Study Guide Exam 1
Study Guide Exam 1 CHM1051
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Popular in Chemistry
This 10 page Study Guide was uploaded by mak15k on Thursday February 4, 2016. The Study Guide belongs to CHM1051 at Florida State University taught by Dr. Albrecht Shmidt/ Dr. Tom in Spring 2016. Since its upload, it has received 108 views. For similar materials see General Chemistry II in Chemistry at Florida State University.
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Date Created: 02/04/16
exam 1: Wednesday, Feb. 10th chm1051 exam 1 study guide 1. Provide the seven Bravais Lattices AND define what a unit cell is. Unit Cell: smallest repeating unit in a structure with no gaps; use math to create symmetrically equivalent objects with unit cells exam 1: Wednesday, Feb. 10th 2. What are the structural differences between amorphous and crystalline solids? Amorphous solids o short range order o no sharp melting point- softened in a range of temperatures o undergo irregular breakage o isotrophic- properties* are independent of the direction in which they are measured o less rigid o Examples: fiberglass, cellophane, polyurethane Crystalline solids o long range order o melt at sharp temperatures o can be cleaved along definite planes o anisotrophic- properties* are in different directions o more rigid o Examples: copper, potassium nitrate amorphous (left) and crystalline (right) 3. The radius of a gold atom is 1.44 Å and elemental silver crystallizes in a face- centered cubic lattice. Calculate the density of silver. a) What we know: FCC= CCP= 4 atoms per unit cell 1 Å= 10 m-10 3 density= g/cm b) Pythagorean theorem indicates: a + b = 4r 2 8 8¿ d= r √ = 1.14 Å( √ = 407pm c) Volume: V= d 3 407pm = 6.74x10 pm 7 3 =6.74x10 cm23 3 d) Mass of unit cell: 4 atoms 1mole 197g Au = 23 = = 1.31x10 -2g/U.C. 1 U.C. 6.022x10 atoms 1 mole Au e) Density: exam 1: Wednesday, Feb. 10th m 1.31x10 −2g perU .C. d= = −23 3 = 19.3 g/cm 3 V 6.74x10 cm 4. Indicate the periodic trends for atomic radii and first ionization energy. s e IE increases a r c Size decreases d E I s IE= ionization energy & size refers to atomic radius (size of atom) s a r 5. Zr in 0.01 Å larger than Hf. Why? What is the name of this phenomenon. The 4f14 electrons are so diffuse that it's almost as if they are not even there, i 2 so She 5d electrons feel the most pressure This is called Lanthenide Contraction: o 5s and 5p orbitals penetrate the 4f-subshell, so the 4f orbital is not shielded from increasing nuclear change, which causes the atomic radii of atoms to decrease throughout the series ALSO... o Why are Eu & Yb larger than expected? They have half-filled electron shells- we want full shells! (1) Eu: [Xe]6s 5d 4f ------------> (2) Eu: [Xe] 6s 4f 2 1 13 2 14 (1) Yb: [Xe] 6s 5d 4f -----------> (2) Yb: [Xe] 6s 4f Since the 5d orbital is the highest, take the electron from here and place in the f orbital. This creates a full s-shell and a partially filled f- shell, as opposed to a full s-shell, partially filled d- AND f-shells...it's more stable now. 6. Discuss the differences between the bonding and chemical behavior of actinides versus lanthanides. While answering this question address why Eu(IO3)3 and Eu2O3 emit the same color light when they are irradiated by long-wavelength UV light. Lanthanides and actinides have primarily ionic bonding properties, which are governed by a charge density. Due to the similarity of the oxidation state and their similar ionic radii (caused by the f-element contractions) separation between the two is difficult. The f sublevel contains seven orbitals, each of which will hold two electrons. Therefore, it is possible to place 14 electrons in the 4f sublevel. Generally exam 1: Wednesday, Feb. 10th speaking, the lanthanides have electron configurations that follow the Aufbau rule, and the 4f sublevel is filled as atomic number increases from cerium (Ce) to lutetium (Lu). However, there are three lanthanide metals that have properties similar to the d block: cerium (Ce), lutetium (Lu), and gadolinium (Gd). All of these metals contain a d electron in their electron configuration. A similar overall trend holds for the 14 elements in the actinide series (numbers 90 to 103): from thorium (Th) to Lawrencium (Lr), the 5f sublevel is progressively filled. The chemistry of the lanthanides differs from main group elements and