Nme Key for Chem 130 Second Exm On the following pges you will find questions tht cover the structure of molecules, ions, nd solids, nd the different models we use to explin the nture of chemicl bonding. Red ech question crefully nd consider how you will pproch it before you put pen or pencil to pper. If you re unsure how to nswer one question, then move on to nother question; working on new question my suggest n pproch to the one tht is more troublesome. If question requires written response, be sure tht you nswer in complete sentences nd tht you directly nd clerly ddress the question. Question 1 /4 Question /8 Question 3 /1 Question 4 /1 Question 5 /1 Question 6 /1 Question 7 /8 Question 8 /1 Totl /100 Some potentilly useful equtions nd constnts re provided here. A periodic tble nd other potentilly useful dt re provided on seprte hndout. c = ln E = hn KE = hn W 1 = 1.09737 10 l æ 1 nm ç - è n1 n - 1 ö ø V Q $Q d AVEE = xie. + yie 1 + zie 3 x + y + z (vlence shell electrons only) FC = B æ EN V - N ö δ = - - ç V N B è EN + EN b ø - c =.998 10 8 m/s h = 6.66 10 34 Js N A = 6.0 10 3 mol 1
Problem 1. For ech of the following, drw one vlid Lewis structure, which need not be the best structure. Next, indicte the three-dimensionl geometry by both drwing its VSEPR structure nd by providing the nme for the geometry round the underlined centrl tom. Then, predict whether the molecule or ion is polr or is non-polr. Finlly, give the idelized bond ngles for the stted bonds bsed on your VSEPR structure; if there is more thn one possible bond ngle, then list ech unique bond ngle nd nnotte your Lewis structure to indicte which is which. Your nswers for these lst two items must be consistent with the bonding geometry you identify. See the figure below for the different wys to drwing the Lewis structures nd the corresponding VSEPR structures. The number of vlence electrons, the bonding geometry, the polrity, nd the idel bond ngle(s) for ech re listed here: SO: 18 electrons; bonding geometry is bent (from trigonl plnr electron domin); polr; idel bond ngle of 10 SO3 : 6 electrons; bonding geometry is trigonl pyrmidl (from tetrhedrl electron domin); polr; idel bond ngle of 109.5 SF5+: 40 electrons; bonding geometry is trigonl pyrmidl; non-polr; idel bond ngles of 90, 10, nd 180 Problem. The triiodide nion, I3, is liner. The corresponding nion of fluorine, F3, however, does not exist. In 1 3 sentences, explin why it is impossible for n nion of F3 to form. A Lewis structure for I3 shows tht it hs five electron domins (see figure below), which requires tht the centrl tom, iodine in this cse, ccommodte more thn n octet of electrons. Fluorine cnnot do this; thus, the F3 ion is not possible.
Problem 3. The element Z, which is n element in the first three rows of the periodic tble, forms the molecule ZFO with trigonl pyrmidl bonding geometry round Z. In ddition, the length of the Z O bond suggests it hs bond order greter thn 1 but less thn. Identify Z nd, in 3 sentences, explin how you rrived t your identifiction. A trigonl pyrmidl structure requires three bonding domins nd one non-bonding domin for totl of four electron domins. A Lewis structure tht is consistent with this (see below) nd tht shows single bond between Z nd O nd double bond between ZO requires totl of 6 electrons. Subtrcting 1 electrons for the two oxygens nd seven electrons for the one fluorine leves Z with seven electrons, which plces it in group 18. As Z hs 10 electrons round it (two single bonds, one double bond, nd lone-pir; Z cnnot be F; thus, it must be Cl. Problem 4. Consider the ion NF+, which hs skeletl structure of N N F. Drw ll possible resonnce structures for this ion nd nnotte ech of your structures with the forml chrge for ech tom. See the figure below for the three possible resonnce structures nd forml chrges. These structures must ccount for the 16 vlence electrons in NF+. Circle the resonnce structure bove tht provides the best overll picture of the bonding for NF+ nd in no more thn two sentences, explin your reson for selecting this resonnce structure. Of the three resonnce structures, the one tht is circled hs the fewest toms with forml chrges, which mkes it the resonnce structure tht best represents the ion s bonding.
