Nme Chem 130 Third Exm On the following pges you will find eight questions covering vries topics rnging from precipittion solubility, cid bse, nd oxidtion reduction rections to metl lignd complexes nd coordintion compounds. 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 /9 Question /9 Question 3 /9 Question 4 /9 Question 5 /4 Question 6 /4 Question 7 /3 Question 8 /4 Totl /100 Potentilly useful equtions nd constnts: c = λν KE = hν W E = hν = hc/λ 1 = 1.09737 10 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 =.998 10 8 m/s h = 6.66 10 34 Js N A = 6.0 10 3 mol 1 Also vilble on seprte hndouts re periodic tble, tble of electronegtivity vlues, nd tbles of K, K b, nd E vlues.
Problem 1. Consider the wek cids H Se, H S, nd HF. Identify which cid is the strongest nd, in two or three sentences, explin your reson for selecting this compound. The strongest cid is the molecule with the wekest X H bond. The fctors tht govern the strength of this bond re the electronegtivity of X nd the size of X, with greter electronegtivity nd lrger size wekening the X H bond. Of the two, the size of X is more importnt thn the electronegtivity of X; thus, H Se is the strongest cid in this group. Problem. Consider the wek bses BrO, BrO, nd O. Identify which bse is the strongest nd, in two or three sentences, explin your reson for selecting this ion. To determine which is the strongest bse we must determine which of the conjugte cids (HBO, HBrO, nd HO ) is the wekest, remembering tht weker cids hve stronger conjugte bses. The strength of the conjugte cids re ffected by the electronegtivity of the centrl element ( more electronegtive centrl element pulls electron density towrd it, wekening the OH bond) nd by the number of oxygens (the greter the number of oxygens, the more positive the oxidtion stte for the centrl element, which pulls electron density towrd it, wekening the OH bond). Bsed on this, HBrO is weker cid thn HO, nd HBrO is weker cid thn HBrO ; thus, HBrO is the wekest cid nd BrO is the strongest bse. Problem 3. The silver ion, Ag +, forms precipittes when rected with NI nd N CO 3. Both precipittes re yellow in color, so it is difficult to use color s unique identifier. One of the two precipittes, however, is soluble in cidic solutions, but the other is not. Write rections showing the precipittion of Ag + with NI nd Ag + with N CO 3. Identify the precipitte tht is soluble in cid nd write rection showing this solubility; use H 3 O + s the cid. In one sentence, explin why this is the precipitte tht is soluble in n cidic solution. The rections forming precipittes re: Ag + (q) + NI(q) " AgI(s) + N + (q) or Ag + (q) + I (q) " AgI(s) Ag + (q) + N CO 3 (q) " Ag CO 3 (s) + N + (q) or Ag + (q) + CO 3 (q) " Ag CO 3 (s) The precipitte tht is soluble in cid is Ag CO 3 (s) becuse crbonte is wek bse; AgI(s) will not dissolve in cid becuse iodide is not wek bse. The rection is or Ag CO 3 (s) + H 3 O + (q) " Ag + (q) + HCO 3 (q) + H O(l) Ag CO 3 (s) + H 3 O + (q) " Ag + (q) + H CO 3 (q) + H O(l)
Problem 4. Metls such s gold nd pltinum re very unrective, which is one reson they re vluble. In 800 A.D. the Islmic lchemist Jbir Ibn Hyyn discovered tht pltinum, which will not dissolve in H or HNO 3 only, will dissolve in mixture of these cids. This mixture is more commonly known s qu regi (or royl wter). The rection responsible for dissolving pltinum is HNO 3 (q) + 6H + (q) + 3(s) + 1 (q) " NO(g) + 3 4 (q) + 4H O(l) Identify the oxidizing gent nd reducing gent in this rection nd the specific chnges in oxidtion sttes tht re occurring. The oxidizing gent is HNO 3 (q) in which the nitrogen picks up three electrons s it chnges from n oxidtion stte of +5 in HNO 3 (q) to n oxidtion stte of + in NO(g). The reducing gent is (s), in which pltinum loses two electrons s it chnges from n oxidtion stte of 0 in (s) to + in 4 (q). Problem 5. Provide the chemicl formul for the coordintion compound hexminecoblt(iii) pentchlorocuprte(ii). The chemicl formul is [Co( ) 6 ][Cu 5 ]. The brckets indicte tht the ch metl-lignd complexes is present s n ion. Hexmine mens tht there re six lignds bound to the Co 3+ ion, nd pentchloro indictes tht there re five ions bound to the Cu + ion. Note tht ech metllignd complex is written so tht the metl ppers first nd the lignds pper second. Problem 6. Provide the nme for the coordintion compound (NH 4 ) 4 [Fe(ox) 3 ] where ox is the oxlte nion, C O 4. The nme of the coordintion compound is mmonium trioxltoferrte(ii). Note tht there is no prefix before simple ctions nd nions (those tht re not metl-lignd complexes) s their number is fixed by the chrge on the metl-lignd complex. NH 4 + is the mmonium ion. The prefix before oxlto (note the chnge from te to to) is tri not tris s there is there is no mbiguity regrding numbers in the lignd s nme. Iron is nmed using its Ltin form when it ppers in metllignd complex tht is n nion; thus, we use ferrte here insted of ironte. The metl-lignd complex hs net chrge of 4 to blnce the four mmonium ions. As ech oxlte hs chrge of, the iron must hve chrge of +.
