Homework before 23rd Sept 2009

Dear CH304,

From KWOK The Chem Teacher

I am certain all of you are aware that pressure affects equilibria which has changes in the number of gaseous molecules. Hence, to ensure a constant pressure to be created is more difficult than to create an environment for constant temperature. Hence, please ponder on the above question. The solution isn't complicated but it is an application of what we have learnt.

Lastly, ensure your answers are clear and concise. This solution does not need excessive elaboration.

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Background

The background of the question came from two different questions found in separate preliminary examination paper. The questions raised a curious thoughts, how is it possible to ensure a constant pressure for gaseous reaction system with temperature being kept constant?

Let's take Haber Process as the example. From its equation, you can tell that the number of particles at equilibrium will be less than the number of particles at the start of the reaction. Hence, if pressure was kept constant (i.e. inital P = final P), with a decrease number of particles, the volume must decrease. Therefore a contraction occurs.

Thus, the curious question. How is it possible for constant pressure, constant volume and constant pressure be maintained for the reaction for Haber Process? Fundamentally, the pressure exerted by the system has to be that exerted by the gas particles.

Hence, if initially we only have the reactants and the total pressure is 200 atm, how can a reduction in gas particles result in the pressure to still remain the same; when volume and temperature are constant? We would actually expect that pressure decreases.

Therefore, my suggestion is to add inert gas such as He into the reaction mixture. The inert gas added will ensure that the total pressure remains, in addition, this addition will not keep causing the equilibrium position to shift.

However, Jing Yew made an excellent suggestion of using a vortex. Personally, I am clueless to what it is. However, a check with a Physics teacher I learnt that it is a machine that can alter the kinetic energy of the gas particles. Hence, if this machine is feasible, it can alternate the speed of the particles at equilibrium to ensure that the total pressure remains the same as the initial pressure.

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The other question

The simplest way is to cool the reaction mixture quickly and separate the species. Quickly because you do not want equilibrium to have any time to respond to the change in temperature. In addition, cooling ensure that the particles do not have enough energy to overcome Ea, hence hardly any forward or backward reaction can take place.

Subsequently, extract a small portion of the sample (to ensure only small amount of NH₃ is present), then titrate against standardised concentration of HCl with a suitable indicator The small amount ensure that you will not need to have to use an exaggerated amount of HCl to attain full neutralisation.

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Misconceptions

Some glaring misconceptions include

(1) Thinking that removing gases (especially NH₃) will do the trick. That is a BIG NEVER. Since, the number of gas particles will decrease as we compare the initial with the equilibrium, removing particles will make the situation worse.

(2) Adding more gas reactants. Will that will ensure that you can get a constant temperature, but can you maintain the equilibrium? If you add a reactant, chances are equilibrium position will shift right. Hence, you will constantly have a situation where equilibrium position shifts and hence you can't achieve the equilibrium.

(3) Using a pH to determine the concentration of ammonia present. That is extremely inaccurate! We never do that!

(4) Apply PV = nRT. I am now sure how are u going to separate the ammonia from the mixture and hence measure the partial pressure of ammonia. Subsequently, using the ideal gas equation to calculate number of moles of ammonia.

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Conclusion

Personally, I am not sure if this is what happen in industries. However, from my extensive search on the web, my sense is probably Jing Yew may be more accurate than I am. Since, the gases in Haber process are actually injected into the reaction vessel at a particular speed to create the desired pressure (Note: More KE the gas particles has, the greater force they exert when the collide of the vessel's wall). The notion of speed causing the pressure gives me that suspicion that it could be a voltex or at least some instructment that can adjust the speed of the particles and yet ensuring the volume, pressure and temperate are kept constant.

Homework - Change in venue.

Dear 1CH304,

There is a new online assignment, please read the following:

Usage of Google Site - Ground Rules

Regards
Mr Kwok

Homework 8 (due Tuesday 2nd June 2009)

Dear CH304,

This application question may be slightly more challenging but it has to do with chemical bonding.

