Archive for the ‘Berry Reference (Notes)’ Category

SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 4)

July 27, 2010

Welcome back to Berry Berry Easy. Today, the Berry Berry Teacher will like to share with all Berry Readers, Part 4 of the SPM Form 5 Chemistry notes on “Oxidation and Reduction”. In the previous part, oxidation and reduction were viewed in terms of changes made to the oxidation numbers. However, we also mentioned that it can also be understood through electron transfer. So yes, you guessed it! This part will focus on redox reaction in terms of electron transfer. Similar to very easy “change in oxidation number”, understanding redox reactions in terms of electron change are just as easy.

Do also try to memorise the table presented at the end of this post. Most of the time, objective questions will ask about the colours of the precipitation when iron (II) or iron (III) ions are present in the various reagents. So don’t forget this tips. Do pop quiz with your friends to memorise the table below.

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction (Part 4)

Oxidation and Reduction in Terms of Electron Transfer

2I^- (aq) –> I₂ (aq) + 2e
Oxidation: Iodide ion, I^- is a reducing agent because it donates/loses electrons to become I₂.

Br₂ + 2e –> 2Br^- (aq)
Reduction: Bromine water, Br₂ is an oxidising agent because it receives/accepts electrons from I^- to form bromide ions, Br^-.

–> Overall reaction: 2I^- (aq) + Br₂ –> I₂ (aq) + 2Br^- (aq)

Conversion of Fe²⁺ Ions to Fe³⁺ Ions and Vice Versa

A) Common oxidising agent (change Fe²⁺ ions to Fe³⁺ ions):

• bromine, Br₂
• chlorine, Cl₂
• hydrogen peroxide, H₂O₂
• concentrated nitric acid, HNO₃
• acidified potassium manganate(VII), KMnO₄ solution
• acidified potassium dichromate(VI), K₂Cr₂O₇ solution

Fe²⁺ (aq) –> Fe³⁺ (aq) + e
Oxidation: Iron(II) ion, Fe²⁺ is a reducing agent because it donates/loses one electron to become Fe³⁺.

Br₂ (aq) + 2e –> 2Br^- (aq)
Reduction: Bromine water, Br₂ is an oxidising agent because it receives/accepts electrons from Fe²⁺ to form bromide ions, Br^-.

–> Observation: iron(II) sulphate solution changes colour from pale green to yellowish-brown.
–> Overall reaction: 2Fe²⁺ (aq) + Br₂ (aq) –> 2Fe³⁺ (aq) +2Br^- (aq)

B) Common reducing agent (change Fe³⁺ ions to Fe²⁺ions):

• zinc powder, Zn
• aluminium, Al
• Magnesium, Mg
• Calcium, Ca
• Sulphur dioxide, SO₂
• Hydrogen sulphide, H₂S
• Sodium sulphide solution, Na₂SO₃
• Tin(II) chloride solution, SnCl₂

Zn (s) –> Zn²⁺ (aq) + 2e
Oxidation: Zinc powder, Zn is a reducing agent because it donates/loses two electrons to form zinc ions, Zn²⁺.

Fe³⁺ (aq) + e –> Fe²⁺ (aq)
Reduction: Iron(III) ion, Fe³⁺ is an oxidising agent because it receives/accepts one electron to become Fe²⁺.

–> Observation: iron(III) sulphate solution changes colour from yellowish-brown to pale green.
–> Overall reaction: 2Fe³⁺ (aq) + Zn (aq) –> 2Fe²⁺ (aq) + Zn²⁺ (aq)

C) Investigate the presence of iron(II) and iron(III) ions

The next Part in this series contains arguably the most important and memorable “series” in your SPM Chemistry studies, namely the Electrochemical Series. It’ll be something that you’ll memorise even after you leave school. So stay tune and log in frequently to Berry Berry Easy.

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SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 3)

July 24, 2010

As promised in the previous post in this series, let Berry Berry Easy present all Berry Readers with Part 3 of the SPM Form 5 Chemistry notes on “Oxidation and Reduction”. The previous two parts were focused on the concepts of redox reactions and oxidation numbers. (You ought to understand how to write a redox reaction from the two half reaction components by now, else do revise from the beginning before starting this part) Now, we are going to show all of you how oxidation and reduction can be understood in terms of changes in oxidation numbers. (Do remember that you can also view it through the gain and loss of electrons, but that’s for another time).

