Structure of Atom PDF Past Paper

Structure of Atom OBJECTIVE TYPE QUESTIONS Multiple Choice Questions (MCQs) 1. The table given below shows the numbers 5. An atom with 3 protons and 4 neutrons will of protons, neutrons and electrons in four ions. have a valency of For which ion i...

Structure of Atom OBJECTIVE TYPE QUESTIONS Multiple Choice Questions (MCQs) 1. The table given below shows the numbers 5. An atom with 3 protons and 4 neutrons will of protons, neutrons and electrons in four ions. have a valency of For which ion is the data correct? (a) 3 (b) 7 (c) 1 (d) 4 Number of Ion 6. The schematic atomic structures of three Protons Neutrons Electrons 40 2+ elements X, Y and Z are given as: 20Ca 20 20 20 19 – 9F 9 10 8 16 2– 8O 8 8 10 X Y Z 23 + Which of the following statements are incorrect? 11Na 11 12 11 40 2+ 19 – I. Z can form ZCl3 and ZCl5. (a) 20Ca (b) 9F II. Y exists in monatomic form. 16 2– 23 + (c) 8O (d) 11Na III. X and Z combine to form X3Z type compound. 2. In a sample of ethyl ethanoate (CH3COOC2H5) IV. X and Y combine to form XY2 type compound. the two oxygen atoms have the same number V. X will gain two electrons to form a stable of electrons but different number of neutrons. compound. Which of the following is the correct reason for (a) I and II (b) I, II and IV it? (c) II, IV and V (d) III, IV and V (a) One of the oxygen atoms has gained 7. The electron distribution in an aluminium electrons. atom is (b) One of the oxygen atoms has gained two (a) 2, 8, 3 (b) 2, 8, 2 neutrons. (c) 8, 2, 3 (d) 2, 3, 8 (c) The two oxygen atoms are isotopes. 8. Select the correct statements. (d) The two oxygen atoms are isobars. (i) An atom is divisible and consists of charged 3. Which pair of molecules has the same particles. number of electrons? (ii) It was known by 1900 that the atom was not (a) N2 and F2 (b) Cl2 and CO2 a simple, indivisible particle but contained at (c) H2O and H2S (d) O2 and C2H4 least one sub-atomic particle – the electron. 4. Atomic models have been improved over the (iii) E. Goldstein in 1886 discovered the presence years. Arrange the following atomic models in of new radiations called canal rays which the order of their chronological order led to the discovery of another sub-atomic (i) Rutherford’s atomic model particle – the proton. (ii) Thomson’s atomic model (iv) Proton had a charge, equal in magnitude but (iii) Bohr’s atomic model opposite in sign to that of the electron and (a) (i), (ii) and (iii) (b) (ii), (iii) and (i) its mass was approximately 2000 times as (c) (ii), (i) and (iii) (d) (iii), (ii) and (i) that of the electron. (v) The mass of an electron is considered to be (a) P - 4, Q - 3, R - 1, S - 2 (b) P - 2, Q - 1, R - 4, S - 3 negligible and its charge is minus one. (c) P - 2, Q - 3, R - 4, S - 1 (d) P - 4, Q - 2, R - 1, S - 3 (a) (i), (ii) and (iii) 14. Which of the following has the same number (b) (iii), (iv) and (v) of electrons as an oxide ion (O2–)? (c) (ii), (iii) and (iv) (a) K+ (b) Mg2+ (c) Cl– (d) S2– (d) All the statements are correct. 15. The given table shows the number of protons, 9. The electronic configuration of elements neutrons and electrons in atoms or ions. Which A, B, C and D are (2, 8, 1), (2, 8, 2), (2, 8, 6) and atom/ion in the table is an isotope of the atom (2, 8, 7) respectively. Which of them can make with the composition of 11p, 11e and 14n? an ion with two negative charges? Number Number Number (a) A (b) B Atoms/ of of of (c) C (d) D Ions protons electrons neutrons (p) (e) (n) 10. The nucleon number of the bromine atom is 79 and its proton number is 35. Which of the P 11 11 12 following is true about the bromine atom? Q 18 18 22 (a) It has 79 neutrons. (b) It has 44 electrons. R 15 18 16 (c) Its electronic configuration contains three S 11 10 14 shells which has 7 electrons in outermost (a) P (b) Q (c) R (d) S shell. 16. Dalton’s atomic theory successfully explained (d) It has similar chemical properties as (i) law of conservation of mass chlorine. (ii) law of constant composition 11. The ion of an element has 3 positive charges. (iii) law of radioactivity Mass number of the atom is 27 and the number of (iv) law of multiple proportion. neutrons is 14. What is the number of electrons (a) (i), (ii) and (iii) (b) (i), (iii) and (iv) in the ion? (c) (ii), (iii) and (iv) (d) (i), (ii) and (iv) (a) 13 (b) 10 17. Which of the following is an accurate (c) 14 (d) 16 illustration of a nitrogen atom? 12. Which of the following do not represent 8n Bohr’s model of an atom correctly? (a) (b) 8p (c) (d) (i) (ii) (iii) (iv) 18. Observe the given figure and answer the (a) (i) and (ii) (b) (ii) and (iii) question that follows : (c) (ii) and (iv) (d) (i) and (iv) Gold atoms 13. Match the column I with column II and select (I) the correct answer by choosing an appropriate a-particles option. Column I Column II P. Mass of proton 1. 9.1 × 10–28 g (III) Q. Charge of electron 2. 1.6 × 10–19 C R. Mass of electron 3. –1.6 × 10–19 C (II) S. Charge on proton 4. 1.67 × 10–27 kg (I) Most of the fast moving α-particles passed straight through the gold foil. (II) Some of the α-particles were deflected by 17p 20p the foil by small angles. 20n 20n (III) Surprisingly one out of every 12000 particles appeared to rebound. Identify the correct observations. (iii) (iv) (a) (I) and (II) (b) (II) and (III) Which of the following options contains the (c) (I) and (III) (d) All are correct. correct pairs? 