If you noticed SiO₂ does not exists like CO₂, where we obtain an SiO₂ with a pair of Si=O bond.

Interestingly, Phosphorous prefer to exists as P₄, with single bonds linking the phosphorous atoms while Nitrogen exists as N₂ where there is a triple bond between the two N atoms.

Please explain why we observe this.

You may wish to read about this (an article which you will learn in JC 2). If you understand that article, you will be able to explain for the above situation.

Regards
Mr Kwok
P.S. This is an interesting question to develop thinking skills. The chemistry concept of this will be largely covered in 2010.

Suggested Answer:

In CO₂, the C=O bond is made up of a sigma bond and a pi bond. If you have read the hybridisation article, you would realise that for carbon has to undergo hybridisation of the 2s and 2p orbitals to form suitable hybrid orbital for the double bond. Since CO₂ contains 2 C=O, the carbon is sp hybridised. Hence, the sp hybridised orbitals overlap with the 2p orbital of oxygen to get the sigma bond, while the 2p orbital carbon overlap in a side-on manner with 2p orbital of oxygen to get the pi bond.

In SiO₂, it is just sigma bonds. Hence, silicon has hybridised its 3s and 3p orbitals to give sp³ orbitals, which are used for head-on overlapping with the p orbital of oxygen.

Hence, it is quite clear that the formation of pi bond is what is different in both cases. As the formation of pi bond between C and O is favourable, hence carbon uses the sp hybridised orbitals. The reason why pi bond between C and O is favourable is because the p orbital of C and that of O are able to effectively overlap with each other.

Usually large period 3 and below elements find it harder to form pi bonding. Their p orbital usually prefer not to have side-on overlap with the 2p orbital of small atoms such as O as the overlapping may not be sufficiently effective for the particular type of hybridisation to take place.

It is this reluctance to form pi bond between atoms that results in phosphorous to exist as P₄ instead of P₂. The former allows each P atom to form 3 bonds (hence satisfying their octet requirements), while the latter forces each p orbtial to contribute two 3p orbitals for side-on overlapping for pi bond. This side-on overlapping is not sufficiently effective and hence the hybridisation to get sp or sp² is not favoured. Instead a sp³ is used to obtain sigma bonds between P atoms.

Comments:

I hope that you found this entry abit interesting. It is probably a "S" paper Chemistry question, but it is really testing us whether we understand why bonding take place between two atoms and why different types of bond can occur between the two atoms.

I shall also publise all your contributions and perhaps you could take time and compare the difference in how the thought processes were formulated.

Commendable Student's answers:

Please read Johanan's and Brandon's input. Jing Yew's contribution is interesting but it did not hit what I feel is the key point. The same should be said about Lyria's answer, which was coherent but I was not convinced that she understood the key point about how easy was it for the pi bond to be formed.

Homework 7 (due 25th May 2009)

Dear CH304,

Please copy/print out the following questions. Answer them on writing paper and submit your scripts on Monday.




Regards
Mr Kwok

Homework 6 (due before 11th May 2009) (part 2)

Dear CH304,
This is the second question.

Suggest an explanation to why XeF₂ exists but NeF₂ does not. [Hint: You are required to use the concept of atomic structure and chemical bonding in your answer]

Answer to Question 2

Xe is a Period 5 element and although it has 8 valence electrons, it has available empty d-orbital which can be used to expand the octet for additional bonding. While Ne is a Period 2 element and does not have available d-orbtial which can be used to expand the octet.

In additional, to expand the octet in Xe, it requires the unpairing of the valence electrons and exciting them to the d orbital, since in a same shell the d orbital has a higher energy than the s orbital and the p orbital. Thus, this rearrangement of electronic configuration requires energy. Noticeably, this energy required is compensated by the Xe-F bond that is formed (bond formation releases energy). Therefore, XeF₂ exist.

