Chem 14A 2010 Fall Midterm Solutions

2pt for final answer 1sf 1units 6 q5a the average

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2pt for final answer. -1sf -1units 6
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Q5A. The average speed of a helium atom at 25°C is 1.23 × 10 3 m·s –1 . What is the average wavelength of a helium atom at this temperature? (6pt) 1 23 1 24 27 4.00 g mol mass of He atom 6.022 10 atoms mol 6.64 10 g or 6.64 10 kg = × = × × From the de Broglie relationship, λ = h p , and p = mv, we can calculate wavelength. 1 34 27 1 34 2 1 27 1 11 ( ) 6.626 08 10 J s (6.64 10 kg)(1230 m s ) 6.626 08 10 kg m s (6.64 10 kg)(1230 m s ) 8.11 10 m h mv λ = × = × × = × = × For partial credit 1pt for each step. -1sf -1units Q5B. The energy required to break a C–C bond in a molecule is 348 kJ·mol –1 . Will visible light be able to break this bond? If yes, what is the color of that light? If not, what type of electromagnetic radiation will be suitable? (6pt) Energy needed to break one bond is: C – C 348 kJ mol C–C bonds × 1 mol C–C bonds 6.022 × 10 23 C–C bonds × 1000 J 1 kJ = 5.78 × 10 -19 J 1pt λ = hc Δ E = 3.44 × 10 -7 m or 344 nm 1pt No visible light cannot break the bond. 2pt 344 nm is in the ultraviolet region of the electromagnetic spectrum. 2pt 7
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Q6. Consider the bonding in CH 2 = CHCHO. Draw the most important Lewis structure. Include all nonzero formal charges. Identify the composition of all the bonds and the hybridization of each lone pair—for example, by writing σ (H1s,C2sp 2 ). (14pt) C C C O H H H H σ ( H1 s ,C2 sp 2 ) σ ( C2 sp 2 ,C2 sp 2 ), π ( C2 p ,C2 p ) σ ( C2 sp 2 ,C2 sp 2 ) σ ( C2 sp 2 ,O2 sp 2 ), π ( C2 p ,O2 p ) σ ( H1 s ,C2 sp 2 ) O sp 2 (H1 s ,C2 sp 2 ) σ 2pt Lewis structure 2pt All atoms have a formal charge of zero. 10pt For correctly identifying each of the 9 bonds and the lone pair. 8
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9 Q7. Determine the following information for the species given. (a) AsF 2 Cl 3 (4pt) VSEPR formula: AX 5 molecular shape: trigonal bipyramidal hybridization of central atom: sp 3 d give one bond angle: 90 o , 120 o , 180 o (b) NH 4 + (4pt) VSEPR formula: AX 4 molecular shape: tetrahedral hybridization of central atom: sp 3 give one bond angle: 109.5 o (c) H 2 CBr 2 (5pt) VSEPR formula: AX 4 molecular shape: tetrahedral hybridization of central atom: sp 3 polar or non-polar: polar give one bond angle: 109.5 o or ~109.5 o (d) H 2 SiO (5pt) VSEPR formula: AX 3 molecular shape: trigonal planar hybridization of central atom: sp 2 polar or non-polar: polar give one bond angle: 120 o or ~120 o 1pt for each correct answer.
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Q8A. Consider a hypothetical species He-H.
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  • Fall '10
  • Lavelle
  • Atom, Electron, TA, Molecule, λ, Constants and Formulas