19. A monovalent anion has 10 electrons and 10 Isotopes Isobars neutrons. The atomic number and mass number (a) (i) and (ii) (iii) and (iv) of the element are respectively ______ and _____. (b) (i) and (iv) (ii) and (iii) (a) 10, 20 (b) 9, 18 (c) (i) and (iii) (ii) and (iv) (c) 11, 20 (d) 9, 19 (d) (ii) and (iv) (i) and (iii) 20. In the Thomson’s model of atom, which of 23. Information about two atoms, X and Y are the following statements are correct? shown below : (i) The mass of the atom is assumed to be Nucleon Proton uniformly distributed over the atom. Atom number number (ii) The positive charge is assumed to be X 14 7 uniformly distributed over the atom. Y 15 7 (iii) The electrons are uniformly distributed in the positively charged sphere. Which of the following is correct about these (iv) The electrons attract each other to stabilise two atoms? the atom. Atom X Atom Y (a) (i), (ii) and (iii) (b) (i) and (iii) (a) Electronic 2, 8, 4 2, 8, 5 (c) (i) and (iv) (d) (i), (iii) and (iv) configuration : 21. Mass spectrum of lead is given as (b) Number of 7 7 60 neutrons : 53.8% 50 (c) Number of valence 5 5 % Abundance 40 electrons : 30 23.6% 22.6% (d) Number of electron 2 3 20 shells : 10 0 24. Match the column I with column II and select 206 207 208 the correct answer by choosing an appropriate Mass option. The average atomic mass of lead is Column I Column II (a) 208 (b) 207.567 P. Neutron 1. Rutherford’s (c) 207.302 (d) 209 atomic model 22. Vivek was given few models of atoms Q. Plum pudding model 2. Bohr’s atomic representing their structures. He was asked to model choose one pair of isotopes and one pair of isobars. R. Mass of the atom is 3. Thomson’s atomic concentrated at the model centre of atom 17p 18p S. Stationary orbit 4. Chadwick 18n 22n (a) P - 1, 4; Q - 1, 2, 3; R - 2; S - 1, 2, 3 (b) P - 1; Q - 1, 2, 3; R - 2; S - 1, 2 (c) P - 1; Q - 3; R - 2; S - 1 (i) (ii) (d) P - 4; Q - 3; R - 1; S - 2 25. The main drawback of Rutherford’s model (a) Na3PO4 (b) Na2CO3 of the atom is that (c) Na2SO4 (d) Na2SO3 (a) it does not explain the stability of atom 29. Which pair of atoms contains the same (b) it does not show the location of protons number of neutrons? (c) it does not explain neutral nature of an atom (a) 114 119 48Cd and 50Sn (b) 59 59 27Co and 28Ni (d) it does not explain existence of a nucleus in 133 132 63 65 an atom. (c) 55Cs and 54Xe (d) 29Cu and 29Cu 26. Rutherford concluded from the a-particle 30. Study the table carefully and select the scattering experiment that correct statement. (i) most of the space inside the atom is empty Element Number of Number of Number of because most of the α-particles passed protons neutrons electrons through the gold foil without getting deflected. U 11 12 10 (ii) very few particles were deflected from their V 20 20 20 path, indicating that the negative charge of W 16 18 18 the atom occupies very little space. X 20 19 18 (iii) a very large fraction of α-particles were Y 14 15 18 deflected by 180°°, indicating that all the Z 10 10 10 negative charge and mass of the gold atom (a) W is a noble gas. were not concentrated in a very small volume (b) X and Y are cations. within the atom. (c) U and V are anions. Identify the incorrect statements. (d) Z is the lightest element while V is the (a) (i) and (ii) (b) (ii) and (iii) heaviest. (c) (i) and (iii) (d) (i), (ii) and (iii) 31. Two particles X and Y have the composition 27. Select the correct statements. as shown in the table. (i) The radius of the nucleus is about 107 times Particle Number Number Number less than the radius of the atom. of of of (ii) There is a positively charged centre in an electrons neutrons protons atom called the nucleus. Nearly all the mass X 10 8 8 of an atom resides in the nucleus. (iii) The electrons revolve around the nucleus in Y 18 18 17 circular paths. The particles X and Y are (iv) The size of the nucleus is very large as (a) metal atoms (b) non-metal atoms compared to the size of the atom. (c) negative ions (d) positive ions. (a) (i) and (iv) 32. A has 9 protons, 9 electrons and 10 neutrons, (b) (ii) and (iii) B has 12 protons, 12 electrons and 12 neutrons. (c) (i), (ii) and (iii) Formula of the compound between A and B is (d) All the statements are correct. (a) BA2 (b) AB2 (c) B2 A3 (d) AB4 28. Schematic atomic structures of three 33. Match the following:- elements are given below : List-I List-II (P) Proton 1. Thomson (Q) Electron 2. Goldstein (R) Neutron 3. Rutherford Which of the following is the correct formula (S) Nucleus 4. Chadwick of the compound formed by the given three (a) P-4, Q-3, R-2, S-1 (b) P-1, Q-2, R-3, S-4 elements? (c) P-2, Q-1, R-4, S-3 (d) P-2, Q-1, R-3, S-4 34. An element L has 9 protons and its valency (a) P-1, Q-2, R-3, S-4 (b) P-4, Q-3, R-2, S-1 is 1. Another element M has valency 3 and 5. (c) P-2, Q-4, R-3, S-1 (d) P-2, Q-3, R-4, S-1 What is the difference in the number of electrons 36. The relative atomic mass of naturally in L and M? occurring chlorine is not a whole number. What (a) 6 (b) 5 (c) 4 (d) 3 is the reason for this? 35. Match the following:- (a) Chlorine atoms can have different numbers List-I List-II of neutrons. (P) Electrons 1. Number of positively (b) Naturally occurring chlorine cannot be charged particles in obtained pure. nucleus (c) Chlorine is unstable. (Q) Carbon dating 2. Negatively charged (d) The mass of the electrons has been included. particles (R) Valence electrons 3. Technique to know 37. Which one of the following is not isoelectronic age of fossils with neon atom? (S) Atomic number 4. Number of electrons (a) 8O2– (b) 11Na+ – in outermost shell (c) 9F (d) 12Mg+ Case Based MCQs Case I : Read the passage given below and Case II : Read the passage given below and answer the following questions from 38 to 40. answer the following questions from 41 to 44. The maximum number of the electrons which The table shows the number of sub-atomic are permitted to be assigned to an energy shell particles in arbitrary elements, A to H. of an atom is called the electron capacity of that Atom Number of Number of Number of shell. The distribution of electrons in different protons electrons neutrons orbits or shell is governed by a scheme known as Bohr-Bury scheme. According to this scheme : A 1 1 0 (i) The maximum number of the electrons that B 3 3 4 can be present in any shell is given by the C 4 4 6 formula 2n2 where, n is the number of energy D 5 5 5 level. E 6 6 6 (ii) The maximum number of electrons that can F 6 6 7 be accommodated in the outermost shell is 8. G 9 9 10 Electrons are filled in the shells in a stepwise manner in increasing order of energy of the H 9 9 11 energy shell. 