5d transition series, this period contains six groups on the extreme right involve the filling of 14 elements which involve filling up of 4f orbitals s-and p-orbitals, respectively. These groups represent beginning from cerium (Z = 58) to lutetium (Z = 71). the main groups of the periodic table and are numbered The series of elements are called inner transition as 1, 2, 13, 14, 15, 16, 17 and 18 corresponding to series or lanthanoid series. general configurations of ns1, ns2, ns2 np1, ns2 np2, ns2 np3, ns2 np4, ns2 np5, ns2 np6, respectively (in the old Elements of Seventh Period system these were denoted as IA, IIA, IIIA, IVA, VA, VIA, VIIA and VIIIA (or zero) respectively corresponding The seventh period, is similar to the sixth to general configurations of ns1, ns2, ns2 np1,..........ns2 period with successive filling up of the 7s, 5f, 6d and 7p np6 respectively. The elements present in these groups orbitals and includes most of the man made radio- are known as normal or representative elements. active elements. This period ends at the element with The first three periods containing 2, 8 and 8 atomic number 118, which would belong to the noble electrons involve the filling of only s-and gas family. Some of the elements of this period have p-subshells. But the fourth period involves the filling of been recently established and named. This period also 3d-subshells also and this results into ten more groups. contains 14 elements which involve the filling of 5f These ten groups lie in between first two and last six orbitals starting from actinium (Z = 89). This is also groups, (see periodic table) i.e., between group 2 (IIA) called 5f– inner transition series or actinoid series. and group 13 (IIIA) because 3d-orbitals are filled after Table 5. Number of elements (corresponding to electrons) in each period. Period (n) Orbitals being filled up Number of electrons or elements in the period First Second (1) 1s 2 =2 Third (2) 2s 2p 2+6 =8 Fourth (3) 3s 3p 2+6 =8 Fifth (4) 4s 3d 4p 2 + 10 + 6 = 18 Sixth (5) 5s 4d 5p 2 + 10 + 6 = 18 Seventh (6) 6s 4f 5d 6p 2 + 14 + 10 + 6 = 32 (7) 7s 5f 6d 7p 2 + 14 + 10 + 6 = 32

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 3/9 4s-orbitals but before 4p-orbitals. These are numbered a systematic nomenclature for naming the new from 3 to 12 (in the old system these were usually elements until their names are officially recognized. labelled as B groups i.e., IB, IIB, IIIB,...etc.). The This nomenclature is based on the Latin words for elements present in these groups are called transition their numbers. elements. The name is derived from the fact that they represent transition (change) in character from reactive The IUPAC Rules for Nomenclature of metals (elements of groups 1 and 2) on one side to the Elements with Z > 100 are : non-metals (elements of group 13 to 18) on the other side. (i) The name of the element should be derived directly from the atomic number using the numerical There are two more rows at the bottom of the roots for their numbers. For example, nil for 0, un for periodic table. These rows consist of fourteen elements 1, bi for 2, tri for 3 and so on as given in Table 6. after lanthanum (Z = 57) and fourteen elements which© follow actinium (Z = 89). These are placed separately inModern Publishers. All rights reserved.Table 6. Notations for numbers (word root) used by the periodic table to save space and avoid undue IUPAC Nomenclature of Elements. sidewise expansion of the periodic table. This arrangement also helps in keeping elements with Digit Root Name Abbreviation similar properties in a single column. The elements in the first row, starting from cerium are called 0 nil n lanthanoids (or lanthanides) and the elements 1 un u present in the second row starting from thorium are 2 bi b called actinoids (or actinides). These lanthanoids 3 tri t and actinoids together are called inner transition 4 quad q elements or rare earth metals and these are built up 5 pent p by filling of f-orbitals. 