Even if Ne can expand its octet, Ne's valence electrons are so close to the nucleus and thus experience strong electrostatic attraction. Thus to unpair them and move it to a higher energy orbital, you will need so much energy that even the formation of the Ne-F bond is unable to compensate the energy required use to rearrange the electrons configuration.

Comments for Question 2
1. If you have submitted the work, a mark is awarded to you. If you have mentioned expansion of octet due to available d orbitals or available orbitals a second mark is given to you. However, if you mentioned 3d orbital, you will not get the mark.

2. All of you had problems in getting the 3rd and 4th mark. I guess one has to realise that in order to expand the octet for bonding, there is a need to rearrange the electronic configuration of Xe, to make the electrons unpaired. This gives the 3rd mark. One of the unpaired electrons will occupy a higher energy orbital such as the d orbital and the unpair electron awaits to be bonded with F. (Note that formation of Xe-F bond makes use of equal sharing of electrons, where Xe and F each contribute one electron. Hence, you will need Xe to unpair its electrons pair.)

3. Finally, because the Xe-F bond is formed (bond formation is exothermic or energy releasing - this was mentioned in an earlier blog post), the energy release can compensate the energy required to rearrange the electronic configuration. This is the 4th mark.

4. Hence, to expand the octet these are the available reasons. Hence, sometimes when we want to consider about expansion of octet, the type of bond formed will determine if the expansion is favourable. Hence, if a strong covalent bond can be formed, we will get expansion of the octet. For this reason, XeBr₂ does not exist as the Xe-Br bond is not sufficiently strong.

Homework 6 (due before 11th May 2009) (part 1)

Dear CH304

There are two questions in today's assignment. Please copy question 1 and answer it on a piece of writing paper. Submit you answers before Tuesday's assembly.

Question 1 - Done in writing paper.
(a) Draw the dot and cross diagrams of (i) N₂O (ii) O₃ (iii) HNO₃ (iv) H₃PO₄.
[Hint: (iii) and (iv) are mineral acids. As they are mineral acids they have at least one O-H.]

(b) Draw the lewis structure of H₂O₂. On this diagram, label all bond angles. In addition, determine the shape of H₂O₂.

(c) XeF₂ is one of the most stable Xenon compounds. (i) Draw the dot and cross diagram of XeF₂. (ii) Predict with explanation which bond is stronger Xe-F or He-F.

(d) Suggest an explanation for the following suitations. (i) MgCl₂ has stronger interatomic bonds than NaCl. (ii) Given that the bond energy of C-Cl is 328 kJmol^-1 and while the C-N is 292 kJ mol^-1.

[Marks allocation for Q1: 3 + 4 + 4 + 4]

Suggested Solutions




Comments
(1) Becareful with your explanation in (c)(ii) - It cannot be a contradiction.

(2) Becareful with your explanation in (d)(ii) - It cannot contradict the given data. Hence, atom size for effective overlap isn't suitable. Lone pair repulsion is irrelevant as C does have that.

Homework 5 (due before 4th May 2009)

Dear CH304,

This week's question is as following observations:
(i) Explain why the bond energy of O-F is smaller than the bond energy of O-Br. (The former is 212 kJmol^-1 and the latter is 217 kJmol^-1.)

(ii)As working, draw the Lewis diagram of N₂ and the CN^- ion. You do not need to submit your Lewis diagram. (Clue: Draw HCN first)

Explain why the triple bond in N₂ is stronger than the triple bond in CN^-. (The former's bond energy is 944 kJmol^-1 and the latter is 890 kJmol^-1

Answers
The following pictures illustrate and explain the questions.
Although, O-Br bond also have lone pair on both O and Br, the repulsion is much less because of Br's large size.

In addition, the size of the atom of the element decreases as we go along the period. While the size of the atom of the element increases when we go down the group. Hence, using this concept, we are able to conclude that the C triple bond N is weaker than the N triple bond N.

Comments

For (i)
Generally (i) has better answers. But, many missed out the key phrase - lone pair on O and the lone pair repulsion on F repel each other because they are so close together (due to O and F being small sized), hence weakening the O-F bond. - I think Samuel's answer for (i) is nice. Succinct, not trying to explain too much and doesn't complicate the situation further.