41. The pair of isotopes from the table is/are 38. What is the maximum electrons capacity of (i) C and D (ii) E and F N shell? (iii) B and C (iv) G and H (a) 24 (b) 8 (c) 18 (d) 32 (a) (ii) only (b) (iv) only 39. Identify the element with the configuration K-2, L-8, M-3. (c) (ii) and (iv) only (d) (i), (ii), (iii) and (iv) (a) Aluminium (b) Magnesium 42. Which of the given elements attains noble (c) Sodium (d) Beryllium gas configuration by gaining an electron? 40. Which of the following configuration (i) A (ii) E represent sodium? (iii) C (iv) H (a) 2, 8, 4 (b) 2, 8, 5 (a) (iii) only (b) (iv) only (c) 2, 3 (d) 2, 8, 1 (c) (i) and (iv) only (d) (i) only 43. Identify pair of isobars from the table. Case IV : Read the passage given below and (a) C and D (b) B and E answer the following questions from 48 to 50. (c) G and H (d) E and F The mass of an atom is due to the masses of 44. The atom _____ has nucleon number 13 and protons and neutrons in the nucleus. The relative atom _____ has valency 3. masses of protons and neutrons are almost equal (a) G and F (b) F and D to one. Therefore, the atomic mass of an element (c) C and E (d) F and B should be nearly a whole number. But in many cases the atomic masses are fractional. The Case III : Look at the diagram given below and main reason for these fractional atomic masses answer the following questions from 45 to 47. is that these elements occur in nature as a The given diagrams show the atomic structures mixture of several isotopes. The atomic mass of of elements X and Y. an element is the average of the atomic masses Proton Neutron of these isotopes in the ratio of their proportion of occurrence. 48. Chlorine occurs in nature in the form of two isotopes with atomic masses 35 u and 37 u in the ratio of 3 : 1 respectively. Atomic mass of chlorine is X Y (a) 35.5 u (b) 34.5 u (c) 35 u (d) 36 u 45. Element X and Y could be _____ and _____ 49. An element occurs in two isotopic forms with respectively. atomic masses 10 and 11. What is the percentage (a) Be and B (b) C and O abundance of two isotopes in the sample having atomic mass 10.80? (c) F and N (d) C and N (a) 20, 80 (b) 50, 50 (c) 25, 70 (d) 60, 40 46. Valency of elements X and Y are respectively, 50. The fractional atomic masses of elements (a) 4 and 3 (b) 2 and 5 are due to the existence of (c) 1 and 4 (d) 3 and 4 (a) isotopes having different masses 47. Elements X and Y are (b) diagonal relationship (a) isotopes (b) isoelectronic (c) equal number of electrons and protons (c) isobars (d) isomers. (d) none of these. Assertion & Reasoning Based MCQs For question numbers 51-60, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Both assertion and reason are true, and reason is correct explanation of the assertion. (b) Both assertion and reason are true, but reason is not the correct explanation of the assertion. (c) Assertion is true, but reason is false. (d) Assertion is false, but reason is true. 51. Assertion : Cathode rays get deflected 53. Assertion : Thomson’s atomic model is towards the positive plate of electric field. known as ‘raisin pudding’ model. Reason : Cathode rays consist of negatively Reason : The atom is visualized as a pudding of charged particles known as electrons. positive charge with electrons (raisins) embedded 52. Assertion : Anions are larger in size than in it. the parent atom. 54. Assertion : The number of electrons gained, Reason : In an anion, the number of protons in lost or shared by the atom of an element so as the nucleus is less than the number of electrons to complete its octet is called the valency of the moving around it. element. Reason : Elements having the same number of 58. Assertion : Bohr’s orbits are called stationary valence electrons in their atoms possess different orbits. chemical properties. Reason : Electrons remain stationary in these 55. Assertion : For noble gases, valency is zero. orbits for sometime. Reason : Noble gases have 8 valence electrons. 59. Assertion : The distribution of electrons in different orbits or shells is governed by a scheme 56. Assertion : The size of the nucleus is very known as Bohr-Bury scheme. small as compared to the size of the atom. Reason : Electrons are filled in the shells in a Reason : The electrons revolve around the stepwise manner in increasing order of energy of nucleus of the atom. the energy shell. 57. Assertion : Electrons moving in the same 60. Assertion : In Rutherford’s gold foil experiment, orbit will not lose or gain energy. very few a-particles are deflected back. Reason : On jumping from higher to lower Reason : The size of the nucleus is very small as energy level, the electron will gain energy. compared to the size of the atom SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Question (VSA) 1. Atomic number is defined in terms of protons 5. What characteristic feature is seen in the and not in terms of electrons. Why? configurations of chemically inactive elements? 2. Given below is a diagram of the nucleus of an 6. What would you conclude from the atom. observation that cathode rays rotate a light 8p paddle wheel placed in their path? 8n 7. An oxide of an element Z has a formula Z2O3. (a) Complete the diagram to show the electronic (a) How many electrons are there in the arrangement of this atom. outermost shell of an atom of element Z? (b) Write the electronic configuration of the (b) Write down the formula for the chloride of Z. element. 8. Neutrons can be found in all atomic nuclei 3. What will be the charge on an atom with except in one case. Which is this atomic nucleus mass number one and atomic number one? and what does it consist of? 4. In Rutherford’s model of an atom, fast 9. Find valencies of the elements having atomic moving alpha (a)-particles were made to fall on a numbers 10 and 15. thin gold foil. State two properties of a-particles. 