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Periodicity Chapter 7 1 Atomic and Ionic Radii 2 Atomic Radii • Increase down the chart ✓ because of increased energy levels, each one further from the nucleus. • Decrease across (L-R) ✓ because of increased effective nuclear charge, Zeff, pulling on electrons that are in the same energy level. 3 Based on atomic radii, which of the following molecules would you expect to have the shortest bonds? 1. F2 2. Cl2 3. Br2 4 Based on atomic radii, which of the following molecules would you expect to have the shortest bonds? 1. F2 • Since fluorine is the smallest of atoms (since it has the fewest energy levels), the bonds of the fluorine molecule would be the smallest 2. Cl2 3. Br2 5 Which of the following is(are) expected to be a correct order of atomic radius from smallest to largest? Select all that are correct 1. 2. 3. 4. 5. 6. H<C<F Mg < Ar < Al S < Si < Na Ne < B < H K < Na < Li P < As < Sb 6 Which of the following is(are) expected to be a correct order of atomic radius from smallest to largest? Select all that are correct. 1. 2. 3. H<C<F Mg < Ar < Al S < Si < Na As you proceed from right to left, • the effective nuclear charge is less pulling the electron cloud in not so tightly. 4. Ne < B < H 5. K < Na < Li 6. P < As < Sb Down the table in a column, • the number of energy levels increase making a longer distance over which the attraction of the protons is acting on the electrons • and each full energy level provides more shielding thus reducing the effective nuclear charge. 7 Which has larger radius: 1. Cl − 2. Cl 8 Which has larger radius: 1. Cl − 2. Cl • All negative ions are larger than their parent atom • an e− has been added and would repel the other e− in the energy level 9 Which has larger radius: 1. Al 3+ 2. Al 10 Which has larger radius: 1. Al 3+ 2. Al 11 Which of the following gives the correct ordering of their ionic size (from smallest to largest)? 1. 2. 3. 4. 5. 6. − F + Na < Ne < − + F < Na < Ne − + Ne < F < Na + − Na < Ne < F + − Na < F < Ne They are all the same size 12 Which of the following gives the correct ordering of their ionic size (from smallest to largest)? F− < Ne < Na+ F− < Na+ < Ne Ne < F− < Na+ Na+ < Ne < F− • These particles are all isoelectronic, but you must consider the number of protons. • Na+ has greater effective nuclear charge pulling in the 2s2 2p6 electrons. 5. Na+ < F− < Ne 6. They are all the same size 1. 2. 3. 4. 13 Order the following according to increasing atomic/ionic radius. 3− + 2− N Li C O 1. 2. 3. 4. 5. 6. 7. 8. + Li 2− O 3− N C< < < 3− 2− + C < N < O < Li + 3− 2− Li < N < O < C + 2− 3− Li < C < O < N 3− 2− + N < O < C < Li 3− 2− + N < O < Li < C 2− 3− + O < N < Li < C 2− 3− + O < N < C < Li 14 Order the following according to increasing atomic/ionic radius. 3− + 2− N Li C O 1. 2. 3. 4. • • • 5. 6. 7. 8. C < Li+ < O2− < N3− C < N3− < O2− < Li+ Li+ < N3− < O2− < C Li+ < C < O2− < N3− Li+ is smallest since it only has a 1st energy level of eC is smaller than the neg ions because when they add electrons, the extra repulsion (p/e- ratio smaller) they become larger in size. O2− is smaller than N3− because it has less protons holding the 10 eN3− < O2− < C < Li+ N3− < O2− < Li+ < C O2− < N3− < Li+ < C O2− < N3− < C < Li+ 15 Size of Atoms and their Ions • Cations ✓ ✓ ✓ are always smaller than their parent atom the atom lost electrons the ion’s outermost e− and the outer electrons now one energy level closer to nucleus • Anions ✓ ✓ are always larger than their parent atom the atom gains e− increasing repulsion 16 Lattice Energy 17 Lattice energy • The enthalpy change (always positive) that accompanies the separation of a solid compound into separate gaseous ions. • Coulomb’s Law shows that the force of attraction between two oppositely charged particles is directly proportional to the magnitude of the charges and inversely proportional to the distance between those charges. + − + − (Q )(Q ) (Q )(Q ) Fattraction α F=k 2 2 d d 18 Lattice Energy + − (Q )(Q ) F=k 2 d • Factors that affect lattice energy ✓ The magnitude of the ionic charge ✓ 3+ / 3− hold tighter than 1+/1−, thus 3+/3− has higher lattice energy ✓ The ionic radius, although less important than the charge factor since ionic radii do not vary greatly. ✓ Smaller ions get closer to each other and the +/− pull felt is stronger thus higher lattice energy 19 Lattice Energy • The energy required to separate a mole of solid ionic compound into its gaseous ions. ScF3 5096 20 Select the correct lattice energy values comparisons. Choose all that apply. 