6 hex h 7 sept s NOMENCLATURE OF ELEMENTS WITH 8 oct o ATOMIC NUMBER MORE THAN 100 9 enn e Today, we have elements upto 118. These elements were earlier named traditionally by its (ii) The names are derived by combining together discoverer or discoverers. However, in recent years, the roots in the order of digits which make up the disputes regarding the original discoveries of some atomic number and ium is added at the end. elements of atomic number 104 and above have arisen. This is because the new elements with very high (iii) In certain cases, the names are shortened. atomic numbers are very unstable and only minute For example, if enn occurs before nil, the second n quantities of elements are obtained. This has led to should be dropped and written as ennil. Similarly, the controversy in deciding the claims of more than one final i of bi and tri should be dropped when these occur scientist for the discovery and names assigned by before ium as bi + ium ⇒ bium, tri + ium = trium them for the new element discovered. (iv) The symbol of the element is obtained from For example, both American and Soviet scientists the first letters of the numerical roots (abbreviations) claimed credit for discovery of element 104. The of the digit which make up the atomic number of the Americans named it Rutherfordium (symbol Rf) while element. Soviets named it Kurchatovium (symbol Ku). To overcome this problem, the IUPAC in 1997 gave the For example, for atomic number 104, the name approved official names for elements with atomic may be written as : numbers 104 to 109. The IUPAC has also recommended un (1) + nil (0) + quad (4) + ium ⎯→ unnilquadium and is assigned the symbol Unq. The IUPAC names for the elements with atomic number more than 100 are given in Table 7. Table 7. Nomenclature of elements with atomic number above 100. Atomic Name Symbol IUPAC IUPAC Number official name symbol 101 Unnilunium Unu Mendelevium Md Nobelium Nb 102 Unnilbium Unb Lawrencium Lr Rutherfordium Rf 103 Unniltrium Unt Dubnium Db Seaborgium Sg 104 Unnilquadium Unq Bohrium Bh Hassium Hs 105 Unnilpentium Unp Meitnerium Mt Darmstadtium Ds 106 Unnilhexium Unh Rontgenium Rg 107 Unnilseptium Uns 108 Unniloctium Uno 109 Unnilennium Une 110 Ununnilium Uun 111 Unununium Uuu

3/10 Ununbium Uub Copernicium MODERN'S abc + OF CHEMISTRY–XI *112 Ununtrium Uut Nihonium Cn* Ununquadium Uuq Flerovium Nh **113 Ununpentium Uup Moscovium Fl* *114 Ununhexium Uuh Livermorium Mc **115 Ununseptium Uus Tennessine Lv* *116 Ununoctium Uuo Oganesson Ts **117 Og **118 © * The element 112 has been named as copernicium with chemical symbol Cn. It is named in honour of scientist astronomerModern Publishers. All rights reserved. Nicolaus Copernicus. The element 114 is named Flerovium with chemcial symbol Fl to honour the Flerov Laboratory of Nuclear Reactions and the element 116 is named Livermorium with chemical symbol Lv to honour the Lawrence Livermore National Laboratory. ** Recently, IUPAC have confirmed the discovery of four new elements with atomic numbers 113, 115, 117 and 118 and named them. Therefore, the seventh row of the periodic table is now complete. DIVISION OF ELEMENTS INTO s, p, d AND General characteristics of s-block elements f-BLOCKS The general characteristics of s-block elements are : The long form of the periodic table can be divided (i) They are soft metals having low melting and into four main blocks. These are s, p, d and f-blocks. boiling points. This division of elements is based upon the electronic configurations of the atoms. In this division, the (ii) They have low ionisation enthalpies. elements which involve the filling of a particular (iii) They are very reactive and readily form orbital (i.e., s, p, d or f) are grouped together as shown in Fig. 2. (These are also shown in Long form of the univalent (alkali metals) or bivalent (alkaline Periodic Table. earth metals) ions by losing one or two valence electrons respectively. Because of their high ns1–2 ns2 np1–6 N reactivity, they are never found pure in nature. O (iv) They act as strong reducing agents. (n – 1)d1–10 ns0–2 B (v) Most of them impart characteristic colours to L the flame. E (vi) They mostly form ionic compounds except lithium and beryllium. G (ii) p-block elements A The elements in which the last electron enters the S p-orbital of their outermost energy level are called E p-block elements. The elements of groups 13 to 18 S involving addition of one (ns2 np1), two (ns2 np2), three (ns2 np3), four (ns2 np4), five (ns2 np5), and six (ns2 np6) (n – 2)f1–14 (n – 1)d0–2 ns2 electrons respectively in p-orbitals and s-orbitals are already filled in their atoms constitute p-block. The Fig. 2. Division of periodic table into s, p, d and f-blocks. general electronic configuration for the atoms of this (i) s-block elements block may be written as : The elements in which the last electron enters the General electronic configuration of p-block s-orbital of their outermost energy level are called elements : ns2 np1–6 s-block elements. It consists of elements of groups 1 General characteristics of p-block elements. and 2 having the ground state electronic configurations The general characteristics of p-block elements are : of outermost shell as ns1 and ns2 respectively (where n (i) They include both metals and non-metals. There stands for outermost energy shell). The elements corresponding to ns1 configuration are called alkali is a regular gradation from metallic to non- metals while those corresponding to ns2 configuration metallic character as we move along a period are called alkaline earth metals. Thus, the general from left to right in this block. The metallic electronic configuration of s-block elements may be character increases as we go down the group. expressed as : (ii) Their ionisation enthalpies are relatively high as compared to s-block elements. General electronic configuration of s-block (iii) They form mostly covalent compounds. elements : ns1–2 (iv) Some of them show more than one oxidation states in their compounds. (v) There occurs a gradation from reducing to oxidising properties as we move across the periods of this block.

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 3/11 © The elements of s- and p-block are collectively Fourth transition series Modern Publishers. All rights reserved.called representative elements. The elements of last Actinium (Z = 89), Rutherfordium (Z = 104) to group (18) having ns2np6 configuration are called noble gases. All the orbitals in the valence shell of the noble Copernicium (Z = 112) gases are completely filled and they have no tendency (7th period : 6d-orbitals are gradually filled) to lose or gain electrons. Therefore, the noble gases General characteristics of d-block elements. exhibit very low reactivity. Preceding the noble gas family are two chemically important groups of non- The general characteristics of d-block elements are : metals. These are halogens (group 17) and (i) They are metals having high melting and chalcogens (group 16). These two groups of elements boiling points. can readily accept one and two electrons respectively (ii) Most of them form coloured compounds. to attain noble gas configurations and form univalent and divalent negative ions. In these elements, the non- (iii) They have a good tendency to form complex metallic character increases as we move from left to compounds. right across a period and metallic character increases as we go down the group. (iv) Their compounds are generally paramagnetic. (v) They exhibit several oxidation states. (iii) d-block elements (vi) Most of the transition elements such as Mn, The elements in which the last electron enters Ni, Co, Cr, V, Pt and their compounds are the d-orbitals of their last but one (called penultimate) used as catalysts. energy level constitute d-block elements. This block (iv) f-block elements consists of the elements lying between s and p blocks starting from fourth period and onwards. They The elements in which the last electron enters constitute groups 3 to 12 in the periodic table. In these the f-orbitals of their atoms are called f-block elements the outermost shell contains one or two elements. In these elements, the last electron is added electrons in their s-orbital (ns) but the last electron to the third to the outermost (called antipenultimate) enters the last but one d-subshell i.e., (n – 1)d. The energy level; (n – 2) f. These consist of two series of general electronic configuration for the atoms of d- elements placed at the bottom of the periodic table : block may be written as : (i) The first series follows lanthanum, La (Z=57) General electronic configuration : (n – 1)d1–10 ns0–2 and the elements present in this series (58Ce – 71Lu) are called