* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download THE UNIVERSITY OF LETHBRIDGE DEPARTMENT OF CHEMISTRY
Bremsstrahlung wikipedia , lookup
Rotational spectroscopy wikipedia , lookup
Coupled cluster wikipedia , lookup
Aromaticity wikipedia , lookup
Electron paramagnetic resonance wikipedia , lookup
History of electrochemistry wikipedia , lookup
Surface properties of transition metal oxides wikipedia , lookup
X-ray photoelectron spectroscopy wikipedia , lookup
Rotational–vibrational spectroscopy wikipedia , lookup
Bose–Einstein condensate wikipedia , lookup
State of matter wikipedia , lookup
Heat transfer physics wikipedia , lookup
Physical organic chemistry wikipedia , lookup
Rutherford backscattering spectrometry wikipedia , lookup
Homoaromaticity wikipedia , lookup
Chemical bond wikipedia , lookup
Electrochemistry wikipedia , lookup
THE UNIVERSITY OF LETHBRIDGE DEPARTMENT OF CHEMISTRY CHEMISTRY 2000 PRACTICE TERM TEST #1 ANSWER GUIDE Winter 2005 Explanations in red text are not necessarily required for full marks on the question. Instructor: Dr. R.T. Boeré Time: 50 minutes Value: 50 points No. of Pages = 4. 1. [10] a) What is the coordination number of the Ti atom in [Ti(NH3)4(OH)2]Cl? Six. There are four N and two O atoms directly bound to Ti b) Write out the electron configuration of neutral Ti 1s22s22p63s23p64s23d2 c) What is the dn configuration of the Ti atom in [Ti(NH3)4(OH)2]Cl? Justify your answer. Ti is in 3+ oxidation state, i.e. [Ar]3d1, so this is a d1 configuration d) Write an acceptable name for the complex salt [Ti(NH3)4(OH)2]Cl. Dihydroxotetramminetitanium(III) chloride List cation before anion. List ligand names before metal Use “o” name for anionic ligands (hydroxo), ammine for coordinated NH3, roman number for oxid. st. e) Predict the number of unpaired electrons in the complex ion in [Ti(NH3)4(OH)2]Cl. Justify your answer with a d-orbital splitting diagram There should be one unpaired electron. The splitting diagram looks like: ↑ The geometry is octahedral (the only one we have considered so far.) The sole u.p.e. must therefore fill in lower orbital with “up” spin. All other electrons are core electrons, which are always paired. 2. [4] What inter-molecular forces must be overcome to: a) Melt ice Hydrogen bonds are the dominant IMF in the structure of ice; these must be partially broken to allow liquid water to form. Water is though to have a structure near the m.p. of “flickering clusters” of the IceI structure. Additional IMF that contribute to the structure are DP-DP and dispersion forces. b) Melt solid I2 I2 is a non-polar molecule. Thus only Induced-DP – Induced-DP forces operate. With the very large, easilly polarizable iodine molecule, these dispersion forces are sufficiently strong to cause it to be a solid. c) Remove the water of hydration from MnCl2.4H2O Ion-DP forces dominate, because the water in this formula is most likely coordinated directly to the Mn, and is not merely water of crystallization as the “bottle label” formula would indicate. Dispersion forces are of much less importance. The true formula for this compound is likely to be either [Mn(OH)4]Cl2 (ionic) or [Mn(OH)4Cl2] (molecular) d) Convert liquid NH3 to gaseous NH3. Hydrogen bonding dominates. General DP-DP and dispersion forces also contribute. 1 3. Cuprite is an oxide of copper with the following cubic structure: [6] Oxide ions are at the cube corners and in the cube center. The central oxide ion is tetrahedrally surrounded by four copper ions. a) What is the formula of cuprite? Justify your answer. 8 corner O2- x 1/8th = 1 1 central O2– = 1 4 internal Cu ions. Therefore the formula is Cu2O, b) What is the oxidation state of the copper in this structure? Explain. Oxygen is invariant in its anionic state, except in peroxide and superoxide, neither of which fit the geometry of this lattice. From the formula, 2 Cu ions to 1 O2– anion, it is Cu(I), not Cu(II) c) Would you expect cuprite to be paramagnetic or diamagnetic? Justify your answer. The electron configuration of neutral Cu is [Ar]4s13d10 Cu+ would lose the s electron to give: [Ar]3d10, which has all subshells filled. Hence it would be expected to be diamagnetic. 4. The graph at the right presents the normal boiling points of 2nd, 3rd, 4th, and 5th period element hydrides of the Group 14 and 16 elements. Boiling Points of Group 14 and Group 16 element hydrides [10] Explain the position of each of the eight points in the graph in terms of the kinetic molecular theory of liquids and the kinds of intermolecular forces that operate between the molecules in pure liquids of the given composition. Discuss all the intermolecular forces that apply in each system. The group 14 compounds follow the series: bp. CH4 < SiH4 < GeH4 < SnH4 i.e. the bp. goes up with mass, as predicted by the kinetic molecular theory. The only forces that operate among these symmetrical (tetrahedral) molecules are induced-DP – induced-DP forces Temperature - deg C 150 100 H2O 50 Period H2Te 0 2 3 -50 H2S -100 5 SnH4 GeH4 SiH4 -150 -200 4 H2Se CH4 Grp14 -250 Grp16 This trend is also seen for the bp. H2S < H2Se < H2Te, but all have higher bp. than their Group 14 analogues SiH4, GeH4 and SnH4. This is due to DP-DP forces operating in the polar (angular shape) Group 16 compounds. H2O has a bp ca. 200 degrees higher than expected for a molecule of its size. This shows the dominant influence of H-bonding in this compound. 2 5. The density of copper metal is 8.95 g cm–3. If the unit cell edge length is 361.5 pm, is the copper unit cell simple cubic, body-centred cubic, or face-centred cubic? [8] We calculate the density for the three types of possible unit cells: The density can always be calculated if you know have a postulate for a possible unit cell contents, as well as the geometry of the unit cell. d= m # atoms per unit cell × 63.546 g mol = = # atoms per unit cell × 2.234 V 6.022 ⋅ 1023 atoms mol × (3.615 ⋅ 10−8 cm)3 CELL TYPE # atoms/unit cell calc’d density SIMPLE CUBIC 1 2.234 g/cm3 BCC 2 4.467 g/cm3 g cm3atom FCC 4 8.935 g/cm3 The density for FCC matches well with experiment. Therefore copper metal is FCC. 6. Determine for each of the following molecule the normal or “average” geometry, and demonstrate for each whether or not it is a polar or non-polar molecule. For polar molecules, show the net molecular dipole moment using a Vector (please point the vector towards the negatively charged end of the molecule.) Central atoms are underlined. Show enough work to justify your choice of geometry. [12] 3