1. LiF > LiI 2. RbF > LiF 3. MgO > NaF 4. CaS > BaS 5. K2S > K2O 6. MgF2 > AlF3 21 Select the correct lattice energy values comparisons. 1. LiF > LiI • F is a smaller size than I, greater force 2. RbF > LiF 3. MgO > NaF • Mg is larger charge, greater force 4. CaS > BaS • Ca is smaller size, greater force 5. K2S > K2O 6. BaF2 > AlF3 22 Each of the processes below are measurable and represent some process. Identify all of the ∆H’s below. • • • • • Na(s) + ½Cl2(g) → NaCl(s) ∆H_ = −411 kJ + − Na(g) → Na (g) + 1e ∆H_ = 496 kJ Cl2(g) → 2Cl(g) ∆H_ = 243 kJ − − Cl(g) + 1e → Cl (g) ∆H_ = −349 kJ Na(s) → Na(g) ∆H_ = 109 kJ 23 Each of the processes below are measurable and represent some process. Identify all of the ∆H’s below. • • • • • Na(s) + ½Cl2(g) → NaCl(s) ∆Hf = −411 kJ + − Na(g) → Na (g) + 1e ∆HIE = 496 kJ Cl2(g) → 2Cl(g) ∆HBDE = 243 kJ − − Cl(g) + 1e → Cl (g) ∆HEA = −349 kJ Na(s) → Na(g) ∆HSub = 109 kJ 24 Lattice energy can not be measured directly, it must be calculated. Calculate the lattice energy for NaCl(s) using Hess’ Law and the following information. NaCl(s) → Na+(g) + Cl−(g) • • • • • Na(s) + ½Cl2(g) → NaCl(s) ∆Hf = −411 kJ + − Na(g) → Na (g) + 1e ∆HIE = 496 kJ Cl2(g) → 2Cl(g) ∆HBDE = 243 kJ − − Cl(g) + 1e → Cl (g) ∆HEA = −349 kJ Na(s) → Na(g) ∆HSub = 109 kJ 25 Lattice energy can not be measured directly, it must be calculated. Calculate the lattice energy for NaCl(s) using Hess’ Law and the following information. NaCl(s) → Na+(g) + Cl−(g) • • • • • NaCl(s) → Na(s) + ½Cl2(g) ∆Hf = +411 kJ + − Na(g) → Na (g) + 1e ∆HIE = 496 kJ ½Cl2(g) → Cl(g) ∆H½BDE = 121.5 kJ − − Cl(g) + 1e → Cl (g) ∆HEA = −349 kJ Na(s) → Na(g) ∆Hsub = 109 kJ 26 Lattice energy can not be measured directly, it must be calculated. Calculate the lattice energy for NaCl(s) using Hess’ Law and the following information. NaCl(s) → Na+(g) + Cl−(g) • • • • • • NaCl(s) → Na(s) + ½Cl2(g) ∆Hf = +411 kJ + − Na(g) → Na (g) + 1e ∆HIE = 496 kJ ½Cl2(g) → Cl(g) ∆H½BDE = 121.5 kJ − − Cl(g) + 1e → Cl (g) ∆HEA = −349 kJ Na(s) → Na(g) ∆Hsub = 109 kJ +788 kJ 27 Magnetism 28 para- and dia- magnetism • paramagnetism ✓ The tendency of a species to be attracted by a magnetic field as a result of having unpaired electrons. • diamagnetism ✓ The tendency of a species not to be attracted (or even slightly repelled) by a magnetic field as a result of all of its electrons being paired. • not to be confused with ✓ ferromagnetism Materials that can maintain a magnetic field in the absence of an external magnetic source (Fe,Co, Ni) 29 Evidence of electron configurations • Paramagnetic studies are used to provide additional evidence of proposed electron configurations ✓ Experimental evidence shows that both Ti and Ti2+ are both paramagnetic, indicating which electrons the atom has lost to beome an ion. ‣ Ti ‣ ‣ ‣ ✓ [Ar] 4s23d2 ⨂ Ti2+ ⊘⊘〇〇〇 [Ar] 3d2 〇 ⊘⊘〇〇〇 If it were the d orbital electrons that were removed, from the Ti atom to produce the Ti ion, then the ion would not be paramagnetic 30 Evidence of electron configurations • Paramagnetic studies are used to provide additional evidence of proposed electron configurations ✓ Iron atoms are paramagnetic ‣ Fe ‣ ✓ ⨂ ⨂⊘⊘⊘⊘ Fe2+ ions do not demonstrate an increase or decrease in paramagnetism ‣ Fe2+ [Ar] 3d6 〇 ⨂ ⊘⊘⊘⊘ ‣ ✓ [Ar] 4s23d6 Fe3+ ions demonstrate increased paramagnetism ‣ ‣ Fe3+ [Ar] 〇 3d5 ⊘⊘⊘⊘⊘ 31 Effective Nuclear Charge 32 The forces on electrons • • • Protons pull on electrons. Higher charges attract more strongly than lower charges. Higher nuclear charge lowers the energy of orbitals making electrons in those orbital more stable. In multi-electron atoms, a second force comes into play, the repulsion from other electrons. This repulsion counteracts the nuclear attractions somewhat, making an electron easier to remove. We call this repulsion shielding and it reduces the full nuclear charge “felt” by an electron. The greatest shielding effect is caused by inner core full energy levels. Some small repulsion, shielding does occur between electrons