transition metals because of the nature of the 4f orbitals. These orbitals are "buried" inside the atom and are shielded from the atom's environment by the 4d and 5p electrons. As a consequence, the chemistry of the elements is largely determined by their size, which decreases gradually with increasing atomic number. This phenomenon is known as the lanthanide contraction. All the lanthanide elements exhibit the oxidation state +3. Actinides are typical metals. All of them are soft, have a silvery color (but tarnish in air), and have relatively high density and plasticity. Some of them can be cut with a knife. The hardness of thorium is similar to that of soft steel, so heated pure thorium can be rolled in sheets and pulled into wire. Thorium is nearly half as dense as uranium and plutonium but is harder than both of them. Unlike the lanthanides, most elements of the actinide series have the same properties as the d block. Members of the actinide series can lose multiple electrons to form a variety of different ions. All actinides are radioactive, paramagnetic, and, with the exception of actinium, have several crystalline phases. All actinides are pyrophoric, especially when finely divided (i.e., they spontaneously ignite upon exposure to air). The melting point of actinides does not have a clear dependence on the number of f electrons. The unusually low melting point of neptunium and plutonium (~640 °C) is explained by hybridization of 5f and 6d orbitals and the formation of directional bonds in these metals. Like the lanthanides, all actinides are highly reactive with halogens and chalcogens; however, the actinides react more easily. Actinides, especially those with a small number of 5f electrons, are prone to hybridization. This is explained by the similarity of the electron energies at the 5f, 7s, and 6d subshells. Most actinides exhibit a larger variety of valence states. Inorganic ions have broad UV absorption bands from non-bonding electrons. Transition metal ions and complexes absorb visible light upon excitation between filled and unfilled d-orbitals. Lanthanide and actinide ion absorptions come from excitation of 4f and 5f electrons. F-electrons are shielded from s, p, and d-orbitals and have narrow absorption bands. Because these bands are so narrow, the same color light can be emitted. exam 1: Wednesday, Feb. 10th 7. For the following compounds indicate the following: Lewis dot structure, hybridization of the central atom, structure type, bond angles, indicate if a dipole moment exists, and, if applicable, resonance structures. If multiple bonding exists between two atoms, indicate the types of bonds present. Dipole moments occur when there is a separation of charge. They can occur between two ions in an ionic bond or between atoms in a covalent bond; dipole moments arise from differences in electronegativity. The larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor into the size of the dipole moment. The dipole moment is a measure of the polarity of the molecule. When atoms in a molecule share electrons unequally, they create what is called a dipole moment. This occurs when one atom is more electronegative than another, resulting in that atom pulling more tightly on the shared pair of electrons, or when one atom has a lone pair of electrons and the difference of electronegativity vector points in the same way. One of the most common examples is the water molecule, made up of one oxygen atom and two hydrogen atoms. The differences in electronegativity and lone electrons give oxygen a partial negative charge and each hydrogen a partial positive charge. (http://chemwiki.ucdavis.edu/Physical_Chemistry/Atomic_Theory/Dipole_Mom ents) A. O3 Lewis Dot: Structure Type: Triangle- bent The bent arrangement of ozone is a result of its electronic structure. And the dipole moment is a result coming from Hybridization: Bond Angle: 120° Dipole Movement: Yes B. XeF4 exam 1: Wednesday, Feb. 10th Lewis Structure: Type: Tetrahedral Dipole Movement: No Bond Angle: The shape is square planar because the lone pairs prefer opposite corners of the octahedron. The bond angles will be 90° and 180° in the plane of the molecule. C. BrCl3 Lewis Structure: Structure and Bond Angles: Dipole Movement: No exam 1: Wednesday, Feb. 10th 8. Illustrate the molecular orbital diagrams starting from atomic orbitals for CN+, CN, and CN–. Predict