# 2 Atomic structure and the periodic table

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1 Page 35 Questions 1 a) Isotopes have the same number of protons but a different number of neutrons. Therefore, B and D are isotopes of element number 17 (chlorine). B is a 35 Cl atom and D is a 37 Cl + ion. Isotopes can be neutral (atoms) or charged (ions). b) Atoms are neutral if the number of electrons is the same as the number of protons. Thus B and F are neutral atoms. [e] B is a 35 Cl atom and F is a 40 Ar atom. c) Cations are positively charged ions because they have more protons than electrons. A, C and D are cations. [e] A is K +, C is Mg 2+ and D is Cl +. d) Anions are negatively charged ions because they have more electrons than protons. Element E (Br ) is the only anion. e) For atoms and ions to have the same electronic configuration, they must have the same number of electrons. Elements A (K + ) and F (Ar) both have 18 electrons, arranged 2,8,8. 2 a) There are two reasons. The more important is that most elements consist of a mixture of isotopes (tin has ten naturally occurring isotopes) and so the average mass is not a whole number. The second reason is that the relative mass of an isotope is not quite a whole number. For example, the relative mass of sodium (which has only one isotope, 23 Na) is This is because protons and neutrons do not have a mass of exactly one and when they join to form a nucleus, some of their mass is changed into energy. b) Let the % of 65 Cu be x. This means that the % of 63 Cu is (100 x). relative atomic mass = 65x + 63(100 x)/100 = x + 63(100 x) = x 63x = = 50 2x = 50 x = 25 The abundances of the two isotopes are 65 Cu = 25.0% and 63 Cu = 75.0%. [e] Note that the answer is given to three significant figures, because the data in the question were given to three significant figures. 3 The relative atomic mass is the sum of the % of each isotope multiplied by its mass, all divided by 100. relative atomic mass of silicon = [( ) + (29 4.7) + (30 3.1)]/100 = 28.1

2 [e] Note that the value given on your calculator is However, this must be changed to the one decimal place required by the question. Always check that your answer agrees with the value given in at the back of the exam paper, but remember that most values in the table have been rounded up to one decimal place. 4 a) The peaks in a mass spectrum are caused by positive ions. Bromine exists as Br 2 molecules and there are two isotopes of bromine, of relative masses 79 and 81. The peak at m/z 158 is caused by the ( 79 Br 79 Br) + ion, the one at 160 by the ( 79 Br 81 Br) + ion and the one at 162 by the ( 81 Br 81 Br) + ion. b) The peaks at 158 (due to the ion containing the Br-79 isotope only) and at 162 (due to the ion containing the Br-81 isotope only) are the same height, so the relative abundance of the two isotopes must be the same. [e] Students studying statistics will know that in a 50:50 mixture of the two isotopes, the chance of a 79 Br 79 Br molecule is: ½ ½ = ¼ It is the same for 81 Br 81 Br. However, the chance of a 79 Br 81 Br molecule is: 2 ½ ½ as there are two ways (head tail and tail head) of obtaining this species. Thus, the peak heights are: ¼:½:¼ or 1:2:1 c) The energy of the bombardment in the mass spectrometer may cause the molecular ion to dissociate into an atom and a monatomic ion. The equation for the formation of the peak at 79 is either: ( 79 Br 81 Br) + 79 Br Br or ( 79 Br 81 Br) + 79 Br Br or ( 79 Br 79 Br) + 79 Br Br [e] Note that the singly charged molecular ion only forms one positive ion. The other particle is a neutral free radical. 5 a) Phosphorus: [Ne] 3s 2 3p 3 b) Cobalt: [Ar] 4s 2 3d 7 The answer [Ar] 3d 7 4s 2 is equally acceptable. 6 Sodium in its ground state is 1s 2 2s 2 2p 6 3s 1. In an excited state, the 3s-electron is promoted to a higher level such as the 3p level, so one answer is 1s 2 2s 2 2p 6 3p 1. Another correct answer is 1s 2 2s 2 2p 6 4s 1.

