Chemistry 65.100 A, V
Midterm Test #2
November 7, 1997
Answers
Part A.
1. There are five charge clouds around the P atom in PCl5. The hybridization must therefore be sp3d.
2. Standard state is the state of an element or compound (e.g. s, l, g, 1M solution) at standard conditions (25oC, 1 atm)
3. A disproportionation reaction is one in which a compound is both reduced and oxidized in the same reaction, or a compound is produced by both oxidation and reduction reactions of two other species.
4. Hess’ law states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes for the steps making up the reaction.
5. H ® Br i.e. the negative end is at the Br atom.
6. If the bond order increases, this means that the electron removed must have been in an antibonding molecular orbital.
7. DH is a state function, and so its value says nothing about the rate of the reaction.
8. Li is a very electropositive element (very low electronegativity). Thus, the reaction Li à Li+ + e- happens very easily, more so than for Zn. Thus, Li is easily oxidized, and thus Li must be a stronger reducing agent.
Part B.
1a. SbF5 valence electrons = 5 + (5 x 7) = 40. Distributing the electrons two to a bond, and putting lone pairs on the F atoms to complete the ectets, we see that there are no lone pairs left to put on the Sb:
This is of the type AX5, and must therefore be trigonal bipyramidal (like PCl5)
1b. SeO3-2 valence electrons = 6 + (3 x 6) + 2 = 26. Distributing the electrons two to a bond, and putting lone pairs on the oxygens to complete their octets leaves one lone pair left, which goes on the central atom (the Se). Note that you could (and should) make double bonds between the Se and each of the O’s, but this would not change the number of charge clouds around the Se, and so would not change the shape of the ion:
This is of the type AX3E, and must therefore be trigonal pyramidal (like ammonia).
1c. SeCl4 valence electrons = 6 + (4 x 7) = 34. Distributing the electrons two to a bond and putting lone pairs on the Cl atoms to complete their octets uses only 32 of these electrons. The remaining pair goes as a lone pair on the Se:
This is of the type AX4E, and must therefore be shaped like a seesaw (like SF4).
2. Cl2(g) à Cl-(aq) + ClO-(aq)
(Note that this is a disproportionation reaction).
The Cl in Cl2 has an oxidation number of 0, while in Cl- it is -1, and in ClO-, it is +1. Thus, the unbalanced oxidation reaction is:
Cl2 à ClO-
to balance for Cl, use 2 ClO-’s:
Cl2 à 2 ClO-
to balance for O, add H2O:
Cl2 + 2 H2O à 2 ClO-
to balance for H, add H+:
Cl2 + 2 H2O à 2 ClO- + 4 H+
to balance the charge, add electrons:
Cl2 + 2 H2O à 2 ClO- + 4 H+ + 2 e-
The unbalanced reduction reaction is:
Cl2 à Cl-
to balance for Cl, use two Cl-‘s:
Cl2 à 2 Cl-
to balance the charge, add electrons:
Cl2 + 2 e- à 2 Cl-
to obtain the balanced reaction, add the two half reactions:
Cl2 + 2 H2O à 2 ClO- + 4 H+ + 2 e-
Cl2 + 2 e- à 2 Cl-
2 Cl2(g) + 2 H2O(l) à 2 Cl-(aq) + 2 ClO-(aq) + 4 H+(aq)
to balance it in a basic solution, add OH- to remove the H+ (as water):
2 Cl2(g) + 2 H2O(l) + 4 OH-(aq) à 2 Cl-(aq) + 2 ClO-(aq) + 4 H+(aq) + 4 OH-(aq)
or,
2 Cl2(g) + 2 H2O(l) + 4 OH-(aq) à 2 Cl-(aq) + 2 ClO-(aq) + 4 H2O(l)
or,
2 Cl2(g) + 4 OH-(aq) à 2 Cl-(aq) + 2 ClO-(aq) + 2 H2O(l)
or,
Cl2(g) + 2 OH-(aq) à Cl-(aq) + ClO-(aq) + H2O(l)
3. We want the following reaction: 3 C(s) + 4 H2(g) à C3H8(g)
Since C3H8 must appear on the right hand side, reverse the first reaction, and change the sign of its DH:
3 CO2(g) + 4 H2O(l) à C3H8(g) + 5 O2(g) DH = +2220 kJ
We need 3 C(s) on the left, so multiply the second reaction and its DH by 3:
3 C(s) + 3 O2(g) à 3 CO2(g) DH = 3(-394 kJ) = -1182 kJ
And we need 4 H2 on the left, so multiply the third reaction and its DH by 4:
4 H2(g) + 2 O2(g) à 4 H2O(l) DH = 4(-286 kJ) = -1144 kJ
Adding these three reactions, the CO2’s and H2O’s and O2’s cancel, leaving:
3 C(s) + 4 H2(g) à C3H*(g) DH = +2220 - 1182 - 1144 = -106 kJ