NAME___Example with Hints______Chem 241 Sp 2006

Helpful information:

NA = 6.022x1023 mol-1, Avogadro’s number R =8.3145 JK-1mol-1, the gas constant

c = 3 x 108 m/s, the speed of light e = 1.602 x 10-19 C, the charge of an electron

m = 9.11 x 10-31 kg, the mass of an electron eo = 8.85 x 10-12 C2J-1m-1, permittivity of vacuum

h = 6.6262 x 10-34 J•s, Planck’s constant 1 J = 1kg•m2•s-2 = 6.24 x 1018eV

Ion-Dipole: Dipole-Dipole Induced Dipole

For full credit, you must show your work, or explain your reasoning.

Pg 1______Pg 2_____ Pg 3_____ Pg 4______Pg 5______Pg 6______Pg 7_____Pg 8_____ Pg 9______

Total: (possible)______


1. Which would you expect to have the higher melting point, dibromine, Br2, or iodine

monochloride, ICl? Explain your reasoning clearly.

Hint: intermolecular forces

2. Slater’s rules allow you to predict atomic/ionic radii.

a). In no particular order the radii of Mn0, Mn2+, and Mn4+ in the gaseous phase are 0.80Å, 0.53Å and 1.40Å. – Which value belongs to which species? Hint: Zeff

b) Slater’s “rules” are approximations. From the above data explain the origin of a major problem with these rules. Hint: 2p and 2s orbitals are different

3. The coordination compound [Ti(H2O)6]3+ absorbs light at 520nm (520´10-9 m).

a) What color is this complex? Hint: what color is not absorbed? This a little harder than a real question.

b) What is the energy for the crystal field splitting(DO)? Hint: DE = hc/l watch your units! From lecture.

c) What is the crystal field stabilization energy (CFSE)? Hint: different from D0.

d) what is the CFSE for Mn(H2O)62+? Hint: it doesn’t depend on Do.

4. Write the ground state electronic configuration for each: B, Al, Zn2+.

Hint: remember the Aufbau principle. See pg. 12.

5. The second ionization energies of some Period 4 elements are (eV, or electron volts):

Ca (11.87) Sc(12.80) Ti(14.15) Cr(16.50) Mn(15.64)

Identify the orbital from which ionization occurs, and account for the trend in values.

Hint: pg 35.

6. Draw all isomers of the following, including any optical isomers; “en” is defined in section 19.4 of your text.

A) Cr(en)33+

B) PtCl2(NH3)2

Hint: pg 492, 490, 500.

7. Give the formulas for:

A) pentaamminechlorocobalt(III) chloride

B) cis-dichlorobis(ethylenediamine)ruthenium(II)

Hint: Co3+ and Ru2+.

8. Name the following:

A. cis-[CrCl2(NH3)4]

B. [Ni(CN)4]2-

Hint: remember the metal oxidation states

9. The ligand SCN− is an ambidentate ligand. Which end of the ligand would you expect to coordinate to Cr3+ and Pt2+?

Hint: HSAB

10. Rank the group I elements in terms of increasing electronegativity.

Hint: related to IEs and Zeff.

11. Using MO theory, predict which of these reactions is more favored. Explain your reasoning. (hint: only consider the MOs resulting from the 2p orbitals).

NO + CN à NO+ + CN−

NO + CN à NO− + CN+

Hint: which reaction gives you a higher sum of bond orders? Higher bond order is correlated with a stronger bond (DGrxn is more negative). And I’m not interested in the crossover of s and p bonding orbitals, as it won’t change your answer (pg 49).

12. What transition-metal group (or groups!) contains members that find uses as oxidizing agents, pigments, and lubricants? This is simply to remind you that your reading is fair game, and mimics content (but not format) of an exam question.

Hint: pigments and oxidants are common, but few metal complexes are lubricants.

13A. How is nickel purified from its compounds?

B) In what way does this resemble Fe purification from Fe-ores?

C) Draw the molecular structure of the key complex for Ni purification, and its d-orbital splitting pattern.

Hints: CO(g). And think about the CFSE for the key intermediate – what structure would give you more CFSE?

14. The highest oxidation state for nickel in a simple complex is [NiF6]2-.

a) Why would you expect fluoride to be the ligand?

b) Is this high-spin or low-spin? Paramagnetic or diamagnetic?

Hint: spectrochemical series

15. Which of the following MO diagrams (showing only the frontier orbitals) is most appropriate for MgO? Explain why.

Hint: HSAB

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