in the same orbital sets and same energy levels 33 What is the approximate Zeff (effective nuclear charge) “felt” by the 3p electrons in the phosphorus atom? • Input a value 34 What is the approximate Zeff (effective nuclear charge) felt by the 3p electrons in the phosphorus atom? • • • • +5 The 10- core electrons blocks 10 of the 15+ from the protons resulting in +5 For atoms in the s and p valence orbitals, we can consider the effective nuclear charge to be equal to the number of valence electrons. sort of..... 35 Zeff • • The effective nuclear charge experienced by the valence electron in sodium depends mostly on the 11+ charge of the nucleus and the 10− charge of the neon core. If the neon core were totally effective in shielding the valence electron from the nucleus, then the valence electron would experience an effective nuclear charge of 1+. But, the 3s1 electron has some probability of being inside the Ne core. As a consequence of this "penetration," the core is not totally effective in screening the 3s1 electron from the nucleus. Thus the effective nuclear charge experienced by the 3s1 electron is actually a teeny bit greater than 1+. 36 In a germanium atom, when comparing electrons in the 3s, 3p and 3d orbitals, which electron(s) will experience the greatest effective nuclear charge? Electron(s) in the 1. 2. 3. 4. 5. 3s orbital 3p orbital 3d orbital 3s and 3p will feel the same 3s, 3p, and 3d will all feel the same 37 In a germanium atom, which electron(s) will experience the greatest effective nuclear charge? Electron(s) in the 1. 3s orbital • You might be inclined to think that 3s and 3p & 3d would “feel” or “see” the same effective nuclear charge, and it is nearly the same, however s orbital electrons feel slightly more nuclear charge since the s orbital penetrates deeper into the inner core of electrons. (check the next slide) 2. 3p orbital 3. 3d orbital 4. 3s and 3p will feel the same 5. 3s, 3p, and 3d will all feel the same 38 Orbital Penetration & Zeff • • • • The radial probability function for the 2s orbital of the hydrogen atom (red curve) shows a "node" of probability close to the nucleus, whereas the 2p orbital (blue curve) does not. As a result, an electron in the 2s orbital for a many-electron atom "sees" more of the nuclear charge than does an electron in the 2p orbital. In other words, the effective nuclear charge experienced by the 2s electron is greater than that for the 2p electron. This difference leads to the observation that in a many-electron atom the orbitals for a given n value increase in energy s < p < d < f, that is, orbital ns is lower energy than np, which is lower energy than nd. 39 Ionization Energy 40 Ionization Energy • The amount of energy required to forcibly remove an electron from a gaseous atom. ✓ ✓ Energy added as heat, light, or electricity Equation: X + IE → + X + − e 41 Which will have the highest first ionization energy? 1. C 2. Si 42 Which will have the highest first ionization energy? 1. C 2. Si • C has fewer energy levels thus the e− is closer and “feels” greater nuclear attraction on the electron, making it more difficult to remove an electron 43 Is the first ionization energy of carbon larger, smaller or the same as nitrogen? 1. larger 2. smaller 3. the same as 44 Is the first ionization energy of carbon larger, smaller or the same as nitrogen? 1. 2. • 3. larger smaller extra proton provides greater nuclear attraction, making the valence electron harder to remove. the same as 45 Is the first ionization energy of oxygen larger, smaller or the same as nitrogen? 1. larger 2. smaller 3. the same as 46 You would think that the trend continues for N and O, however the ionization energy of oxygen is lower than for nitrogen. Why? 47 You would think that the trend continues for N and O, however the ionization energy of oxygen is lower than for nitrogen. Why? Even though there is one added proton, the added repulsion of 4th electron that is paired in the p orbital is greater than the increase in proton pull from the extra proton, this repulsion makes the IE required to remove that p electron lower in oxygen compared to nitrogen N ⨂ ⊘⊘⊘ O ⨂ ⨂⊘⊘ In fluorine, the added proton overshadows the increased repulsion. F ⨂ ⨂⨂⊘ 48 Which will have the highest first ionization energy? 