Modern ABC Chemistry XI

3/68 MODERN'S abc + OF CHEMISTRY–XI D10. The electronic configuration of the element which (c) 1s2 2s2 2p6 3s2 3p6 3d5 4s2 4p6 is just above the element with atomic number (d) 1s2 2s2 2p6 3s2 3p6 3d5 4s2 43 in the same group is (a) 1s2 2s2 2p6 3s2 3p6 3d7 4s2 (b) 1s2 2s2 2p6 3s2 3p6 3d6 4s2 © D11. Which of the elements whose atomic numbers Modern Publishers. All rights reserved. are given below, cannot be accommodated in the present set up of the long form of the periodic table? (a) 118 (b) 107 (c) 102 (d) 126 Passage-II D10. (d) D11. (d) Matrix Match Type Questions 2. Match the property given in Column I with the property given in Column II Each question contains statements given in two columns, Column I Column II which have to be matched. Statements in Column I are labelled as A, B, C and D whereas statements in Column II (A) Ionisation enthalpy (p) decreases down the group are labelled as p, q, r and s. Match the entries of Column I (B) Negative electron (q) increases down the group with appropriate entries of Column II. Each entry in Column gain enthalpy I may have one or more than one correct option from Column (C) Metallic character (r) increases from left to right II. The answers to these questions have to be appropriately in a period (in general) bubbled as illustrated in the following example. (D) Atomic size (s) decreases from left to right in a period (in general) If the correct matches are A-q, A-r, B-p, B-s, C-r, C-s 3. Electronic configuration of some elements is given in Column I and their first ionization and D-q, then the correctly bubbled matrix will look like as enthalpies are given in Column II. Match the electronic configuration with first ionization shown: p qr S enthalpy. A p qr S B p qr S Column (I) Column (II) C p qr S D p qr S Electronic First ionization configuration enthalpy/kJ mol–1 1. Match the elements/ions in Column I with the (A) 1s2 2s2 2p6 (p) 495 characteristics in Column II from left to right. (B) 1s2 2s2 2p6 3s1 (q) 1681 (C) 1s2 2s2 2p5 (r) 1314 Column I Column II (D) 1s2 2s2 2p4 (s) 2080 (A) Cl, Br, I (B) B, C, N (p) ionisation enthalpy increases (q) negative electron gain enthalpy (C) O, O–, O2– (D) Cl, F, P decreases (r) atomic size decreases (s) belong to same period (1): (A) – (q) (B) – (p), (r), (s) (C) – (q) (D) – (q) (2) : (A) – (p), (r) (B) – (p), (r) (C) – (q), (s) (D) – (q), (s) (3): (A) – (s) (B) – (p) (C) – (q) (D) – (r) Hints & Explanations for Objective Type Questions Difficult A. Topicwise Multiple Choice Questions A6. (d) : The statements (a), (b) and (c) are correct but statement (d) is incorrect because it belongs to A3. (d) : 3(1s22s1), 12(1s2 2s2 2p6 3s2) have electron in 7th period (outermost shell). outermost s– subshell. A7. (c) : Mt : [Rn] 5f14 6d7 7s2 : d-block because the A5. (c) : d-block elements have maximum tendency to form outer most electron is in d-subshell. complexes. A8. (a) : All p-block elements have 6 elements in a period.