In addition, using electronegativity to explain (i) is NOT correct.

Interestingly, only Nicholas mentioned that overlapping between small atoms is weak because orbtial overlap is not effective. This is highly incorrect.

For (ii)
I think the biggest problem is that many of you tried to explain that lone pair repulsion, hence making the bond weaker. That is not true as illustrated by the picture above, where both CN^- and N₂ have lone pairs of electrons.

Hence, in short. The general strategy to explain strength of covalent bond is as follow.

(A) Direct application of the definition of covalent bonds.

If the two examples you are given cannot be concluded using (A), then either one of the two exceptions will be employed.

(B1) A polar covalent bond. A partial positive and partial negative charge, this is an added electrostatic attraction that strengthens the bond between the two atoms.

(B2) Lone pair repulsion. This causes the bond to be weaker.

Homework 4 (due 27th April)

Dear CH304,

Predict with suitable explanation which compound has stronger inter-atomic bond, BeBr₂ or AlBr₃?

Answer
BeBr₂ and AlBr₃ covalent compounds. If both were ionic compound, both compound contains a very small and highly charge cation and a large anion. This results in the cation to polarise the anion's electron cloud, hence making electron clouds to overlap.

Since both BeBr₂ and AlBr₃ are covalent compounds, Be is a smaller atom than Al. Hence, the extent of overlapping of atomic orbital between Be and Br is greater than in Al and Br. Hence, Be-Br single bond is stronger than in Al-Br single bond.

Comments
The most comprehensive answer goes to Xing Ting, Johanan, Wei Jie and Dung. =)

The following are the misconceptions made. I have included the names so that we can all learn from each other. I think it wld be excellent if you would like to take this question into further discussion.

1. Stating that BeBr₂ and AlBr₃ are dative covalent compounds. Please don't say that. Covalent compound is good enough. - Eileen's answer

2. Not applying to the definition of covalent bond (key words missing out) completely. - Eileen's, Jo's, Daniel's

3. Thinking that BeBr₂ are AlBr₃ ionic compounds - Xiao Min's, Lyria

4. Incomplete sentence expression (with ambiguous word(s)). - Sherlyn (what is it?), Jing Yew (Am i going to assume that you meant covalent bonds are formed?)

5. Totally missing the point. Thinking that I am trying to account for dative bond or why covalent bond, when I am actually trying to determine strength of bond. - Peck Fen, Wen Han, Brandon

6. Writing irrelevant information (e.g. "Bond pair replusion", "more bonds bonds formed around Al, does not mean the bond is stronger", "VSEPR" and "Charges of Be and Al") - Renaldy, Benjamin, Crystal, Kevin, Nicholas

7. Trying to answer too much. Uses both definition of covalent bonding. (can get abit confusing). - Samuel

Homework 3 (due before 20th April 2009)

Dear CH304,

Question

You have learnt about the oxidising agent KMnO₄. This reagent is made up of potassium and permanganate ions. In an ion of MnO₄^-, there is covalent bond between Mn and O. By using your knowledge from these posts, (i) suggest plausible reason(s) to why covalent bond exists between Mn and O and (ii) to why a dimer or polyer is not formed (which is what happen in AlCl₃ and BeCl₂ respectively.).

Suggested Answers

(i) If ionic bonding occurs in MnO₄^-, the Mn must exist as a Mn⁷⁺. This cation will be very small and will have a very large change, hence highly polarising. O^2- is generally quite small, but when placed next to a Mn⁷⁺, it will polarise the electron cloud of O and hence resulting in an overlapping of electron cloud between Mn and O. Therefore, covalent bond exists between Mn and O instead.

(ii) Usually such structures, we will expect to see that the metal atom will contain empty orbitals while the non-metal atom will contain available lone pair of electrons of donation. This results in a formation of dimer (Al₂Cl₆) or a polymer (e.g. BeCl₂). In the situation of MnO₄^-, it is possible to suggest that O being a highly electronegative element is reluctant to donate its lone pair of electrons to form dative bond. Hence, there is no dimer formed nor polymer.

