Chemistry 104B - Spring 2002 - JohnArnold - Midterm 1

Chemistry 104B - Spring 2002 - JohnArnold - Midterm 1 - 1(5...

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Unformatted text preview: 1. (5 points) (a) Fill in the missing elements in the following table. Periodic Table (b) (5 points) Identify three elements that tend to form square-planar complexes, show their common oxidation states in this geometry, and draw structures of two examples with different metals. RhU), Ir(I), Ni(H) (depending on ligands), Pd(II), PtUI) «t 7) flew-“ls fly 0’. 5 . Pth/[III'Rh‘n‘fiCI H3N/lnr-Ptl“\\\\C| Pthr ‘Ppha H3N( \CI 4* 2' (c) (I point) Identify a biologically relevant metal from the 2nd row of transition elements. MO t t (d) (3 points) Briefly describe what is meant by the term ‘lanthanide contraction’. ZIP f The outermost electrons experience an increasin Z ff due to the poor shieldin of f , and are therefore drawn in to slightly closer to the nucleus, leadingutothe‘decrease in atomic radii as the atomic number increases (as you go across the period). ---‘ Fifi V.” 2. (6 points). For the following octahedral complexes, determine the number of unpaired electrons, spin—Qmagnetic moment, and ligand field stabilization energy: ‘ _ - “W F—_er_‘—'—~—._._m (a) [RuC16]3 '4 I M Ru(III) (15 low spin, 1 unpaired e- LFSE = —2.0 A0 1 + \ tt,:(1(1+2))A(1/2) = 1.73 14131 H Wm (b) [Cr(NH3)6r+ Cr(III) d3 3 unpaired c" L LFSE : -12 A0 1 5 1.15 : (3(3+2))A(1/2) = 3.87 pa Wei—fr; 3. (14 points) For each of the following, determine the: (i) number of valence d electrons on the metal ion \ Vi, (ii) oxidation state of the metal ion (iii) total electron count at the metal ion (iv) structures of (a) and (b) (a) 0s3(C0)12 (b) (n6-06H6)2(:r (1“, 03(0), 186's d6, 0(0), 18e‘s so 00/111,, I .\\\\CO '05" | (c) Potassium hexacyanoiron (H) d6, Fe(II), 186's (‘1) C's Re\ CI3RBL—R6CI3 d4,Re(1H), 126's 4. (5 points) Using a simple crystal field splitting diagram (and reference to any of the character tables supplied if you think them useful), show what would happen to the d orbital energies if an octahedral complex is stretched along the z axis. (Note that at the limit, the common square— planar geometry would result.) a d m u. r x __ n 3r 3% (if, lg, mu d’a'yz ,9; aw “U 7i xx £3 ’ v':-—“'--- dx I / 3—“ —— — — —4':'-;::::_'_' ' ‘ ' ' ' ' ’4’ spherical field ‘*~._ ""*— dag m {ma W; 5- ,’II 5“—— dyz dxz _ 2:66; — square planar field hisluv/ [1 it} + 2. A {6417, lW at?“ "F 2, [W (11L {'Ori dfimjmfl, ,OIY} out +1 &¥v WVWL-B (MM "42.. I21) A? WM] aw?” 6‘ flit“: a“ 5. (8 points) Sketch all geometric isomers of the following complexes. Indicate which would exhibit optical activity: (a) [CO(OCH2CH2NH2)3] A = H2NCH20H20- N ‘ o A mer / \ / N W" Br EL NH Br FL NH3 Jr L QJ< fac mer (c) Diaquodiiododinitroplatinum(IV) | ONO ONO ONO/,1,“ \\\\OH2 ONO/1,1," l ‘\\\\OH2 Um““ I “‘\\\\OH2 I ‘ “Pfu .Ptu '. pt 1C _ “1,3,5; -. / \ / \ If \ rt w m6 \ T C.“ ONO 0H2C6 H20 I 0H2 51} h“ ( l “h l ONO A) \ ML " dwtm L. h #01110 I has” L L 0H2 E 0H2 {‘3‘ D‘- Ilflln.,F|)t__u\\\OH2 HQO/I'fn.,Plt..\\\‘\ONO i ONO/UII.,FL...