End of chapter exercises Problem 1: Write down only the word/term for each of the following descriptions. 1. The sum of the number of protons and neutrons in an atom 2. The defined space around an atom's nucleus, where an electron is most likely to be found Practise more questions like this Answer 1: a) atomic mass number b) electron orbital Problem 2: For each of the following, say whether the statement is true or false. If it is false, re-write the statement correctly. 1. 2010Ne and 2210Ne each have 10 protons, 12 electrons and 12 neutrons. 2. The atomic mass of any atom of a particular element is always the same. 3. It is safer to use helium gas rather than hydrogen gas in balloons. 4. Group 1 elements readily form negative ions. Practise more questions like this Answer 2: 1. False. They both have 10 protons and 10 electrons, but 2010Ne has 10 neutrons and 2210Ne has 12 neutrons. 2. True. 3. True. 4. False. Group 1 elements readily form positive ions. Problem 3: The three basic components of an atom are: 1. protons, neutrons, and ions 2. protons, neutrons, and electrons 3. protons, neutrinos, and ions 4. protium, deuterium, and tritium Practise more questions like this Answer 3: d. protons, neutrons, and electrons Problem 4: The charge of an atom is: 1. positive 2. neutral 3. negative 4. none of the above Practise more questions like this Answer 4: c. neutral Problem 5: If Rutherford had used neutrons instead of alpha particles in his scattering experiment, the neutrons would: 1. not deflect because they have no charge 2. have deflected more often 3. have been attracted to the nucleus easily 4. have given the same results Practise more questions like this Answer 5: d. not deflect because they have no charge Problem 6: Consider the isotope 23492U. Which of the following statements is true? 1. The element is an isotope of 23494Pu 2. The element contains 234 neutrons 3. The element has the same electron configuration as 23892U 4. The element has an atomic mass number of 92 Practise more questions like this Answer 6: 3. The element has the same electron configuration as 23892U Problem 7: The electron configuration of an atom of chlorine can be represented using the following notation: 1. 1s22s83s7 2. 1s22s22p63s23p5 3. 1s22s22p63s23p6 4. 1s22s22p5 Practise more questions like this Answer 7: d. 1s22s22p63s23p5 Problem 8: Give the standard notation for the following elements: 1. beryllium 2. carbon–12 3. titanium–48 4. fluorine Practise more questions like this Answer 8: 1. Beryllium has 4 protons and 5 neutrons. So the number of nucleons is 9. The standard notation is: 94Be 2. Carbon has 6 protons and 6 neutrons. So the number of nucleons is 12. The standard notation is: 126C 3. Titanium has 22 protons and 26 neutrons. So the number of nucleons is 48. The standard notation is: 4822Ti 4. Fluorine has 9 protons and 10 neutrons. So the number of nucleons is 19. The standard notation is: 199F Problem 9: Give the electron configurations and Aufbau diagrams for the following elements: 1. aluminium 2. phosphorus 3. carbon 4. oxygen ion 5. calcium ion Practise more questions like this Answer 9: a) Aluminium has 13 electrons. So the electron configuration is: b) Phosphorus has 15 electrons. So the electron configuration is: c) Carbon has 6 electrons. So the electron configuration is: d) Oxygen has 8 electrons. The oxygen ion has gained two electrons and so the total number of electrons is 10. The electron configurationis: e) Calcium has 20 electrons. The calcium ion has lost two electrons and so the total number of electrons is 18. The electron configuration is: Problem 10: For each of the following elements give the number of protons, neutrons and electrons in the element: 1. 19578Pt 2. 4018Ar 3. 5927Co 4. 73Li 5. 115B Practise more questions like this Answer 10: a) Z = 78 and A = 195. So the number of protons is 78 and the number of neutrons is N=A−Z=195−78=117 . Since the element is neutral the number of electrons is also 78. b) Z = 18 and A = 40. So the number of protons is 18 and the number of neutrons is N=A−Z=40−18=22 . Since the element is neutral the number of electrons is also 18. c) Z = 27 and A = 59. So the number of protons is 27 and the number of neutrons is N=A−Z=59−27=32 . Since the element is neutral the number of electrons is also 27. d) Z = 3 and A = 7. So the number of protons is 3 and the number of neutrons is N=A−Z=7−3=4 . Since the element is neutral the number of electrons is also 3. e) Z = 5 and A = 11. So the number of protons is 5 and the number of neutrons is N=A−Z=11−5=6 . Since the element is neutral the number of electrons is also 5. Problem 11: For each of the following elements give the element or number represented by x: 1. 10345X 2. 35xCl 3. x4Be Practise more questions like this Answer 11: a) This element has 45 protons and atomic mass number of 103. Looking on the periodic table we find that the element with 45 protons and atomic mass number of 103 is Rhodium (Rh). b) We are given that chlorine has 35 protons and neutrons. We need to find out how many protons it has. From the periodic table we find that x = 17 c) We are given that beryllium has 4 protons. We look at the atomic mass given on the periodic table and see that it is 9. So x = 5. Problem 12: Which of the following are isotopes of 2412Mg: 1. 1225Mg 2. 2612Mg 3. 2413Al Practise more questions like this Answer 12: b) is the only correct answer. An isotope has the same number of protons, but a different number of neutrons. c) is a different element and a) has 25 protons, but -13 neutrons, which is not chemically possible. Problem 13: If a sample contains 69% of copper–63 and 31% of copper–65, calculate the relative atomic mass of an atom in that sample. Practise more questions like this Answer 13: The contribution from copper-63 is: 69100×63=43,47 The contribution from copper-45 is: 31100×65=20,15 Now we add the two values to get the relative atomic mass: 43,47+20,15=63,62u This value is slightly higher than the one on the periodic table which is mainly due to rounding errors. Problem 14: Complete the following table: Element or ion Electron configuration Core electrons Valence electrons Boron (B) Calcium (Ca) Silicon (Si) Lithium ion (Li+) Neon (Ne) Table 1 Practise more questions like this Answer 14: We first write down the number of electrons for each element: B: 5 electrons, Ca: 20 electrons, Si: 14 electrons, Li: 2 electrons, Ne: 10 electrons For the electron configuration we use the filling rules to write the electron configuration. The core electrons are the electrons in the innermost shell. The valence electrons are the electrons in the outermost shell. Element or ion Electron configuration Core electrons Valence electrons Boron (B) 1s22s1 2 3 Calcium (Ca) 1s22s22p23s23p64s2 18 2 Silicon (Si) 1s22s22p23s23p2 10 4 Lithium ion (Li+) 1s2 2 2 Neon (Ne) 1s22s22p6 10 8
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