Common errors and comments

There some nice answers given by your classmates. I reckon that Jing Yew provides the best answer. Joey gave me a very detailed analysis to why the bonding between Mn and O is covalent for MnO₄^-. Wei Jie also provided me with a good answer to why the bond between Mn and O is covalent.

Basically, I required you to make use of the ideas to suggest why a metal and non-metal produces a covalent bond between them. In addition, to account for formation of dative bond. If you notice, generally covalent compounds are metal and non-metal combination, it is capable of dative bond. - This is not the case for MnO₄^-

1. Suggesting that a Mn²⁺ will exist in MnO₄^-. This is not true.

2. Mn has no empty orbital - Actually I do not think I can conclude that. I think the d orbital is available. And since Mn has the electronic configuration of 1s²2s²2p⁶3s²3p⁶3d⁵4s², it should still have empty orbital available for lone pairs of electrons donated to it.

3. Some fail to conceptualise why AlCl₃ exists as a dimer. That idea is discussed in tutorial and explained over here! There is a weakness in conceptualising the phenomenon of AlCl₃.

4. I don't understand what has octet got to do with the type of bond formed between Mn and O.

5. The only two ways to account for why MnO₄^- has covalent bonds between Mn and O: (i) The above answer. (ii) Using IE. It takes up too much energy to form Mn⁷⁺, hence the ion prefers to have covalent bond instead.

6. One of you mentioned about orbital overlap, since Mn 3d and 4s are relatively close in energy. I would think that by saying these orbitals overlap presupposes that covalent bond is formed. I am actually curious then to why this doesn't occur to the other metals.

7. Some mentioned that O lacks lone pair. That is not true.

8. Mass of the atoms is irrelevant to the charge density.

Homework 2 (due before 13th April 2009)

Dear CH304,

This weekend's assignment. Last week's assignment can be found here.

In the current course of Chemistry, you have learnt that MgCl₂ exists as a giant ionic lattice structure. While, H₂O is a simple discrete molecule. This implies that ionic bond exist in MgCl₂, while covalent bond exists in H₂O. Suggest an explanation to why MgCl₂ is not covalent. In addition, suggest why H₂O is not ionic. You are highly encouraged to use the following articles as a stepping stone.

1) Atomic structure - Introduction
2) Chemical bonding - Interatomic Bonds

Suggested Answer

Mg has two valence electrons while Cl has 7 valence electrons. In order for Cl to obtain the octet configuration it needs 1 more electron. There are two possible ways this may be done: (1) gaining an electron from a donor. (2) Through sharing of electrons, Cl gains the 8th electron. The third way will be for Cl to lose all its 7 valence electron and this is not possible because it would require too much energy.

If Cl obtain its 8th electron through sharing of electrons with Mg, thus resulting in a Mg-Cl covalent bond to be formed between Mg and Cl. This results in Mg to have only 4 electrons; 2 of its own and 2 derived through sharing. This is a far short of octet.

Hence, Mg will prefer to lose 2 electrons. The energy required to remove 2 electrons is compensated by the ionic bond formed between Mg²⁺ and Cl^-. Therefore, MgCl₂ exist as an ionic compound.

While in the situation of H₂O, the sharing of electrons will allow both H and O to obtain the full shell configuration. Hence, water has the covalent bond, O-H, and it exists as a simple discrete molecule.

If H₂O exists as an ionic compound, it is not possible for O to ionise all its valence electrons as that is energetically not feasible. In addition, for H to ionise its electron to give H⁺ will not be favourable too. This is because H has only 1 quantum shell. The close proximity between the valence electron of H and the nucleus requires a huge amount of energy to ionise the atom H. Therefore, these scenarios explain why ionic bonding cannot occur between H and O in H₂O.

