Chem 130 Second Exam

Transcription

1 Nme Chem 130 Second Exm On the following pges you will find seven questions covering vries topics rnging from the structure of molecules, ions, nd solids to different models for explining 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 question, move to nother; 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. Prtil credit is willingly given on ll problems so be sure to nswer ll questions! Question 1 /3 Question /10 Question 3 /10 Question 5 /10 Question 6 /10 Question 7 /18 Question 4 /10 Totl /100 Potentilly useful equtions nd constnts: c = λν KE = hν W E = hν = hc/λ 1 = λ 1 nm n1 n 1 FC B = V N V Q!Q d OX = V N B (0if lest EN;1if most EN) δ = V N B EN EN + EN b c = m/s h = Js N A = mol 1 Also vilble on seprte hndouts re tble of electronegtivity vlues bsed on verge vlnce electron energies, tble of pcking possibilities bsed on reltive sizes of ctions nd nions, nd periodic tble.

2 Problem 1. For ech of the following molecules or ions, drw ny one vlid Lewis structure of your choosing (it need not be the best structure). Annotte your structure by indicting the forml chrge on ech tom. Finlly, give the nme for the bonding geometry round the underlined centrl tom, predict whether the molecule or ion is polr (P) or non-polr (NP), nd provide the idelized bond ngle for the stted bonds. Molecule or Ion CCl F Lewis Structure There re 4 + (7) + (7) = 3 electrons in the molecule nd four substituents round the centrl tom, requiring four electron domins. The only possible LS hs single bonds from C to Cl nd from C to F. All forml chrges re zero. Bonding Geometry tetrhedrl Polr or Non-Polr? polr Idel Bond Angle for n F C Cl IO F There re 7 + (6) + (7) + 1 = 34 electrons nd four substituents round the centrl tom, requiring five electron domins with the lone pir nd two oxygens (lrger thn the fluorines) in xil positions. Using single bonds to O nd F gives forml chrges of zero for F nd 1 for O nd +1 for I. see-sw polr n O I O 10 ICl 4 There re 7 + 4(7) + 1 = 36 electrons nd four substituents round the centrl tom, requiring six electron domins. The only possible LS hs single bonds from I to Cl. All forml chrges on Cl re zero, leving forml chrge of 1 on I. squre plnr non-polr Cl I Cl 90 XeOF 4 There re (7) = 4 electrons nd five substituents round the centrl tom, requiring six electron domins. With double bond between Xe nd O, nd single bonds between Xe nd F, ll forml chrges re zero. squre pyrmidl polr n F-Xe-F bond is 90 or 180

3 Problem. The element Z forms the moleculr compound ZBr 3 with trigonl pyrmidl electron domin geometry. Identify one exmple of n element tht could be Z. In no more thn three sentences, clerly explin your reson for picking this element. A trigonl pyrmidl geometry mens tht there re three bonds between Z nd Br, nd lone-pir of electrons on Z. The structure, therefore, hs totl of 6 electrons. Ech bromine provides seven electrons, for totl of 1, leving five electrons for Z. Possibilities for Z re nitrogen (N), phosphorous (P), rsenic (As), ntimony (Sb), nd bismuth (Bi). Problem 3. Consider the molecules nd ions listed below. The underlined centrl toms in three of these species use the sme type of hybrid orbitls to form bonds with the remining toms; the fourth species uses different set of hybrid orbitls. Circle the species tht is different nd indicte the type of hybrid orbitls it uses. In no more thn three sentences, clerly explin your reson for picking this species. NO CO 3 BF 3 NH 3 Lewis structures for the ions NO nd CO 3, nd the molecule BF 3 hve identicl geometries consisting of three electron domins. Ech hs trigonl plnr electron domin geometry nd, therefore, ech hs the sme hybridiztion. Ammoni, NH 3, hs four electron domins (three of which re bonding), which requires tetrhedrl rrngement of electron domins nd sp 3 hybridiztion.

4 Problem 4. Shown to the right is the moleculr orbitl digrm for the molecule BO. Complete the moleculr orbitl digrm by filling in the electrons for ech tom nd for the molecule. Bsed on your moleculr orbitl digrm, wht is BO s bond order? Is BO prmgnetic or dimgnetic compound (circle your nswer below)? In no more thn three sentences, clerly explin the resons for your nswers. The bond order for BO is.5. BO is: prmgnetic dimgnetic The bond order is the difference between the number of bonding nd ntibonding electrons, divided by two; thus (7 )/ or.5. There is single unpired electron in the σ p orbitl, so BO is prmgnetic. Problem 5 The following compounds re generlly considered covlent: HCl, ICl, nd SCl. As we hve seen, pure covlent bond is rre. Rnk these compounds from the one showing the lest ionic chrcter to the one showing the gretest ionic chrcter. Plce your nswers in the tble nd show ny relevnt work nd/or explntion in the spce below the tble. lest ionic chrcter gretest ionic chrcter SCl ICl HCl Ionicity is given by the difference in electronegtivities (ΔEN) between the two elements. The gretest ΔEN is for HCl (0.57), so it hs the most ionic chrcter, nd the smllest ΔEN is for SCl (0.8), so it hs the lest ionic chrcter. Problem 6. The following compounds hve similr ionic chrcters, but differ in the degree of metllic nd covlent chrcter: MgH, BSi nd ZnS. Rnk these compounds from the one showing the most metllic chrcter to the one showing the most covlent chrcter. Plce your nswers in the tble nd show ny relevnt work nd/or explntion in the spce below the tble. most metllic chrcter most covlent chrcter BSi MgH ZnS Covlency is given by the verge electronegtivity of the elements. The lrgest verge electronegtivity is ZnS (.095), so it hs the most covlent chrcter, nd the smllest verge electronegtivity is for BSi (1.40), which it hs the most metllic chrcter.

5 Problem 7. Shown below re three cross-sections through the unit cell of clcium titnte, lso known s the minerl perovskite. Wht is the empiricl formul for clcium titnte bsed on this unit cell? Be sure to clerly explin how you rrived t this formul. Ech of the eight clcium ions is on corner nd contributes 1/8 th ech to the unit cell for totl of one clcium ion. The six oxygen ions re on fces nd contribute ½ ech to the unit cell, for totl of three oxygen ions. The single titnium ion is in the middle of the unit cell nd contributes itself wholly to the unit cell. The empiricl formul for clcium titnte, therefore, is CTiO 3. The titnium ion cn be considered to sit in two types of holes hole in lttice defined by clcium ions nd hole in lttice defined by oxygen ions. For ech, stte the type of hole in which the titnium ion sits nd wht percentge of these holes re filled. For lttice of clcium ions, titnium sits in cubic hole nd occupies 100% of these holes. For lttice of oxygen ions, titnium sits in n octhedrl hole nd occupies 100% of these holes. To how mny oxygen ions is ech clcium ion coordinted? Briefly explin how you rrived t your nswer in no more thn three sentences nd/or with n pproprite sketch. Ech clcium is coordinted to 1 oxygen ions. To see this, consider the figure on the left, which shows cross-section through the xy-plne of eight unit cells. The blck circle is clcium ion on the corner of these eight unit cells. Ech open circle is n oxygen ion sitting on the fce shred by two unit cells. We cn see here tht the coordintion number within the xy-plne is four. We cn lso drw cross-sections through the xz-plne nd the yz-plne, ech of which lso will contin four oxygen ions, giving totl of 1 oxygen ions coordinted to this single clcium ion.