Lesson 1:1
- A reaction where Br (or, Cl, F, H, O, S) is added to an element or a compound.
- Yes. Back
Lesson 2:1
Material with high specific heat makes good coolant in heat exchangers, coupled with a large liquid range it should be a good liquid to remove heat for systems with temperature of 200�C to 1200�C. Back
Lesson 2:2
- Fluorine, chlorine, bromine, and iodine are collectively known as halogens. Their anions F‾, Cl‾, Br‾, and I‾ are then known as halides.
- Although barium sulphate is said to be insoluble, what it meant was its solubility is very very low. Still there will be a little barium sulphate dissolved in water.
- LiO (from reaction with oxygen in the air) amd LiOH (from reaction with water in the air). Back
Lesson 4:1
- Be(H2O)4Cl2
- Be(H2O)4Cl2
- Be(OH)2 + 2 HCl + 2 H2O � Be(H2O)4Cl2
- Be(OH)2 + 2 Na(OH) � Na2[BeO2] + 2 H20 Back
Lesson 5:1
Be do not form a cation ion, so it is not possible to determine the charge-to-radius ratio. Back
Lesson 6:1
No, B(OH) is acidic. Back
Lesson 6:2
The chlorine atoms can be attached to either B or N. In this case they were attached to B. Back
Lesson 8:1
RADIOCARBON DATING: 12C, the commonest type of carbon, has a mass of twelve. 13Carbon has six protons and seven neutrons, and 14Carbon has six protons and eight neutrons. Carbon 14 is produced by high-energy electromagnetic radiation in the upper atmosphere of the earth and is radioactive (that is unstable).
Carbon 14 can become part of the carbon dioxide in our atmosphere and eventually made its way into plants and animals. It stayed in plants (as the carbohydrates of the wood), and animals (the calcium carbonate in bones and skeletons). Once the organisms died they can no longer take up carbon 14. So the amount of carbon 14 in the remains is that left behind from their decay into the more stable carbon 12.
If science can know what is the proportion of carbon 14 during the formation of the wood and skeleton, then it would be possible to determine how long the wood or skeleton has been around by analysing the proportion of carbon 14 in the remains. The rate of decay of carbon 14 has been determined as a reduction of half the amount after every 5570 years. This technique of finding out how old the remains are is known as radiocarbon dating, or carbon dating.
General the proportion of carbon 14 that existed in ancient time was assumed to be that of the amount present in our present atmosphere. That means the proportion in our plants and our skeleton. This was based on the consideration that the only source of carbon 14 is from the upper atmosphere undisturbed by human activities, and the rates of production and decay of carbon 14 are equal.
It must be noted that nuclear explosion can also be a source of carbon 14. Back
Lesson 8:2
- C : [He]: 4 (sp�) + H : [He]: 2s + 3 Cl : [Ne]: 3s�, 3p5 �
1 C-H σ-bond + 3 C-Cl σ-bond
The molecule is tetrahedral in shape.
The C-Cl bonds are all equal in length and slightly longer, the C-H is shorter and stronger.- C : [He]: 3 (sp�), 2p + 2 O‾ : [He]: 2s�,2p5 + O : [He]: 2s�,2p4 �
3 C-O σ-bond + 2 C-O π-bond
The molecule is planar in shape.
Owing to resonance of the electrons in the C-O bonds, all C-O bonds have the same length and strength. Back
Lesson 8:3
6 C : [He]: 3 (sp�), 2p � 6 C-C σ-bond forming a hexagonal planar ring
The remaining p-orbitals are perpendicular to the plane. They can overlap alternating with the one next. Consequently the electrons can move throughout the plane. So graphite is an electrical conductor. Back
Lesson 8:4
Carbon dioxide dissolves in water to give carbonic acid which can then react with metallic cations present in the sea water to precipitate out as carbonates. Or use by sea creatures to form corals or other carbonate compounds. Back
Lesson 8:5
CO form bonds with transition metals: O≡C→M. Oxygen holds onto its electron more firmly than carbon. Back
Lesson 9:1
Stannane and Plumbane. Back
Lesson 10.1
Using the molecular orbital theory of bonding, the molecular electron configuration for nitrogen is; N2 : [(σ)�(σ*)�]; (σ1)�(σ1*)�(σ2)�(π)4
So the bond order is 3. This is a very strong bond. Back
Lesson 10.2
pKb = - log10[ a(NH4+) x a(OH‾) / a(NH3) ]. Back
Lesson 10.3
Sodium sulphate is soluble whereas barium sulphate will precipitate from the final product. The nitrous acid can be easily separated by filtration. Back
Lesson 12.1
white orbital has no electronThe oxygen underwent a sp� hybridisation. The remaining p-orbitals are all perpendicular to the plane. Their electrons can then move throughout the entire length of the molecule. Such resonance gives ozone the needed stability to exist. Of course it is not as stable as O2. The lone electron pair in the centre oxygen exert a greater electron repulsion on the σ-bond as compared with that between σ-bond-σ-bond repulsion. So the O-O-O bond is less than 117� while the -O-O angle is more than 121�. When the centre oxygen has only one electron in its p-orbital it can form a π-bond with the other oxygen atom with also one electron in its p-orbital. So the O-O bonds are of equal length and strength, and is between that of a σ-bond and a π-bond. Back
Lesson 12.2
Sulphuric acid can hold on strongly to water molecules, thus preventing them reacting to reform the acetic acid. Back
Lesson 15.1
As the halide ions increases in size, its surface charge density decreases (every ion has only one negative charge). So the ionic bond with a particular metallic cation becomes weaker. Back.
Lesson 15.2
- Chlorine oxide as oxygen is more electronegative than chlorine.
For F2O it will be oxygen fluoride.- The barium sulphate will precipitate out making it easy to isolate the required product. Back.
Lesson 15.3
First identify the elements W, X, Y, Z. W is sulphur, X is Ar, Y is potassium, and Z is calcium. Oxides on the right hand side of the Perioic Table are basic while those on the right handside are acidic. Inert elements do not form oxides. So the answer is W. Back.