10. Why are anode rays also called canal rays? Short Answer Type Question (SA I) 11. (a) If the number of electrons in an ion is 12. Draw Bohr ’s model of an atom with 10 and the number of protons is 9, then three shells. How many electrons L-shell can (i) what would be the atomic number of the ion? accommodate? (ii) what is the charge on the ion? (b) An ion M 2+ contains 10 electrons and 12 13. The atom of an element has 9 protons, 9 neutrons. What is the atomic number and electrons and 10 neutrons. mass number of the element M? Name the (a) What is the atomic number of the element? element. (b) What is the mass number of the element? (c) Name the element and give its electronic = 3). Compare the structures and state one configuration. similarity and one difference between them. (d) Predict the valency of the element. 17. Give two points of differences between 14. Element X has a proton number of 7. It also isotopes and isobars. has seven neutrons. 18. An element has two electrons in N-shell. (a) Deduce the number of electrons and the nucleon number of X. Identify the element. (b) Represent X by writing the chemical symbol, 19. Justify the given statements : including the proton and nucleon numbers. (a) Most of the space in an atom is empty. 15. Justify the statement ‘atomic number of an (b) The elements are identified by their atomic element is equal to the number of electrons in numbers and not by their mass numbers. a neutral atom only and not in anion’. 20. Given that the percentage abundance of the 20 22 16. Draw the electronic structures of sodium isotope 10 Ne is 90% and that of the isotope 10 Ne (proton number = 11) and lithium (proton number is 10%, calculate average mass of neon. Short Answer Type Question (SA II) 21. (a) Electronic configuration of a neutral atom D 17 20 17 ‘X’ is 2, 8, 6. What is the electronic configuration E 18 22 18 of X2– ? F 19 21 19 (b) What is a valence shell? How many electrons can be present in valence shell? Making use of these data, find (a) a pair of ions 22. Atoms of elements R, S and T have 8, 9 and 11 protons respectively. Neon has 10 protons. (b) an atom of a noble gas (a) What is the chemical formula of the (c) a pair of isobars compound formed between? (d) a pair of isotopes. (i) R and T (ii) S and T 26. (a) On the basis of Thomson’s model of an (b) What is the formula of a molecule of R? atom, explain how the atom is neutral as a whole. 23. (a) Which fact is proved by the following (b) What are canal rays? observation in Rutherford’s scattering experiment 27. For the following statements, write T for ‘very few alpha particles are deflected back’? True and F for False. (b) How will you find the valency of nitrogen, (a) J.J. Thomson proposed that the nucleus of oxygen and fluorine? an atom contains only nucleons. 24. (a) Explain, why 23He and 13H are not (b) A neutron is formed by an electron and considered isotopes. a proton combining together. Therefore, it is (b) What are octet and duplet rules? How do neutral. elements attain octet? 1 (c) The mass of an electron is about times 25. Elements from A to F have in them the 2000 that of proton. distribution of electrons, neutrons and protons (d) An isotope of iodine is used for making as follows : tincture iodine, which is used as a medicine. Atoms/ Number of Number of Number of 28. (a) Helium atom has an atomic mass of Ions electrons neutrons protons 4 u and two protons in its nucleus. How many A 4 10 3 neutrons does it have? B 10 12 11 (b) Write the distribution of electrons in carbon C 17 18 17 and sodium atoms. 29. Summarise the rules for writing the (b) Which of these elements have a complete distribution of electrons in various shells for the outermost shell? first eighteen elements. (c) Which of these elements have 5 valence electrons? 30. (a) What is the number of valence electrons (d) Which of these elements have 6 valence in the atom of an element A having atomic electrons? number 20? Name the valence shell of this atom. (e) Which of these elements have 2 valence (b) The atom of an element has 9 protons, 9 electrons? electrons and 10 neutrons. (f) Write the valencies of each of the elements. (i) What is the atomic number of the element? 32. (a) Explain the formation of a cation. Give (ii) What is the mass number of the element? its main characteristics. (iii) Name the element and give its electronic (b) What is electronic configuration of Al3+? configuration. 33. The average atomic mass of sample of an (iv) Predict the valency of the element. element X is 16.2 u. What are the percentages of 31. Use the information to answer the following isotopes 168X and 188X in the sample? questions : 34. (a) Who discovered electron, proton and neutron? Element P Q R S T U V (b) Compare the properties of electrons, protons Proton number 7 8 10 12 15 18 19 and neutrons. Write five properties. (a) Which of these elements have only four filled 35. How will you find the valency of chlorine, electron shells? sulphur and magnesium? Long Answer Type Question (LA) 36. Observe the given figure and answer the (b) An ion M 3+ contains 10 electrons and 14 questions that follow : neutrons. What is the atomic number and mass Gold atoms number of the element M? Name the element. a-particles 38. Give reasons for the following : (a) Isotopes of an element are chemically similar. (b) An atom is electrically neutral. (c) Noble gases show least reactivity. (d) Nucleus of an atom is heavy and positively charged. (e) Ions are more stable than atoms. 39. (a) An element X has an atomic number 12 and a mass number 26. Draw a diagram showing (a) Which experiment is represented by the the distribution of electrons in the orbits and the given figure? nuclear composition of the neutral atom of the (b) List three observations of this experiment. element. What is the valency of the element and (c) S t a t e c o n c l u s i o n s d r a w n f r o m e a c h why? observation of this experiment. (b) If this element X combines with another (d) Write the features of the nuclear model of element Y whose electronic configuration is 2, an atom. 8, 7. What will be the formula of the compound (e) What were the drawbacks of this model of formed? an atom? 40. The table shows the numbers of electrons, 37. Answer the following : neutrons and protons in some atoms and ions of (a) Are there elements with the same number elements. of electrons, protons and neutrons? Give few (The letters used in the table are not the chemical examples. symbols of the elements.)  