Fifth Exam CHEM 1A Summer 2013

Transcription

1 ifth Exam CEM 1A Summer 2013 Name: Last KEY irst Instructions: Read every problem carefully. Gauge your time. Use the proper number of significant figures on your results. Don t just believe the calculator, estimate of your numerical answer; points will be deducted if you give an impossible answer(s). If you need me to clarify what I am asking, let me know. D NT ask me if your answer is correct or if you are solving a problem correctly, unless you are ready to turn in the exam. Write your final answer in the box provided for every problem, and show your work clearly for partial credit. Useful Information: kj/mol D(C C) 348 D(C ) 485 D(C ) 413 D(C ) 358 D( ) 155 D( ) 436 D(N ) 391 D( ) 190 D( ) 463 D(C=) 799 D(N N) 941 D(=) 495 1

2 1. (6) Consider the A 2 4 molecule depicted below, where A and are elements. The A A bond length in this molecule is d 1, and the four A bond lengths are each d 2. In terms of d 1 and d 2 define the distance of the bond length. (hint: the bond length is 2 r.) d 2 r A = d 1 /2. A A Since d 2 = r A + r, Then d 2 = d 1 /2 + r The bonding length is: 2 (r = d 2 d 1 /2) = 2 d 2 d 1. d 1 2. (6) a. Estimate the volume of an atom of gold. The element has a density of g/cm 3. ( ) ( ) b. ow many atoms of gold are there in five drops of water? (int: 1 ml = 20 drops.) 1 ml = 1 cm 3 3. (6) Arrange these ionic compounds in order of increasing magnitude of lattice energy: Ca, KBr, KCl, Sr. Explain your reasoning thoroughly. No reasoning, no points! KBr KCl Sr Ca KBr and KCl should have lattice energies of smaller magnitude than Ca and Sr because of their lower ionic charges (1+, 1 compared to 2+, 2 ). When we compare KBr and KCl, we expect KBr to have a lattice energy of lower magnitude due to the larger ionic radius of the bromide ion relative to the chloride ion. Between Ca and Sr, we expect Sr to have a lattice energy of lower magnitude due to the larger ionic radius of the strontium ion relative to the calcium ion. 18 2

3 4. (4) Write the electron configuration for the following ions (use the kernel notation): a. Au + [e]4f 14 5d 10 b. Rh 3+ [Kr]4d 6 5. (6) Using the following template, show the directions in which the atomic radius increases along a group and a period. Draw an arrow to show the tendencies, and briefly explain, using complete sentences, your reasoning. (No explanation, no points.) The atomic radius decreases as we move from left to right on a period since the electrons are inserted into the same energy level (i.e. same distance from the nucleus), but the number of protons is increasing so the electrons feel a stronger pull towards the nucleus and the atomic radius decreases. The atomic radius increases as we move from top to bottom on a family. The electrons are inserted into a higher energy level (i.e. they are farther from the nucleus), therefore, the atoms are more voluminous. 6. (4) What is electronegativity? Define it. Using the following template, point out where you find the element with the highest electronegativity; give its name and symbos. Electronegativity refers to the relative ability of an atom to attract the bonding electrons towards it. luorine () ighest electronegativity 7. (6) Generally, as one goes across a period from left to right the ionization energy increases; nevertheless, there are discontinuities to the trend. Write the full electron configuration of Mg and Al; then explain the discontinuity on ionization energy between periods IIA and IIIA. IIA Mg 1s 2 2s 2 2p 6 3s 2 IIIA Al 1s 2 2s 2 2p 6 3s 2 3p 1 The e- is removed from a p orbital, which is further from the nucleus. There is also a small amount of repulsion by the s electrons. 20 3

