Question 1
There are many useful benefits that we can gain from the development of MRI imaging technology. An argument for continuing the research of atomic structure is basically due to the greater understanding of the universe. As we study the interaction of very small particles, we can gain a better understanding to how they work together to ultimate create a world we live in. The data which we have on atom and electrons interact is based on observation alone, and the reason is unknown. When we involve in a larger system we cannot know the prediction accurately. For example, the properties of hydrogen are distinguished between the different muscles and tissues within the human body because a human body consists of 63% of hydrogen atoms and we learn all this data through the understanding of atomic particles. If we continue this research we can discover many ideas such new elements, their properties, amount of energy and how we use that to our advantage.
Atomic –structure research is very costly and it is always risky that the money that can be put into the research may not be worth it since there is a chance that it will not result in new discoveries. A study shoes that well off Ontarians were 38% more likely to have MRIs done than the remaining percentage that are less fortunate. If a lot of money was to be invested in more advanced technologies, at the end of the day, not a great percentage of the overall population would have access to it since only a small percentage of the population is well off, therefore only the rich would benefit.
Question 2
The visible line spectrum is that portion of the spectrum which is visible by human eye and is produced when electron jumping from higher energy level to the lower energy level orbit thus releasing energy and the electromagnetic waves produces longer wavelengths, this mean that these lines are lower in energy. The ultraviolet light is invisible to human eye but visible to several insects and birds and it indicates that electron is jumping from highest energy level orbit to lowest energy level orbit thus releasing shorter wavelengths and more energy as compared to the visible portion of the spectrum.
Question 3
a)
Zr = 1s2, 2s2, 2 p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d2
b) The Zr4+ ion exists because the electrons from 4d and 5s orbitals are lost to form a more stable electron configuration. The Zr electron loses 4 electrons to become more stable and less reactive. It will have the configuration like Krypton the nearest Nobel gas.
c) The Zr4+ ion is more stable in comparison to the Zr atom because the valence shell is completely full which makes it less likely to react. The Zr atom does not have a filled valence shell, making it less stable and more reactive.
Question 4
1s
H
1s 1s
N 2s 2p
H
H
Question 5
The Lewis structure of sulphite molecule:
Total number of valence electrons
= 6e + 3*6e +2e
= 26e
The Lewis structure for the sulphate ion has a formal charge of -1 on all the oxygen atom and +1 charge on sulfur atom.
The S is the central atom because there is only one Sulphur atom and three oxygen atoms.
Question 6
a)
• After the hybridization, 2s and 2p orbitals are mixed to produce 4 sp3 hybrid orbitals. And one of the hybrids orbital is occupied by a lone pair of electrons.
b) The type of hybrid orbitals found in NH3 is sp3 hybrid orbitals. Three of the hybrid orbitals take part in bonding, where one of the hybrids is occupied by a lone pair of electrons.
c) In energy level diagram of C atom, it can have two single electrons in its 2p orbitals but it is very unstable mainly due to the empty 2p orbital. But when orbital hybridization of the C atom it can promote its electron to the empty 2p orbital. Now the 2s and 2p orbitals of carbon mix together to form four new identical orbitals called sp3 hybrid orbital. C atom now make 4 bonds like it can make bond with 4 hydrogen atoms to complete its valence orbit to make stable.
In the energy level diagram of N atom, it can have three single electrons in its 2p orbitals and 2 electrons in its 2s orbital. After the orbital hybridization of N atom, it has four hybrid orbitals called sp3 hybrid orbital atom can make 3 bonds like it can make bond with 3 hydron atom to complete its valence orbit to make stable.
Question 7
a) i) CaBr2 ΔEN= 2.9-1.0=1.9
ii)Na3N ΔEN= 3-0.9=2.1
iii)CH4 ΔEN= 2.5-2.2=0.3
b) CaBr2 is considered as ionic compound since ΔEN > 1.7
Na3N is considered as ionic compound since ΔEN > 1.7
CH3 is considered as molecular compound since ΔEN < 1.7
c) The bonds in Na3N would have the greatest ionic character of the three mentioned here, if Na3N existed, which is unlikely due to its instability. Next will be the bonds in CaBr2, and the bonds in CH4 will be the most covalent. So, the ionic character in decreasing order are:
1. Na3N (most ionic character)
2.CaBr2 (iconic character
3.CH4 (least ionic character)
Question 8
a)
• Since iodine is surrounded by one lone pair and three bond pairs with oxygen, this ion is a trigonal pyramidal. It makes an angle of 107 degree.
b) The shape is pyramidal because the central iodine atom is surrounded by 3 bond pairs and 1 lone pair.
c) It’s a polar molecule because its bond dipole didn’t cancel out each other. The iodine lone pair repelled the bonding oxygen electron pair downwards which gives the ion a pyramidal shape. The iodine end of the molecule is slightly more positive and the oxygen end of the molecule is slightly more negative.