Problem 7. The tble below contins dt for series of coordintion compounds. Fill in the missing blnks. You my wish to use the boxes below the tble to explin your resoning, nd the spce t the bottom of the pge for scrtch work. empiricl formul # ions #free oxidtion stte for coordintion number for moleculr formul 4 6 5 4 +4 6 [( ) 6 ] 4 b 4 4 3 +4 6 [( ) 4 ] c 4 0 0 +4 6 ( ) 4 d 4 K 3 0 +4 6 K [ 6 ] e K 0 + 4 K[( ) 3 ] f 4 0 +/+ 4/4 [( ) 4 ][ 4 ] Ammoni,, is lignd in ll cses; chloride,, cn be either lignd or n nion; potssium, K +, does not hve lone-pir of electrons to donte nd is present s ction only. The number of chlorines minus the number of potssiums in the empiricl formul gives pltinum s primry vlency (oxidtion stte); thus, (), (b), nd (d) hve oxidtion sttes of +4. Remember tht mmoni does not hve chrge nd does not, therefore, help to stisfy pltinum s primry vlency. If there re no ions, then there re no free chlorines. If there re ions, then the number of free chlorines is equl to the number of ions minus the number of potssiums, which re never lignds, minus one for the metl-lignd complex ion; thus, () hs 5 0 1 = 4 free chlorines; (c) hs no free chlorines becuse there re no ions; nd (d) hs 3 1 = 0 free chlorines. Pltinum s secondry vlency (coordintion number) is the number of lone-pirs bonding to pltinum, which is equl to the number of molecules plus the difference between the number of chlorines in the empiricl formul nd the number of free chlorines; thus the coordintion number for () is 6 + (4 4) = 6; the coordintion number for (b) is 4 + (4 ) = 6; the coordintion number for (c) is + (4 0) = 6; nd the coordintion number for (d) is 1 + (3 0) = 4. Remember tht potssium cnnot serve s lignd s it does not hve lone pir of electons to donte to pltinum. The number of ions for (e) is becuse there is potssium ion, K +, nd, s result, metl-lignd complex tht is n nion. The chemicl formuls follow from the informtion on the primry nd secondry vlencies of pltinum, nd from the number of ions nd the number of free chlorines. The use of brckets round the metl-lignd complexes in ll but (c) indictes tht the metl-lignd complexes re ions. Another possible nswer for (f) is [( ) 3 ][( ) 3 ]; however ( ) ( ) is not possible nswer s the two metl-lignd complexes re neutrl nd will not produce ions when dissolved.
Problem 8. The lignd nitrite my bind through its nitrogen ( NO ) or through one of its oxygens ( ). In the boxes below, drw ll possible geometric isomers nd linkge isomers for the octhedrl coordintion compound ( ) (NO ) 3. Do not include opticl isomers in your nswer. Be creful to drw ech unique isomer only once. There my be more boxes thn nswers. You my wish to use the use the bottom of this pge or the reverse side of this pge for your initil work; however, be sure to plce your finl nswers here. As shown below, there re six possible geometric nd linkge isomers. The three chlorines cn dopt either meridionl or fcil geometry. The two mmonis cn dopt cis or trns geometry when the chlorines re meridionl, but cis geometry only when the chlorines re in fcil rrngement. O N meridionl chlorine cis mmoni meridionl chlorine trns mmoni O N meridionl chlorine trns mmoni NO fcil chlorine cis mmoni meridionl chlorine cis mmoni fcil chlorine cis mmoni