1. 2. 3. 4. 5. C N O Al Si 49 Which will have the highest first ionization energy? 1. 2. • 3. 4. 5. C N While the O as it has greater nuclear charge; the extra repulsion caused by the paired p4 electron makes it a bit easier to remove than one of the p3 electron in N O Al Si 50 Ionization Energy • The amount of energy required to forcibly remove an electron from an atom. ✓ ✓ Energy added as heat, light, or electricity Equation: X + IE → X+1 + e-1 • Increases across the chart (L-R) ✓ ✓ Due to increased effective nuclear charge No increase in shielding • Decreases down the chart ✓ ✓ due to outermost electrons further from the nucleus increased shielding F ir st Ion iza t io n En er gie s ( kJ/mole ) H 1 1311 Li Be 2 521 899 Na Mg 3 496 737 K Ca 4 419 590 Rb Sr 5 403 550 Cs Ba 6 376 503 7 Fr Ra He 1 2370 B C N O F Ne 2 799 1087 1404 1314 1682 2080 Al Si P S Cl Ar 3 576 786 1052 1000 1245 1521 Ga Ge As Se Br Kr 4 579 762 944 941 1140 1351 In Sn Sb Te I Xe 5 558 709 832 869 1009 1170 Tl Pb Bi Po At Rn 6 589 716 703 812 1037 51 7 Ionization Energy F ir st Ion iza t io n En er gie s ( kJ/mole ) H 1 1311 Li Be 2 521 899 Na Mg 3 496 737 K Ca 4 419 590 Rb Sr 5 403 550 Cs Ba 6 376 503 7 Fr Ra He 1 2370 B C N O F Ne 2 799 1087 1404 1314 1682 2080 Al Si P S Cl Ar 3 576 786 1052 1000 1245 1521 Ga Ge As Se Br Kr 4 579 762 944 941 1140 1351 In Sn Sb Te I Xe 5 558 709 832 869 1009 1170 Tl Pb Bi Po At Rn 6 589 716 703 812 1037 7 52 Zeff • • The effective nuclear charge experienced by the valence electron in sodium depends mostly on the 11+ charge of the nucleus and the 10− charge of the neon core. If the neon core were totally effective in shielding the valence electron from the nucleus, then the valence electron would experience an effective nuclear charge of 1+. But, the 3s1 electron has some probability of being inside the Ne core. As a consequence of this "penetration," the core is not totally effective in screening the 3s1 electron from the nucleus. Thus the effective nuclear charge experienced by the 3s1 electron is actually a teeny bit greater than 1+. 53 Orbital Penetration & Zeff • • • • The radial probability function for the 2s orbital of the hydrogen atom (red curve) shows a "bump" of probability close to the nucleus, whereas that for the 2p orbital (blue curve) does not. As a result, an electron in the 2s orbital for a many-electron atom "sees" more of the nuclear charge than does an electron in the 2p orbital. In other words, the effective nuclear charge experienced by the 2s electron is greater than that for the 2p electron. This difference leads to the observation that in a many-electron atom the orbitals for a given n value increase in energy with increasing l value, that is, orbital ns is lower energy than np, which is lower energy than nd. 54 Which ionization energy will be the largest? (Choose more than one if the same.) 1. 2. 3. 4. 5. 1st IE of Na 2nd IE of Na 1st IE of Mg 2nd IE of Mg 3rd IE of Mg 55 Which will be the largest? (Choose more than one if the same.) 1. 2. 3. 4. 5. • 1st IE of Na 2nd IE of Na 1st IE of Mg 2nd IE of Mg 3rd IE of Mg 6 The same 2p electron is being removed, however the Mg has one extra proton creating greater nuclear charge. 56 Successive Ionization Energy • The amount of energy required to repeatedly remove electrons. ✓ Energy could be added in the form of heat, light, or electricity. X + IE → X+ + e− Second: X+ + IE → X2+ + e− Third: X2+ + IE → X3+ + e− Etc, etc, etc. There is a very large increased value when the electron being removed is one more than the number of valence electrons (in red on the next side) ✓ Compared to the previous, the removal of this electron is from a full energy level that is one level closer to the nucleus than the previous electron removed. • First: • • • • 57 Successive Ionization Energy H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar 1st 1311 23 70 521 899 799 1087 1404 1314 1682 20 80 496 737 576 786 1052 1000 1245 1521 2nd 3rd 4th 5220 73 04 11752 1756 14 8 49 20899 2422 3657 25 0 19 2393 4622 6223 2856 4573 7468 3396 5297 7468 3367 6050 8423 3946 6165 9301 45 64 6918 9542 1447 77 38 10546 1814 2750 11 5 78 1582 3232 4361 1901 2914 4959 2258 3387 4544 2287 3850 5162 2653 3927 5886 5th 6th 7th 8th 32660 37 8 22 9446 10990 11028 12138 