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 3/69 A9. (c) : The elements having similar valence electronic B15. (a) : Electronic configuration of the element, Z = 114 is configuation belong to same group of the periodic [Rn]86 5f14 6d10 7s2 7p2 table. Therefore, A and B having 3 electrons in the It belongs to group 14 i.e. carbon family. valence shell belong to same group. B16. (d) : X having ground state configuration 1s22s22p3 is N A13. (a) : P3–, S2– and Cl– have same number of electrons and its valency is –3. (18) and size decreases as nuclear charge Formula: Mg+2X–3 or Mg3X2 increases. Cl– < S2– < P3– B18. (c) : The configuration corresponds to that of Cl, which has the highest negative electron gain enthalpy. A17. (c) : It corresponds to Cl and has largest negative electron gain enthalpy. B19. (e) : Alkali metals have lowest value of ionisation enthalpy in a period. On moving down the group A28. (c) : Li (g) + I.E ⎯⎯→ Li+(g) + e–(g) from top to bottom, the ionisation enthalpy decreases. Hence from the graph, M has least ionisation enthalpy. B20. (b) : Ionic radius increases from top to bottom and decreases on moving from left to right in a periodic table. © Moles of Li in 7 mg = 7 × 10–3 = 1 × 10–3 mol Modern Publishers. All rights reserved. 7 B22. (c) : The most reactive metal will be the one having lowest ionisation enthalpy and the least reactive Amount of energy needed to convert 1 × 10–3 mol of non-metal will be the one having lowest or positive Li atoms electron gain enthalpy. 520 × 103 J × 1 × 10–3 = 520 J A33. (d) : Amount of energy released for 1 × 106 atoms B23. (b) : For isoelectronic ions, the ionic radius decreases = 4.9 × 10–13 J Amount of energy released for 6.022 × 1023 (1 mol) atoms = 4.9 × 10–13 × 6.022 × 1023 with increase in nuclear charge. 1 × 106 Correct order is : O2– (+8) > F– (+9) > Na+ (+11) > Mg2+ (+12) >Al3+ (+13) = 29.5 × 104J = 295 kJ mol–1 B24. (b) : The configurations of atoms and ions after removing A34. (c) : Chlorine has the highest negative electron gain enthalpy. It belongs to period 3 and group 17. one electron are : Li+ : 1s2 Li : 1s2 2s1 A35. (c) : The fourth ionisation enthalpy is abruptly Be : 1s2 2s2 Be+ : 1s2 2s1 increased, which means that the electron is B : 1s2 2s2 2p1 B+ : 1s22s2 removed from noble gas configuration. C : 1s2 2s2 2p2 C+ : 1s22s22p1 B5. (c) : 117 : [Rn] 5f146d107s27p5 : It will belong to halogen Hence the order of IE2 is Li > B > C > Be family. B25. (d) : Size decreases down the group. B7. (d) : Group 17 > Group 16 B26. (c) : Phosphorus belongs to group 15 and third period. Group 17 → Cl > F > Br > I Group 16 → S > Se > Te > Po > O B35. (d) : Larger is the negative charge and size of anion, ∴ The correct order is : O < S < F < Cl higher is its polarising power. B8. (a) : For isoelectronic species, smaller the positive charge B36. (b) : In isoelectronic ions, with increase in nuclear charge, on the cation, larger is the ionic radius. size decreases, B10. (c) : For option (a), the order is H– > H > H+, for option Therefore, Ca2+ < K+ < Cl– < S2–. (b) the order is O2– > F– > Na; for option (d), the order is N3– > Mg2+ > Al3+. B37. (c) : Isoelectronic species are those which have same Hence, (c) is correct: O2– > F– > Na+ correct. number of electrons. K+, Ca2+, Sc3– and Cl– have 18 B13. (a) : With increase in nuclear charge, size decreases in electrons. isoelectronic species. Ca2+ (Z = 20) < K+ (Z = 19) < Ar (Z = 18) B38. (d) : Among these neon has the highest positive electron B14. (a, d) : The correct order of increasing –ve electron gain gain enthalpy. enthalpy value among halogens is I < Br < F < Cl option (a) is wrong. B39. (d) : In general, ionization enthalpy decreases from top to Nitrogen has the electronic configuration, 2s22p3 (i.e., half filled 2p configuration). Therefore, it has bottom (Ba < Ca), (Se < S) and increases from left to higher first ionization enthalpy than oxygen. Option (d) is wrong. The correct order is : B < C < O < N. right in a period (Ca < Se), (S <Ar). Therefore, correct order of increasing first ionization enthalpy is : Ba < Ca < Se < S < Ar B40. (c) : Electron gain enthalpy = – Ionisation potential ∴ = – 5.1 eV B43. (a) : With increase in nuclear charge, size decreases in isoelectronic ions. N3–(Z = 7) > O2–(Z = 8) < F–(Z = 9) B44. (d) : The outer electronic configuration of the element ‘X’ is : 3d10 4s2 4p3. It has half filled p-orbitals and completely filled s-and d-orbitals.

