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Hi All A-Level/JC/H2/H1 Chemistry Students

Continue to work hard and stay focus for the coming papers

See you on 23 Nov or 24 Nov or 25 Nov for P1 preparation

All the best
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FAQ : Why alkanes and alkenes are less dense than water?

Alkanes and alkenes have idid whereas water has hydrogen bonding. Stronger inter molecular forces and tighter packing causes water to be more dense.

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FAQ : Why alkanes and alkenes are less dense than water?

Alkanes and alkenes have idid whereas water has hydrogen bonding. Stronger inter molecular forces and tighter packing causes water to be more dense.

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Qns : Why is gibbs free energy = 0 for phase change reactions?

Ans : During a phase change, there is an equilibrium between the two phases. The two phases can interchange and hence delta G is zero at equilibrium

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Hi Rachel

Q14 JJC 09/3
No of protons in X + 84 = 115
No of protons in X = 115 – 84 = 31
Atomic number of X = 31
Hence X is Ga
Ans (B)

Q19 NJC 09/2
X and Xn+ has the same no of protons as they are the same element.
Xn+ has fewer electrons than X.
Xn+ is smaller than X due to higher effective nuclear charge (same nuclear charge since same no. of protons but weaker shielding effect).
Xn+ and X have different number of electrons and hence different electronic configuration.
Ans (C)

Q21 HCI 09/5 H1
Discuss on Monday

Q22 NYJC 09/4
Discuss on Monday

Q24 AJC 09/3
N: 1s2 2s22p3
In GaN, N is N3-
Therefore 3 more electrons added to 2p orbital.
Ans (D)

Q28 SAJC 09/4
Element X is in Group III while element Y is in Group VI. Hence the formula formed between X and Y is X2Y3.
Ans (B)

Q37 NJC 09/10 H1
Second I.E. is energy taken in to remove the 2nd outermost electron from one mole of gaseous singly charged cation to form one mole of gaseous doubly charged cation.
Ans (B)

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Hi All A-Level/JC/H2/H1 Chemistry Students

JC2 Chemistry Lesson Plan for the month Jan and Feb

– Introduction to Organic Chemistry
– Alkanes
– Alkenes
– Arenes
– Halogen Derivatives

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Hi All A-Level/JC/H2/H1 Chemistry Students

New Class J1 H2 Chemistry starting this Sunday 3 Feb 2012 at 12pm to 2pm

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GCE A Level 2012 Paper 1 Ans

1 A 6 A 11 D 16 D 21 D 26 A 31 A 36 B
2 C 7 D 12 C 17 D 22 B 27 C 32 D 37 D
3 D 8 C 13 B 18 C 23 B 28 B 33 A 38 A
4 B 9 A 14 B 19 C 24 D 29 B 34 B 39 B
5 C 10 D 15 C 20 C 25 C 30 A 35 B 40 B

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Please be inform that J1 H2 Chemistry will start on 3 Feb Sunday 12pm to 2pm

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Introduction to Organic Chemistry Summary

Classification of Organic Compounds Hydrocarbons can be classified as:

• aliphatic (straight or branched chain) eg hexane C6H14
• alicyclic (closed ring) eg cyclohexane C6H12
• aromatic eg benzene C6H6

Functional Group

• Each type of organic compound contains the same reactive group of atoms that governs the chemical properties. This reactive group of atoms is called a functional group.
• Compounds in a homologous series with the same functional group and general formulae are called homologues. They have similar chemical properties due to the same functional group.

Naming compound from given structure

1. Identify most important Functional Group (FG). Halogen assigned as Substituent.
2. Find longest chain that contains FG
3. Identify and assign number to FG and Substituent
a) smallest number assigned to FG
b) if no FG, smallest number assigned to Substituent
c) if more than 1 Substituent, sum of numbers as small as possible
4. Arrange Substituent in alphabetical order
a) no spacing between letters
b) dash “-” between letter and number
c) comma “,” between numbers
d) numbers represent position of FG and Substituent
e) if have 2, 3 or 4 of the same Substituent, add prefix “di”, “tri”, or “tetra” respectively

Drawing structural formula from name

1. Draw and number parent skeleton
2. Draw Functional Group
3. Draw all Substituents from Left to Right
4. Fill in remaining Hydrogen

Isomerism

• different compounds possessing the same molecular formula but existing in different forms because they have different arrangement of atoms.
◦ Structural Isomerism – same molecular formula but different structural formula
◦ Stereoisomerism – same structural formula but different spatial arrangement of atoms
▪ Total number of stereoisomers = 2n, where n is the number of stereocentres (chiral C or alkene with cis-trans isomerism)

For complete summary please contact Mr Ong @ 9863 9633

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Introduction to Organic Chemistry Summary

Classification of Organic Compounds Hydrocarbons can be classified as:

• aliphatic (straight or branched chain) eg hexane C6H14
• alicyclic (closed ring) eg cyclohexane C6H12
• aromatic eg benzene C6H6

Functional Group

• Each type of organic compound contains the same reactive group of atoms that governs the chemical properties. This reactive group of atoms is called a functional group.
• Compounds in a homologous series with the same functional group and general formulae are called homologues. They have similar chemical properties due to the same functional group.