Atom/Ion Electrons Neutrons Protons Using only the letters from the table, identify S 10 10 10 (a) a negative ion. T 17 20 17 (b) a positive ion. U 10 12 11 (c) two atoms of the same element. V 10 8 8 (d) a noble gas. W 17 18 17 (e) an atom with a nucleon number of 20. OBJECTIVE TYPE QUESTIONS V. X will lose two electrons to form a stable compound. It shows the valency of +2. 1. (c) : 7. (a) : Aluminium has 13 electrons. Its electron distribution Number of is 2, 8, 3. Ion Protons Neutrons Electrons 8. (d) 40 2+ 20Ca 20 20 18 9. (c) : The two negative charges can be shown by that 19 – element, which needs two electrons for its stability. 9F 9 10 10 16 2– Therefore, it should contain 6 electrons in its valence shell. 8O 8 8 10 23 + 10. (d) : The nucleon number of bromine atom is 79 and its 11Na 11 12 10 proton number is 35. 2. (c) : Two isotopes have same number of electrons and It has 35 electrons. protons but different number of neutrons. It has (79 – 35) 44 neutrons. 3. (d) : Number of electrons in O2 = 2 × 8 = 16 Its electronic configuration contains four shells which has 7 Number of electrons in C2H4 = (2 × 6) + (4 × 1) = 16 electrons in outermost shell. Number of electrons in N2 = 2 × 7 = 14 It has similar chemical properties as chlorine as it has same Number of electrons in F2 = 2 × 9 = 18 valency 1 as chlorine. Number of electrons in Cl2 = 2 × 17 = 34 11. (b) : Mass number of the atom = 27 Number of electrons in CO2 = 6 + (2 × 8) = 22 Number of neutrons = 14 Number of electrons in H2O = (2 × 1) + 8 = 10 Number of protons = 27 – 14 = 13 Number of electrons in H2S = (2 × 1) + 16 = 18 Number of electrons in the atom = 13 4. (c) : Thomson’s atomic model, followed by Rutherford’s Number of electrons in ion with 3 positive charges model which is followed by Bohr’s model.  = 13 – 3 = 10 5. (c) : For the atom n = 4, p = 3, hence e = 3 12. (c) : First shell can accommodate maximum of two Distribution of electrons = 2, 1 electrons and second shell can accommodate maximum of 6. (d) : Number of electrons in X = 12 eight electrons. Number of electrons in Y = 10 13. (a) Number of electrons in Z = 15 14. (b) : O2– has 10 electrons. I. Z has five valence electrons, thus, it can form ZCl3 and K+ has 18 electrons. ZCl5 as it shows valency of +3 and +5. Mg2+ has 10 electrons. II. Y is neon and exists in monatomic form. Cl– has 18 electrons. III. X Z S2– has 18 electrons. Valency 2 3 15. (a) : Isotopes are the atoms of the same element with Compound X3Z2 different mass numbers i.e. , they have same number of IV. Y has a complete octet. Thus, it does not combine with protons but different number of neutrons. any element. Mass number = No. of protons + No. of neutrons Mass number of the given atom = 11 + 14 = 25 Number of neutrons in atom Y = 15 – 7 = 8 Atoms/Ions p e n Mass number Number of electron shells for both X and Y = 2 P 11 11 12 11 + 12 = 23 24. (d) 25. (a) : The revolution of the electron in a circular orbit is Q 18 18 22 18 + 22 = 40 not expected to be stable. The revolving electron would lose R 15 18 16 15 + 16 = 31 energy and finally fall into the nucleus. S 11 10 14 11 + 14 = 25 26. (b) : Very few particles were deflected from their path, 16. (d) : Dalton’s atomic theory successfully explained law indicating that the positive charge of the atom occupies very of conservation of mass, law of constant composition and little space. A very small fraction of α-particles were deflected law of multiple proportion. by 180°, indicating that all the positive charge and mass of 17. (a) : 147N the gold atom were concentrated in a very small volume within the atom. No. of protons = 7, No. of electrons = 7 No. of neutrons = 7 27. (b) : The radius of the nucleus is about 105 times less than the radius of the atom. The size of the nucleus is very Electronic configuration of N = 2, 5 small as compared to the size of the atom. 18. (d) 28. (b) : Three elements have the electronic configuration 19. (d) : As the monovalent anion has 10 electrons hence (2,4), (2,6) and (2,8,1) respectively. Hence, the elements are the element has (10 – 1) = 9 protons. carbon (C), oxygen (O) and sodium (Na). The charge on Na Hence, atomic number of the element = 9 is +1 and on carbonate (CO3) is –2. Mass number of the element = 9 + 10 = 19 So, 20. (a) : Thomson’s model could be compared with a raisin pudding model according to which the mass of atom is uniformly distributed over the atom in the form of positive Thus, the formula of the compound formed by these elements charge and electrons are uniformly distributed over the atom. is Na2CO3. Electrons do not attract each other. 29. (c) : Number of neutrons = 21. (c) : The average atomic mass of lead  Mass number – Atomic number 114 119 59 59  23.6 22.6 53.8  48 Cd : 66   50Sn : 69   27Co : 32   28 Ni : 31  = 206× + 207× + 208×   100 100 100  133 132 63 65 55Cs : 78   54 Xe : 78   29 Cu : 34   29 Cu : 36  = (48.616 + 46.782 + 111.904) = 207.302 u 30. (d) : W is not a noble gas, it is an anion ; W 2– 22. (c) : Atomic no. = No. of protons (Q no. of protons = 16 and no. of electrons = 18) Mass no. = No. of protons + No. of neutrons X is dipositive cation ; X 2+ (i) ⇒ atomic no. = 17 and mass no. = 17 + 18 = 35 (Q no. of protons = 20 and no. of electrons = 18) (ii) ⇒ atomic no. = 18 and mass no. = 18 + 22 = 40 Y is not a cation, it is an anion ; Y 4– (iii) ⇒ atomic no. = 17 and mass no. = 17 + 20 = 37 (Q no. of protons = 14, no. of electrons = 18) (iv) ⇒ atomic no. = 20 and mass no. = 20 + 20 = 40 U is monopositive cation ; U + Out of these four atoms, (i) and (iii) have same atomic no. i.e., 17 but different mass numbers. Hence, they are isotopes. (Q no. of protons = 11, no. of electrons = 10) Atoms (ii) and (iv) have same mass no. i.e., 40 but different V is a neutral element, atomic numbers. Hence, they are isobars. (Q no. of protons = no. of electrons). 23. (c) : The electronic configuration for both atoms X and Mass no. of U is 23 (11 + 12). Y is 2, 5. Hence, the number of valence electrons = 5 Mass no. of V is 40 (20 + 20). The electronic configuration of an atom is based on its proton Mass no. of W is 34 (16 + 18). number and not on its nucleon number. Mass no. of X is 39 (20 + 19). Number of neutrons in atom X = 14 – 7 = 7 Mass no. of Y is 29 (14 + 15). Mass no. of Z is 20 (10 + 10). 