Chemistry 65.100 A, V
Midterm Test #2
November 6, 1998
Answers
Part A.
1. BF3 is a trigonal planar molecule. The B atom must therefore be using sp2
hybrid orbitals.
2. The term "standard state" refers to the state of a substance under standard
conditions (25oC, 1 atm).
3. PCl5: P (+5) Cl (-1)
H3AsO3: H (+1) As (+3) O (-2)
4. PCl3 is a trigonal pyramidal molecule. Cl is more
electronegative than P, so the dipole in each bond points from P to Cl.
The overall dipole therefore points from
the P atom downward between the Cl atoms:
5. We can say nothing about the value DH, since it is
a state function. The path and rate are irrelevant.
6. The d-orbitals of As are close in energy to the p orbitals. Electrons can thus occupy
these d-orbitals, resulting in an expanded octet.
Part B.
VSEPR - only the name of the shape is required for full marks.
(a) GeCl4
Valence electrons = 4 + (4 x 7) = 32
The Lewis structure is thus:
Which is of the form AX4. The molecule is thus tetrahedral.
(b) ICl3
Valence electrons = 7 + (3 x 7) = 28
The Lewis structure is thus:
Note the two lone pairs in the I atom. This is of the form AX3E2, which is T- shaped.
(c) SeO2
Valence electrons = 6 + (2 x 6) = 18
The Lewis structure is thus:
Which is of the form AX2E. The shape is thus bent.
2. The S is being oxidized and the Cl is being reduced. The oxidation reaction is balanced first:
S2O6-2(aq) à
2 SO4-2(aq)
S2O6-2(aq) + 2 H2O(l) à 2 SO4-2(aq)
S2O6-2(aq) + 2 H2O(l) à 2 SO4-2(aq) + 4 H+(aq)
S2O6-2(aq) + 2 H2O(l) à 2 SO4-2(aq) + 4 H+(aq)
+ 2 e-
Then the reduction reaction:
2ClO2- (aq) à Cl2(g)
2ClO2- (aq) à Cl2(g)
+ 4 H2O(l)
2ClO2- (aq) + 8 H+(aq) à Cl2(g) + 4 H2O(l)
2ClO2- (aq) + 8 H+(aq) + 6 e-
à Cl2(g) + 4 H2O(l)
Multiply the oxidation reaction by 3 to make the electrons cancel and add the two half reactions:
3 S2O6-2(aq) + 6 H2O(l)
à 6 SO4-2(aq) + 12 H+(aq)
+ 6 e-
2ClO2- (aq) + 8 H+(aq) + 6 e-
à Cl2(g) + 4 H2O(l)
___________________________________________
3 S2O6-2(aq) + 2 H2O(l) + 2ClO2- (aq) à 6 SO4-2(aq) + 4 H+(aq) + Cl2(g)
To convert to basic solution, add OH-(aq) to both sides:
3 S2O6-2(aq) + 2 H2O(l) + 2ClO2- (aq) + 4 OH-(aq) à 6 SO4-2(aq) + 4 H+(aq) + Cl2(g) + 4 OH-(aq)
and cancel the resultant waters:
3 S2O6-2(aq) + 2ClO2- (aq) + 4 OH-(aq) à 6 SO4-2(aq) + Cl2(g) + 2 H2O(l)
3. Ba(s) à Ba(g)
+175.4 kJ mol-1
Ba(g) à Ba+(g) + e-
+502 kJ mol-1
Ba+(g) à Ba+2(g)
+ e-
+963 kJ mol-1
H2(g) à 2 H(g)
+436 kJ mol-1
2 H(g) + 2 e- à 2 H-(g)
2(-73) = -146 kJ mol-1
Ba+2(g) + 2 H-(g) à
BaH2(s) -U (U = lattice energy)
___________________________________________________________
Adding all of the above reactions gives:
Ba(s) + H2(g) à BaH2(s)
which has an energy of -40.9 kJ mol-1 (given)
Thus, -U = -40.9 – (175.4+502+963+436-146) = -1971.3
Thus, U = +1971.3 kJ mol-1 (yes, lattice energy is
positive!)
4. For the C2 molecule:
(a) The bond order = (6 – 2)/2 = 2
(b) To ionize C2 to C2+, an electron is removed from the p2p orbitals. Since these are bonding orbitals, removing the electron results in less bonding character, and so the bond length of C2+ will be greater than that of C2.
(c) To make C2-, the extra electron will be added to the s2p orbital. Since this is a bonding orbital, the extra electron will give C2- more bonding character, and so C2- will have a greater bond energy than that of C2.
(d) C2+ and C2- are paramagnetic, since each will contain one unpaired electron.