Naming compound from given structure

1. Identify most important Functional Group (FG). Halogen assigned as Substituent.
2. Find longest chain that contains FG
3. Identify and assign number to FG and Substituent
a) smallest number assigned to FG
b) if no FG, smallest number assigned to Substituent
c) if more than 1 Substituent, sum of numbers as small as possible
4. Arrange Substituent in alphabetical order
a) no spacing between letters
b) dash “-” between letter and number
c) comma “,” between numbers
d) numbers represent position of FG and Substituent
e) if have 2, 3 or 4 of the same Substituent, add prefix “di”, “tri”, or “tetra” respectively

Drawing structural formula from name

1. Draw and number parent skeleton
2. Draw Functional Group
3. Draw all Substituents from Left to Right
4. Fill in remaining Hydrogen

Isomerism

• different compounds possessing the same molecular formula but existing in different forms because they have different arrangement of atoms.
◦ Structural Isomerism – same molecular formula but different structural formula
◦ Stereoisomerism – same structural formula but different spatial arrangement of atoms
▪ Total number of stereoisomers = 2n, where n is the number of stereocentres (chiral C or alkene with cis-trans isomerism)

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HYDROCARBONS – SUMMARY

ALKANES

• saturated hydrocarbons with the general formula of CnH2n+2 for aliphatic alkanes and CnH2n for cyclic alkanes.
For example: hexane, C6H14 and cyclohexane, C6H12
• non-polar molecules and their physical properties, such as melting point and solubility is affected by induced dipole – induced dipole interactions (chemical bonding)
• C atoms are sp3 hybridised and forms 4 single bonds with other atoms.

Reactivity

• generally unreactive as they are non-polar and all the bonds (C-C and C-H) are saturated.
• much of the chemistry of alkanes involves free radical chain reactions, which take place under vigorous conditions and usually yield a mixture of products.
• preparation of alkanes is not suitable for the lab preparation of halogenalkanes due to low yield and difficulty in separating the mixture of products.

Cracking / Pyrolysis of Alkanes

• In order to produce smaller hydrocarbons which are more useful and in greater demand as fuels for example, larger alkanes can be broken into smaller alkanes, alkenes and hydrogen by the process
known as cracking or pyrolyisis, which can be done either at very high temperatures (thermal cracking) or using catalyst (catalytic cracking)

For complete summary please contact Mr Ong @ 9863 9633

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Atomic structure – FAQ Part 1

Q: Why is the angle of deflection directly proportional to charge of the particle?
A: The greater the charge of the particle, the greater would be the attractive force exerted on it from the oppositely charged plate. Hence greater would be the deviation from its original direction of motion.

Q: Why is the angle of deflection inversely proportional to the mass of the particle?
A: If two particles are moving at the same speed but one is more massive than the other,the heavier particle would have a greater kinetic energy. Thus it requires more energy to be exerted on the heavier particle to cause it to deflect. Since the applied electric field is exerting the same amount of force on these two different particles with different
masses, the heavier particle would be deflected to a lesser extent.

Q: Why do isotopes react similarly?
A: This is because in chemical reaction, it is the electrons that are transferred between different atoms; atoms either gain or loose or share electrons. The nucleus remain intact

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Atomic structure – FAQ Part 2

Q: What is the meaning of this symbol 8O2–?
A: This represents an oxide ion. The subscript on the left is the atomic number whereas
the superscript on the right indicates the extra electrons the oxygen has acquired. For this 8O2– species, the number of protons (8) is not equal to the number of electrons (10). The number of electrons would be the same as the number of protons for a neutral atom only.

Q: What is the meaning of this symbol O2 2–?
A: This represents a peroxide ion. The subscript on the bottom right indicates that there are 2 oxygen atoms in this species covalently bonded to each other. The superscript on the right indicates the number of extra electrons this species have acquired.