lanthanoids or lanthanides. These are Thus, it is the d-orbital which is progressively also called rare earth elements. being filled in the elements of this block. These elements are also called transition elements because (ii) The second series follows actinium, the properties of these elements are midway between Ac (Z = 89) and the elements present in this series those of s-block and p-block elements. In a way, a(9r0eThof–r1a0d3Liora)catriveecanlaletduraec.tinoids or actinides. These transition metals form a bridge between the chemically active metals of s-block elements and the less active The general electronic configuration of f-block elements of Groups 13 and 14. Therefore, they elements may be written as : represent transition (change) in behaviour and take their familiar name “transition elements.” General electronic configuration : (n – 2)f1–14 (n – 1)d0 – 2 ns2 The d-block comprises three complete rows of ten elements and one incomplete row. These rows are For example, for lanthanoids, the general called first, second and third transition series which electronic configuration is 4f1—14 5d0—1 6s2 while for involve the filling of 3d, 4d and 5d-orbitals respectively. actinoids, it is 5f1—14 6d0 – 2 7s2. These series are also called transition series. The elements included in these two series are First transition series called inner transition elements, because they form Scandium (Z = 21) to Zinc (Z = 30) transition series within the transition elements of (4th period : 3d-orbitals are gradually filled) d-block. Second transition series Yttrium (Z = 39) to Cadmium (Z = 48) General characteristics of f-block elements (5th period : 4d-orbitals are gradually filled) The general characteristics of f-block elements are : Third transition series Lanthanum (Z = 57), Hafnium (Z = 72) to (i) They are heavy metals. Mercury (Z = 80) (6th period : 5d-orbitals are gradually filled) (ii) They generally have high melting and boiling points. (iii) They exhibit variable oxidation states. (iv) They form coloured ions. (v) They have the tendency to form complex compounds. (vi) Actinoids are radioactive in nature. Many of the actinoid elements have been made only in nano gram quantities or even less by nuclear reactions

3/12 MODERN'S abc + OF CHEMISTRY–XI and their chemistry is not fully studied. The in the Periodic Table. It may be noted that the change elements after uranium are called from metallic to non-metallic character is not abrupt transuranium elements. as shown by zig-zag line in Fig 3 ahead. The elements bordering this line (silicon, germanium, arsenic, On the basis of above four blocks, the elements antimony, tellurium, etc.) show characteristic can be classified into the following four general types : properties of both metals and non-metals. These are © called metalloids or semimetals. Modern Publishers. All rights reserved.Classification of Elements into four General Types Advantages of the Long Form of the Periodic Table 1. Noble gases. The noble gases are found at the end of each period in group 18. Except for helium, The important advantages of the long form of the elements have completely filled s- and p-orbitals the periodic table are given below : of the outermost shell i.e., ns2 np6. Helium has 1s2 configuration. All these elements are highly stable and (i) This classification is based on the atomic number chemically inert under ordinary conditions. which is a more fundamental property of the elements. 2. Representative elements (s and p-block elements). All the elements of s and p-block with (ii) Since this classification is based on the atomic exception of noble gases are called representative number and not on the atomic mass, the position elements. They represent two groups 1 (alkali metals) of placing isotopes at one place is fully justified. and 2 (alkaline earth metals) on the extreme left and five groups from group 13 to 17 on the right hand side (iii) The position of elements in the periodic table is of the periodic table. governed by the electronic configurations, which determine their properties. 