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction (Part 3)

Oxidation and Reduction in Terms of Changes in Oxidation Numbers

Redox reactions – oxidation number of all elements change.

Rusting of iron, combustion, displacement of metal from its salt solution, displacement of halogen from its halide solution and electrolysis are redox reaction.

-10 …. -3  -2  -1  0  +1  +2  +3  …  +10

<———-  Reduction || Oxidation ———->

• H₂ (g) + CuO (s) –> H₂O (l) + Cu (s)
  Hydrogen: 0 –> +1 (Oxidised to water & Hydrogen is a reducing agent)
  Copper oxide: +2 –> 0 (Reduced to copper & Copper oxide is a oxidising agent)

• 2Zn (s) + O₂ (g) –> 2ZnO (s)
  Zinc: 0 –> +2 (Oxidised to zinc ion & Zinc is a reducing agent)
  Oxygen: 0 –> -2 (Reduced to oxide ion & Oxygen is an oxidising agent)

• 2Mg (s) + CO₂ (g) –> 2MgO (s) + C (s)
  Magnesium: 0 –> +2 (Oxidised to magnesium ion & Magnesium is a reducing agent)
  Carbon dioxide: +4 –> 0 (Reduced to carbon & Carbon dioxide is an oxidising agent)

• Br₂ (l) + 2HI (aq) –> 2HBr (aq) + I₂ (s)
  Hydroiodic acid / Hydrogen iodide: -1 –> 0 (Oxidised to iodine & Hydroiodic acid is a reducing agent)
  Bromine: 0 –> -1 (Reduced to hydrobromic acid & Bromine is a oxidising agent)

Non-redox reactions – oxidation number of all elements remain unchanged.

Precipitation, decomposition and neutralisation are not redox reaction (non-redox reaction)

Precipitation:

• AgNO₃ (aq) + NaCl (aq) –> AgCl (s) + NaNO₃ (aq)
  +1 +5 3(-2)      +1  -1              +1  -1        +1 +5  3(-2)

No change in the oxidation numbers.

Decomposition:

• ZnCO₃ (s) –> ZnO (s) + CO₂ (g)
  +2 +4  3(-2)    +2 -2       +4  2(-2)

No change in the oxidation numbers.

Neutralisation:

• NaOH (aq) + HCl (aq) –> NaCl (aq) + H₂O (l)
  +1 -2 +1          +1 -1             +1  -1             2(+1)  -2

No change in the oxidation numbers.

Revision time:

Questions to ask yourself at this point:

1. Do you understand what is a redox reaction?
2. Can you write the two half reactions out of a redox reaction?
3. Do you understand the concept of oxidation number?
4. Give three examples of an oxidising agent and the example of the reaction involved.
5. Give three examples of a reducing agent and the example of the reaction involved.
6. Can you differentiate a redox reaction with a non-redox reaction?
7. What is the characteristics of a non-redox reaction?

If you can answer all these questions, it shows that you understand the basics of oxidation and reduction.

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SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 2)

July 22, 2010

Berry Berry Easy would like to present all with Part 2 of SPM Form 5 Chemistry “Oxidation and Reduction”. In the previous part, the concept of redox reaction was introduced to all students. By now, all of you should be comfortable in identifying the “two reactions” (reduction and oxidation) that make up a redox reaction. In this part, the intention is to convey the concept of “oxidation number“. Ample of examples on it and sample calculations are also given. However, do take note also that some metals might break convention and show different oxidation numbers (as shown in the tables towards the end).

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction(Part 2)

Oxidation Number – is the charge that the atom of the element would have if complete transfer of electron takes place.

Oxidation number

(i) Free elements have an oxidation number of zero.

Na = 0
Mg = 0
C = 0
H₂ = 0
Br₂ = 0

(ii) Oxidation number of a simple monoatomic ions is the same as its charge.

Na⁺ ion = +1
Mg²⁺ ion = +2
O^2- ion = -2
Cl^- ion = -1
H⁺ ion = +1

(iii) Sum of the oxidation number for a neutral compound is zero.

CaH₂
(+2) + 2(-1)
= 0
Sum of oxidation number is 0

Al₂O₃
2(+3) + 3(-2)
= 0
Sum of oxidation number is 0

Iodine, Bromine, Chlorine, Nitrogen, Oxygen, Fluorine

—> Electronegativity increase

Cl₂O
2(+1) + (-2)
= 0
Sum of oxidation number is 0.