Problem 5. The underlined centrl tom for two of the nions below use identicl hybrid orbitls to form bonds with the nion s other toms; the remining nion uses different set of hybrid orbitls. Circle the nion tht is different nd, in 3 sentences explin the reson for your choice. As prt of your nswer, identify the specific hybrid orbitls used by ech nion s centrl tom. NO CO3 BF4 The BF4 ion is the one tht is different. See the figure below for the Lewis structures for these three ions, which show tht the two ions on the left both contin double bond tht requires sigm bond nd pi bond; the ion on the right, however, hs only single, sigm bonds. As result, the two ions on the left re trigonl plnr nd hve centrl toms tht re sp hybridized; the ion on the right, however, is tetrhedrl nd hs centrl ion tht is sp 3 hybridized. Problem 6. Moleculr orbitls form when tomic orbitls interct with ech other. Consider the three p-orbitls on nitrogen. When these orbitls interct with the three p-orbitls on nother tom of nitrogen to form N, the resulting moleculr orbitls re of four types: sigm bonding, sigm ntibonding, pi bonding, nd pi ntibonding. For ech of these four types of moleculr orbitls, drw picture in the tble below tht shows how the electrons re distributed in spce reltive to the two nitrogen toms. The pictures below show the different moleculr orbitls. The key fetures your pictures needed to mke cler re (1) tht the electron density for sigm bond (both bonding nd ntibonding) is ligned long the xis tht connects the two nitrogens, but tht the electron density for pi bonds (both bonding nd ntibonding) is locted bove nd below the xis tht connects the two nitrogens; nd () tht the ntibonding orbitls (both bonding nd ntibonding) hve node tht cuts cross the xis tht connects the two nitrogens. Precise drwings were not necessry. sigm bonding sigm ntibonding pi bonding pi ntibonding
Problem 7. Consider the led hlides PbF, PbCl, nd PbBr. Which of these compounds likely hs the highest melting point? Explin your resoning in 1 sentences. Answers here needed to be cler tht Coulomb s lw is importnt, which could be mde directly or inferred by noting tht these re ionic, not covlent, compounds. As the ction is the sme in ll three cses, we focus on the nions only. The nions ll hve the sme chrge, so the distnce between the ions is the key. The fluoride nion is the smllest of the three; thus, s distnce is inversely proportionl to the energy of ttrction, PbF will hve the highest melting point. Problem 8. The figure below shows three cross-sections through the lttice structure for solidstte compound tht consists of rhenium ions nd oxide ions. From left-to-right, these cross-sections re t z = 0, 0.5, nd 1. Wht is the empiricl formul for rhenium oxide. Be sure tht how you rrived t your formul is cler by nnotting the figure bove to show the contribution of ech ion. The chemicl symbol for rhenium is Re. For rhenium, there re eight ions, ll on corners t one-eighth ech; thus, there is one full rhenium ion. For oxide, there re 1 ions, ll on edges t one-fourth ech; thus, there re three full oxide ions. The empiricl formul is ReO3. Bsed on your empiricl formul, wht is the chrge on the rhenium ion? In one sentence, explin how you rrived t your nswer. Ech oxide hs chrge of for totl chrge of 6; thus, rhenium hs chrge of +6. Wht kind of lttice structure do the rhenium ions exhibit? In one sentence, explin how you rrived t your nswer. The rhenium ions re on the corners of the unit cell, which mens they hve lttice structure tht is simple cubic. In wht kind of holes within the oxide lttice re rhenium ions found? In one sentence, explin how you rrived t your nswer. The coordintion number for rhenium within the oxide lttice is six, which mens tht the rhenium ions re in octhedrl holes. See the picture to the right for more detiled view. Wht is coordintion number for the oxide ions reltive to the rhenium ions? In one sentence, explin how you rrived t your nswer. Ech oxide is between two rhenium ions, which gives them coordintion number of two.