1. (15 marks) Consider the oxides of the following ions: Co²⁺, S⁶⁺, Ti⁴⁺, Cs⁺, Zn²⁺. Answer the following questions. Note that a particular oxide may be used more than once or may not be used at all in answer to these questions.

(a) Give the formula of the oxide most likely to give an acidic solution when dissolved in water. Write an equation for the reaction.

SO₃ - has the most acidic cation. It is the acid anhydride for sulfuric acid.

SO₃ + H₂O ---> H₂SO₄

CoO - Co²⁺ is a d⁷ cation, so the transitions within the split d-orbitals will make it colored.

ZnO (or CoO). Amphoterism is found in oxides of +2 ions such as Be²⁺, Zn²⁺, and Sn²⁺, or small +3 ions such as Al³⁺

ZnO + 2 HCl ---> ZnCl₂ + H₂O

ZnO + H₂O + 2 NaOH ---> Na₂[Zn(OH)₄]

Cs₂O. Cs⁺ has no acid character, and the oxide ion is a strong base. The Group 1A oxides are anhydrides to the common strong bases such as KOH and NaOH.

Cs₂O + H₂O ---> 2 CsOH

Molecular oxides are generally found in compounds between and a nonmetal, whereas metal oxides exhibit ionic bonding and are network structures. SO₃ is a gas.

The oxygens on phosphate, each of which carries an average charge of -3/4 (total charge of -3 spread over 4 oxygens), are hydrogen-bond acceptors from water molecules. The size of phosphate is large enough (about 220 pm in radius) to accomodate a dozen water molecules about itself. A schematic drawing of a hydrated phosphate ion is shown below.

The phosphate ion attracts the hydrogen from a water molecule to produce the hydrogen phosphate ion and hydroxide, a process known as hydrolysis.

PO₄^3-(aq) + H₂O ---> HPO₄^2- + OH^-

Metal cations behave as Bronsted acids in water. For example, solutions of aluminum chloride are acidic because the hydrated aluminum ion exerts an inductive effect on the O-H bonds, resulting in hydrolysis and production of H₃O⁺ ions.

Al(H₂O)₆³⁺ + H₂O ---> Al(H₂O)₅(OH)²⁺ + H₃O⁺

Cation acidity is determined primarily by the charge Z and size (radius) r. There is an approximate linear relationship between the pKa of the hydrated ion and Z²/r of the cation. The electronegativity of the cation also affects the acidity, but to a lesser extent than the charge, and only if the electronegativity is greater than about 1.8.

In order of Z²/r, with Zn²⁺ being rated more acidic than Mg²⁺ because of its higher electronegativity: C⁴⁺ > Al³⁺ > Zn²⁺ > Mg²⁺ > K⁺.
 
 

When an ionic compound melts, the forces that must be overcome are the ionic bonds between cations and anions. The energy of attraction between them is proportional to the product of the charges on the ions divided by the contact distance in the crystal lattice:

E_ionic proportional to (Z⁺)(Z^- )/d

CaF₂ will melt higher than CaCl₂ because the Ca²⁺-F^- distance is shorter than the Ca²⁺-Cl ^- distance. Both compounds will melt higher than NaF because of the higher charge on the calcium ion compared to the sodium ion.
 
 

In NaCl, the sodium ions and the chloride ions are both octahedral 6-coordinate. The unit cell has chloride ions at all the corners and in the center of every face. The sodium ions lie in the center of each of the 12 edges, and in the center of the cell.

In CsCl, the chloride ions form a simple cubic packing at the corners of the cell, and there is a cesium ion in the center. The cesium ions and the chloride ions are both cubic 8-coordinate.

The CsCl structure is different from the NaCl structure because cesium is so much larger, and can accomodate 8 chlorides about itself.

Isomorphous replacement occurs when ions of a similar size and the same charge replace each other in a crystalline material without changing the structure of the crystal. Albite and anorthite form a continuous series in which gradually increasing amounts of [Na⁺ + Si⁴⁺] are replaced by [Ca²⁺ + Al³⁺]

One Si⁴⁺ in albite is replaced by Al³⁺, which is about 20% larger than Si⁴⁺. However, the charges aren't balanced now, so Na⁺ must be replaced by Ca²⁺, which is about the same size as Na⁺. Now the charges [Ca²⁺ + Al³⁺] are the same as the charges [Na⁺ + Si⁴⁺] in the original compound. This process can continue until every Si⁴⁺ and Na⁺ in albite are replaced by Al³⁺ and Ca²⁺.

It doesn't occur in molecular compounds because, e.g., replacing 10% of Mn in Mn₂O₇ by Tc gives a mixture that contains Tc₂O₇, Mn₂O₇, and TcMnO₇. These three compounds could be separated by chromatography or some other method, unlike solid structures where the individual compounds that comprise the "end members" cannot be separated.

Cr³⁺ is a d³ ion. In octahedral coordination, the electron configuration is t_2g³e_g⁰. The colors are different because the crystal field splitting energy is different.

The low splitting energy for Cr³⁺ in beryl results in the red part (low energy) of the spectrum being absorbed, so the transmitted color appears green. In corundum, the splitting energy is higher so a higher-energy part of the visible spectrum is absorbed, so the low-energy part of the spectrum is transmitted (red).