13373 13624 14820 16 0 07 6272 6947 6542 7526 46988 53 2 50 13325 15167 15148 16644 18033 18361 19693 21 5 16 8500 9359 8587 63970 71 3 12 17869 19972 20175 21767 23465 23658 25858 27 1 12 11028 11964 83652 91 9 50 22963 25501 25742 27575 29110 30489 31734 33 4 42 13778 58 Electron Affinity 59 Electron Affinity • The energy change accompanying addition of electron to gaseous atom: ✓ ✓ ✓ ✓ This is NOT the opposite of ionization energy. Cl + e− → Cl− It’s a measure of the attraction of an atom for an electron. The greater the attraction, the more negative its affinity value will be. ‣ remember energy is released when bonds form...and an incoming electron is like forming a bond between that electron and the atom. 60 Electron Affinity 61 PES Photon Electron Spectroscopy Evidence for the Validity of the Shell Orbital Model 62 PES − Measuring Binding Energy • X-Rays of a particular energy (Ex-ray) are “shot” at atoms • Electrons from each energy level of that atom leap away with varying amounts of excess kinetic energy from the X-ray • The charged hemisphere bends that spray of electrons ✓ faster (more KE) electrons bend less, slower (Less KE) electrons bend more • Detector tells us how many of each electron of a particular energy (KEdetected). • Binding energy (BEelectron) can be calculated • BEelectron = Ex-ray − KEdetected Photo Electron Spectra Relative Number of Electrons for Scandium 1000 Scandiu m Electron Detector 100 10 Energy 1 0.1 Negatively Charged Hemisphere (Constant V) X-Rays 63 Which element is represented by the PES black peaks shown below. lithium carbon oxygen neon Photo Electron Spectra Relative Number of Electrons 1. 2. 3. 4. 100 10 1 0.1 Energy 64 Which element is represented by the PES black peaks shown below. lithium carbon oxygen neon Photo Electron Spectra 2p4 Relative Number of Electrons 1. 2. 3. 4. 100 2s2 2 1s 10 1 0.1 Energy • Turn to your mate and decide what element is represented by the red peaks. 65 Which element is represented by the PES black peaks shown below. lithium carbon oxygen neon Photo Electron Spectra 2p4 Relative Number of Electrons 1. 2. 3. 4. 100 2s2 2 1s 10 1 0.1 Energy • Explain to your mate why the first 2 peaks of the red element are to the right of the first two peaks of the black element. 66 Which elements are represented by the PES black peaks shownbelow. Photo Electron Spectra Relative Number of Electrons O e C o 100 10 1 0.1 Energy • Explain to your mate why the first 2 peaks of the red element are to the right of the first two peaks of the black element. 67 Isotopes Percent Abundance 68 Determine the average atomic mass for a recently discovered element, Biggsium, Bg. Its isotopes are found in nature according to the chart below. Calculate the average atomic mass. • Report your answer to the nearest 10ths place. isotope % abundance 300Bg 65% 30% 5% 303Bg 304Bg 69 Determine the average atomic mass for a recently discovered element, Biggsium, Bg. Its isotopes are found in nature according to the chart below. Calculate the average atomic mass. • Report your answer to the nearest 10ths place. [(300 × 0.65) + (303 × 0.3) + (304 × 0.05)] = 301.1 isotope % abundance 300Bg 65% 30% 5% 303Bg 304Bg 70 You would like to calculate the % abundance found in nature of the newly discovered element Maxogen, Mx. There are only two naturally occurring isotopes: 324Mx and 327Mx. The average atomic mass is 325.2 • ths 10 Report the % to the nearest place for the isotope that occurs in highest quantity. 71 You would like to calculate the % abundance found in nature of the newly discovered element Maxogen, Mx. There are only two naturally occurring isotopes: 324Mx and 327Mx. The average atomic mass is 325.2 • ths 10 Report the % to the nearest place for the isotope that occurs in highest quantity. 324x + 327(1 − x) = 325.2 324x + 327 − 327x = 325.2 −3x = −1.8 • x = 0.6 60% is the 324Mx 72 Cesium is the most reactive of all the alkali metals (Francium has not been studied extensively due to its radioactivity.) What atomic property is most responsible for its high reactivity? 1. Cs is the softest of the alkali metals. 2. Cs has the lowest ionization energy of the alkali metals and the greatest metallic character. 3. Cs has the greatest density of the alkali metals. 