magnetic properties and relative bond lengths based upon the MO’s. C and N each have four valence orbitals: one 2s and three 2p. The 2s are at a lower energy and the 2p orbitals are equal in energy to each other. So the atomic orbitals for C and for N each look like this: 2p 2s The orbitals on the N are slightly lower in energy than those on the C, but that makes little difference here. The s orbitals are spherical in shape, so they can directly overlap and therefore sigma bond every time. The s orbitals on C and N are also very close together in energy since they are both 2s, again leading to a nice, strong overlap and a good, strong bond. So the s atomic orbitals overlap to form two molecular orbitals like this: 2s 2s The orbital of higher energy is antibonding; any electrons in this one will kill bonding, not help it. Orbitals of lower energy than the original orbitals contribute to bonding. On to the p orbitals. Of the three p orbitals on each C and N, one on each atom will point right at each other, leading to direct overlap. Again, the energy match is excellent, leading to a good strong interaction. So two of the p orbitals will overlap like this: 2p 2p This produces another sigma bonding set: one bonding and one antibonding. The other two p orbitals on each atom are orthogonal, meaning they are parallel and do not point right at each other. They do, however, match up energetically and point along like axes, so they can form pi bonding interactions: exam 1: Wednesday, Feb. 10th 2p 2p With the s orbitals being lower in energy and any sigma bonding orbitals being lower in energy, the overall MO diagram will look like this: * * * Where the * indicates an anitbonding molecular orbital that is higher in energy than the atomic orbitals that combine to form it. Now fill the MO diagram and find bond orders. In neutral CN, there are nine valence electrons (five from N, four from C). Filling these according to Hund's rule, all the orbitals are full up to the pibonding one, which has two electrons in it. First, add up the electrons in the bonding and antibonding orbitals. In this case, there are seven bonding electrons and two antibonding electrons. Bond order is (72)/2, or 2.5 Anionic CN has an extra electron. This goes into a bonding orbital. That makes the bond order (82)/2 = 3. Cationic CN has one less electron than neutral CN. That comes out of a bonding orbital, making the bond order (62)/2=2 exam 1: Wednesday, Feb. 10th 9. Explain the difference between p-type and n-type semiconductors. How does this doping affect conductivity? "Figure 1.1- Typical junction alignment (+) and (-) represent major holes (h) and electrons (e)" exam 1: Wednesday, Feb. 10th N-Type: Atoms with one more valence electron than silicon are used to produce "n-type" semiconductor material. These n-type materials are group V elements in the periodic table, and thus their atoms have 5 valence electrons that can form covalent bonds with the 4 valence electrons that silicon atoms have. Because only 4 valence electrons are needed from each atom (silicon and n-type) to form the covalent bonds around the silicon atoms, the extra valence electron present (because n-type materials have 5 valence electrons) when the two atoms bond is free to participate in conduction. Therefore, more electrons are added to the conduction band and hence increases the number of electrons present. P-Type: Atoms with one less valence electron result in "p-type" material. These p-type materials are group III elements in the periodic table. Therefore, p-type material has only 3 valence electrons with which to interact with silicon atoms. The net result is a hole, as not enough electrons are present to form the 4 covalent bonds surrounding the atoms. In p-type material, the number of electrons trapped in bonds is higher, thus effectively increasing the number of holes. In doped material, there is always more of one type of carrier than the other and the type of carrier with the higher concentration is called a "majority carrier", while the lower concentration carrier is called a "minority carrier." Additionally, memorize the following equation for problems concerning Bragg's Law: nλ d= 2sinϴ Solve for d= distance between spacing of atoms, where n= 1 (index of refraction of air), λ= wavelength, and ϴ= angle of incidence.
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