43. (a) : Elements C and D have different atomic numbers Therefore, Z is lightest element and V is the heaviest. 4 and 5 respectively but same mass number (10) so, these 31. (c) : Particle X is O2– and particle Y is Cl–. They both are are isobars. negative ions. 44. (b) : The atom F has nucleon number (6 + 7) 13 and 32. (a) : Symbol 24 19 element D will attain noble gas configuration after gaining 12 B 9A 2, 8, 2 2, 7 3 electrons so, its valency is 3. 45. (d) : Element X has proton number 6 (Z = 6) so, it is Valency 2 1 carbon (C) and element Y has proton number 7 (Z = 7) so it Formula is BA2. is nitrogen (N). 33. (c) 46. (a) : Valency of element X is 4 as it has electronic 34. (a) :  K L Distribution configuration   and element Y is 3 as it has electronic At. no.= No. of  2 4 Element protons (= No. of electrons Valency of electrons) K L M N  K L configuration  . L 9 2 7 – – 1  2 5 M 15 2 8 5 – 3, 5 47. (c) : Elements X and Y are isobars as they have different atomic numbers 6 and 7 but same mass number 14. Hence, the element L is fluorine (F) and M is phosphorus (P). 35 × 3 + 37 × 1 The difference between the number of electrons of P and F 48. (a) : Atomic mass = = 35.5 u 3 +1 is (15 – 9) = 6. 49. (a) : Let the percentage of isotope with atomic mass 10 35. (d) =x 35 37 36. (a) : Chlorine has two isotopes : 17 Cl and 17 Cl. They \ Percentage of isotope with atomic mass 11 = 100 – x both have different number of neutrons. Average atomic mass = 10.80 37. (d) : Isoelectronic species have same number of x × 10 + (100 − x ) × 11 10.80 = electrons. 100 No. of electrons in Ne = 10 10.80 × 100 = 10x + 1100 – 11x No. of electrons in O2– = 8 + 2 = 10 x = 1100 – 1080 = 20 + Abundance of isotope – 10 = 20% No. of electrons in Na = 11 – 1 = 10 – Abundance of isotope –11 = 100 – 20 = 80% No. of electrons in F = 9 + 1 = 10 + No. of electrons in Mg = 12 – 1 = 11 50. (a) : The fractional atomic masses of elements are due to 38. (d) : Maximum number of electrons present in a shell is the existence of isotopes having different masses, the atomic given by 2n2. mass of the element is the average value which comes out \ Number of electrons in N (4) = 2 × 42 = 32 to be fractional. 39. (a) : 2, 8, 3 electronic configuration means atomic 51. (a) : In electric field, negatively charged particles always number of the element is 13. Element with atomic number move towards positive plate and vice - versa. 13 is aluminium. 52. (a) 53. (a) 23 54. (c) : Elements having the same number of valence 40. (d) : 2, 8, 1 is configuration of 11Na. electrons in their atoms possess similar chemical properties. 41. (c) : E and F has same atomic number (Z = 6) and different mass numbers 12 and 13 respectively so, these are 55. (a) 56. (b) isotopes. Similarly, G and H has same atomic number (Z = 9) 57. (c) : Electrons moving in the same orbit will not lose or and different mass number 19 and 20 respectively so these gain energy. On jumping from higher to lower energy level, are isotopes. the electron will lose energy. 42. (c) : Elements A and H will attain noble gas configuration 58. (c) : Electrons in different orbits have fixed energies. after gaining one electron. 59. (b) 60. (b)  SUBJECTIVE TYPE QUESTIONS 12. 1. An atom may lose or gain electrons, but the number of + Nucleus protons remains constant till the same atom exist. K-shell (n = 1) L-shell (n = 2) M-shell (n = 3) 2. (a) 8p Number of electrons in L-shell = 2n2 = 2 × 22 = 8. 8n 13. (a) The atomic no. of element = No. of protons = 9 (b) The mass no. of element = No. of protons + No. of neutrons (b) Electronic configuration = 2, 6 = 9 + 10 = 19 3. The atom will not carry any charge because the atom (c) The element with Z = 9 is fluorine (F). Its electronic contains one unit negative charge in the form of electron and configuration : 2, 7. one unit positive charge in the form of a proton. (d) The valency of fluorine is 1 and is calculated as 8 – 7 = 1. 4. (a) Alpha particles are positively charged particles. 14. p = 7, n = 7 (b) They are doubly charged helium ions having a mass (a) Number of electrons = Number of protons = 7 number of 4 (consists of 2 protons and 2 neutrons). Number of nucleons = number of p + number of n 5. Chemically inactive elements have 8 electrons in their = 7 + 7 = 14 valence shell except helium which has 2 electrons in its 14 (b) 7X valence shell which is the maximum capacity of K shell. 15. In neutral atom, No. of protons = No. of electrons 6. When a light paddle wheel is placed in the path of = Atomic number cathode rays, the blades of the paddle wheel begin to rotate. An anion is formed by gain of one or more electrons by an It shows that cathode rays consist of material particles having atom. Therefore, anion contains more electrons than neutral mass and velocity. atom or in other words number of electrons in anion is greater 7. (a) An ion of element Z has the formula Z 3+. Hence, it than atomic number (the number of protons). has 3 valence electrons. 16. Sodium (proton number = 11), Electronic configuration (b) ZCl3  = 2, 8, 1 8. In case of hydrogen atom, there is no neutron. It consists Lithium (proton number = 3), Electronic configuration = 2, 1 of only one proton. 9. Atomic number = 10 Atomic number = 15 Electronic configuration Electronic configuration = 2, 8 = 2, 8, 5 Valency = 0 Valency = 3 Sodium Lithium 10. Canal rays are positively charged anode rays. The canal Similarity : Both have same valency, which is equal to one. rays are called so because they pass through the holes or the Difference : They have different number of shells. canals in the cathode. 