3 7 The boxes must show electrons with their spins or. The answer is: 8 Periodicity is the term used to describe a regular pattern of physical or chemical properties. An example is that the first ionisation energies reach a maximum value at regular intervals (at each noble gas). 9 Boron has a nuclear charge of +5 and the electronic configuration 2,3. Thus, it has two inner shielding electrons. Its effective nuclear charge is approximately +3 (+5 2). Magnesium has 12 protons in its nucleus and the electronic configuration 2,8,2. Therefore, there are ten inner shielding electrons, giving it an approximate nuclear charge of +2 (+12 10). Bromine has 35 protons and the electronic configuration 2,8,18,7. The nucleus is shielded by 28 inner electrons, so the approximate effective nuclear charge is +7 (+35 28). Bromine has the largest effective nuclear charge and magnesium the smallest. [e] Remember that the effective nuclear charge is approximately equal to the nuclear charge (number of protons) minus the number of inner (shielding) electrons. The argument used in this answer ignores the repulsion between electrons in the outer orbit, which reduces the effective nuclear charge from that calculated. 10 Negative ions are bigger than their parent neutral atoms; positive ions are smaller. Thus the order of increasing radius is K + < Cl < Cl. 11 Helium has two electrons that are totally unshielded and so are extremely difficult to remove. Neon has ten protons in the nucleus and the 2p-electrons are shielded by two inner (1s) electrons; argon has 18 protons and the 3p-electrons are shielded by ten inner electrons; and lastly krypton has 36 protons with the 4p-electrons shielded by 28 electrons. So although the nuclear charge increases, so does the number of shielding electrons. These two effects cancel each other out. The ionisation energy decreases from neon to krypton because the radii of the outer electron shells increase, which results in the electrons being held less firmly and, therefore, being easier to remove. 12 There is a big jump in ionisation energy between the sixth and the seventh values. The seventh electron is therefore being removed from an inner shell. Therefore, element X has six outer electrons and is in group 6 of.

4 13 TOF: this is very accurate in finding the molar mass and as some of the molecules fragment, clues can be obtained about the molecule s structure. However, it is no use for complex molecules that fragment considerably under the high energy of the process. ESI: because low energies are used it is useful for finding the molar masses (and hence identity) of complex molecules such as those in the pharmaceutical industry. However, as the molecular ions do not fragment, it cannot be used to elucidate structure. 14 The nuclear charge increases by 8 from fluorine to chlorine, but by 18 from chlorine to bromine. This causes the outer electrons to be drawn much further in as the 4s and 4p electrons penetrate towards the nucleus and so feel the pull of the nucleus strongly. Page 36 Exam practice questions 1 a) The first ionisation energy is the energy required to remove one electron ( ) from each atom of a mole of gaseous atoms ( ) (in their ground state). b) i) C ( ) ii) A ( ) c) i) Electron affinity is the energy change when 1 mol of electrons is added ( ) to 1 mol of gaseous atoms ( ) of the element. ii) B ( ) 2 a) i) The gaseous element is bombarded by fast moving electrons, which causes it to lose an electron and form a positive ion ( ). ii) The positive ion is accelerated across an electric potential ( ). iii) It is deflected by a magnetic field ( ). b) ( ) + ( ) + ( ) ( ) = ( ) c) D ( ) [e] There is no 80 Br isotope (as there is no peak at 159), so A is wrong. d) ESI can be used to find the molar mass of a fragile molecule such as a pharmaceutical ( ). The determination of 4 decimal place molar masses ( ) can be used to identify the molecular formula of a substance ( ). 3 a) The general upward trend from sodium to argon is caused by an increase in the number of protons (the nuclear charge) ( ) thus increasing the pull on the outer electrons ( ). This means that more energy is required to remove it ( ). The electronic configuration of sulfur is [Ne] 3s 2 3p 2 x 3p 1 y 3p 1 z whereas that of phosphorus is [Ne] 3s 2 3p 1 x 3p 1 y 3p 1 z ( ). The two electrons in the 3p x orbital of sulfur repel each other and so make the removal of one of them easier ( ) (lowering the ionisation energy). b) Marked slightly below the value for sodium ( ).

5 c) The second electron has to come from the 2p orbital/inner orbit ( ). As it is only shielded by the two 1s electrons, it is held on much more strongly ( ).

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