4 8. (6) A molecular compound is composed of 58.8% e, 7.2%, and 34.0% by mass. The approximate molecular mass is g. a. What is the molecular formula of the compound? b. Draw the best Lewis Structure for this compound. ( ) ; ( ) ; e 4 MM = ( ) ; #V. e - s = (4* 7) = 42 e 9. (8) N 2 has a linear, unsymmetrical structure that may be thought of as a hybrid of two resonance forms. owever, three Lewis Structures can be drawn. a. Draw the three Lewis Structures for dinitrogen oxide. (Label them as A, B and C.) C N N N N N N A B C Nitrogen is the central atom because is both: the least electronegative and the one that forms more bonds. b. Which of the three structures is the most important? Explain. Structure A is the most important. Structures A and B have the same formal charges, but on structure A the negative charge is on the most electronegative atom (oxygen) as expected. 10. (4) Which will experience the greater effective nuclear charge, the electrons in the n = 3 shell in Ar or the n = 3 shell in Kr? Which will be closer to the nucleus? Explain thoroughly. Krypton has a larger nuclear charge (Z = 36) than argon (Z = 18). The shielding of electrons in the n = 3 shell by the 1s, 2s and 2p core electrons in the two atoms is approximately equal, which means the electrons in the n = 3 shell of krypton experience a greater effective nuclear charge due to the presence of a higher number of protons, and are thus closer to the nucleus. 18 4

5 11. (6) a. Arrange the following set of bonds from less covalent to more covalent. Explain your reasoning. C,, Be. Be < C < The closer the atoms are on the periodic table, the smaller the electronegativity difference between them and the more covalent the bond. b. Arrange the following set of bonds from less ionic to more ionic. Explain your reasoning. C S, B, N. N < C S < B The farther apart two atoms are on the periodic table, the larger the electronegativity difference between them and the more ionic the bond they form. 12. (8) Give the skeleton structures of the following compounds. Include lone pairs on those elements that require them. a. C 2 CCCNC 2 C 2 CC C C C C N C C C C b. CCCCCC 2 NC C C C C C C N C 13. (8) Using bond enthalpies, estimate the Δ for the combustion of ethanol (C 2 5 ). Balance the equation (all substances are gases) and draw the structural formulas so that you can visualize bonds. C 2 5 (g) (g) 2 C 2 (g) (g). C C 3 2 C Δ = [D(C C) + 5 D(C ) + D(C ) + D( ) + 3 D(=)] [4 D(C=) 6 D( )] Δ=[(348 kj/mol) + 5 (413 kj/mol) + (358 kj/mol) + (463 kj/mol)+3(495 kj/mol)] [4(799 kj/mol)+6(463kj/mol)] Δ = (348 kj/mol kj/mol kj/mol ) (5974 kj/mol) = 1,255 kj/mol 22 5

6 14. (6) Circle the best choice in the list and explain your reasoning clearly: a. Largest ionic radius Cr 6+, P 3-, Cl -, K +, or Ti 4+ This is an isoelectronic series, the more protons the smaller the radius, and the more electrons the larger the radius; the one with less protons and more electrons is selected. b. Lowest second ionization energy: Na, Mg or Al. Na has only one valence electron, so it s too hard to remove two. Mg and Al have enough electrons to be removed, but Al has more protons so it s harder to remove the second electron form it. Mg is the choice! 15. (4) In general, electron affinity becomes more exergonic (more favorable) as we go from left to right across a row on the periodic table; however, there are some exceptions. ne of them happens between groups IVA and VA (the process is less exergonic for VA, but it should be more exergonic). Explain this exception using the arrow diagrams for the electron configurations of C and N. or C: or N: rom the diagrams we see that adding an electron to C half-fills the p subshell, whereas, in N the electron is added to an already occupied orbital so there is repulsion and the process is less favorable (less exergonic). 16. (12) Complete the following table: ormula Lewis Structure (Show Resonance Structures) Electron Domain Geometry Molecular Geometry Bond Angle Molecular Geometry (Use wedges and dashed lines) Polar? C 3 C Tetrahedral on both, C and. Tetrahedral around C Bent around on C Less than 109 n C <109 Yes Se4 Se Trigonal Bipyramidal Seesaw Less than 90 and less than 120 Se <90 <120 Yes 22 6