17. Differences between isotopes and isobars: 11. (a) (i) Atomic number (Z) = No. of protons = 9 Isotopes Isobars (ii) Charge on the ion = –1 (i) These are the atoms of (i) These are the atoms Here, one electron is more than proton. So this one extra the same element. of different elements. electron attains –1 charge on the ion. (ii) These have same atomic (ii) These have different number. atomic numbers. (b) No. of electrons in M2+ ion = 10 (iii) These have different (iii) These have same Atomic number of atom M = 10 + 2 = 12 mass numbers. mass numbers. No. of protons in atom M = 12 (iv) The chemical properties (iv) These have different Mass number of atom M = No. of protons + No. of neutrons of isotopes are similar but chemical and physical = 12 + 12 = 24 their physical properties properties. The element M with atomic number 12 is magnesium (Mg). are different. 18. An element can be identified with its atomic number (Z) (ii) Oxygen has 6 electrons in valence shell, hence its valency which is equal to the number of electrons in its neutral atom. is 8 – 6 = 2. If there are 2 electrons in N shell, it means K, L and M shells (iii) Fluorine has 7 electrons in valence shell, hence its are completely filled. Two electrons can be accommodated valency is 8 – 7 = 1. in K shell, eight in L shell and 8 in M shell since outermost 24. (a) 23He and 13H are not considered as isotopes because orbit cannot have more than 8 electrons, 2 electrons go to they have different atomic numbers and are different elements. N orbit. (b) Octet rule was proposed by G.N.Lewis. According to this Hence total number of electrons = 20 rule “The atom of an element combines with another atom to K L M N have eight electrons in its outermost shell”. An atom having 2 8 8 2 8 electrons in its outermost shell is least reactive or most The element is calcium. stable. If there is only one shell, then stability is attained by 19. (a) This can be justified by Rutherford’s a-scattering having 2 electrons in the shell and this is called duplet rule. experiment. Since most of the alpha particles could pass Element attains octet in the following ways : through the sheet, made up of atoms of gold, undeflected, (i) by losing or gaining electrons. this means that they did not come across any obstruction. (ii) by sharing electrons with other atoms. Thus, most of the space in an atom is empty or hollow. 25. (a) A pair of ions is A and B. (b) In the study of the atomic structure, we have seen that Because A has 4 – 3 = 1 negative charge and B has 11 – 10 = 1 the isotopes of an element have different mass numbers. In positive charge. isobars, the atoms of different elements have same mass (b) An atom of noble gas is E. numbers. However, the atomic numbers of no two elements Electronic configuration of E is 2, 8, 8. can be the same. Therefore, the elements are identified by (c) A pair of isobars is E and F. their atomic numbers and not by their mass numbers. Mass number of E = 22 + 18 = 40 20 20. Abundance of isotope 10 Ne = 90% and of Isotope Mass number of F = 21 + 19 = 40 22 10Ne = 10% (d) A pair of isotopes is C and D. 20 × 90 + 22 × 10 Average atomic mass of neon = = 20.2 u. Atomic number of C = 17 100 Atomic number of D = 17 21. (a) X = 2, 8, 6 Mass numbers of C and D are 35 and 37 respectively. No. of electrons in neutral atom = 2 + 8 + 6 = 16 26. (a) According to Thomson’s model of an atom : X + 2e– → X2– (i) An atom consists of a positively charged sphere and the No. of electrons in X2– = 16 + 2 = 18 electrons are embedded like the seeds in a watermelon. Electronic configuration of X2– = 2, 8, 8 (ii) The negative and positive charges are equal in magnitude. (b) The outermost shell of an atom is called valence shell. So, the atom as a whole is electrically neutral. The number of electrons that can be present in valence shell (b) The beam of rays which travel in a direction away from is 1 - 8. anode towards cathode when a gas taken in a discharge tube 22. R(p = 8), Electronic configuration = 2, 6 is subjected to the action of high voltage under low pressure S(p = 9), Electronic configuration = 2, 7 are known as canal rays. It is also called anode rays. It was T(p = 11), Electronic configuration = 2, 8, 1 discovered by E. Goldstein in 1886. (a) (i) RT2 (ii) ST 27. (a) F : Because it was not proposed by J.J. Thomson. (b) R2 (b) F : Because neutron is an independent sub-atomic 23. (a) This shows that in the centre of atom a very small particle. positively charged body called nucleus is present. (c) T : Because it is a fact known from experiments. (b) (i) Nitrogen has 5 electrons in valence shell, hence its (d) F : Because tincture iodine is a solution of ordinary iodine valency is 8 – 5 = 3. in alcohol. 28. (a) Mass number of helium = 4 (c) P(2, 5) and T(2, 8, 5)   (d) Q(2, 6) Number of protons = 2 (e) S(2, 8, 2) Number of neutrons (n) (f) P(3), Q(2), R(0), S(2), T(3), U(0), V(1) = Mass number (A) – No. of protons (p) 32. (a)A cation is formed when an atom loses one or more =4–2=2 than one electrons from valence shell. Thus, no. of neutrons = 2 M → M+ + e– (b) Atomic number of carbon = 6 For example, Hence first shell (K-shell) have 2 electrons and the remaining Na → Na+ + 1e– 4 electrons will be present in the second shell, i.e. L-shell. Mg → Mg2+ + 2e– Thus the distribution will be Characteristics of cations : K L (i) Cations are positively charged. 2 4 (ii) Cations are formed when an atom loses electrons from Atomic number of sodium = 11. Hence, first shell (K-shell) its valence shell to attain octet. will have 2 electrons and second shell (L-shell) will have 8 (iii) Cations are smaller in size than parent atom. electrons and third shell (M-shell) will have 1 electron. Thus, (iv) The charge acquired by a cation is equal to the number the distribution will be of electrons lost by the valence shell. K L M (b)Atomic number of aluminium is 13 and its electronic 2 8 1 configuration is 2, 8, 3. Al 3+ is formed by removing 3 29. The distribution of elements in different orbits is electrons from aluminium atom. governed by a scheme called Bohr-Bury scheme. There are Al – 3e– → Al3+ following rules : Hence, electronic configuration of Al3+ is 2, 8. (i) The maximum number of electrons present in any shell 16 18 33. Let the % of isotope 8X = x and % of isotope 8X = is given by the formula 2n2. Where n = no. of orbit. 