(Chlorine, bromine and iodine usually have the oxidation number of -1 except when combine with a more electronegative element.)

HClO
(+1) + (+1) + (-2)
= 0
Sum of oxidation number is 0.

(Chlorine, bromine and iodine usually have the oxidation number of -1 except when combine with a more electronegative element.)

(iv) Polyatomic ion, the sum of the oxidation numbers of all the atoms equals the charge on the ion.

SO₄ ^2-
(+6) + 4 (-2)
= +6 + (-8)
= -2
Sum of oxidation number is -2

Cr₂O₇^2-
2(+6) + 7(-2)
= -2
Sum of oxidation number is -2

(v) Calculating the oxidation numbers of elements in compounds or ions.

K₂Cr₂O₇
2 (+1) + 2x + 7 (-2) = 0
x = +6
Oxidation number of chromium in K₂Cr₂O₇ is +6

NO^3-
x + 3(-2) = -1
x = +5
Oxidation number of nitrogen in NO^3- is +5

Hydrogen peroxide, H₂O₂
2(+1) + 2x = 0
x = -1
Oxidation number of oxygen in H₂O₂ is -1 (and not -2)
(Usually oxidation number for combined oxygen usually is -2 except in peroxides)

F₂O
2(-1) + x = 0
x = +2
Oxidation number of oxygen in F₂O is +2 (and not -2)
(Usually oxidation number for combined oxygen usually is -2 except in fluorine compounds)

NaH
(+1) + x = 0
x = -1
Oxidation number of hydrogen in NaH is -1 (and not +1)
(Usually oxidation number for combined hydrogen usually is +1 except in metal hydrides)

AlH₃
(+3) + 3x = 0
x = -1
Oxidation number of hydrogen in AlH₃ is -1 (and not +1)
(Usually oxidation number for combined hydrogen usually is +1 except in metal hydrides)

MgH₂
(+2) + 2x = 0
x = -1
Oxidation number of hydrogen in MgH₂ is -1 (and not +1)
(Usually oxidation number for combined hydrogen usually is +1 except in metal hydrides)

(vi) Some metals show different oxidation numbers.

(vii) Usually non-metals have negative oxidation numbers but Cl, Br & I can have positive or negative oxidation number.

Stay tune for the next installment in the “Oxidation and Reduction” series with focus on the difference between redox reaction and non-redox reactions.

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STPM Chemistry Form 6 Notes – Terminology and Concepts: Liquid and Solid States (Part 5 – Final)

July 15, 2010

Berry Berry Easy would like to present the final part on long-running topic of “Liquid and Solid States” for STPM Form 6 Chemistry. Despite being the final part in the series, this is certainly the most important part in the whole chapter due to the berry important phase diagram, triple point (uniqueness of three phases coexisting), the uniqueness of water (be thankful that water had the properties it had or else we would be living today). The aforementioned subtopics are rather popular in topical tests and major examinations. So make sure you understand them fully. Do also make use of the super summary table at the of the post. (Remember to draw out the phase diagrams)

STPM Chemistry Form 6 – Terminology and Concepts: Liquid and Solid States (Part 5)

Summary: Definition of the states of matter

Phase – refers to a single homogeneous physical state of a heterogeneous system. There are three phases with the same composition solid, liquid and gas.

Triple point – the point of a condition of temperature and pressure at which the solid, liquid and vapour phases exist simultaneously at equilibrium.

Critical point – is the highest temperature and highest pressure at which there is a difference between liquid and vapour states. At either a temperature or a pressure over the critical point, only a single fluid state exists, and there is a smooth transition from a dense, liquid-like fluid to a tenuous, gas-like fluid/or pressure that is required to liquefy a gas at its critical temperature.

Supercooling – metastable condition where a liquid can exist below its freezing point.

Phase Diagrams

In laboratory, experiments are being carried out on two environmental factors which is temperature and pressure (referred to as independent variables).

A) The Phase Diagram of Water
- ice (solid), water (liquid) and water vapour / steam (gas)

• Vapour Pressure Curve
  - critical point = critical temperature (374˚C) and critical pressure (200 atmospheres)
  - temperature above 374˚C and critical pressure 200 atmospheres, the vapour and liquid are indistinguishable (no longer two separate phases) because the densities of the gas and liquid are equal (meniscus separating a liquid from its vapour disappears).