4. Cs is the most conductive of the alkali metals. 5. Cs has the most electrons of the alkali metals. 73 Cesium is the most reactive of all the alkali metals (Francium has not been studied extensively due to its radioactivity.) What atomic property is most responsible for its high reactivity? 1. Cs is the softest of the alkali metals. 2. Cs has the lowest ionization energy of the alkali metals and the greatest metallic character. • Remember that IE gives us a direct insight to the ease with which electrons leave an atom, and the easier an electron leaves, and the more metallic character it has. 3. Cs has the greatest density of the alkali metals. 4. Cs is the most conductive of the alkali metals. 5. Cs has the most electrons of the alkali metals. 74 The electron configuration of germanium is 2 10 2 [Ar]4s 3d 4p This element is known to form more than one oxide. Which pair of formulas include the two most likely formulas? 1. 2. 3. 4. 5. GeO and GeO2 GeO2 and Ge2O7 Ge2O3 and GeO2 GeO and Ge2O3 Ge2O3 and Ge2O7 75 The electron configuration of germanium is 2 10 2 [Ar]4s 3d 4p This element is known to form more than one oxide. Which pair of formulas include the two most likely formulas? 1. GeO and GeO2 • Ge like (Sn and Pb) forms oxidation states +2 and +4 2. GeO2 and Ge2O7 3. Ge2O3 and GeO2 4. GeO and Ge2O3 5. Ge2O3 and Ge2O7 76 How does the electron structure of a phosphorus atom differ from a phosphorus ion? 1. The phosphide ion has more electrons. 2. The phosphorus atom has more unpaired electrons. 3. The phosphide ion has more core electrons. 4. The phosphorus atom is smaller than the phosphorus ion. 77 How does the electron structure of a phosphorus atom differ from a phosphorus ion? • 1. 2. 3. 4. Phosphorus forms a −3 oxidation state, gaining 3 electrons The phosphide ion has more electrons. The phosphorus atom has more unpaired electrons. The phosphide ion has more core electrons. The phosphorus atom is smaller than the phosphorus ion. 78 Which radiation below has the highest energy? 1. 2. 3. 4. 5. infrared microwaves ultraviolet visible x-rays 79 Which radiation below has the highest energy? 1. 2. 3. 4. 5. • infrared microwaves ultraviolet visible x-rays in order from lowest to highest it would be: mw < IR < visible < UV < x-rays 80 Which radiation below can be used to analyze the concentration of colored solutions? 1. 2. 3. 4. 5. infrared microwaves ultraviolet visible x-rays 81 Which radiation below can be used to analyze the concentration of colored solutions? 1. 2. 3. 4. infrared microwaves ultraviolet visible • Since color is a manifestation of visible light. We used the spectrophotometer to measure +1 the intensity of color of a red FeSCN solution to measure its concentration. 5. x-rays 82 Which radiation below is used in the analysis of bond length and atomic arrangement in crystal structures. 1. 2. 3. 4. 5. • infrared microwaves ultraviolet visible x-rays x-ray crystallography is used to measure bond lengths and radii of atoms since the nucleus’s show up as more dense on the xray image 83 Which radiation below is used as a detection beam across doorways and windows? 1. 2. 3. 4. 5. infrared microwaves ultraviolet visible x-rays 84 Which radiation below is used as a detection beam across doorways and windows 1. infrared • IR is also used for motion detectors 2. microwaves 3. ultraviolet 4. visible 5. x-rays 85 The value of Plank’s constant is 6.63x10-34 J 8 -1 sec. The velocity of light is 3x10 m sec . Which value in joules is closest to the energy of 15 -1 a photon with frequency of 8x10 sec 1. 2. 3. 4. 5. -20 10 1x -20 5 x 10 -19 5 x 10 -18 1 x 10 -18 5 x 10 86 The value of Plank’s constant is 6.63x10-34 J 8 -1 sec. The velocity of light is 3x10 m sec . Which value in joules is closest to the energy of 15 -1 a photon with frequency of 8x10 sec 1. 2. 3. 4. 5. • -20 10 1x -20 5 x 10 -19 5 x 10 -18 1 x 10 -18 5 x 10 E = hv, so the speed of light is a distractor and not necessary to calculate the energy. 87 Calculate the wavelength of light that is released when a hydrogen electron transitions from n=4 to n=1 • E= • wavelength = 88 Calculate the wavelength of light that is released when a hydrogen electron transitions from n=4 to n=1 • Pull equations off the blue sheet. 89 Which is a list of elements in order of increasing first ionization energy? Select all that apply. 