100 – x (ii) The maximum number of electrons that can be accommodated in the outermost shell is 8. Average atomic mass of an element (X) = 16.2 (iii) Electrons in an atom do not occupy a new shell unless Average atomic mass all the inner shells are completely filled. Massof 168 X × percentange of 168 X + 30. (a) The electronic configuration of element A is Massof 18 18 8 X × percentage of 8 X = K L M N 100 2 8 8 2 x × 16 + (100 − x ) × 18 or 16.2 = Therefore, the N shell is the outermost shell or the valence 100 shell. Number of valence electrons 16x + 1800 − 18x or 16.2 = = number of electrons in the outermost shell 100 = 2. or 1620 = 16x + 1800 – 18x (b) (i) The atomic no. of element = No. of protons = 9 or 1620 = –2x + 1800 or 2x = 1800 – 1620 (ii) The mass no. of element or 2x = 180 = No. of protons + No. of neutrons 180 \ x= = 90 = 9 + 10 = 19 2 (iii) The element with Z = 9 is fluorine (F). Thus % of isotope 168X = 90% Its electronic configuration : 2, 7. % of isotope 188X = (100 – 90) = 10% (iv) The valency of fluorine is 1 and is calculated as 8 – 7 = 1. 34. (a) J.J. Thomson discovered electron. E. Goldstein 31. (a) V(2, 8, 8, 1)   (b) R(2, 8) and U(2, 8, 8) discovered proton. Chadwick discovered neutron. (b) the number of such a-particles is very small, thus the space Property Electron Proton Neutron occupied by the heavy positive centre must be very small. 1. Nature Negatively Positively Neutral (d) The nuclear model of atom had following features : charged charged (i) There is a positively charged centre in an atom called the 2. Charge –1 unit or +1 unit or Zero nucleus. Nearly whole the mass of an atom resides in the –19 –1.6 × 10–19C +1.6 × 10 C nucleus. 3. Mass 9.1 × 10–31 kg 1.67 × 10–27 kg 1.675 × 10–27 (ii) The electrons revolve around the nucleus in circular paths. kg 4. Location Present around Present in Present in (iii) The size of the nucleus is very small as compared to the the nucleus nucleus nucleus size of the atom. 5. Relative (e) Drawbacks of Rutherford’s atomic model : 1 mass u 1u 1u (i) It has been found that, if an electrically charged particle 1838 revolves around the circular path, then it always radiates out 35. Valency of an atom is the number of electrons gained, energy. Thus, if an electron moves around the nucleus, it must lost or shared so as to complete the octet of electrons in the continuously radiate out energy and hence, gradually move valence shell. towards nucleus in a spiral path, till it collides with nucleus. Valency of chlorine: It has electronic configuration = 2, 8, 7 However, we know that atom is very stable. Rutherford’s Thus, one electron is gained to complete its octet and so its model cannot explain this stability. valency is 1. (ii) Rutherford’s model of atom does not say anything about Valency of sulphur: It has electronic configuration = 2, 8, 6 the arrangement of electrons in an atom. Thus, two electrons are gained to complete its octet and 37. (a) Yes, there are elements with the same number of hence its valency = 2 electrons, protons and neutrons. For example Valency of magnesium : It has electronic configuration = Element No. of No. of No. of 2, 8, 2 electrons protons neutrons Thus, it can lose two electrons to attain octet and hence its 4 2He 2 2 2 valency = 2 12 6C 6 6 6 36. (a) Rutherford’s a-particle scattering experiment 14 (b) Rutherford observed that : 7N 7 7 7 16 (i) Most of the a-particles (nearly 99%) passed through 8O 8 8 8 the gold foil undeflected. 20 10Ne 10 10 10 (ii) Some of the a-particles were deflected by small angles. 24 12Mg 12 12 12 (iii) A very few a-particles (1 in 12,000) were either deflected 28 by very large angles or were actually reflected back along 14Si 14 14 14 their path. 32 16S 16 16 16 (c) Rutherford explained his observation as follows : 40 20Ca 20 20 20 (i) Since most of the a-particles passed through the foil undeflected, it indicates that the most of the space in an (b) Number of electrons in M3+ ion = 10 atom is empty. Number of electrons in the neutral atom = 13 (ii) a-Particles being positively charged and having (M – 3e– → M3+) considerable mass, could be deflected only by some heavy, For a neutral atom, number of electrons = number of protons positively charged centre. The small angle of deflection of = atomic number a-particles indicated the presence of a heavy positive centre \ Atomic number = 13 in the atom. Rutherford named this positive centre as nucleus. Mass number = Number of protons + number of neutrons (iii) a-Particles which make head-on collision with heavy = 13 + 14 = 27 positive centre are deflected through large angles. Since The element with atomic number 13 is aluminium. 38. (a) Isotopes of an element have same atomic number as Thus, number of neutrons = 26 – 12 = 14 well as electronic configuration. Since the chemical properties Its electronic configuration = 2, 8, 2 of elements are related to their electronic configuration, i.e., Thus, number of electrons present in the outermost shell is the elements with similar configuration will have similar 2, so its valency is 2. properties. Thus, the isotopes of an element are chemically similar. (b) In an atom, the number of protons in the nucleus is equal 12p to the number of electrons in the extra-nuclear portion. Since 14n each proton and each electron has the same charge but with opposite magnitude, the atom is electrically neutral. (c) The atoms of noble gas elements have complete outermost shells. Hence, they are least reactive. (b) Electronic configuration of Y = 2, 8, 7 (d) Nucleus of an atom is made up of protons which are So, valency of Y = 8 – 7 = 1 positively charged and neutrons that are neutral. Hence net charge on nucleus is positive. The total mass of neutron and Thus, formula of compound formed by X and Y = XY2 proton makes it heavy. One atom of X combines with two atoms of Y to achieve (e) When an atom changes into an ion (cation or anion) the stable noble gas configuration. valence shell of the ion has a complete octet or duplet which 40. (a) V makes ions more stable than atoms. (b) U 39. (a) Atomic number of X = 12 (c) T and W Thus, number of protons = number of electrons = 12 (d) S Mass number of X = 26 (e) S