Problem Set #10 PDF

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Problem set 10...

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Problem(Set(10(–(Periodic(Table(&(Trends! Chem!105! 1. (a) What is the difference between a group and period in the Periodic Table? Groups are columns and contain elements with similar valence electron configurations and thus similar chemical properties. Periods are rows and contain elements with valence electrons in the same energy level (n).

(b) The Periodic Table can be divided into roughly 4 blocks according to the type of subshell being filled. State the subshell being filled in each of these blocks. Orange on left= s, blue= d, red= p, yellow= f 2. (a) There are also different ‘families’ of elements in the Periodic Table. List a transition metal, an alkali metal, a halogen, a noble gas, an alkaline earth metal, a nonmetal, a lanthanide, an actinide, and a metalloid in that order. Ni, Li, Br, Ar, Ca, C, Ce, Th, Si (b) In which set of elements on the periodic table are the p orbitals being filled? a. alkali metals b. halogens c. actinides d. lanthanides e. transition metals 3. Size, ionization energy, and electron affinity are all properties that show definite trends across the Periodic Table. (a) Describe the trends for each of these properties going down a group and across a period.

(b) Explain the reason for the change in atomic radii across a period as well as down a group. The z effective nuclear charge increases from left to right across the periodic table, atomic radii increase down columns because valence electrons become further and further away from the nucleus (nuclei become larger) Identify each statement as true or false. If false, re-write the statement so it is true: (c) Ionization energies are always negative. False, positive (d) Oxygen has a larger first ionization energy than fluorine. False, smaller

(e) The second ionization energy of an atom is always larger than its first ionization energy. True 4. (a) When an atom forms an anion, its radius increases. When an atom forms a cation, its radius decreases. Explain why this is so. Adding an electron will decrease z effective nuclear charge, taking away an electron will increase z effective nuclear charge (b) Arrange the following atoms and ions in order from smallest to largest radius. a. Sr, Kr, Br, Rb, Ar Ar, Kr, Br, Sr, Rb b. Ne, Ne+, Ne– Ne+, Ne, Nec. Al, Cs, Ne, Rb, S Ne, S, Al, Rb, Cs d. Cl, Cl–, Cs, Cs+ Cl, Cl-, Cs+, Cs e. K+, Ca2+, S2-, Cl-, Ar+ Ar+, Ca2+, K+, Cl-, S2-, 5. (a) The compound, NaCl, consists of sodium cations and chlorine anions. The sodium and chlorine ions are not the same size as they are in the elemental state. Explain how their relative sizes change (larger or smaller) in relationship to their size in the elemental state. Na will decrease in size, Cl will increase in size. Losing electrons makes a cation smaller because the positively charged nucleus can more strongly attract the remaining negatively charged electrons. Fewer electrons also leads to less electron repulsion between neighboring electrons. Gaining electrons makes anions larger because there are more repulsions between the electrons, which also shield each other from the attraction of the nucleus. (b) In the depiction of the crystal structure of salt to the right, which ions are the purple ions, and which are the green? Purple is Na+, green is Cl6. Which of the following atoms has the smallest first ionization energy? a. Al b. P c. Sr d. Ga e. Rb 7. (a) The first ionization energies (IE) of hydrogen and helium are about 1300 kJ/mol and 2300 kJ/mol respectively. Yet, the electron removed from both of these originated in a 1s orbital. Explain in 1-3 sentences the large IE difference observed between these elements. Helium has twice as many protons in its nucleus as Hydrogen. Because the electrons in He and H are in the same orbital (same distance away from nucleus), the electrons are more tightly bound to the nucleus of He (Z eff is larger). Thus, it is more difficult to ionize (pull an electron away from) He. (b) Another particularly interesting exception to the trend in 1st ionization energy is found by comparing nitrogen and oxygen. The 1st IE is higher for N than for O, even though O has 1 more proton in its nucleus. Offer an explanation.

It is helpful to look at the electron diagrams of N and O. The 2p orbitals of nitrogen are exactly halffilled, an energetically favorable state that makes the first ionization energy of nitrogen lower than

expected based on the general trend. In contrast, ionizing oxygen (removing the 4th electron in the 2p orbitals) gives oxygen the energetically favorable half-filled orbitals. Thus, the 4th 2p electron in oxygen is more easily removed than one of the three 2p electrons in nitrogen. 8. An electron can be removed from Na or Rb by electromagnetic radiation (the photoelectric effect). Which element, Na or Rb, would require the shortest wavelength to remove the electron? Na. Removing an electron would ionize the atom, so the question is indirectly asking which atom has the greatest ionization energy since the shortest wavelength corresponds to highest energy (inversely proportional). The valence electrons of Rb are in the n=5 shell, much further from the nucleus than the valence electrons of Na in the n=3 shell. The attraction to the positive nucleus is thus felt more strongly by the electrons in Na (they are in a lower energy state) so it will take more energy to remove them. 9. Based on their positions in the periodic table, predict which atom of the following pairs have the most negative electron affinity: a. Cl, Ar b. K, Co c. S, Ge d. Sn, Te 10. “Photo-gray” lenses for eyeglasses darken in bright sunshine because the lenses contain tiny, transparent AgCl crystals. Exposure to light removes electrons from Cl– ions, forming a chlorine atom in an excited state (indicated below by the asterisk): Cl– →hn (UV light) Cl* + e– + The electrons are transferred to Ag ions, forming silver metal: Ag+ + e– → Ag Silver metal is reflective, producing the photo-gray color. How might substitution of AgBr for AgCl affect the light sensitivity of photo-gray lenses? In answering this question, consider whether more or less energy is needed to remove an electron from a Br– ion than from a Cl– ion. The lenses would be more sensitive because it takes less energy to remove an electron from Br- than Cl- due to their difference in size (Br is larger than Cl so Zeff is smaller)....