• Melting Temperature Curve
  - melting temperature point decrease with pressure- supercooling is the cooling of a liquid to below its freezing point without a change taking place from the liquid to the solid state. A phenomenon (metastable condition) shows the vapour pressure of water below its freezing point.

• Triple Point
  - Water triple point is at temperature 0.01˚C and pressure 0.006 atm (610 N m^-2). All the three phases (ice, water and water vapour) coexist at equilibrium.

• Normal Melting Temperature Point
  - the temperature at which both the solid and the liquid states of the substance exist in equilibrium at a pressure of 1 atm (101 kNm^-2)

• Normal Freezing Temperature Point
  - the temperature at which both the liquid and the solid states of the substance exist in equilibrium at a pressure of 1 atm (101 kNm^-2)

Unsual Behaviour of Water

• i) Why ice can float?
  - the volume of water increase when the change of phase from liquid to solid.
  Reasons: Ice (solid) has an open structure (hydrogen bond).

• ii) Why the melting temperature curve slopes to the left (melting point decreases with pressure)?
  - (In most of substances (except water), an increase in pressure will push the molecules even closer / Increase in pressure favours the physical state which is higher density)
  Reasons: Increasing the pressure favours the formation of liquid water due to the latent heat of fusion is absorbed from the surroundings during melting.

B) The Phase Diagram of Carbon Dioxide
- solid carbon dioxide (dry ice), liquid carbon dioxide and gas carbon dioxide

• Vapour Pressure Curve
  critical point = critical temperature (374˚C) and critical pressure (217 atmospheres)
  - temperature above 374˚C and critical pressure 217 atmospheres, the vapour and liquid are indistinguishable because the densities of the gas and liquid are equal. At this point, carbon dioxide gas can be liquefied.

• Melting Temperature Curve
  - melting temperature point increase with pressure
  - melting temperature curve slopes to the right
  - density of dry ice (solid carbon dioxide) is higher than the density of liquid carbon dioxide. It is because the carbon dioxide molecules are held closer together (smaller volume).
  - Increasing the pressure favours the formation of solid carbon dioxide due to the latent heat of fusion is liberated (given out) to the surroundings.

• Triple Point
  - Carbon dioxide triple point is at temperature -57˚C and pressure 5.1 atm. All the three phases (solid, liquid and gas) coexist at equilibrium.

• Normal Sublime Temperature Point
  - the temperature at which both the solid and the gas states of the substance exist in equilibrium at a pressure of 1 atm (101 kNm^-2)
  - at atmospheric pressure, solid carbon dioxide (dry ice) sublimes to form carbon dioxide gas at -78˚C.

Berry Important Points of Water Phase Diagram and Carbon Dioxide Phase Diagram

Now we have reached the conclusion of this topic. Be sure to understand this topic in full if you want to score in chemistry. Do also learn the Berry Important Points table of the water and carbon dioxide phase diagram by heart.

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SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 1)

July 9, 2010

Berry Berry Easy would like to present all Berry Readers with one of the most interesting topic (as voted by most students) in SPM Form 5 Chemistry. Without prolonging the suspense, the interesting topic as mentioned by most chemistry takers is “Oxidation and Reduction“. So why is this interesting? Based on a straw poll, it is simply because of the coolness of the word Redox, which is a shorthand for REDuction/OXidation = REDOX!!! Teachers only need to mention this once, and students can remember this virtually forever. Its concept are also rather simple, but at the hardest end (not covered by the syllabus), it can be hard to identify a redox reaction.

One obvious tip to understand redox reactions is to think of them as components of a bigger reaction. So a redox reactions is best imagined as two smallers parts of oxidation and reduction. (If you would like to impress your friends, you can also tell them that, non-redox reactions are called methathesis reactions.)

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction (Part 1)

Redox reaction – chemical reactions in which both oxidation and reduction occur simultaneously.

1) Oxidation

• gain of oxygen, O₂ by a substance
• loss of hydrogen, H₂ from a substance
• a loss of electrons
• occurs when there is an increase in oxidation number

2) Reduction

• loss of oxygen, O₂ by a substance
• gain of hydrogen, H₂ from a substance
• a gain of electrons
• occurs when there is an decrease in oxidation number

Oxidation Number – is the charge that the atom of the element would have if complete transfer of electron takes place.