1. Si < P < Cl 2. N < P < As 3. Sr < Ca < Mg 4. Cl < Br < I 5. F < Ne < Na 90 Which correctly list elements in order of increasing first ionization energy? 1. Si < P < Cl due to greater effective nuclear charge as you proceed L>R across the chart 2. N < P < As 3. Sr < Ca < Mg due to less energy levels as you proceed up the chart. 4. Cl < Br < I 5. F < Ne < Na 91 The overall electron configuration of the sulfide ion is most similar to (isoelectronic with) the electron configuration of the 1. 2. 3. 4. 5. oxide ion chlorine atom oxygen atom sodium ion potassium ion 92 The overall electron configuration of the sulfide ion is most similar to (isoelectronic with) the electron configuration of the 1. 2. 3. 4. 5. • oxide ion chlorine atom oxygen atom sodium ion potassium ion -1 Since potassium ion has 18e (since it -1 lost 1e ) as does the sulfide ion (since it -1 gained two e ) 93 Which describes the behavior of potassium metal during a chemical reaction? Select all that apply. 1. K atoms become ions with a 1+ charge. 2. K atoms take on protons. 3. K atoms become ions with a corresponding decrease in radius. 4. When K atoms bond with sulfur, the bond is stronger than the bond that forms when K atoms bond with selenium. 94 Which describes the behavior of potassium metal during a chemical reaction? 1. K atoms become ions with a 1+ charge. • due to the loss of 1e-1 2. K atoms take on protons. 3. K atoms become ions with a corresponding decrease in radius. • cations lose e-1 and the p/e ratio increases, and the valence e-1 are one energy level closer. 4. When K atoms bond with sulfur, a stronger ionic bond results than when K atoms bond with selenium. • Coulombs law says that when distance is shorter, the force is greater (and the bond is stronger). S is smaller than Se. 95 -34 10 The value of Planck’s constant is 6.63 x 17 -1 Jsec. The velocity of light is 3.0 x 10 nm sec . Which value is closest to the wave length in nanometers of a quantum of light with frequency 15 -1 of 6 x 10 sec ? 1. 2. 3. 4. 5. 10 25 50 75 100 96 -34 10 The value of Planck’s constant is 6.63 x 17 -1 Jsec. The velocity of light is 3.0 x 10 nm sec . Which value is closest to the wave length in nanometers of a quantum of light with frequency 15 -1 of 6 x 10 sec ? 1. 10 2. 25 3. 50 • c = wavelength*frequency • Planck’s constant is a distractor 4. 75 5. 100 97 What is the number of half-filled orbitals in an atom of phosphorus? 1. 2. 3. 4. 5. none one three five seven 98 What is the number of half-filled orbitals in an atom of phosphorus? 1. 2. 3. 4. 5. none one three five seven 99 Which description of electron configuration applies to atoms of potassium, chromium, and copper? 1. 2. 3. 4. 5. only one half-filled orbital only one filled sublevel only three half-filled orbitals only nine filled orbitals only five filled sublevels 100 Which description of electron configuration applies to atoms of potassium, chromium, and copper? 1. only one half-filled orbital • they all have a half filled s1 orbital (sublevel) 2. only one filled sublevel 3. only three half-filled orbitals 4. only nine filled orbitals 5. only five filled sublevels 101 The most abundant isotopes of hydrogen and oxygen are 11H, 21H, 168O, and 178O, respectively. Using these isotopes only, what is the number of different possible values for the molar mass of water in grams? • input your numeric answer 102 The most abundant isotopes of hydrogen and oxygen are 11H, 21H, 168O, and 178O, respectively. Using these isotopes only, what is the number of different possible values for the molar mass of water in grams? • 4 • • • there are 6 combinations, but 2 of them repeat 18(16+1+1), 19(16+1+2), 20(16+2+2) 19(17+1+1), 20(17+1+2), 21(17+2+2) 103 Which set of formulas or symbols best illustrates the Law of Multiple Proportions? 12 14 1. 6C and 6C 14 14 2. 6C and 7N 3. CO and CO2 4. CH4 and CCl4 5. C2H5OH and CH3OCH3 104 Which set of formulas or symbols best illustrates the Law of Multiple Proportions? 12 14 1. 6C and 6C 14 14 2. 6C and 7N 3. CO and CO2 4. CH4 and CCl4 5. C2H5OH and CH3OCH3 105 All of the following can be inferred from the Lewis dot diagram of a neutral atom below EXCEPT X 1. 2. 3. 