3. Transition elements. The elements of d- block elements are called transition elements. These (iv) It is easy to remember and reproduce. include elements of group 3 to 12 lying in between the (v) The systematic grouping of elements into four representative elements (between s and p-blocks). blocks; s, p, d and f has made the study of the 4. Inner transition elements. The elements elements more simple. of f-block are called inner transition elements. These (vi) The position of some elements which were misfit comprise two series of 14 elements called lanthanoids on the basis of atomic mass is now justified on and actinoids. the basis of atomic number. For example, argon proceeds potassium because argon has atomic Classification as Metals and Non-metals number 18 and potassium has 19. (vii) The lanthanoids and actinoids which have In addition to classifying the elements into properties different from other groups are placed s-, p-, d- and f–block, the elements can be classified on separately at the bottom of the periodic table. the basis of their properties as : Defects of the Long Form of the Periodic Table (i) Metals and (ii) Non-metals Although the Long Form of the periodic table has helped in systemising the study of chemistry of Metals comprise more than 78% of all known elements, yet it has certain defects. The main defects of this table are : elements and appear on the left hand side of the (i) The position of hydrogen is not settled. It Periodic Table. They have high melting and boiling resembles with alkali metals as well as halogens. points and are generally solid at room temperature (ii) Lanthanoids and actinoids have not been (mercury is an exception which is a liquid at room accommodated in the main body of the periodic temperature). Gallium and caesium also have very low table. melting points, 303 K and 302 K respectively. They Prediction of Period, Group and Block of an Element are good conductors of heat and electricity. They are malleable (can be flattened into thin sheets) and ductile The period, group or block of an element can be predicted from the electronic configuration of (can be drawn into wires). They have bright lustre. the elements. On the other hand, non-metals are present on 1. Period of the element corresponds to the the right hand side of the periodic table. They are principal quantum number of the valence shell. usually solids or gases at room temperature with low melting and boiling points (boron and carbon are 2. Block of the element corresponds to the subshell which receives the last electron. exceptions). The only liquid non-metal is bromine. Most non-metals are brittle and are neither malleable nor ductile. In general, the elements become more metallic as we go down a group and non-metallic character increases as we move from left to right across a period

© Modern METALLIC CHARACTER INCREASES GroupPuIst Period1 18 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES bl2nd Period1 2 2 H 13 14 15 16 17 He ish3rd Period3 4 Transition Elements 5 6 7 8 9 10 Li Be B C N O F Ne 11 12 3 4 13 14 15 16 17 18 22 Na Mg 5 6 7 8 9 10 11 12 Al Si P S Cl Ar e4th Period Ti 19 20 21 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr r5th Period 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 sRb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe . A6th Period 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La* Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn ll7th Period 87 88 89 104 105 106 107 108 109 111 110 112 113 114 115 116 117 118 Fr Ra Ac** Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og riMETALLIC CHARACTER INCREASES ghMetals tMetalloids 67 68 58 59 60 61 62 63 64 65 66 69 70 71 Ho Er Ce Pr Nd Pm Sm Eu Gd Tb Dy Tm Yb Lu sNon-metals Lanthanoids re*Discoveries of these elements have been claimed 90 91 92 Actinoids 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr served.Fig 3. Position of metals, non-metals and metalloids in the periodic table. 3/13

