Chemistry Answers Chapter 2

   Chapter 2 3 1 C 6 8 6 F 9 10 10 Fe3+ 26 30 23 (c) 14 (d) (e) 19 − 56 5 No. of electrons No. of protons Species 18 Ar 18 22 18 K+ 19 20 18 (c) 35 17 Cl– 17 18 18 C H 1 1 1 5 6 5 C 11 5 16 2− 8 O 8 8 10 D 19 − 9 9 10 10 B 7 A B 6 Ca2+ No. of electrons 4 0 20 40 18 39 19 40 No. of neutrons 3 1 1 20 (a) (b) No. of protons Li H 7 (a) (b) 4 Species 3 No. of electrons Tellurium has a greater proportion of heavier isotopes (with more neutrons). No. of neutrons 2 No. of protons Examples are density (related to mass) or, for gases, rate of diffusion. Species 1 No. of neutrons Exercises 2 1 B F B 6 10:01 8 Let x atoms be 20Ne atoms. The remaining atoms are 22Ne. 16 Sub-level No. of orbitals 4s 1 4p 3 4d 5 4f 7 number of 22Ne atoms = 100 – x total mass = 20 × x + (100 – x) × 22 = 2200 – 2x 2200 – 2x average mass = 100 From the Periodic Table we see that the relative atomic mass of neon = 20.18 2200 – 2x 20.18 = 100 2018 = 2200 – 2x 17 1s22s22p63s23p64s2 18 1s22s22px22py22pz23s23px13py13pz1, so three unpaired electrons 19 C 21 (a) V is 1s22s22p63s23p63d34s2 (c) Se is 1s22s22p63s23p63d104s24p4 (d) Sr is 1s22s22p63s23p63d104s24p65s2 x = 91; abundance 20Ne = 91% probability of 35Cl = 43 probability of 37Cl = 14 3 4 probability of Cl– Cl (M = 70) = × 56.25% 35 35 3 4 = 9 16 = 22 D 25 (a) O2− is 1s22s22p6 1 = probability of 37Cl–37Cl (M = 74) = 41 × 14 = 16 6.25% relative abundance/% 26 30 20 10 Ion (a) Ti2+ (b) Fe2+ (c) Ni2+ (d) Zn2+ 3d 4s 6.25 71 72 73 mass charge 74 75 Let the abundance of 25Mg be x. Consider 100 atoms. (78.90 × 24) + (x × 25) + (100 − 78.90 − x) × 26 24.31 = 100 = 1893.6 + 25x + 2600 – 2051.4 – 26x 27 28 Mg 11.20% and 26Mg 9.90% B C 15 4s < 4p < 4d < 4f (a) Cl is 1s22s22p63s23p5 (b) Nb is 1s22s22p63s23p63d104s24p64d35s2 (c) Ge is 1s22s22p63s23p63d104s24p2 (d) Sb is 1s22s22p63s23p63d104s24p64d105s25p3 x = 11.20 11 (a) Ne is 1s22s22p6 (b) Negatively charged ions would be F−, O2− or N3−; positively charged ions would be Na+, Mg2+ or Al3+. = 2442.2 – x 12 B 37.5 70 25 24 (d) Cu2+ is 1s22s22p63s23p63d9 56.25 0 10 B (c) Ti3+ 1s22s22p63s23p63d1 50 40 23 (b) Cl− is 1s22s22p63s23p6 probability of 35Cl–37Cl/37Cl–35Cl (M = 72) = 2× 43 6 = 37.5% × 14 = 16 60 C (b) K is 1s22s22p63s23p64s1 2x = 2200 – 2018 = 182 9 20 13 A 14 A 29 (a) Si (b) Mn (c) Sr 30 11 31 20 33 B 34 B 35 (a) C has the electronic configuration 1s22s22p2. The 4th electron is removed from a 2s orbital, the 5th electron from the 1s 32 [Kr]4d (d) Sc 10 7 10:01 Answers orbital. Electrons in 1s orbitals are closer to the nucleus and experience a stronger electrostatic force of attraction. (b) The 2nd electron is removed from a 2p orbital, the 3rd electron from the 2s orbital. Electrons in 2s orbitals are closer to the nucleus and so experience a stronger electrostatic force of attraction. was explained in 1905 by Albert Einstein. He realized that the pollen was being jostled by something even smaller; water molecules. Einstein didn’t just base this theory on his observations – he used complex mathematics to show that an atomic model could explain Brownian motion. 