Q: It seems that the relative isotopic mass is a whole number. Is it really so?
A: The word ‘relative’ means that the value obtained is measured with respect to some other thing, hence this number has no physical unit. In this case here, the relative isotopic mass of an isotope is the mass measured with respect to 1/12 of the mass of a 12C atom, which has a value of 1 unit (1/12 × 12). As the mass of an atom arises mainly from the nucleus and since the total number of nucleons in 35Cl is 35, the relative
isotopic mass seems to have a value of 35 too. But in reality, the actual relative isotopic mass is less than the mass of all nucleons added up. This phenomenon is known as mass defect. The differences in the two masses (less than 1%) arise because part of the mass has been converted to binding energy (according to E = mc2) which is necessary to hold the nucleons together.

Q: Why does the energy of an electron increase as we move away from the nucleus?
A: By convention, when an electron is ‘free’, that is, not subjected to any other electrostatic interactive forces (attractive or repulsive), it has an assigned zero energy. This is when the electron is infinitely away from the nucleus. But now if you want to bring an electron from n=1 Principal Quantum Shell to infinity, you got to do work against the attractive force of the nucleus, you got to ‘break’ the ‘bond’ between the
electron and the nucleus. Breaking bond needs energy. The energy that you put in while doing work (energy is conserved from Law of Conservation of Energy) is gained by this electron, hence its energy has increased. When an electron moves from infinity and is attracted by the nucleus, a ‘bond’ is formed, and energy will be released.

Q: Why does the 4s sub-shell have a lower energy than the 3d sub-shell given that n=4 Principal Quantum Shell should have a higher energy than n=3 Principal Quantum Shell ?
A: n=4 should have a higher energy than n=3. But the s sub-shell has a relatively lower energy than the d sub-shell for the same Principal Quantum Shell. These 2 different opposing factors counteract each other, resulting in the 4s sub-shell having a lower energy than the 3d sub-shell. The same explanation account for the relative energies of the 5s and 4d sub-shells.

Q: Does that mean that the 4s sub-shell is now closer to the nucleus than the 3d sub-shell?
A: No. On the average, n=4 Principal Quantum Shell is still further away from the nucleus than n=3. So although the energy of 4s sub-shell is lower than the 3d sub-shell, it does not imply that the distances have changed. Remember a Principal Quantum Shell is actually a band, not a single discrete line.

Q: Why do we need to know the electronic configuration of an element?
A: Knowing the correct electronic configuration would enable us to know which electron is to be removed and which orbital does it reside in. This is important as the removal of different electrons from different orbitals need different amounts of energy.

Q: What is electron spin?
A: You can imagine an electron like the Earth, rotating at a particular axis.

Q: Why can’t two electrons in the same orbital have the same spin?
A: Well, when an electron spins, this spinning charged-particle creates a magnetic field. If two electrons spin in the same direction, the magnetic field created would be repulsive in nature and the energy level of these two electrons would be higher as compare to if they spin in opposite directions to create an attractive magnetic field. An analogy can
be used here: Picture a spinning electron as moving in one particular direction. Two such spinning electrons will be moving in opposite directions and the chances of them ‘meeting’ will be lower. This results in lesser inter-electronic repulsion.

Q: Why do we need to first place electrons in empty orbitals of the same sub-shell before pairing them in an orbital?
A: Well, electrons repel each other. By occupying different orbitals, the electrons remain as far apart as possible from one another, thus minimizing electron-electron repulsion. Take note that each orbital represent a particular region of space, hence two different orbitals would be two different region of space separated from each other. Take for
instance the three p orbitals in a p sub-shell, each is oriented erpendicularly from each other, occupying different regions in space.

Q: Why is the electronic configuration of 25Mn not 1s2 2s2 2p6 3s2 3p6 3d7?
A: The 4s orbital is filled first before the 3d orbitals. This is because the 4s orbital has a lower energy level as compared to the 3d orbitals.

Q: If the 4s is filled before the 3d, why the electronic configuration of 25Mn not 1s22s2p63s23p63d54s23d5?
A: The electronic configuration is always written in the order of increasing Principal Quantum Number. This also indicates the order of increasing energy level of the various sub-shells.

Q: Why is symmetrical distribution of similar charge preferred?
A: If similar charges are distributed symmetrically, this would mean that all charges are spread out evenly and as far apart as possible. Such situation would result in a similar amount of electrostatic repulsion at each point in space. Consequently, such state would have a lower energy as compared to a state of asymmetrical distribution.

Q: Why must the atom be in the gaseous state?
A: When we carry out ionization, the species must be gaseous atoms. In the gaseous state, the atoms have very minimal interaction with each other. Thus the energy input would solely be responsible for removing the electron and not in overcoming other types of bond. So remember that the gaseous state symbol is very important here.