3/14 MODERN'S abc + OF CHEMISTRY–XI 3. Group is predicted from the number of Solution: For n = 4, the electronic configuration is electrons in the outermost shell or 3d14s2. The element belongs to d-block elements. penultimate (last but one, i.e., n–1) shell as Group number = No. of electrons in (n – 1)d subshell follows : + No. of electrons in nth shell For s-block element, group number is equal to number of =1+2=3 valenceelectrons(nselectrons). Example 6. (a) Elements A, B, C and D have atomic numbers For p-block element, group number is equal to 12, 19, 29 and 36 respectively. On the basis of 10 + number of valence electrons (ns and np). electronic configuration, write to which group of the periodic table each element belongs. For d-block element, group number is equal to total (b) Predict the blocks to which these elements can number of electrons in (n–1)d and ns subshells. be classified. Also predict their periods and groups. (c) Which of these are representative elements ? Solution: (a) ©Example 1. Modern Publishers. All rights reserved. What would be the IUPAC name and symbol of the element with atomic number 120 ? Element At. No. Electronic Group configuration Solution: The roots for 1, 2 and 0 are un, bi and nil respectively. A 12 1s2 2s2 2p6 3s2 2 ∴ Name of element : Unbinilium B 19 1s2 2s2 2p6 3s2 3p6 4s1 1 Symbol : Ubn C 29 1s2 2s2 2p6 3s2 3p6 3d10 4s1 11 D 36 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 18 Example 2. How would you justify the presence of 18 elements [According to old system, these elements belong to in the 5th period of the periodic table ? groups : IIA (A), IA (B), IB (C), VIIIA or zero (D)] Solution: The 5th period involves the filling of 5th (b) Prediction of blocks shell (n=5). This period has 4d, 5s and 5p available orbitals. The total number of available orbitals are 9 and therefore (i) A receives the last electron in 3s-subshell, therefore, it the maximum number of electrons that can be accommodated belongs to s-block. is 18. Thus, 5th period contains 18 elements. Period = 3rd Example 3. Group = No. of valence electrons = 2 (ii) B receives the last electron in 4s-orbital, therefore, it The elements Z = 107, 108 and Z = 109 have been belongs to s-block. made recently. Indicate the group in which you will place the above elements. Period = 4th Solution: The electronic configurations of these Group = No. of valence electrons = 1 elements are : (iii) C receives the last electron in 3d-orbital, therefore, it Z = 107 [Rn] 5f14 6d5 7s2 belongs to d-block. Z = 108 [Rn] 5f14 6d6 7s2 Period = 4th Z = 109 [Rn] 5f14 6d7 7s2 Group = No. of electrons in ns and (n—1) d subshells = 11 These elements will be placed in d-block in groups 7th, 8th and 9th respectively. (iv) D receives the last electron in the 4p-orbital, therefore, it belongs to p-block. Example 4. Arrange the following elements in the increasing Period = 4th order of metallic character : Group = 10 + valence electrons = 10 + 8 = 18 Si, Be, Mg, Na, P. Solution: We know that metallic character increases (c) The elements A and B are representative elements. down a group and decreases along a period as we move from left to right. Example 7. Hence the order of increasing metallic character is P < Si < Be < Mg < Na. The electronic configurations of some elements are given below : Example 5. (i) 1s2 2s2 2p6 3s1 Predict the position of the element in the periodic (ii) 1s2 2s2 2p6 3s2 3p6 4s2 3d1 table satisfying the electronic configuration (iii) 1s2 2s2 2p3 (n-1)d1ns2 for n = 4

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 3/15 (iv) 1s2 2s2 2p6 3s2 3p6 4s2 (v) 1s2 2s2 2p6 3s2 3p5 (vi) 1s2 2s2 2p6 3s2 3p3  1. Write the name and deduce the atomic numbers of the following atoms : (vii) 1s2 2s2 2p6 (viii) 1s2 2s2 2p6 3s2 3p4 © (i) The third alkali metal. Modern Publishers. All rights reserved.Name the elements. Out of these which (ii) Second transition element. 1. is an alkaline earth metal 2. has lowest chemical reactivity 3. belong to group 15 of the periodic table (iii) The fourth noble gas. 4. is a transition element 5. is halogen 6. belong to p-block 7. belong to second period. (iv) Fourth element in the second period. Solution:  2. The outer electronic configurations of some elements are : Names of elements: (i) 3s2 3p4 (ii) 3d10 4s2 (iii) 3s2 3p6 4s2 (iv) 6s2 4f 3 (i) Sodium (Na) (ii) Scandium (Sc) (iii) Nitrogen (N) (iv) Calcium (Ca) State to which block in the periodic table each of these elements belong. (v) Chlorine (Cl) (vi) Phosphorus (P) (vii) Neon (Ne) (viii) Sulphur (S).  3. How many elements are present in the third period of the p-block elements ? 1. Calcium (iv) is an alkaline earth metal. 2. Neon (vii) has lowest chemical reactivity.  