a. The anion (more electronegative) component in an ionic compound is typically a nonmetallic element and designated as X. The cation (more electropositive) component is a metallic element and designated as M. A vacant lattice site is designated as V. Atoms or ions can occupy cation (M) sites, anion (X) 0r interstitial (i) sites. To facilitate the description, Kroger-Vink notation is widely accepted.
Kroger-Vink notation
X=the entity occupying the defect site (M, X, V or substitutional elements).
Y=type of site occupied (M, X, i).
Z=the excess charge associated with the site
(• = positive, ’ = negative, X = neutral)
Kroger-Vink notation for Frenkel defect can be written as
since all X atoms remain on anion sites while M atoms are distributed over cation and interstitial sites. Furthermore, each Frenkel defect consists of one vacancy.
Preservation of regular site ratio
The ratio between the numbers of regular cation and anion sites must remain constant and equal to the ratio of the parent lattice. Thus if a normal lattice site of one constituent is created or destroyed, the corresponding number of normal sites of the other constituent must be simultaneously created or destroyed so as to preserve the site ratio of the compound.
b. The introduction of defects increases entropy ΔS and decreases free energy ΔG. A minimum value for ΔG is reached for an optimum concentration of defects. The structure with defects is more stable.
ΔG = ΔH – T(ΔS)
It is important for you at this point to review the information on isomers. Most organic compounds have isomers. Because they are composed of the same number and kind of atoms, isomers have similar properties. In general, however, the boiling points of the unbranched compound are higher than that of their highly branched isomers. It is also true for their melting points. There is no known way of predicting exactly how many isomers most compounds can form. Each of these two structures is an isomer of butane.
Pentane, C5H12, the next member of the alkane family, has three isomers. Hexane, C7H16, has nine. It is estimated that triacontane, C30H62, has over four billion isomers.
c. The reduction Cu2+ to Cu2 by Ag
Cu2++ 2e– → Cu = 0.159v
Ag++ e = -0.799 volts
From the equations above, the silver ions have greater attraction forces for electrons (more positive voltage) than ions so that they will be converted to atoms. Meaning copper will give up its electrons.
The oxidation of Cu to Cu2+ by Fe3+ (Assume that the Fe3+ reduced to Fe2+ )
Cu2++ 2e– → Cu = 0.340v
Fe2+ + 2e– → Fe = – 0.44v
Fe3+ + 3e– → Fe = – 0.04v
From the equations above, the copper ions have greater attraction forces for electrons (more positive voltage) than iron II ions, so they will attract electrons from iron, thus turning copper II ions to copper and converting iron to Fe2+. Meaning iron will give up its electrons to copper ions; therefore, oxidation of Cu to Cu2+ by Fe3+ is impossible because iron will be oxidized as copper will be reduced when the required condition is are available.
The oxidation of Fe2+ to Fe3+ by Cl2 (g)
Iron (II) ions can be oxidized to iron (III) ions by bubbling chlorine gas in a solution, but not in a mare gas only. The number of electrons lost by the iron ll ions must be equal to the number of electrons gained by the chlorine. So, one chlorine molecule will oxidize two iron ll ions. So, the oxidation is possible.
The oxidation of H2O to O2 (g) by Fe2+ (under acid conditions)
H2O (L) + 2e– → O2 (g)
It is impossible to oxidize H2O to O2 by iron because, in order for the reaction to take place, it requires water in the presence of oxygen to form rust but not produce oxygen. But iron can be reduced to iron(II) then iron(III) in water and in the presence of oxygen. Again, in the presence of acidic solutions, oxygen can be formed when a metal is oxidized in the presence of water.
The reduction of Fe2+ to Fe3+ by Zn
Fe3+ + 3e– → Fe = – 0.04v
Zn2+ + 2e– → Zn = – 0.7626v
From the equations above, the iron ions have greater attraction forces for electrons (more positive voltage) than zinc ions, so iron ions will be converted to iron atoms. Meaning zinc will give up its electrons to iron ions; therefore, reduction of Fe2+ to Fe3+ by Zn is impossible because zinc will be oxidized as the iron will be reduced.
The reduction of Mn2+ to Mn by Tl
From the equations above, the Tl ions have greater attraction forces for electrons (more positive voltage) than Mn ions; also, Tl ions will be converted to Tl atoms. Meaning Mn will give up its electrons to Tl ions. Therefore, the reduction of Mn2+ to Mn by Tl is possible because Mn will be oxidized as TI will be reduced.