Refer to the diagram in Question 2a. Basicity of an oxo anion increases as the negative charge on the ion increases. The greater the negative charge on the oxo ion (e.g. in the series ClO₄^-, SO₄^2-, PO₄^3-) , the more the proton in the water molecule will be attracted to the ion, leading to hydrolysis. Thus ClO₄^- is the least basic and PO₄^3- the most basic ion.

Basicity decreases with increasing number of oxo groups if the charge is fixed. The more oxo groups there are for a given charge, the more the negative charge can be spread around. Comparing two ions with the same charge, but different numbers of oxo groups [ClO^- and ClO₄^- ], there is only one oxo group in ClO^- so it carries the full negative charge (O being more electronegative than Cl). However, in ClO₄^- there are four oxo groups over which to spread the charge, so each has a -1/4 charge and will be less able to attract a proton from a water molecule. [Can also argue that the more oxo groups, the higher the oxidation number of the central atom and thus the greater the inductive effect: as the central atom attracts more electrons to itself from the oxygen, there is less electron density on the oxygen to bond to a proton so the base will get weaker with increasing oxidation number of the central atom.]

The more electronegative the central atom, the weaker the base. This is because a more electronegative central atom a greater its ability to attract electron density from the oxygens, leaving less electron density to bond to protons. Thus ClO^- is a stronger base than IO^-.

Method A. Charge distribution per oxo group.

ion	# oxo groups	charge per oxo group
pyrophosphate	6	-4/6 = -0.667
phosphate	4	-3/4 = -0.75
cyclic triphosphate	6	-3/6 = -0.50

Based on the charge distribution per oxo group, phosphate is the most basic and cyclic triphosphate the least. NOTE: oxo groups are oxygens bonded to only one other atom.

Method B. Cancelling 2 oxo groups with one negative charge

ion	# oxo groups	charge	after cancellation
pyrophosphate	6	-4	-1 spread over 2 P = -1/2 per P
phosphate	4	-3	-1 for P
cyclic triphosphate	6	-3	0

Same results as Method A, you just have to remember that the negative charges or excess oxo groups have to be spread over all the phosphorus atoms.

Ions: Cs⁺, Bu₄N⁺, Mg²⁺, Fe³⁺, PO₄^3-, MnO₄^-, SO₄^2-

(ii) Which cation(s) will be electrostatic structure makers? Fe³⁺, Mg²⁺

(iii) Which cation(s) will disrupt the icebergs in liquid water? Cs⁺

(iv) Which anion is an electrostatic structure breaker? MnO₄^-

(v) Which ion is a hydrophobic structure maker? Bu₄N⁺

(vi) Which cation(s) are likely to give insoluble salts with the perchlorate ion? Cs⁺, possibly Bu₄N⁺

(vii) Which cation(s) are likely to give insoluble salts with the phosphate ion? Fe³⁺, Mg²⁺

(viii) Write the formula of any salt containing just these ions that will precipitate for reasons connected with an entropy change. FePO₄ or Mg₃(PO₄)₂

(ix) Write the formula of any salt containing just these ions that will precipitate for reasons connected with an enthalpy change. CsMnO₄
 
 

7. (10 marks). Draw and briefly discuss the structures of typical silicates and aluminosilicates, showing how complex silicates are built up from the simple orthosilicate anion. You may wish to use the following silicates as examples in your answer: Na[AlSi₃O₈] (albite), CaMg[SiO₃]₂ (diopside), (Mg,Fe)₂[SiO₄] (olivine), Be₃Al₂[Si₆O₁₈] (beryl), and Mg₃[Si₄O₁₀](OH)₂ (talc).

This is well-covered in your text. Just make sure that you calculate the O:Si or O"(Si+Al) ratio in the silicate or aluminosilicate anions (enclosed in square brackets). Practice drawing the chains and sheets!

Enthalpy of hydration for the sulfate ion = -1115 kJ/mol

Enthalpy of hydration for the barium ion = -1304 kJ/mol

Lattice energy of BaSO₄ = -2423 kJ/mol

(i) What is the definition of the lattice energy?

Energy released when gaseous cations and anions come together to form 1 mole of the crystalline salt

Ba²⁺ (g) + SO₄^2- (g) ---> BaSO₄ (s)

SO₄^2- (g) ---> SO₄^2- (aq)

Precipitation of BaSO₄: Ba²⁺ (aq) + SO₄^2- (aq) ---> BaSO₄ (s). The enthalpy of precipitation is the enthalpy change for this reaction, so put the reactions in an order that adds to give this equation.

Reaction	Enthalpy change
SO42- (aq) ---> SO42- (g)	1115 kJ/mol x 1 mol = 1115 kJ
Ba2+ (aq) ---> Ba2+ (g)	1304 kJ/mol x 1 mol = 1304 kJ
Ba2+ (g) + SO42- (g) ---> BaSO4 (s)	-2423 kJ/mol x 1 mol = -2423 kJ
Ba2+ (aq) + SO42- (aq) ---> BaSO4 (s)	Answer: -4 kJ for the enthalpy of precipitation

This page is http://chemiris.labs.brocku.ca/~chemweb/courses/chem232/Midterm_Exam_2_2001.html
Created April 6, 2001 by M. F. Richardson
© Brock University, 2001