3/70 MODERN'S abc + OF CHEMISTRY–XI B45. (d) : Sc has the highest ionization enthalpy. This is B57. (d) : 119: un (1) + un (1) + ennium = ununennium or uue because, it is difficult to remove an electron from 4s B58. (c) : Second electron gain enthalpy is always positive orbital of Sc as compared to 3s orbital of Na due to for an element because energy is required to add the second electron to negatively charged ion. poor shielding of d-orbitals and high effective B59. (d) : For isoelectronic ions, the size is affected by nuclear charge. nuclear charge. B46. (a) : F has the highest negative electron gain enthalpy. B60. (a) : In case of Be electron is to be removed from ©B48. (a) : Mg2+, Na+, O2– and F–, all have 10 electrons each and filled 2s subshell while in case of B, the electron Modern Publishers. All rights reserved. is to be removed from outer 2p subshell (2s2 are therefore, isoelectronic. 2p1). Hence, Be has higher ionization enthalpy than B though its nuclear charge is less. B50. (c) : For isoelectronic ions, higher the positive charge, B61. (b) : K has the electronic configuration : [Ar]4s1. After smaller will be the ionic radius. the removal of one electron, it acquires stable noble gas configuration. Hence, second ionisation B51. (c) : Noble gases have high ionization potential. energy will be high and therefore, there is a large difference between first and second ionisation B52. (d) : In general, electrons are added one at a time to the enthalpies. (n–1)d orbitals in transition elements. However, C1. (a,c) : (a) all have 10 electrons, (c) all have 18 electrons. because of extra stability of d5 and d10 configurations, C2. (c,d) : 14 elements of 6th period are called lanthanoids, two electrons appear to have entered the d-subshell 4th period begins with potassium. C3. (b,c) : O > S and N > O. due to shifting of one electron from ns to (n–1)d C5. (a,b) : C < N > O, F– > Na+ > Mg2+ C6. (b,c) : These belong to f-block orbital. B54. (d) : C N Si < C ; Si < P ; C < N ; P < N 2.5 3.0 Si P 1.9 2.2 So, correct order is Si < P < C < N B55. (b) : Electronegativity decreases as we go down the group and atomic radius increases as we go down the group. B56. (c) : Atomic no = 15 : 1s2 2s2 2p6 3s2 3p3 Group no = 15, Valence electron = 2 + 3 = 5. Valency = 8 – 5 = 3

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 3/71 ©Time Allowed: 1½ Hr. Maximum Marks: 25 Modern Publishers. All rights reserved. Q. 1 and 2 are assertion reason 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. 1. Assertion: The electron gain enthalpy of F is less negative than that of Cl. Reason: The atomic size of F is less than that of Cl. (1) 2. Assertion: Nitrogen has higher ionization enthalpy than oxygen. Reason: Oxygen has stable half filled configuration (1) 3. Which of the following is correct set of ionisation enthalpy? (a) Si < P < S (b) F < Cl > Br (c) B < C > N (d) N > O < F (1) 4. Which of the following electronic configuration has highest negative electron gain enthalpy? (a) 1s2 2s2 2p6 3s2 3p5 (b) 1s2 2s2 2p5 (c) 1s2 2s2 2p6 (d) 1s2 2s2 2p6 3s2 (1) 5. The correct set of decreasing ionic size is (a) Be2+ > Mg2+ > S2– > Cl– (b) S2– > Cl– > Na+ > Mg2+ (c) Cl– > S2– > Mg2+ > Be2+ (d) S2– > Cl– > Mg2+ > Na+ (1) 6. Which among I, I+ and I– has smallest size ? (1) 7. Arrange the elements : B, Al, Mg and K in the decreasing order of metallic character. (1) 8. How would you account for the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium? (2) 9. (a) Would you expect the second electron gain enthalpy of oxygen as positive, more negative or less negative than the first? Explain. (b) Give the general electronic configurations of (i) p-block (ii) actinoids (2) 10. Among the elements of second and third period, predict the following : (i) element (except noble gases) having largest ionization enthalpy and element having lowest ionisation enthalpy. (ii) element (except noble gases) having smallest atomic size and element having largest atomic size. (iii) elements having six electrons in the valence shell. (3) 11. Consider the ground state electronic configuration given below : A : 1s22s22p5 B : 1s22s22p4 C : 1s22s22p63s2 3p5 D : 1s22s22p63s1 E : 1s22s22p4

3/72 MODERN'S abc + OF CHEMISTRY–XI Which of the above configuration corresponds to element having (3) (a) maximum negative electron gain enthalpy? (3) (b) largest atomic size. (5) (c) having highest ionisation enthalpy. 12. Explain the following : (a) Oxygen has lower ΔiH than N and F. (b) Elements in the same group have similar physical and chemical properties. (c) Size of an anion is larger than the corresponding atom but the size of cation is smaller than the corresponding atom. 13. (a) What is the basic difference in approach between the Mendeleev’s periodic law and modern periodic law ? (b) Explain the variation in acidic and basic character of oxides of second row elements. (c) Account for the variation in metallic or non-metallic character in third row elements. © 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.