Practice Exam Final Answer Key 2021 - Inorganic Chemistry PDF

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Practice final exam - Spring 2021...

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Practice Final Exam Key 2021 1) Write the chemical formula for the following a) Pentaaquabromochromium(III) [Cr(H2O)5Br]2+

b) Hexacarbonylmanganese(I) perchlorate [Mn(CO)6]ClO4 c) Ammonium tetrachlororuthenate NH4[RuCl4] d) Trisethylenediaminecopper(II) sulfate [Cu(en)3]SO4 2) Use the Born-Haber cycle to calculate the enthalpy of formation of aluminum hydride. The following table of data is available for your calculations. (e = 1.602 x 10-19 C, vacuum permittivity = 8.85 x 10-12 J-1 C2 m-1, constant for repulsion between ions at short range = 34.5 pm). Use the Born-Mayer equation to calculate the lattice enthalpy of aluminum hydride. Term Value ΔHBDE[H2(g)] 436 kJ/mol ΔHsub[Al (g)] 293.4 kJ/mol [IE1(Al)] 577 kJ/mol [IE2(Al)] 1816 kJ/mol [IE3(Al)] 2744 kJ/mol [EA(H)] 72 kJ/mol A 1.748 3+ 53 pm Al 140 H-

N A Z +Z − e 2  d *   1 − HL =  A(kJ / mol ) 4od  d  HL = 2766 kJ/mol 3) Use the Born-Haber cycle to calculate the enthalpy of formation of aluminum hydride. Based upon your calculated value will this compound form? ΔHf = ΔHsub[Al (g)] +[IE1(Al)] + [IE2(Al)] + [IE3(Al)] + 3/2 ΔHBDE[H2(g)] + 3[-EA(H)]- HL

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ΔHf = 2552 kJ/mol Positive ΔH f, ΔH f will be similar to ΔG f so the formation is not favored 4) How many NbO formula units are present per unit cell? 0, 1

1/2

1/8 *8 = 1 O and one Nb in center = NbO 5) Calculate the density of NbO if a = 291.5 pm. 7.3 g/cm3

6) S2, the major component of sulfur vapor above 720 °C, has a sulfur-sulfur distance of 189 pm. This is significantly shorter than the sulfur-sulfur distance of 206 pm in S8. Use molecular orbital theory to create a homonuclear diatomic diagram (assume similarities to O2) and calculate the bond order in S2.

BO = (8-4)/2 = 2

O2

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7) What does that suggest about the bonding in S8? From the MO diagram we know that the bond order is 2 and this is a double bond. So the S-S distance of 189 is a double bond. Within S8, the bond distance is longer (206 pm) so that suggests that this is a weaker bond (most likely a single bond).

8) On the basis of VSEPR, predict the structures of XeOF2, XeOF4, XeO2F2, and XeO3F2 and indicate the point group for each. (12 points) XeOF2 T-shaped (C2v) XeOF4 Square pyramid (C4v) XeO2F2 Seesaw (C2v) XeO3F2 trigonal bipyramid (D3h)

9) Permanganate can be reduced to the Mn(II) cation when the sulfurous acid is oxidized to sulfuric acid. Provide a balanced equation for this reaction. 2 MnO4- (aq) + 5 H2SO3 (aq) + H+(aq) →2 Mn2+(aq) + 5 HSO4- (aq) + 3 H2O (l)

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10) Consider the redox equation from question 9. The reduction of permanganate to form Mn(II) has a standard reduction potential of +1.51 V. The standard reduction potential of sulfuric acid to form sulfurous acid is +0.158 V. Calculate the Gibbs free energy of the reaction (F = 9.648 x 104 C/mol) and indicate if the reaction is favorable. MnO4- (aq) + 8 H+ (aq) + 5 e- → Mn2+ (aq) + 4 H2O (l) HSO4- (aq) + 3 H+ (aq) + 2 e- → H2SO3 (aq) + H2O (l)

E = +1.51 V E = +0.158 V

ΔG = -1304.4 kJ/mol Reaction is favorable because ΔG is negative

11)

The Latimer diagram for Manganese is shown below:

Create a Frost diagram based upon the Latimer diagram provided above.

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12) The 1st, 2nd, 3rd, and 4th ionization energy for an element is 577, 1816, 2744, 11574 kJ/mol. Suggest which group on the periodic table this element belongs to and explain your reasoning. (5 points) Likely belongs to the Group 13 (3A). Energy difference between IE1, IE 2, and IE 3 are only about 1500 kJ/mol and it is around 8000 for IE 4. That means it take relatively less energy to remove the first 3 electrons to form a +3 cation and create the noble gas configuration.

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