Assignment 7 and Practice Third Exam Solutions

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1 Assignment 7 and Practice Third Exam Solutions 1. Draw the structures of the following molecules, and don t forget their lone pairs! (a) thiocyanide anion, SC Isoelectronic analogies: S is like, and is like,, too; so, SC is like C, or C 2 : [ S C ] or [ S = C = ] (b) acetone, (C 3 ) 2 C A methyl group, C 3, is a methane molecule with one removed; methyl groups make single bonds, and thus acetone is like formaldehyde, 2 C, with methyl groups in place of formaldehyde s hydrogens: C C C (c) thioacetamide, C 3 CS 2 Start with acetone, change the to S, and change one methyl group to 2 (with a lone pair on the ): S C C (d) oxalic acid, 2 C 2 4 (which will make more sense to you if written CC) xalic acid is simply two carboxylic acid groups ( C) joined through a single C C bond: C C 2. Some short answer or fill-in-the-blank questions: (a) Why is chlorine pentafluoride polar but phosphorus pentafluoride isn t? 5 has a steric number of six and an octahedral electron pair geometry, but one of those pairs is a lone pair, leading to a square-based pyramidal molecular geometry (just like I 5 shown in igure 3.12 on page 84 in the text). P 5 is a symmetric trigonal bipyramid. 1

2 (b) Why is the P 3 dipole moment less than that for 3? The molecular structures of P 3 and 3 are the same: trigonal pyramidal. ut the P bond dipole moment is smaller than the bond dipole moment because P is less electronegative than. (c) There are 3 distinctly different molecules with the empirical formula C 2 2. And here they are: C = C C = C C = C (d) The molecule P 2 3 could have a dipole moment. There are three isomers, one non-polar, and two polar: P P non-polar polar P polar 2 (e) The molecular orbital electron configuration of the carbide ion, C 2, is σ 2 1s σ1s * 2 σ2s 2 σ2s * 2 π2p 4 σ2pz 2. C 2 2 is isoelectronic to 2, with a triple bond: C C The molecule C 3 is a lovely pale blue gas that decomposes when irradiated according to the net reaction 2 C 3 C 3 C forming hexafluoroethane, C 3 C 3, and nitric oxide,. Draw the Lewis electron dot structure of C 3, and then answer the following questions. C or C 2

3 Chem 6, 9 Section Spring 2002 (a) C 3 has one π bond. (It s in the = double bond.) (b) The hybridization scheme used by the atom is sp 2. (c) The hybridization scheme used by the C atom is sp 3. (d) The hybridization scheme used by the atom is sp 2. (e) The molecule has 12 lone pair(s) of electrons. (f) The C bond is not linear. (because is sp 2 hybridized) (g) The formal charge on the atom is zero. (5 valence e s 2 nonbonded electrons (6/2) bonded electrons = 0) (h) The strongest bond in C 3 is most likely between the atoms and. (double bond) (i) The following are paramagnetic: (dd number of electrons) 4. Each of the following Lewis dot structures has one or more deficiency that makes it somewhere between less than optimal or dead wrong. Draw the optimum structure and state what is bad about the structures below. (a) 3 This is dead wrong too many electrons! oron has only three valence electrons, making 3 an electron-deficient molecule: (b) 2 There are 9 e s on. ptimum (resonance, with the odd electron on the less electronegative atom): (c) CS C S ad formal charges on ( 1) and S (+1) compared to optimal structure with no formal charges: C S 3

4 (d) C 2 C o octet around. ptimal: C 5. Some dipole moment questions: (a) The dipole moment of is D = C m, and the bond is 11.3% ionic. What is the bond length? We use the expression relating the dipole moment, µ, to the bond length, R, and the fractional ionicity δ (11.3% ionic means δ = 0.113): (µ/d) (0.2082) (0.8881) (0.2082) R/Å = = = 1.64 or R = 1.64 Å δ Equivalently, we can express these quantities in their fundamental SI units: R = µ δe = C m (0.113) ( C) = m or R = 1.64 Å (b) Given the following bond dipole moments: C 0.40 D C 1.46 D C I 1.19 D predict the magnitude and direction for the dipole moments of the linear molecules: (i) C C (ii) I C C (iii) I C C irst, we use electronegativity values (which will be provided on the exam if needed) to deduce the direction of each of these bond dipole moments. The relevant electronegativities are: C, 2.55;, 2.20;, 3.16; and I, Using these values and remembering that the dipole moment vector points from the negative (more electronegative) to the positive end of the dipole, we can write C C C I This lets us write bond dipole vectors in the three molecules of interest and combine them to form the total dipole moment magnitudes and directions: C C I C C I C C 1.46 D 0.40 D 1.19 D 0.40 D 1.19 D 1.46 D net: 1.86 D net: 1.59 D net: 0.27 D 4

5 (c) The molecule 2 C, phosgene, has a net dipole moment that points one way or another along one bond in the molecule. Which bond is it, and how can you tell? (Don t worry about its direction along that bond.) 2 C is a planar molecule (like formaldehyde) with the central C atom double-bonded to the and single-bonded to the two atoms so that the C= bond is a symmetry axis. The dipole points along this bond, because the C bond moment components perpendicular to this direction cancel each other, while their components along this direction combine with the C= bond moment to produce a net moment that is non-zero. (d) There are many compounds that contain Pt with four things bonded to it in a square planar arrangement. (The chemotherapy drug called cisplatin is one such.) Consider the two generic types of compounds with empirical formulas PtA 3 and PtA 2 2 in which A and are atoms bonded to the Pt. Let s assume A is more electronegative than. Draw all possible isomers of these compounds, and indicate the direction of the net dipole moment (if there is one). There is only one PtA 3 isomer, but two PtA 2 2 isomers, one of which is non-polar: A A A A Pt Pt Pt non-polar A 6. Give the electron pair geometry and the molecular geometry of each of the following oxides. Draw molecular structures for each, showing all single bonds, multiple bonds, and lone pairs to support your geometry assignments. (a) C 2 electron pair geometry: linear molecular geometry: linear = C = (b) S 3 electron pair geometry: trigonal planar molecular geometry: trigonal planar S (plus two resonance forms in which this structure is rotated 120 first clockwise and then counterclockwise) 5

6 (c) Xe 4 electron pair geometry: tetrahedral molecular geometry: tetrahedral Xe 6

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