Homework Assignment #2 Key

Transcription

1 Homework Assignment #2 Key Chapter (a) 16 protons 18 neutrons 15 electrons (b) 41 protons 52 neutrons 41 electrons (c) 13 protons 14 neutrons 13 electrons (d) 29 protons 34 neutrons 28 electrons (e) 26 protons 30 neutrons 24 electrons (f) 26 protons 29 neutrons 23 electrons

2 73. (a) n = 1 l = 0 (b) n = 4 l = 0 (c) n = 3 l = 1 (d) n = 3 l = 2 (e) n = 4 l = (a) [Ne]3s 1 n = 3 (b) [Ne]3s 3 3p 1 n = 3 (c) [Ar]4s 2 n = 4 (d) [Kr]5s 2 n = 5 (e) [Xe]6s 2 n = 6 (f) [Ar]4s 2 3d 10 4p 5 n = Pauli Exclusion Principle: No two electrons in the same atom may have identical sets of 4 quantum numbers. (a) 1s 2?? this does not violate the principle (PEP) (2 different m s values)?? this would violate the PEP (same m s values) the former arrangement is implied. (b) 1s 2 2p 7 -????????? 1s 2 2p 7 violates the PEP (3 rd p orbital has 3 electrons, 2 with the same m s value) this also violates the Aufbau principle (filling the 2p orbitals before filling the 2s orbitals (c) 1s 3 -??? violates the PEP (orbital has 3 electrons, 2 with the same m s value) 95. (a) Na: 1s 2 s2 2 2p 6 3s 1 or [Ne]3s 1 K: 1s 2 s2 2 2p 6 3s 2 3p 6 4s 1 or [Ar]4s 1 (b) They are similar in that they have the same number of outer shell electrons (1) in the same kind of orbital (s) (c) They are different in that they have a different number of core electrons (10 vs. 18) and that their outer electrons are in different shells (n = 3 vs. n = 4) 97. (a) Ti (b) Bi (c) Si (d) Ra

3 Chapter Ionization energy the amount of energy required to remove the outermost electron the first ionization energy is the removal of an electron from an atom the second ionization energy is the removal of an electron from a cation It is generally more difficult to remove an outer electron from a cation. The nuclear charge is the same as the atom, but the # electrons has decreased; hence, the attraction to the nucleus is greater. 33. using the general trends that you have been taught: least negative (least affinity) most negative (greatest affinity) (a) Fr Cs Rb K Na Li (b) At I Br Cl F (c) Li Be B C N O F (d) K Li C Cl F in actuality (consulting Table 6.2), Cl has a greater affinity than F, hence the positions of Cl and F should be reversed in b) and d). 41. As one moves down a group, the number of filled shells of electrons increases, which results in greater shielding of the outermost electrons from the nucleus by the inner shells filled with electrons, and a lower effective nuclear charge on the outermost electrons. Hence, the atom (or ion) is larger as one moves down the group. 45. Definitions such as electronegativity, electron affinity, ionization energy, etc., are generalized descriptions of the behavior of the elements based upon scientists observations. Statement (b) is better because it reflects the fact that electronegativity is our description of the behavior of the element, not the cause.

4 Chapter 7 7. There are covalent bonds (electron pair sharing) between the N and the H atoms in the ammonium cation. The entire ammonium cation is engaged in ionic bonding with the chloride ion. 21. two cations that are isoelectronic with Kr: Rb +, Sr 2+ two anions that are isoelectronic with Kr: Br -, Se 2- examples of compounds they could form: RbBr, SrSe, Rb 2 Se, SrBr 2

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6 47. For some substances it is possible to draw more than one Lewis structure. In this case, neither drawing fully represents the bonding in the structure. The structure is better approximated as an average, or resonance hybrid, of the Lewis structures.

7 71. In covalent compounds, the strongest forces are the covalent bonds between the atoms within a particular molecule. The intermolecular forces between covalent molecules are relatively weak. In ionic compounds, the forces between atoms are the ionic bonds between the ions, which are very strong. Melting disrupts the intermolecular forces between molecules in covalent compounds, and the ionic bonds between ions in ionic compounds. To overcome an ionic bond takes a lot more energy than it takes to overcome an intermolecular force. 73. We show only partial charges on the atoms of polar molecules because no atom takes an electron completely from the other.

8 Uncollected problems: Chapter Hybridized atomic orbitals are combinations of atomic orbitals (s, p, sometimes d) on the same atom which give the right orientations (and energies) to correctly describe the bonding within a molecule, particularly in terms of the geometry (shape). The theory of hybrid orbitals is useful for explaining the bonding in cases when bonding arrangements or bond angles are significantly different than the angles between non-hybridized atomic orbitals. 19. assume that A is the central atom. (a) AB 4 4 things sp 3 (b) AB 2 U 3-5 things sp 3 d (c) AB 3 U - 4 things sp 3 (d) ABU 4-5 things sp 3 d (e) ABU 3-4 things sp 3

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