\\\\\OH2 Q I H20( I \I I/ ‘ \ouo 5 ONO/ l \t O"K V B ONO“ I : _l l‘ / \Vi ‘7 0tle ’iWH/J .‘ _ '. Mex picxfg (Li/3 enantlomers / l 3/ W0 L 6. (6 points) A compound with the empirical formula Fe(H20)4(CN)2 has a magnetic moment corresponding to 2 2/3 unpaired electrons per iron. (a) How do you explain this odd result? (b) Calculate the average spin only magnetic moment per iron. (Hint: Tw M 5 octahedral Fe(II) species are involved, each containing a single type of ligand.) & Z, " l I; l {3 [L a) The compound contains both high spin and low spin Fe(II). The molecular formula is JO” 5 2/ [Fe(H20)6]2[Fe(CN)61 The aquo complex is low spin Fe(II), therefore, contributes 2* 4 unpaired electrons. The cyano complex is high spin Fe(II) has no unpaired electrons. Overall, the molecule has a total of 8 unpaired electrons for 3 Fe center, thus 2 2/3 unpaired electrons per Fe. N b) p5 = (2 2/3*(2 2/3+2))"(1/2) = 3.53 it]; “9‘ R \v see) tale) “a? W M wM¢® \ 7. (4 points) Give electronic configurations for: (1) Hr” [Xe]4f” 4'1 in- (ii) Ni2+ [Ar]3d8 (iii) Tc” [Kr] (iv) Zn2+ [Ar]3d‘° 8. (4 points) A pink solid with the formula CoC13WH3)5H20 dissolves in water to give a pink solution that gives 3 mol equivalents of AgCl on titration with AgN03. When the pink solid is heated, it loses 1 mol equivalent of water to give a purple solid with an unchanged ratio of NH32C1:Co. The purple solid dissolves to give a solution that quickly forms 2 mol equivalents of AgCl on titration with AgNO3, then, more slowly, further reacts to from another mol equivalent of AgCl. Draw the structures of the two complexes. - - 7 - an; [m 5 '- i "AL The first compound has the formula [Co(NH3)5(H20)]C13 *1 pW/Y UlThe second compound has the formula [C0(NH3)5C1]C12 t L 9. (5 points) In lecture, we discussed the claim that one~coordinate complexes of Cu(l) and Agfl) had been prepared. a. Say why you think one-coordinate complexes, in general, are unlikely to be stable. b. What mistakes were made in the case of the Cu(I) and Ag(I) species described in your textbook? c. What characterization techniques would you have used to confirm their true identity. +1 a. One coordinate complexes are rare because positive metal ions prefer to be surrounded by mulitiple ligands (electrons) or surrounded by bulky ligand which prevent dimerization. .L b. The Cu(I) and Ag(I) complexes described in the textbook behave liked the aryl bromide 4 starting material, which suggests that the metal one—coordinate complexes were not isolated. c. Halide test to determine if the “product” is the starting material. Mass spectrometry to Jr 1 determine the molecular weight. 10. (6 points) For which member of the following pairs of complexes would And be the larger? Explain your reasoning: a. [Cr(H20)5]2+ and [amazon/3* i l Larger oxidation stat, larger Am 4 \ b. [CrF6]3' and [Cr(NH3)6]3+ 1n spectrochemical, NH; is a stronger field ligand than F' c. [MnF6]2‘ and [Reth 2"d and 3rd row metal complexes has a larger Am than 1St row metal complex. ...
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Chemistry 104B - Spring 2002 - JohnArnold - Midterm 1 - 1(5...

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