2 Potash, soda, magnesia and barytes are compounds of Group 1 and 2 elements. These compounds were later broken down into their component elements by electrolysis. 3 The Schrödinger model: ionization energy 36 ● ● 1 2 3 4 5 6 7 ionization number ● 37 (a) The ionization energy rises from Na to Ar because the charge of the nucleus increases but the number of inner ‘shielding’ electrons remains the same. The increase in effective nuclear charge makes it progressively more difficult to remove an outer shell electron. (b) Mg has the electron configuration [Ne]3s2, Al has the electron configuration [Ne]3s23p1. The 3p electron, removed from Al, has more energy and is further away from the nucleus than the 3s electron removed from Mg. (c) P has the configuration [Ne]3s23p1x3p1 y3p1z, S has the configuration [Ne]3s23p2x3p1 y3p1z. The electron removed from S comes from a doubly occupied 3p orbital, which is repelled by its partner and is easier to remove than the electron removed from P which comes from a half-filled orbital. 4 does not have well-defined orbits for the electrons does not treat the electron as a localized particle but gives a probability wave description predicts the relative intensities of various spectral lines. (a) [Rn]7s25f146d7 (b) The first g block element would be [Rn]7s25f146d107p6 8s28g1 Z = 86 + 2 + 14 + 10 + 6 +2 +1 = 121 5 1s22s22p63s23p64s23d104p65s24d105p66s24f145d10 6p65f4 6 (a) There would be two types of p orbital and two types of d orbitals. (b) Four groups in the p and d blocks. 7 1290 kJ mol−1 8 The convergence limit corresponds to the transition n = 2 to n = ∞. To obtain the ionization energy, we have to add the energy which corresponds to the n = 2 to n = 1 transition. This is the first line in the Lyman series. I.E. = 1310 kJ mol−1 Challenge yourself 1 8 In 1827 Robert Brown dropped grains of pollen into water and examined them under a microscope. The pollen moved around erratically in the water. This so-called ‘Brownian motion’ Practice questions 1 D 2 A 3 A 4 B 5 D 6 A 7 C 8 B 9 B 10 D (b) 27 protons and 25 electrons [1] (c) 1s22s22p63s23p63d7 or [Ar] 3d7 [1] 14 B 15 1s22s22p63s1 [1] Do not accept [Ne]3s1 11 (54 × 5.95) + (56 × 91.88) + (57 × 2.17) = 55.90 [2] 100 Award [2] for correct final answer. Large increase between 1st and 2nd I.E. as electron now removed from n = 2. Answer must be to 2 d.p. 12 (a) the electron configuration (of argon) or 1s22s22p63s23p6 [1] (b) x = 1 and y = 5 [1] (c) [1] 4s 3d Accept all six arrows pointing down rather than up. 13 First electron easy/easiest to remove or 1 electron in outermost/n = 3 energy level/furthest from nucleus. [1] (a) Cobalt has a greater proportion of heavier isotopes (OWTTE) or cobalt has a larger number of neutrons. [1] [1] Next eight electrons slightly more difficult to remove or show (relatively) small increase as these electrons are in the same energy level/ second energy level/n = 2. [1] Large increase between 9th and 10th I.E. as electron now removed from n = 1 or last two electrons very hard/most difficult to remove or innermost/lowest/closest to the nucleus/energy level (OWTTE). [1] Electron 11 also comes from 1s, so shows a small increase. [1] max [4] 10:01