NOTE: Chem 1A03/1E03 students who are Lab Week O students will have a formal tutorial during the week of October 16. Solutions will be posted.
1. Decide whether the following configuration represents S, S- , S+, S2-, or S2+: 1s2 2s2 2p6 3s2 3p3. Is this configuration a ground state configuration of the species you have chosen, or an excited state configuration?
The configuration shown represents the ground state configuration for S+.
2. a) Which of the following gaseous atoms is smallest: P, Cl, Ca, Si?
The atoms P, Cl, and Si are in period 3; Cl is the smallest of these three, since size decreased across a period (effective nuclear charge increases). Ca is in period 4, and is larger than any of the above-named three atoms. Thus Cl is the smallest.
b) Which of the following gaseous atoms has the highest ionization energy: Li, Ne, Na, Ar?
Ne; the noble gases have very high ionization energies; also, I.E.(Ne) > I.E.(Ar), since the valence electrons of Ne are closer to the nucleus. In both Ne and Ar the core charge is +8.
3. Arrange the following gaseous ions in the order of decreasing ionic radius: S2-, K+, Ca2+, Cl-.
S2- > Cl- > K+ > Ca2+, where ">" means "larger than". All four of these species have 8 valence electrons, and all are [Ne] 3s2 3p6. Recall we calculated core charge to approximate the effective nuclear charge. For these four species, since core charge = full nuclear charge Z - total # electrons in inner shells, the core charges are +6, +7,+9, +10 for the order shown. The higher the core charge in this set, the smaller the size.
4. Which of the following orders of the sizes of halogen atoms and
ions is/are correct?
a) I- > I > Br;
b) Cl > F > I;
c) F- > Cl- > I-;
d) Br- > Cl- > F;
e) Cl > F > F-.
a and d are correct.
a is correct since I- is larger than I, and slso I is larger
than Br.
d is correct since anion size increases as go down a group. Also, since
Cl > F, and Cl- > Cl, Cl- must be larger than F.
5. The following problem includes the concepts of Hess's Law
in Chapter 6 and the enthalpy changes associated with several processes
mentioned in Chapter 8.
a) Define the tterm "lattice energy".
b) Given that the enthalpy of sublimation of cesium metal is 78.2
kJ/mol, the enthalpy of formation of solid CsCl is -443.0 kJ/mol, and the
ionization energy of cesium is 377 kJ/mol, by using approprate data from
the text, calculate the lattice energy for CsCl (in kJ/mol).
a) Lattice energy is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous ions. The word "energy" is used loosely in the above definition, since actually it is the enthalpy change for the process.
b) The target reaction is CsCl(s) --> Cs+(g) + Cl-(g) DH =?
Five steps need to be added to yield this lattice energy. (Note that the data will be given in kJ).
CsCl(s) --> 1/2 Cl2(g) + Cs(s) DH1 = -DHof(CsCl,s) = +443.0
Cs(s) --> Cs(g) DH2 = DHSUBL = +78.2
Cs(g) --> Cs+(g) + e- DH3 = I.E. = +377
1/2Cl2(g) --> Cl(g) DH4 = 1/2 BECl-Cl = +121.4
Cl(g) + e- --> Cl-(g) DH5 = EA = -349 (Table 8.4 in Chang actually lists -EAs.)
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CsCl(s) --> Cs+(g) + Cl-(g) DH = + 671 kJ/mol (where DH = DH1 + DH2 + DH3 + DH4 + DH5)6. The electron affinity of chlorine atom can be determined by using a laser light to ionize the chloride ion in the gas phase.
Cl(g) + e- --> Cl-(g)
b) By referring to Table 8.4 in Chang, calculate the wavelength, in nm, of laser light that corresponds to the electron affinity for chlorine.
Energy of one photon = [349 kJ x 103 J/kJ ]/ 6.022 x 1023 photons/mol = 5.80 x 10-19 J
Since E = h x n = h x c/l,
l = h x c/E = [6.626 x 10-34
J s x 3.00 x 108 m/s]/ [5.80 x 10-19 J]
l = 3.43 x 10-7 m x 109
nm /m = 343 nm
c) To what region of the electromagnetic spectrum does this wavelength
correspond?
This is ultraviolet light.