Comments
There are some decent attempts in trying to answer, but there are some of the glaring misconceptions or inabilities to account:

(1) Unable to do a proper compare and contrast to why some compounds exhibit ionic bonding while other exhibit covalent.

(2) Those who tried to do the compare and contrast gave very superficial comparisons, like using electronegativity difference, or electronegativity etc, these highlight that one did not really understood how these two interatomic bonds were derived. Electronegativity difference shows us a pattern to why certain bonds between particles are ionic and others are covalent; BUT they are not the reason to why those interatomic bonds are formed.

(3) Did not highlight why it is improbable for covalent bond for MgCl₂ and ionic bond for H₂O.

(4) Using polarising power to determine bonding is a poor use of principle. This principle is used to explain anomalies. It is not use to explain for the bonding in the first principle.

(5) This activity does highlight that critical analysis of data and principles is something that you would like to work on. The combination of Johanan (for MgCl₂ and Sherlyn's (for H₂O) answers give a indication of what I consider as critical analysis.

In conclusion, I think these set of assignments will start to allow us to think carefully about a statement and think more carefully about the fundamentals and hence facilitate the learning of this topic.

Homework 1 (due before 6th April 2009)

Dear 1CH304,

Critically comment the following quote:

"A An atom is a very small particle. It is able to be divided into smaller particles, known as protons, neutrons and electrons. These particles are sub-atomic particles and are the smallest particles of an atom. In addition, the atoms of an element are always identical. While the valence electrons of an atom are always degenerate. The chemical property of an atom is largely determined by the protons of the atom."

There are 5 errors in the above quote. Quote the relevant sentences and explain the error(s).

You may wish to this and/or that as resource materials to answer this question.

Suggested Answer
.... are the smallest particles of an atom

(1) This is not true as we have learned that protons, electrons and neutrons can be subdivided. However, we don't need to know the details.

the atoms of an element are always identical

(2) The existence of isotopes refute this statement. For example, the element hydrogen has 3 isotopes, ¹H, ²H and ³H.

..valence electrons of an atom are always degenerate

(3) Degenerate means that the valence electrons are found on the same energy level. That is not true. Using oxygen as an example, it has 6 valence electrons. It's electronic configuration is 1s²2s²2p⁴. The 6 valence electrons are found on 2s and 2p. Clearly, 2s and 2p are of DIFFERENT energy levels.

..chemical property of an atom is largely determined by the protons of the atom

The above statement has two errors:
(4) It is the valence electrons of an atom that determines its chemical property.

(5) The protons and neutrons (hence the nucleus) affects the physical property.

Comments

Generally there are many good answers! I am pleased and I am confident we can expect better ones in the future. However, the accuracy of the explanation was quite weak.

(1) Atoms are very small particles. The atomic radius is close to 10^-9 m, it is indeed very small!

(2) There are a few who struggled in their explanation of degenerate. The orbital in a subshell are degenerate, however two subshells are not degenerate. Interestingly, there are a few who did not know the meaning of degenerate.

(3) Some have left out the explanation and hence their answers were marked down as it did not adhere to the requirement of the question.

Welcome CH304 aka SCones

Welcome CH304,

This week is Orientation week. Please do the following by 30th March 2009.

Please click here to leave behind a name which I will be able to recognise. In future homework posts, you will need to click "comment" and enter this name that you have picked and then submit your assignment.

I would like to highlight that all comments will be screened by me before they are published. Hence, please do not panic when you do not see your comment immediately.

In the meantime, this week's set of homework also includes reading the articles on

(1) Atomic Structure.
(2) Redox equation.

You will be pleased to realise that the following blog labels are relevant to you:

(1) Content labels.
(2) CH304(10)
(3) Poll answers - when it is JC1 work.

As I try to keep the blog less cluttered, generally there are only a maximum two posts available on the main page and hence you need to know how to interpret the labels well.

If you have feedback, you can receive me at the messenge board, the discussion forum and even by simply using the comment function of each post.

So please remember to enter your name in this entry first!

Regards

Mr Kwok