4. Complete the following statements : 3. Nitrogen (iii) and phosphorus (vi) belong to group 15 of the periodic table. (a) There are.........................periods in the long form of the periodic table. 4. Scandium (ii) is a transition element. 5. Chlorine (v) is halogen. (b) The s block element having highest atomic mass is ..................... . 6. Nitrogen (iii), chlorine (v), phosphorus (vi), neon (vii) and sulphur (viii) belong to p-block. (c) The elements Cu, Ag and Au are called ....................... metals. 7. Nitrogen (iii) and neon (vii) belong to second period of the periodic table.  5. An element X belongs to the third period of the p-block elements. It has 4 electrons in the outermost Example 8. shell. Name the element. Element A, B, C, D and E have the following  6. Which family of elements has the electronic electronic configurations : configuration ns2 np4 ? A : 1s22s22p1 B : 1s22s22p63s23p1  7. The element 118 has not yet been discovered. What C : 1s22s22p63s23p3 is its IUPAC name and symbol ? D : 1s22s22p63s23p5  8. Arrange the following elements in the increasing order of metallic character : E : 1s22s22p63s23p64s2 Which among these will belong to the same group B, Al, Mg, K. in the periodic table ?  9. Arrange the following elements in the increasing Solution: We know that the elements having similar order of non-metallic character : valence electronic configuration belong to the same group of the periodic table. Therefore, elements A and B having three B, C, Si, N and F. electrons in the valence shell belong to the same group i.e., group 13 of the periodic table.   10. Rn (Z = 86) is the last noble gas discovered. Predict what will be the atomic number of the next noble gas to be discovered. Write its symbol. 1. (i), K, 19 (ii) Ti, 22 (iii) Kr, 36 (iv) C, 6. 6. Oxygen family. This family is also called chalcogens. 2. (i) p (ii) d (iii) s (iv) f. 7. Ununoctium, Uuo 3. Six. 8. B < Al < Mg < K 4. (a) 7 (b) Ra (c) Coinage. 9. F < N < Si < C < B 5. Silicon. 10. Atomic number of next noble gas to be discovered will be 118. Symbol : Uuo

3/16 MODERN'S abc + OF CHEMISTRY–XI 1 ©Q.1. On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements. Modern Publishers. All rights reserved. Ans. The sixth period of the periodic table begins with n=6. It involves the filling of 6s, 4f, 5d and 6p subshell in the increasing order of energy. The total number of orbitals available are 16 and therefore, the maximum number of electrons that can be accommodated is 32. Thus, the sixth period of the periodic table should have 32 elements. Q.2. In terms of period and group where would you locate the element with Z = 114 ? Ans. The electronic configuration of the element with Z=114 would be [Rn] 5f14 6d10 7s2 7p2. Since it has n=7 for the valence shell, it belongs to 7th period. It receives the last electron in p-orbital. Therefore, it belongs to p-block. The group number will be 10 + 4 (No. of electrons in ns and np orbitals) = 14 So, it belongs to 7th period and 14th group. Q.3. The elements Z = 117 and 118 have been recently discovered and named. In which family group would you place these elements and also give electronic configuration in each case. Name these elements. Ans. The electronic configuration of the element with Z = 117 would be [Rn] 4f14 5d10 7s2 7p5. It has outermost ns2np5 configuration and therefore, it belongs to halogen family or Group 17. IUPAC name: Tennessine, Ts The electronic configuration of the element with Z=118 would be [Rn] 4f14 5d10 7s2 7p6. It has outermost ns2np6 configuration and therefore, belongs to noble gases or group 18. IUPAC name: Oganesson, Og. Q.4. Write the atomic number of the element present in the third period and sixteenth group of the periodic table. Give its name. Ans. Since it belongs to 3rd period, it will have outershell as n = 3. Its configuration would be [Ne] 3s2 3p4. So its atomic number = 16. It is sulphur. Q.5. Give the general electronic configurations of (i) p-block (ii) actinoids Ans. (i) ns2np1 –6 (ii) 5f1–14 6d0–2 7s2 Q.6. An element ‘X’ with atomic number 112 has been recently predicted. Its electronic configuration is : [Rn] 5f 14 6d10 7s2. Predict (i) group and (ii) block in which this element would