IUPAC nomenclature – name inorganic compounds in order to avoid confusion due to elements have variable oxidation numbers.

Oxidation and Reduction in Terms of Gain and Loss of Oxygen

2CuO (s) + C (s) –> 2Cu (s) + CO₂ (g)

• Reduction:
  CuO loses its oxygen to form copper. CuO (oxidising agent) is being reduced.
• Oxidation:
  Carbon gains oxygen to form CO₂. Carbon (reducing agent) is said to be oxidised.

PbO (s) + CO (g) –> Pb (s) + CO₂ (g)

• Reduction:
  PbO loses its oxygen to form lead. PbO (oxidising agent) is being reduced.
• Oxidation:
  Carbon monoxide (CO) gains oxygen to form CO₂. Carbon monoxide (reducing agent) is said to be oxidised.

H₂ (g) + CuO (s) –> H₂O (l) + Cu (s)

• Reduction:
  CuO loses its oxygen to form copper. CuO (oxidising agent) is being reduced.
• Oxidation:
  Hydrogen (H₂) gains oxygen to form H₂O. Hydrogen (reducing agent) is said to be oxidised.

Oxidation and Reduction in Terms of Gain and Loss of Hydrogen

H₂S (g) + Cl₂ (g) –> S (s) + 2HCl (g)

• Reduction:
  Cl₂ gains hydrogen to form hydrogen chloride. Cl₂ (oxidising agent) is being reduced.
• Oxidation:
  Hydrogen sulphide loses hydrogen to form sulphur. Hydrogen sulphide (reducing agent) is said to be oxidised.

2NH₃ (g) + 3Br₂ (g) –> N₂ (g) + 6HBr (g)

• Reduction:
  Bromine gains hydrogen to form hydrogen bromide. Br₂ (oxidising agent) is being reduced.
• Oxidation:
  Ammonia loses hydrogen to form nitrogen. Ammonia (reducing agent) is said to be oxidised.

Do drop by Berry Berry Easy soon again for the continuation of this interesting chapter.

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STPM Chemistry Form 6 Notes – Terminology and Concepts: Liquid and Solid States (Part 4)

July 7, 2010

Berry Berry Easy would like to continue on the interesting topic of “Liquid and Solid States” Part 4 for STPM Form 6 Chemistry. In this part, we’ll be focusing on allotropes which are the different forms which an element exist. (The key point to bring home is “element”, which means that they are homogeneous) The concept of this is easy to grasp but students tend to confuse it with isotopes for unknown reasons. So make sure you know the differences. In short, allotropes of an element varies in the structure of the sane atom, while isotopes of an element varies on the number of neutrons of the atom. (Think of allotropes as how atoms are arranged, while isotopes are how the composition of subatomic particles in an atom varies).

STPM Form 6 – Terminology and Concepts: Liquid and Solid States (Part 4)

The concept of allotropy was proposed in 1841 by the Swedish scientist Baron Jöns Jakob Berzelius (1779-1848). The term is derived from the Greek άλλοτροπἱα (allotropia; variation, changeableness). By 1912, Ostwald proposed that the terms allotrope and allotropy be abandoned and replaced by polymorph and polymorphism but IUPAC and most chemistry texts still favour the usage of allotrope and allotropy for elements only.

Allotropy – existence of elements in two or more different forms (allotropes).

Elements with variable of coordination number or oxidation states tend to exhibit greater numbers of allotropic forms and typically more noticeable in non-metal (excluding the halogens and the noble gases) and metalloids.

Example:

i) Different molecular configuration

Oxygen – O₂ dioxygen (colourless), O₃ trioxygen / ozone (blue), O₄ tetraoxygen, O₈ octaoxygen (red)

ii) Different crystal structures in the solid

Group 14, Group 15, Group 16 of the periodic table

• Group 14: Carbon – graphite, amorphous carbon (soot/coal), diamond, fullerenes C₆₀ (buckyball), Ionsdaleite / hexagonal diamond (meteorites containing graphite strike to the Earth) and carbon nanotubes (buckytubes) is carbon with a cylindrical nanostructure.
• Group 15: Phosphorus – red phosphorus (polymeric solid), white phosphorus (crystalline solid P₄), scarlet phosphorus, violet phosphorus, black phosphorus (semiconductor) and diphosphorus P₂.
• Group 16: Sulphur – rhombic sulphur (large crystals composed of S₈ molecules), monoclinic sulphur (fine needle-like crystals), plastic (amorphous) sulphur (polymeric solid) and other ring molecules S₇ and S₁₂.

