CHEM 306 Inorganic Chemistry Problem Set V Answers

Size: px

Start display at page:

Download "CHEM 306 Inorganic Chemistry Problem Set V Answers"

Jordan Bryant
6 years ago
Views:

1 CEM 306 Inorganic Chemistry Problem Set V Answers 1) I made the (perhaps oversimple) assumption that the values given by the author in Table 12.6 were the values for the reactions, and not just for the products. Given this, the two equilibria we are asked to examine are: i [i( 3 ) 6 ] 2+ G = kj/mol and i en [i(en) 3 ] 2+ G = kj/mol It seems to me that we can use the relation G = -RTln(K), which leads to: i [i( 3 ) 6 ] 2+ K = 1.20 x 10 9 and i en [i(en) 3 ] 2+ K = 6.99 x We are also asked to find the equilibrium constant for the equilibrium: [i( 3 ) 6 ] en [i(en) 3 ] which we can do either by a) G - RTln(K), or b) by remembering that the equilibrium constant for linear combinations of equilibria is the product of the appropriate equilibrium constants. These approaches yield: a) 5.81 x 10 8 and b) 5.83 x 10 8, which is pretty good agreement. (From: Bowser, J.R.; Inorganic Chemistry ; Brooks-le Publishing; California: 1993; p. 547, this K eq is found to be 1.1 x 10 9, so we are a bit off.)(from: McMurry, J; Fay, R.C.; Chemistry ; 2nd Edition; Prentice all; ew Jersey: 1998; p. A-15 (Table C.6), which find K f for i( 3 ) 6 2+ to be 5.6 x 10 8, so, again, we are a bit off.) [A simple calculation, treating the G as that only for i( 3 ) 6 2+, and looking up the G for both 3 and i 2+ (or calculating them) yields an overall G of kj, which leads to a ridiculous K f for the constant.] (My values above are from: McMurry, J; Fay, R.C.; Chemistry ; 2nd Edition; Prentice all; ew Jersey: 1998; p. A-10,A-17.) 2) This is simply a matter of brute force in drawing all the isomers. For [(en) 2 2 ] + :

2 achiral (trans) For [(en) 2 ( 3 )] enantiomers (optical) (cis) achiral (trans) enantiomers (optical) (cis) 2 2 For [(en)( 3 ) 2 2 ] + : 'bis' cis and enantiomers (optical) achiral # achiral #2 3) We are asked to determine (and draw) the structure of the two platinum products, and compound C, and decided whether or not compound C will have a dipole moment. a) The two platinum products are as shown. The first reaction yields the polar cisdiamminodichloroplatinum(ii) 3 Pt 3 whereas the second reaction yields the non-polar trans-diamminedichloroplatinum(ii)

3 3 Pt 3 b) i 2+, a d 8 metal, usually forms square planar complex, which, as we learned last problem set, is diamagnetic. aving a paramagnetic complex tells us that we have a tetrahedral geometry. (This geometry would have an electron configuration (e) 4 (t 2 ) 4, with two unpaired electrons.) ere is the picture: PPh 3 i PPh 3 c) We have i 2+ here, so we have a d 8 complex. It is tetra-coordinate, paramagnetic, and soluble in polar organic solvents. This last statement tells us that we have a polar molecule, so the answer to the dipole moment question is yes, it will. 4) Plutonium is in the actinide series and so potentially can be octacoordinate (that is, have a coordination number of 8.) Proceeding from this assumption, one could draw a picture that looks something like either one of these (note these are not the only possibilities; I will evaluate your answer on its own merits): Pu 4+ plane) (all M- bonds above the

4 r Pu 4+ the plane and two below (two M- bonds above 5) ere are two possibilities for answers (there may well be others.) It should be noted that cyclopentadienyl is a -1 ligand, so the metal needs to be at least 1+. Cr + is d 5 ; Cr 2+ is d 4 ; Cr 3+ is d 3. a) [Cr(η 5 -cyclopentadienyl) 2 () 2 ], Cr

5 b) [Cr(η 5 -cyclopentadientyl)(c) 2 ()], Cr C C 5) Remember that the IR-active bands transform as x, y, and z in the character tables. So, having determined to which point group the complex belongs, determine the Γ r for the vectors representing the C- stretch, extract the Γ i for these, and determine which, if any, of the Γ i are the representations to which x, y, and/or z belong. (A check was done using: ousecroft, C.E.; Sharpe, A.G.; Inorganic Chemistry, 3 rd Edition; Pearson/Prentice- all, Y: 2008; p. 110.) Γ r : a) Mo(C) 5 PR 3 : this molecule belongs to point group C 4v, and has the following C 4v E 2C 4 C 2 2σ v 2σ d Γ r which leads to: 2A 1 + B 1 + E. f these the A 1 s and E are IR-active, so there should be 3 bands, two of symmetry A 1 and one of symmetry E. b) cis-mo(c) 4 (PR 3 ) 2 : this molecule belongs to point group C 2v, and has the C 2v E C 2 σ v (xz) σ v (yz) Γ r which leads to: 2A 1 + B 1 + B 2., all of which are IR-active and yield four IR-active bands. c) trans-mo(c) 4 (PR 3 ) 2 : this molecule belongs to point group D 4h, and has the D 4h E 2C 4 C 2 2C 2 2C 2 i 2S 4 σ h 2σ v 2σ d Γ r which leads to: A 1g + B 1g + E u, of which only the E u is IR-active, giving one band.

6 d) fac-mo(c) 3 (PR 3 ) 3 : this molecule belongs to the point group C 3v, and has the C 3v E 2C 3 3σ v Γ r which leads to A 1 + E, both of which are IR-active, and would give rise to two bands. e) mer-mo(c) 3 (PR 3 ) 3 : this molecule belongs to the point group C 2v, and has the C 2v E C 2 σ v (xz) σ v (yz) Γ r which leads to 2A 1 + B 2 ; both symmetries are IR-active, so there should be three bands.

Structure of Coordination Compounds

Chapter 22 COORDINATION CHEMISTRY (Part II) Dr. Al Saadi 1 Structure of Coordination Compounds The geometry of coordination compounds plays a significant role in determining their properties. The structure