be placed (iii) IUPAC name and symbol. Ans. The configuration of the element is : [Rn] 5f14 6d10 7s2 (i) It belongs to 12th group. (ii) It belongs to d block. (iii) IUPAC name is : Copernicium; Symbol : Cn. Q.7. The element 119 has not been discovered so far. What would be the IUPAC name and the symbol for this element ? On the basis of periodic table, predict the electronic configuration of this element and the formula of its most stable chloride and oxide. (NCERT Exemplar Problem) Ans. IUPAC name : Ununennium Symbol : Uue Electronic : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 configuration 4d10 4f14 5s2 5p6 5d10 5f14 6s2 6p6 6d10 7s2 7p6 8s1 or [At] 8s1 Formula : Chloride : ACl : Oxide : A2O

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 3/17 Q.8. Eka-aluminium and eka-silicon were named given by Mendeleev for the then unknown elements gallium and germanium respectively. A recently discovered element has first named as eka-mercury. What is its atomic number? Write its group number, electronic configuration, IUPAC name, official name and symbol. Ans. The element which comes after mercury in the periodic table is eka-mercury. Its atomic number = 80 + 32 = 112 Electronic configuration: [Rn] 5f14 6d10 7s2 © Modern Publishers. All rights reserved.IUPAC name : Uub Official name: Copernicium, symbol : Cn Q.9. Write the IUPAC names, official names and symbols for the elements having atomic numbers 108, 114 and 116. Ans. Z = 108 Unniloctium Uno ; Hassium Hs Z = 114 Ununquadium Uuq ; Flerovium Fl Z = 116 Ununhexium Uuh ; Livermorium Lv CAUSES OF PERIODICITY Table 8. Electronic configurations of alkali metals. The recurrence of similar properties of the Element At. No. Electronic configuration elements after certain regular intervals when they are arranged in the order of increasing Li 3 1s22s1 atomic numbers is called periodicity. Na 11 1s22s2 2p6 3s1 K 19 1s22s2 2p6 3s2 3p6 4s1 The classification of the elements in certain Rb 37 1s22s2 2p6 3s2 3p6 3d10 4s2 4p6 5s1 families was based upon the observed similarities Cs 55 1s22s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s25p6 in the physical and chemical properties. But, why do certain elements exhibit similar properties and 6s1 why do the similar properties recur at regular Fr 87 1s22s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s25p6 intervals of atomic numbers, were some important questions yet to be answered. 6s2 4f14 5d10 6p6 7s1 Thus, the cause of periodicity of the We know that an atom consists of a nucleus properties of elements is surrounded by electrons. In ordinary chemical reactions, the repetition of similar electronic the nuclei of the atoms do not undergo any change and, configuration of their atoms in the outermost therefore, it is the distribution of the electrons in the energy shell (or valence shell) after certain various shells of their atoms that determine the physical regular intervals. and chemical properties of the elements. Further, it For example, all the alkali metals have a great has been also found that the properties of atoms do not tendency to lose the single electron in order to acquire depend so much on the arrangement of electrons in the a stable noble gas configuration. Consequently, all of inner shells. Thus, it must primarily be the them are very reactive metals. arrangement of electrons in the outermost shell (called Similarly, all the members of the halogen family the valence shell) that determines the properties of the have two electrons in the s-orbital and five electrons atoms. For example, the electronic configurations of in the p-orbital of the valence shell (Table 9). The the atoms of alkali metals group (Table 8) shows that general electronic configuration of halogens may be they all have one electron in the s-orbital of the expressed as ns2 np5. outermost valence shell preceded by the noble gas configuration (ns2 np6, but for Li, ns2). The general Table 9. Electronic configurations of halogens. electronic configuration of these may be written as (noble gas) ns1 : where n stands for principal quantum Element At. No. Electronic configuration number. F 9 1s22s2 2p5 Cl 17 1s22s2 2p6 3s2 3p5 Br 35 1s22s2 2p6 3s2 3p6 3d10 4s2 4p5 I 53 1s22s22p63s23p63d104s24p64d105s25p5