Enantiotropy – the allotropes are stable over a temperature range, with a definite transition point at which one changes into the other.

Example:

i) Tin has three allotropes / two enantiotropy:

• alpha tin is white (metallic) tin stable above 13.2 ˚C.
• beta tin is grey (nonmethallic) tin below 13.2 ˚C.
• gamma tin is rhombic tin.

ii) Iron has four allotropes / four enantiotropy:

• ferrite (alpha iron) stable below 770°C (BCC) and the iron becomes magnetic.
• beta iron stable below 912°C (BCC).
• gamma irons stable below 1394°C (FCC) crystal structure.
• delta irons stable from cooling down molten iron below 1538°C and has a (BCC) crystal structure.
• *BCC – Body-centred cubic
• *FCC – Face-centred cubic

Allotropes of Carbon

i) Graphite – used as lubricant (powder or oily suspension)

• layered lattice structure
• hexoganal for the crystal system
• density is 2.25 g cm^-3
• each carbon atom is bonded by strong covalent bonds (sp² hybridisation / trigonal planar) with three other carbon atoms to formed hexagonal ring.
• the layer are held together by weak van der Waals forces.
• graphite is soft and slippery due to weak van der Waals forces allow the layer to slide over one another.
• graphite is a moderate conductor of electricity along its layer (in the direction parallel but not perpendicular to the laver) due to a free electron (per carbon atom) which can move throughout the solid lattice. (Each carbon atom has one outer shell electron (unhybridised p electron) which is not used to form covalent bonds.)

ii) Diamond – used as abrasives (high velocity cutting tools) and ornaments (high refractive index)

• crystallises in a face-centred cubic structure.
• single giant molecule.
• density is 3.50 g cm^-3
• each carbon atom is bonded by strong covalent bonds (sp³ hybridisation / tetrahedral) with four other carbon atoms to formed three-dimensional giant structure.
• diamond has great hardness and high melting point due to the strong covalent bonds in the 3-D structure.
• diamond is a non-conductor of electricity due to all the four valence electrons of the carbon atoms are involved  with covalent bonding, therefore no free/delocalised electrons.

iii) Fullerene / Buckyball / Buckminsterfullerene – used as lubricant, semi-conductor, superconductors and catalyst

• Molecular formulae of fullerene are C₂₀ (smallest member), C₃₂, C₆₀ (most common member), C₇₀, C₇₆, C₇₈, C₈₄ and C₉₀.
• spherical molecules of 20 – 90 carbon atoms (32 sides, 12 pentagons and 20 hexagons).
• simple molecular solid.
• each carbon atom is bonded by strong covalent bonds (sp² hybridisation / trigonal planar) with three other carbon atoms. It also contains delocalised π electrons which does not exhibit “superaromaticity” that the electrons in the hexagonal rings do not delocalise over the whole molecule.
• Fullerene is a superconductor when it mixed with other metals.

Allotropes of Sulphur (different molecular arrangement)

i) alpha sulphur / rhombic sulphur (large crystals composed of S₈ molecules)

• lemon yellow colour
• shape of an octahedron.
• crystallises with the orthorhombic lattice.
• more stable at room temperature (formed in temperature below 95.6˚C).
• melting point at 113˚C.
• density is 2.07 g cm³.

ii) beta sulphur / monoclinic sulphur (fine needle-like crystals of S₈ molecules)

• deeper yellow colour
• shape of long, narrow and thin needle.
• crystallises with the monoclinic lattice.
• stable at temperature above 95.6˚C.
• melting point at 119˚C.
• density is 1.94 g cm³.

So by now you should know about allotrophy. So your diamonds are just allotropes of carbon.

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SPM Biology Form 4 Notes – Terminology and Concepts: Movement of Substances Across the Plasma Membrane (Part III – Final)

July 2, 2010

Ever wondered why you need to drink isotonic drinks (100 Plus as an example) after doing sports? Okay, mainly because it tastes good as any sweetish cold drink after much exercise is appreciated. However, isotonic drinks does more than that as it replenish fluids after you lose some fluid through sports. Well, Berry Berry Teacher thinks that most of our knowledgeable Berry Readers knows about the isotonic part, but how about hypotonic and hypertonic? Hypertonic do not make you any more hyper, while hypotonic do not have anything to do with hippopotamus. So let us ride through the journey of substances across plasma membrane with Part 3 (final part) of this series.

