Chemistry F.A.Q.
Links/ References
You were asked to deduce a number of Lewis structures in class and then the hybridization of the central atom, several of which we did not have time to discuss. For that reason I have reproduced them here.
All the anions are resonance structures with the negative charge spread between all the atoms in the structure. This is done through overlapping p-orbitals which become available if each atom is sp2 hybridized. It is therefore this low energy configuration which dominates and allows electrons to 'spread' through the pi-system.
All images are off the Web. The format of the nitrite ion is probably the best format to use, with the overall charge on the outside of the square brackets. For those who like dot and cross structures I'm preparing some examples of these too.
The carbonate anion
Lewis structures:
Three charge centres indicate that the central atom is sp2 hybridized. If the oxygen atoms adopt a similar hybridization the two negative charges can be easily delocalised across the four atoms by pi-bonding. (Incidentally, because the double bonds are partial this ion is said to have a 'C-O bond order of 4/3')
Nitrate (NO3-)
Lewis Structures (there are three equivalent structures)
N.B. below is an example of how NOT to write these Lewis structures (ALL non-bonding electrons must be shown):
Nitrite (NO2-)
Lewis Structures (there are two equivalent structures)
Ozone (O3)
Ozone is an interesting structure because it has been shown to be bent (the molecule has a dipole and a measured bond angle of 116° ) Furthermore, both O-O bonds have been shown to be identical in length. This strongly suggests that the valence shell of the central atom is sp2 hybridized and that 4 electrons are delocalised across the three atoms using p-orbitals. p-orbitals are only available if the three oxygens are sp2 hybridised. If the central atom were sp hybridised the molecule would be linear like CO2. The question remains, why isn't it linear or why isn't CO2 bent? These questions are beyond the scope of IB chemistry but most certainly have answers in terms of molecular orbital energies.
In IB just remember that if you have several possible resonance structures then the atoms with three charge centres are going to be sp2 hybridized!
1.http://en.wikipedia.org/wiki/Ozone#Structure
2..http://www.newton.dep.anl.gov/askasci/chem03/chem03464.htm
VSEPR Theory: Who needs to know what? The IB says:
SL need to:
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Deduce the Lewis (electron dot) structures of molecules and ions for up to four electron pairs on each atom. |
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A pair of electrons can be represented by dots, crosses, a combination of dots and crosses or by a line. Note: Cl–Cl is not a Lewis structure. |
SL Need to:
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Predict the shape and bond angles for species with four, three and two negative charge centres on the central atom using the valence shell electron pair repulsion theory (VSEPR). |
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Examples should include CH4, NH3, H2O, NH4+, H3O+, BF3, C2H4, SO2, C2H2 and CO2. |
Predict whether or not a molecule is polar from its molecular shape and bond polarities.
In additon HL need to:
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Predict the shape and bond angles for species with five and six negative charge centres using the VSEPR theory. |
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Examples should include PCl5, SF6, XeF4 and PF6–. |
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Explain hybridization in terms of the mixing of atomic orbitals to form new orbitals for bonding. (We will study hybridization more when we do organic chemistry) |
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Students should consider sp, sp2 and sp3 hybridization, and the shapes and orientation of these orbitals. |
Electron configurations of Transition Metal (TM) ions - clarified?
N.B. In I.B. you are supposed to say that Scandium in not considered a transition element because it doesn't have any compounds in which it has a partially filled d-subshell.
BUT Sc2+ should be written as [Ar] 3d1 from the other rule you have learned (4s empties before 3d to give TM ions)!
99% of scandium compounds are Sc3+, so if they ask you the configuration of Sc2+ they are being mean!
Further reading shows that the existence of Sc2+ hydrides is highly controversial.*
At least one site says a Sc2+ compound exists and boldly says it has a 3d1 configuration.
However, the scandium could also be bonded to another scandium in addition to two hydrogens, in which case it would be trivalent.
*Greenwood and Earnshaw "Chemistry of the Elements" First Edition p.1107
Mass Spectrometer (a small, but important point)
What really happens at the accelerator plates?The ions are accelerated until they all have the same kinetic energy (moving energy=1/2 mv2). Don't forget: a high mass ion will have the same moving energy as a low mass ion, but since it has more mass it will be moving more slowly. After this point the magnet DOES NOT separate the ions based on their speed, but on their charge/mass ratio (as mentioned in class).
At the detector only certain ions get through, the rest are brought 'into focus' by changing the magnetic field.
The detector measures how much charge is needed to neutralize those arriving, which gives a measure of abundance.
1. The Mass Spectrometer
Significant Figures (S.F.)
We must not confuse significant figures and uncertainties!1. In a given number, the significant figures (figures presented) , are those digits that are certain and the first uncertain digit.
2. It is confusing to the reader to see data or values reported without the uncertainty reported with that value.
So, on a two figure balance with a last number which fluctuates by +/- 0.05 we should write masses as:
10.01 +/- 0.05g students are also required to express the uncertainty as a percentage = +/- 0.5%
When we add or subtract numbers it is the number with the fewest decimals that determines the figures in the final result:
10.005 + 0.2 = 10.2
5.789 + 105 = 111
BUT,
in multiplication and division, the number with the least number of significant figures determines the number of significant figures in the result:
10.005 x 0.2 = 2 (1 S.F.)
1.4589 x 1.2 = 1.8 (2S.F.)
If several different operations are involved we apply the last rule AT THE END OF THE CALCULATION
1. Significant figures and uncertainty
2. Another easier? explanation
3. Practice significant figures
Orbitals and the problems with the Bohr theory (HL)
Although the Bohr model of the hydrogen atom is very successful for hydrogen it doesn't explain- the fact that when certain spectral lines are examined closely they are not solid lines of emitted light but made up of a number of hyperfine spectral lines.
- the behaviour of atoms with more than one electron.
Orbitals, as we shall see later help to explain a number of facts about atoms which the more primitive Bohr model cannot.
There is an interesting online simulation of how the different atomic models cope with the absorption and emission of light packets (quanta or photons): http://phet.colorado.edu./new/index.php
De Broglie
Schrodinger
How to calculate the pH of a solution of the sodium salt of a weak acid:
e.g. What is the pH of a 0.100 M solution of sodium acetate (NaAc)?i) Find Kb
ii) Calculate [OH-]
iii) Calculate pH
Ka x Kb = Kw
Ka we know from our databook = 1.74 x 10-5 mol.dm-3 (from pKa = 4.76 for ethanoic acid)
Kw = 1 x 10-14 mol2 dm-6
So, Kb = 5.75 x 10¯10
Here is the chemical reaction for the hydrolysis of NaA:
A¯ + H2O <==> HA + OH¯
Here is the Kb expression for A¯:
| [HA] [OH¯] | |
| Kb = | ---------------- |
| [A¯] |
We can then substitute values into the Kb expression in the normal manner:
| [OH-]2 | |
| 5.65 x 10¯10 = | ---------------- |
| 0.100 |
solve for the hydroxide ion concentration:
[OH¯] = square root of [(5.65 x 10¯10)/ (0.100)]
[OH¯] = 7.52 x 10¯6 M
We then calculate the pH. Since this is a base calculation, we need to do the pOH first:
pOH = - log 7.52 x 10¯6 = 5.124
pH = 14 - 5.124 = 8.876