Question 9
a)
b) Carbon with four groups bonded to it forms a tetrahedral shape. One with hydrogen and the three with chlorine.
c) The bond dipole didn’t cancel each other out therefore it’s a polar molecule. It is more negative on the chlorine side and more positive on the hydrogen side.
Question 10
In methane, each hybrid orbitals of carbon are bonded to hydrogen and bonds are arranged in a tetrahedral geometry, which result in 109.5-degree bond angle. In water molecules, they also have 4 sp3 hybrid orbitals, two of the orbitals are occupied by lone pairs and two are bonded to hydrogen atoms. The lone pair electrons have greater repulsion forces to repel the two bonding hydrogen electron pairs downwards, which distort the normal tetrahedral angle, and result in a bent shape molecule. The sp3 hybrids make a 109.5-degree angle. The angle is decreased because of extra repulsion from the two lone pairs making the bond angle become 104.5 degrees due to the distortion caused by lone pair electrons repulsion. This results in these molecules not being identical.
Question 11
a) Carbon is very good solvent because it has the property of high catenation. Use liquid carbon dioxide (CO2) dry cleaning as a nontoxic professional dry cleaning alternative. CO2 cleaners use the same process as standard dry cleaning except that liquid carbon dioxide is used as the solvent, which eliminates the need for toxic cleaning chemicals. Liquid CO2 can be used as an environmentally friendly replacement for perchloroethylene (PCE), which is one of the most common chemical dry cleaning solvents used by dry cleaners. It is a non-polar solvent used by dry cleaners can lift oil and grease off fabrics and the non-polar nature of the carbon dioxide molecules makes it an ideal cleaning solvent.
b) Pros
• The Carbon dioxide dry cleaning machines have low operating costs and high productivity and the liquid carbon dioxide solution used in its machines can be recycled and reused.
• Clothing cleaned by carbon dioxide won’t change shape, shrink or stretch
Cons:
• Carbon dioxide dry cleaning is less efficient in removing polar stains compare to other solvents.
• Investment in the technology can be a lot costlier than the traditional dry cleaning machines
Question 12
a) As you progress down halogen group intermolecular forces increase because of increasing size of molecule hence increase in observed boiling point. All the halogens exist as diatomic molecules: F2, Cl2, and so on. The intermolecular attractions between one molecule and its neighbors are van der Waals dispersion forces. As the molecule get bigger there are obviously more electrons which can move around and set up the temporary dipoles which create these attractions. The boiling point increases with increasing molar mass so for this specific problem that’s all the information needed, thus Fluorine<Chlorine<Bromine<Iodine in molar mass and intermolecular forces. Halogens only exhibit London forces; they do not exhibit hydrogen bonding and dipole-dipole attraction. You can see that both melting points and boiling points rise as you go down the Group. The stronger intermolecular attractions as the molecules get bigger means that you must supply more heat energy to turn them into either a liquid or a gas so there melting and boiling point increase. If you explore the graphs, you will find that fluorine and chlorine are gases at room temperature, bromine is a liquid and iodine a solid. Nothing very surprising there.
Figure 1 Boiling Points vs Number of Electrons
b)
• The intermolecular attraction, London dispersion forces, increases with the increasing number of electrons. The Halogens exists as a discrete, non-polar, diatomic molecule. The strength of the London dispersion forces determines whether the halogen will exist as a gas, liquid or solid at room temperature. F2 and Cl2 are gases, Br2 is a liquid and 12 is a solid. The strength of the London dispersion forces is proportional to the polarizability of the molecules, which in turn depends on the number of electrons, and the volume of which they are spread.
c) By predicting the graph the boiling point of astatine will be 335-degree C
(It could be 30 degree higher or lower from my estimation)
Question 13
Boiling points are based on the strength of intermolecular forces. Ethane is a non-polar molecule, therefore it doesn’t have dipole-dipole force but a london force. The strength of the London force in ethane is almost the same as the London force in CH3 OH since they have the same number of electrons and similar molar mass. CH3 OH is a polar molecule and the intermolecular forces present are London force, dipole-dipole forces and Hydrogen bonding, which are much stronger than the intermolecular forces present in ethane which gives it a much higher boiling point than ethane
Question 14
a) Two physical properties that the elevator should have are: the cable should have are great strength and very light in molecular weight since it will come into contact with extreme temperatures and complications. The cable material must be chemically stable so that it doesn’t react with any other elements in space, making it able to adapt to the environment and not break.
b) The elevator cable needs to be 23000 miles long so the material needs a strength to weight ratio over a thousand times better than steel. Glass fiber, Polymer, carbon fiber and aramid fiber are all MUCH better than steel but not good enough.
I would suggest the use of carbon nanotubes or diamond fiber for this purpose because, when linked in a nanotube, a pure carbon structure will be nonreactive up to thousands of degrees due to its extremely powerful covalent bonds with the surrounding carbon atoms. It is several times stronger than steel and have no edges. These bonds also give it an enormous tensile strength capable of supporting the full weight of the cable if properly woven.