SPM Form 4 – Terminology and Concepts: Movement of Substances Across the Plasma Membrane (Part 3 – Final)

Type of Solution

1. Hypotonic
2. Isotonic
3. Hypertonic

1) Hypotonic

• Solute concentration in the external solution is lesser than solute concentration inside the cell.
• Water concentration outside the cell is higher than the water concentration inside the cell.

2) Isotonic

• Solute concentration in the external solution is equal to the solute concentration inside the cell.
• Water concentration inside and outside of the cell is the same.

3) Hypertonic

• Solute concentration in the external solution is greater than solute concentration inside the cell.
• Water concentration outside the cell is lower than the water concentration inside the cell.

Types of solutions:

Application

1. Food is soaked in a concentrated salt solution to prevent bacteria and fungus to survive.
2. Chemical fertiliser (dissolved ions) increases solute concentration (decrease water molecules) in soil. Therefore, water leaves from the cell sap of the plant which result the plant wither.

Finally, the end of the interesting substance movement over plasma membrane. Do keep note that this might be a popular essay question. Easy to answer but hard to score, so take note of the terminologies and concepts shown above.

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STPM Chemistry Form 6 Notes – Terminology and Concepts: Liquid and Solid States (Part 3)

June 23, 2010

This is Part 3 of the series of notes on “Liquid and Solid State” from Berry Berry Easy. In the previous part, the crystal systems were discussed in tabulated form. With that knowledge in hand, it is time to look at the packing of it. It is absolutely important that you draw the following structure at least once and try to have a 3D mental image of it. This action of visualising will help you to understand the information more easily. And when you understand, you don’t have to memorise it anymore, it will come natural to you. (Nonetheless, at least for the beginning, try to memorise them. Subsequent topics will require understanding from this topic, so do not skip this topic)

STPM Form 6 – Terminology and Concepts: Liquid and Solid States (Part 3)

Four types of lattice points:

1. Lattice point at the corner of the unit cell (1/8)
2. Lattice point on the edge of the unit cell (1/4)
3. Lattice point on the face of the unit cell (1/2)
4. Lattice point in the centre of the unit cell (1)

Coordination number – the number of atoms, molecules or ions (called the nearest neighbours) that surrounds a given atom, molecule or ion in a crystal lattice.

A) Simple cubic cell

Example: Caesium chloride & Polonium

• Sphere touches six other spheres.
• Four sphere in its own layer, one sphere above the layer and one sphere below the layer.
• Coordination number = 6
• Unit cell contains in total one atom (8 corners x 1/8 = 1)

B) Body-centre cubic lattice

Example: Sodium, Barium, Potassium, Iron, Manganese, Chromium & Vanadium

• Sphere touches eight other spheres.
• Second layer are placed in the hollows between the spheres in the first layer.
• Each sphere atom is in contact with four atoms in the layer above and four atoms in the layer below.
• Coordination number = 8
• Unit cell contains in total of two lattice points per unit cell (8 corners x 1/8 + 1 = 2)

C) Close-packed structures

Example: Sodium chloride

• Unit cell contains in total of four atoms per unit cell (8 corners x 1/8) + (6 faces x 1/2)

i) Cubic close packing (ABCABCABC) / Face-centered cubic / Simple cubic close packing

Thomas Harriot (1585) first pondered the mathematics of the cannonball arrangement or cannonball stack, which has a face-centered cubic lattice.

• Sphere touches twelve other spheres.
• First layer of spheres is packed as closely and each sphere atom is in contact with six other atoms.
• Second layer of spheres is placed on top of the first layer, so that each sphere in the second layer rests on the hollows between the spheres in the first layer.
• Each sphere atom is in contact with six atoms in its own layer, three spheres (atoms) in the layer above and three spheres (atoms) in the layer below.
• Coordination number = 12

ii) Hexagonal close packing (ABABABABA)

• Sphere touches twelve other spheres.
• First layer and the second layer of spheres are packed in the same way as cubic close packing.
• (Difference = the third layer of spheres is placed on top of the first layer)
• Coordination number = 12

Stay tune to Berry Berry Easy for the next part (Part 4) in this series covering allotropy and enantiotropy.

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