1. Estimate the lattice energy for KBr, which exists in the rock salt structure and has a Born exponent of 9.5.
Use the Born-Landé equation: Elat = –NoZ+Z–e2A(1 – 1/n)/4πεodo, where (in SI units)
No = 6.02×1023
Z+ = +1
Z– = –1
e = 1.602×10–19
A = 1.748
n = 9.5
4πεo = 1.113×10–10
do = 138 + 196 = 334 pm = 3.34×10–10 m
so that Elat = –(6.02×1023)(+1)(–1)(1.602×10–19)2(1.748) (1 – 1/9.5)/(1.113×10–10)(3.34×10–10) = 650000 J/mol = 650.0 kJ/mol.
2. Despite having a filled electron configuration (s2d10), Zn is a metal. Explain.
In the solid state, the 4s and 3d orbitals will create bands that should be filled, which would lead to a nonmetallic substance. However, the 4p orbitals in Zn are relatively close in energy to the 4s and 3d orbitals so that upon band formation the bottom of the p-band is lower energy than the top of the s or d band. This allows electrons to move from the s and d orbitals into the p orbitals, which leaves all three type of bands partially filled. A partially filled band is the requirement for a metal.
3. Write the reaction between iodine monochloride and methylamine, showing the Lewis structure for the product(s). Estimate the enthalpy of reaction.
This is a simple Lewis complex formation reaction, given by:
ICl + CH3NH2 → Cl–I–NH2–CH3
In ICl, the Cl is more electronegative so the I, which is the positive end of the molecular dipole, must be the acidic portion of the molecule.
The enthalpy can be found from the Drago-Wayland equation:
–ΔH = EAEB + CACB, where
EA = 10.4
EB = 2.66
CA = 1.70
CB = 12.00
so that ΔH = –[(10.4)(2.66) + (1.70)(12.00)] = –48.1 kJ/mol
4. For each of the following pairs of aqueous solutions, indicate which will have the lower pH and explain your reasoning: a) 0.1 M Co(NO3)2 or 0.1 M Cu(NO3)2; b) 0.1 M Mn(NO3)2 or 0.1 M KMnO4; c) 0.1 M NaNO3 or 0.1 M AgNO3.
a) 0.1 M Co(NO3)2 or 0.1 M Cu(NO3)2
0.1 M Cu(NO3)2 is expected to have a lower pH. Cu2+ has a larger Z* than Co2+ so should be a stronger aqua acid. Since the concentrations of the two solutions is the same, the stronger acid should give a lower pH.
b) 0.1 M Mn(NO3)2 or 0.1 M KMnO4
0.1 M Mn(NO3)2 is expected to have a lower pH. Mn2+ is an aqua acid while K+ is not. The permanganate ion is basic.
c) 0.1 M NaNO3 or 0.1 M AgNO3
0.1 M AgNO3 is expected to have a lower pH. Ag+ has a larger Z* than Na+ because of the poor shielding of d electrons so should be a stronger aqua acid. Since the concentrations of the two solutions is the same, the stronger acid should give a lower pH.
5. Complete and balance the following reaction and estimate the standard potential:
MnO4–(aq) + Mn(s) → Mn2+(aq) + Mn3+(aq)
This problem has two possible answers, depending upon how the redox couples are partitioned.
Answer 1:
MnO4–(aq) → Mn2+(aq) giving:
MnO4–(aq) + 8 H+(aq) + 5 e– → Mn2+(aq) + 4 H2O(l)
From the Latimer Diagram, the standard potential is E°red = 1.51 V
Mn(s) → Mn3+(aq) giving:
Mn(s) → Mn3+(aq) + 3 e–
From the Latimer Diagram, the standard potential is E°red = [1.5 + 2(–1.18)]/3 = –0.3 V
Net reaction: 3 MnO4–(aq) + 5 Mn(s) + 24 H+(aq) → 3 Mn2+(aq) + 5 Mn3+(aq) + 12 H2O(l)
E° = 1.51 – (–0.3) = 1.8 V
Answer 2:
MnO4–(aq) → Mn3+(aq) giving:
MnO4–(aq) + 8 H+(aq) + 4 e– → Mn3+(aq) + 4 H2O(l)
From the Latimer Diagram, the standard potential is E°red = [0.90 + 1.28 + 2.9 + 0.95]/4 = 1.5 V
Mn(s) → Mn2+(aq) giving:
Mn(s) → Mn2+(aq) + 2 e–
From the Latimer Diagram, the standard potential is E°red = –1.18 V
Net reaction: MnO4–(aq) + 2 Mn(s) + 8 H+(aq) → 2 Mn2+(aq) + Mn3+(aq) + 4 H2O(l)
E° = 1.5 – (–1.18) = 2.7 V