4. X belongs to the same family as sulfur X has two half filled p-orbitals X has at least ten kernel (core) electrons X can accept two electrons to become an ion with a charge of 2- 5. X has four electrons with l quantum number = 1 106 All of the following can be inferred from the Lewis dot diagram of a neutral atom below EXCEPT X 1. 2. 3. 4. X belongs to the same family as sulfur X has two half filled p-orbitals X has at least ten kernel (core) electrons X can accept two electrons to become an ion with a charge of 25. X has at least four electrons from l quantum number of 1 107 Which gives a list of species with the same electron configuration; that is, species that are isoelectronic? 1. Mg, Ca, Sr 2+ 2+ 2+ 2. Mg , Ca , Sr 2− 3− 3. F , S , As 4. Fe, Co, Ni 2− + 5. S , Cl , K 108 Which gives a list of species with the same electron configuration; that is, species that are isoelectronic? 1. Mg, Ca, Sr 2+ 2+ 2+ 2. Mg , Ca , Sr 2− 3− 3. F , S , As 4. Fe, Co, Ni 2+ 5. S , Cl , K 109 Which color in the visible spectrum is associated with the lowest frequency? 1. 2. 3. 4. 5. blue green violet yellow red 110 Which color in the visible spectrum is associated with the lowest frequency? 1. 2. 3. 4. 5. blue green violet yellow red 111 According to quantum mechanics, what is the maximum number of electrons that can occupy the third energy level in a ground state atom? • input your numeric answer 112 According to quantum mechanics, what is the maximum number of electrons that can occupy the third energy level in a ground state atom? • • 18 s, d, p electrons only (no f’s etc) 113 Which pair of atoms represents nuclei that have the same number of neutrons? 1. 2. 3. 4. 5. 55Co and 58Co 57Mn and 57Fe 58Ni and 57Fe 58Ni and 57Co 59Ni and 56Fe 114 Which pair of atoms represents nuclei that have the same number of neutrons? 55Co 1. 2. 3. 4. • 5. and 58Co 57Mn and 57Fe 58Ni and 57Fe 58Ni and 57Co both have 30 neutrons 59Ni and 56Fe 115 Quantum Numbers Quantum Numbers are no longer part of the AP curriculum. We will review them before the SAT subject test in June. 116 Describe a possible set of quantum numbers for the electron removed for either the 2nd IE of Na or the 3rd IE of Mg 117 Describe a possible set of quantum numbers for the electron removed for either the 2nd IE of Na or the 3rd IE of Mg • • 2, 1, 1, +½ OR 2, 1, (or -1, or 0), +½(or -½) 118 What is the number of electrons in 23V that have an L quantum number of 2? • Input your answer 119 What is the number of electrons in 23V that have an L quantum number of 2? • Quantum number L = 2 represents d electrons, and Vanadium has d3 or 3 electrons in the d orbitals 120 What is the number of electrons in 16S that have an L quantum number of 1? • Input your answer 121 What is the number of electrons in 16S that have an L quantum number of 1? • Quantum number L = 1 represents p electrons, and sulfur has p4 or 4 electrons in the 3p orbitals AND p6 or 6 electrons in the 2p orbitals 122 Consider the set of quantum number numbers: 3, 2, -1, -1/2 Which set of quantum numbers represents an electron with the same energy (degenerate orbitals) but different angular orientation in space as the electron represented above? 1. 2. 3. 4. 5. 3, 2, -1, +½ 3, 1, -1, -½ 3, 2, 0, +½ 4, 1, 0, +½ 2, 2, -1, -½ 123 Consider the set of quantum number numbers: 3, 2, -1, -1/2 Which set of quantum numbers represents an electron with the same energy but different orientation in space as the electron represented above? 1. 3, 2, -1, +1/2 2. 3, 1, -1, -1/2 3. 3, 2, 0, +1/2 • both d electrons, but different orientation. 4. 2, 1, 0, +1/2 5. 2, 2, -1, -1/2 124 Consider the set of quantum number numbers: 3, 2, -1, -1/2 Which set of quantum numbers represents a different electron in the same (exact) orbital. 1. 2. 3. 4. 5. 3, 2, -1, +1/2 3, 1, -1, -1/2 3, 2, 0, +1/2 2, 1, 0, +1/2 2, 2, -1, -1/2 125 Consider the set of quantum number numbers: 3, 2, -1, -1/2 Which set of quantum numbers represents a different electron in the same orbital. 1. 2. 3. 4. 5. 3, 2, -1, +1/2 3, 1, -1, -1/2 3, 2, 0, +1/2 2, 1, 0, +1/2 2, 2, -1, -1/2 126 Write three possible groups of quantum numbers 4 that could describe any one of the 4f electrons in the Neodymium atom. 127 Write a set of three possible sets of quantum numbers to describe the ground state all three of the 4p3 electrons in the Arsenic atom. 128