Fifth Exam CHEM 1A Summer 2017

Transcription

1 ifth Exam EM 1A Summer 2017 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( ) 348 D( ) 485 D( ) 413 D( ) 358 D( ) 155 D( ) 436 D(N ) 391 D( ) 190 D( ) 463 D(=) 799 D(N N) 941 D(=) 495 1

2 1. (10) (a) Which factors impact the lattice energy? Write the equation to which the lattice energy is related. Both, the charges of the ions involved and the distance between the ions impact the lattice energy. E el = k Q 1Q 2 d (b) Between lithium chloride and potassium bromide, which one has the highest melting point and why? Explain. Since usually high lattice energies lead to higher melting points, we need to find out which of the two compounds has the higher lattice energy between: Lil vs KBr. Since lithium and chloride ions are both smaller than potassium and bromide ions, we know the lattice energy of lithium chloride is higher and that its melting point will be higher as well. (c) Lithium iodide and sodium bromide lattice energies are nearly equal. In terms of the factors that affect the lattice energy, what can you conclude from this observation? Since the charges are the same, this is telling us the inter-ionic distances are also similar. Na is larger than Li, but Br smaller than I; therefore, the effect of distance almost cancels out. 2. (4) Give the electron configuration, using the kernel notation, for the following ions: a. Zn 2+ [Ar]3d 10 b. Au 3+ [e]4f 14 5d 8 3. (8) (a) Electron affinities are exergonic for the most part; however, all noble gases have endergonic electron affinities. learly explain why and give an example. aving the outmost energy level totally filled with electrons, inserting an extra electron in noble gases means the electron must go in a higher E orbital; since this is unfavorable, EA of noble gases are endergonic. Take, for instance, e: 1s 2 the extra electron would go in the second energy level, which is further from the nucleus and less attracted by it. (b) There is a discontinuity on the ionization energy between families VA and VI A. learly explain why using phosphorous and sulfur as an example. Ionization energies increase as we move to the right on a period. The discontinuity between families VA and VIA arises due to two facts: the electron to be removed comes from a double occupied orbital, and also, removing an electron leaves the element with a half-occupied p subshell, which is particularly stable. VA P 1s 2 2s 2 2p 6 3s 2 3p 3 VIA S 1s 2 2s 2 2p 6 3s 2 3p

3 4. (10) (a) What is electronegativity? A measure of the attraction an atom has for shared electrons. (b) ut of P, S, As and Se, give the elements with the lowest and highest electronegativities. Explain your reasoning. Electronegativity increases: from left to right across a row, and from the bottom to the top of a column. Lowest: As ighest: S. (c) Arrange the following set of bonds from less ionic to more ionic. Explain your reasoning.,, Be. < < Be 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. (d) Arrange the following set of bonds from less covalent to more covalent. Explain your reasoning. S, B, N. B < S < N The closer the atoms are on the periodic table, the smaller the electronegativity difference between them and the more covalent the bond they form. 5. (10) (a) Using bond enthalpies, estimate the rxn for the combustion of acetone ( 3 3 ). Write the balance equation using structural formulas for all compounds involved rxn ( kj mol ) = 1619 = [2D( ) + 6D( ) + D( = ) + 4D( = )] [6D( = ) + 6D( )] = [2(348) + 6(413) + (799) + 4(495)]kJ [6(799) + 6(463)]kJ = 5953 kj kj kj 7572 = 1619 mol mol mol 6. (8) onsider the A 2 4 molecule depicted here, 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. (a) In terms of d 1 and d 2, how could you define the bonding atomic radii of atoms A and? (b) In terms of d 1 and d 2, give the length of in the 2 molecule. A d 2 A Since d1 = 2A, it follows that: ra = d1/2. Since d2 = ra + r, Then d2 = d1/2 + r r A = d 1 2 r = d 2 d 1 2 d 1 The bonding radius of : r = d2 d1/2. distance is 2 r = 2(d2 d1/2) = 2d2 d1 d = 2d 2 d

4 7. (10) Drawing the expanded molecular structures (show multiple bonds and lone pairs) that belong to the following condensed structural formulas of organic compounds: a. 2N ircle All sp 3 atoms. ow many are there? 4 N b. 32NN223 ircle All sp 2 atoms. ow many are there? 6 N N 8. (16) (a) Draw the Lewis structure for the thiocyanate ion: SN. If more than one structure can be drawn, circle the one that you think is the dominant structure and explain your choice clearly. # Val. e s = = 16 Skeleton S N S N S N The structure circled is the dominant because it not only has the lowest formal charges (as the one on the right), but it also leaves the negative charge on the most electronegative atom (nitrogen), which makes more sense. (b) Draw the Lewis Structure for the phosphate ion. ybridization on P? # Val. e s = 6 + 4(6) +2 = 32 Skeleton and s. ptimizing the structure: ybridization on P: sp 3 d. 25 4

5 9. (12) Use the templates provided below to give the trend (draw an arrow) on the following properties. Give a thorough explanation of your selection. a. Direction in which the ionization energy increases on a family. Going from bottom to top on a given family the electrons are found closer and closer to the nucleus (smaller n ). Trying to remove an electron from elements at the top of a given family would require more energy due to the stronger attraction by the nucleus. b. Direction in which the electron affinity is less exothermic on a period. Since the number of protons increases from left to right on a period, and the new electron is inserted in the same n energy level, it is easier to insert the electron where there are more protons; therefore, the electron affinity is less exothermic at the left of a given period. c. Direction in which the atomic size increases on a family. As we move down on a family, the valence electrons are on higher and higher energy levels (farther from the nucleus); this means that the size of the atoms increases as we move from top to bottom on a family. 10. (5) What is the chemical behavior of metal oxides? (a) Write the reaction of calcium oxide with water and (b) calcium oxide with sulfuric acid. a (s) + 2 (l) a() 2 (aq) a (s) + 2 S 4 (aq) as 4 (aq) + 2 (l) 11. (8) (a) What is the shape of the following molecule? What is the hybridization of the central atom? Draw the shape, identify and give the values of the expected bond angles. Se sp 3 d Se <90 <120 (b) Is this molecule polar or nonpolar, why? Polar, the lone pair on the central atom increases polarity and the dipole it creates is not cancelled by the fluorine atoms. 25 5