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The solutions to Assignment 1 for Chem 612, a university course offered in the Fall of 2008. The assignment involves calculating thermochemical properties of MgCl2, including enthalpies of formation, sublimation, ionization, dissociation, and electron affinity, as well as the lattice energy. The document also discusses the differences between ionic and polar covalent bonds, and provides Lewis structures for various anions and molecules.
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Chem 612 โ Fall, 2008 Assignment 1 - Solutions
This assignment is due at the beginning of class on Wednesday, September 10, 2008.
a. Write thermochemical equations, including state designations, corresponding to all of the data given above and also the lattice energy, U.
Mg( s ) + Cl 2 ( g ) 6 MgCl 2 ( s ) โ H of = โ642 kJ/mol Mg( s ) 6 Mg( g ) โ H osub = 146 kJ/mol Mg( g ) 6 Mg+( g ) + e โ^ I 1 = 737.7 kJ/mol Mg+( g ) 6 Mg2+( g ) + e โ^ I 2 = 1451 kJ/mol Cl 2 ( g ) 6 2 Cl( g ) D = 243.4 kJ/mol Cl( g ) + e โ^6 Clโ( g ) A = โ349 kJ/mol Mg2+( g ) + 2 Clโ( g ) 6 MgCl 2 ( s ) U = 2522 kJ/mol
b. Sketch the Born-Haber cycle for MgCl 2 ( s ), similar to the cycle shown in class for NaCl( s ).
c. Calculate the lattice energy, U , for MgCl 2 ( s ) to verify the value of 2522 kJ/mol given in the table shown in class. Show your work.
โ H of = โ H osub + I 1 + I 2 + D + 2 A โ U
U = โ H osub + I 1 + I 2 + D + 2 A โ โ H osub โ โ H of = [146 + 737.7 +1451 + 243.4 + (2)(โ349) + 642]kJ/mol = 2522 kJ/mol
d. Explain why the lattice energy of MgCl 2 is so much greater than that of NaCl ( kJ/mol).
Mg2+^ is twice as highly charged and much smaller. Both factors increase the force of attraction between oppositely charged ions, resulting in a much greater potential energy
e. The values of the lattice energy for MgCl 2 , CaCl 2 , and SrCl 2 are 2522 kJ/mol, 2253 kJ/mol, and 2127 kJ/mol, respectively. Why is this the expected trend?
The cations are progressively larger through the series, increasing the separation between oppositely charged ions. Potential energy decreases with separation of the charges.
f. The difference in lattice energies between MgCl 2 and CaCl 2 is 269 kJ, but the difference between CaCl 2 and SrCl 2 is only 126 kJ/mol. Why?
The ionic radii (Shannon CN6 values) are 72 pm, 100 pm, and 110 pm, respectively for Mg2+, Ca2+, and Sr2+. The lattice energy trend reflects the cation size trend. Both Mg2+ and Ca2+^ have noble-gas configurations of the type ns^2 np^6 ( n = 2, 3), whereas Sr2+^ has a configuration 3 d^104 s^24 p^6. The size difference between Ca2+and Sr2+^ is so much less than between Mg2+^ and Ca2+, because Ca and Sr are separated by the intervening first transition series, through which sizes decline more gradually owing to poor shielding by 3 d electrons. As a result the cation-anion separation in SrCl 2 is not that much greater than in CaCl 2 , and the two salts have very similar lattice energies.
All simple ionic compounds, particularly those composed of monatomic ions, are crystalline solids at room temperature. HF, by contrast, is a gas at room temperature. Moreover, it has a dipole moment (ฮผ = 5.93 x 10โ30^ C@m), characteristic of a polar covalent molecule. Ionic compounds contain no molecules of the compound, and therefore have no measurable dipole moments. Clearly difference in electronegativity is not a reliable guide to deciding the nature of the bonding between two atoms. The fault lies in assuming that electronegativities are constant from case to case. This is particularly liable to break down in the case of hydrogen. Indeed, as Gillespie and Popelier assert (p. 15), โalmost all bonds are polar bonds, whether they are approximately described as covalent or ionicโ. โCovalentโ and โionicโ cannot be precisely and quantitatively defined, except to note that there are purely covalent bonds (e.g., homonuclear diatomic molecules) but no purely ionic bonds.
ClF 2 +^ electron count = Cl (7) + 2F (14) + cation charge (-1) = 20
I 3 โ^ electron count = 3Cl (21) + anion charge (1) = 22
ClF 3 electron count = Cl (7) + 3F (21) = 28
Cl is hypervalent.
IF 5 electron count = I (7) + 5F (35) = 42
I is hypervalent.
XeO 3 electron count = Xe (8) + 3O (18) = 26
Cl 2 O 7 electron count = 2Cl (14) + 7O (42) = 56
3+
O Cl
O
O
O Cl
O
O
O 3+
SCl 4 electron count = S (6) + 4Cl (28) = 34
S is hypervalent.
Cl
Cl S Cl
Cl
B(OH) 3 electron count = B (3) + 3O (18) + 3H (3) = 24
B B is electron deficient.
O
O O
H
H
H
B 3 N 3 H 6 electron count = 3B (9) + 3N (15) + 6H (6) = 30
B N N
B B N
H
H
H
H
H
H B N N
B B N
H
H
H
H
H
H