Q: What is a valence electron?
A: Valence refers to the outermost. Thus a valence electron ‘sits’ in the outermost Principal Quantum Shell and is furthest from the nucleus. The Principal Quantum Number for the valence shell corresponds to the Period Number of the element. All other Principal Quantum Shell of electrons before the valence Principal Quantum Shell are known as the inner core electrons.

Q: What is the Effective Nuclear Charge ENC on each of the valence electrons of an oxygen atom?
A: The electronic configuration of an O atom is 1s22s22p4. The nuclear charge consists of 8 protons and the number of inner core electrons is 2 (since there is only one Principal Quantum Shell of electrons before the n=2 valence shell), therefore the ENC is ≈ 6.

Q: Does that mean that each of the 6 valence electrons is attracted by 1/6 of the ENC, which is 1 proton?
A: No. Each of the 6 valence electrons is attracted by 6 protons. This is because the 6 valence electrons are moving round the nucleus within the same distance from the nucleus and the nucleus is considered a point charge. Therefore, the ENC is the same on each of the valence electrons.

Q: Does that mean that the ENC on each the two 1s electrons is equivalent to 8 protons?
A: Yes. There are no other inner core electrons before the n=1 Principal Quantum Shell,therefore there is no shielding effect. The ENC on each of the electron in the n=1 subshell is the same, i.e. equivalent to 8 protons.

Q: The ENC of an O atom is greater than the N atom. So shouldn’t this cause the O atom to have a higher 1st I.E.?
A: Yes, the ENC of O atom is greater than the N atom and this should cause it to have a higher 1st I.E. But experimentally, it has been found that O atom has a lower 1st I.E that what was expected. There must be some other factors that have yet to be considered. The only reason we can use to explain the data collected would be to employ the interelectronic
repulsion factor. As shown here, there is the interplay of two opposing
factors but it seems that the inter-electronic repulsion is a more dominant factor than the ENC effect.

Q: Why is the domineering effect of the inter-electronic repulsion over the ENC factor
being observed in the oxygen versus nitrogen case but not in the beryllium-lithium scenario?
A: Remember we mention before that a p orbital is 1/3 the size an s orbital from the same Principal Quantum Shell? Thus because of this different in size, the inter-electronic repulsion is more prominent in a p orbital than the much bigger s orbital.

Q: The ENC of a B atom is higher than that of a Be atom but there is inter-electronic repulsion faced by the electrons in the 2s sub-shell of Be. Are these two factors less dominant than the ‘difference in energy level’ factor?
A: You are right. Based on the ENC factor, B is expected to have a higher 1st 1.E. than Be. But this not what experimental data makes of it. If we are to consider inter-electronic repulsion factor, Be should have a lower 1st I.E but observed data proves otherwise. Hence, the only factor that could be used to account for the observed experimental data would be because the 2p electron is at a higher energy level than the 2s electron.

Q: Why would attractive force be weaker, the further the electron is from the nucleus?
A: The electrostatic force (F) between the nucleus and electron can be approximated to F ∝ 1/r2, where r is the distance of separation between the charges. Thus, as r increases, the strength of the electrostatic force decreases drastically.

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Chemical Bonding Part 1 FAQ

What are Chemical bonds?

– Binding forces of attraction between particles (atoms, ions or molecules) resulting in a lower energy arrangement.
– The formation of a bond involves the re-distribution of the outer electrons of the atoms concerned.

What is the Octet Rule?

Atoms tend to lose, gain or share electrons until they are surrounded by eight valence
electrons. Atoms try to achieve the same number of electrons as the noble gases closest to
them in the Periodic Table.
All noble gases (with the exception of helium) have eight valence electrons. They have very
stable electronic arrangements. Evidence of stability of noble gases:
– high ionisation energy
– low affinity for additional electrons
– general lack of reactivity

However, there are many exceptions to the octet rule. Nevertheless, it provides a useful framework for introducing many important concepts of bonding.

Why is it that both NCl3 and PCl3 exist, but only PCl5 exist and not NCl5?

Such expansion of octet is observed in some compounds formed by elements of Period 3 (and beyond) This is due to the availability of vacant, low-lying orbitals. The energy required to promote an electron from 3s or 3p to 3d is not very large.

However elements in Period 2 (e.g. O and N) do not have low-lying vacant orbitals for expansion of octet. Promotion of electrons to the next quantum shell requires too much energy and hence Period 2 elements can accommodate only a maximum of eight valence electrons.

Explain which bond is stronger, C—H or Si—H.

C—H bond is stronger since C is smaller than Si so that valence
orbital of C is less diffuse and overlap of its valence orbital with that
of H is more effective.

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