AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Unit 1 - Atomic Structure & Properties
1.1 Moles and Molar Mass 1.2 Mass Spectra of Elements 1.3 Elemental Composition of Pure Substances 1.4 Composition of Mixtures 1.5 Atomic Structure & Electron Configuration 1.6 Photoelectron Spectroscopy 1.7 Periodic Trends 1.8 Valence Electrons & Ionic Compounds ISPS Chemistry Aug 2024 page 1 Atomic Structures & Properties AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. This logo shows it is a Topic Question - it
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ISPS Chemistry Aug 2024 page 2 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.1 Moles & Molar Mass
Formulae ? Surprisingly, AP Chemistry does not require you to memorise or even work out the formulae of the substances needed in mole calculations - the question should include any formulae needed. However, you need to understand formulae - for example, when calculating the oxygen content in a substance you consider O atoms (formula mass = 16, one mole = 16g) but if calculating for oxygen gas you need to consider O2 molecules (formula mass = 32, one mole = 32g). ISPS Chemistry Aug 2024 page 3 Atomic Structures & Properties AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. The Atomic Mass Scale Scientists of the nineteenth century were aware that atoms of different elements have different masses and were able to determine formulae from weights of reacting elements.
They found, for example, that each 100.0 g of water contains 11.1 g of hydrogen and 88.9 g of oxygen. Thus, water contains 88.9 / 11.1 = 8 times as much oxygen, by mass, as hydrogen.
Once scientists understood that water contains two hydrogen atoms for each oxygen atom, H2O, they concluded that an oxygen atom must have 2 x 8 = 16 times as much mass as a hydrogen atom.
Hydrogen, the lightest atom, was arbitrarily
assigned a relative mass of 1 (no units).
Atomic masses of other elements were at first
determined relative to this value. Thus, oxygen
was assigned an atomic mass of 16.
The development of more accurate means of determining atomic masses - the Mass Spectrometer - led to more accurate system based on comparison with the mass of a proton - the Atomic Mass Unit (amu).
ISPS Chemistry Aug 2024 page 4 Atomic Structures & Properties AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Formula and Molecular Weights The formula weight of a substance is the sum of the atomic weights of the atoms in the chemical formula of the substance. Using atomic weights, we find, for example, that the formula weight of sulfuric acid (H2SO4) is 98.1 amu:*
*For convenience, the atomic weights have been rounded off to one decimal place. In the AP exam all calculations should be done using the Atomic Weights provided with the Formula Sheet.
Molar Mass, M
The easiest way to 'scale up' to real world quantities is to stick
with the same number but move from measuring mass in amu to
measuring mass in grammes.
This involves using a very large
number that we refer to as 1 mole
(1 mol) and it is the number of
particles needed to convert from amu
to grammes.
Molar Mass = AW or FW in grammes
A packet of an artificial sweetener powder
contains 40.0 mg of saccharin (C7H5NO3S),
which has the structural formula:
Given that saccharin has a molar mass of 183.18 g/mol, how many moles of saccharin molecules are in a 40.0 mg (0.0400 g) sample of powdered saccharin? How many moles of carbon are in the same sample?
moles of saccharin, n = m ÷ M, n = 0.0400 / 183.18 = 2.2 x 10-4 mol 7 carbon atoms per molecule, so 7 x 2.2 x 10-4 = 1.5 x 10-3 mol
Saccharin tablets contain 5.5 x 10-4 mol of saccharin. What mass of saccharin is in each tablet? mass of saccharin, m = n x M, m = 5.5 x 10-4 x 183.18 = 0.10 g = 100 mg
ISPS Chemistry Aug 2024 page 5 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Molar Number - Avogadro's Number
While molar mass provided us with the tool
to move into real world quantities, it
was not long before scientists found ways to
calculate the actual number of particles
required to move from, for example,
1 H2O molecule = 18.0 amu
to 18 g (1mole) = 6.02 x 1023 molecules
There are many ways in which we can think about the quantity we call 1 mole, but the first two would be:
1 mole is the amount of a substance that contains 6.022 x 1023
elementary entities (atoms, molecules, or other particles)
1 mole is the amount of a substance whose mass in grammes is
numerically equal to its formula weight in atomic mass units
Mole calculations will often require a mathematical understanding of these relationships and the use of simple formulae linking them.
n = m / M and n = number of particles / NA
ISPS Chemistry Aug 2024 page 6 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.1 Practice Problems
( ) ( )
Mass Number
Not surprisingly, AP Chemistry will assume that you already know that the mass of an atom is determined by the number of protons and neutrons - the mass of electrons are considered to be negligible:
Mass Number = number of protons + number of neutrons
Mass Spectrometer and Mass Spectroscopy ?
AP Chemistry probably assumes that you will have met the machine used to determine masses of atoms present (Mass Spectrometer) and have an idea of how the strength of magnetic and electric fields along with the masses of isotopes present work together (Mass Spectroscopy) to produce the final Mass Spectrum.
However, the AP exam will normally only test you on your ability to understand and use the Mass Spectrum. It is important also that you make use of the Periodic Table provided in the exam to help identify possible elements from the mass spectra in a question.
ISPS Chemistry Aug 2024 page 9 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Mass Spectroscopy
Since each proton and each neutron contribute approximately one amu to the mass of an atom, and each electron contributes far less, the atomic mass of a single atom is approximately equal to its mass number (a whole number).
Over the years, the atomic mass unit (amu) has been redefined so that, for example, the mass of a proton is now only approximately 1 amu - it is actually 1.00727 amu. Similarly, the neutron is actually 1.00866 amu.
More importantly, when protons and neutrons come together to form a nucleus there is an 'energy cost' referred to as the binding energy. As Einstein demonstrated, nuclear energy is linked to mass, E = mc2, so the actual mass of the nucleus is always less than what would be calculated.
The significance of this is that we cannot always assume that the atomic mass is the same as the mass number. For example, there are two Chlorine isotopes,
Isotope 1: 17 protons 18 neutrons mass number = 35
17 x 1.00727 18 x 1.00866 atomic mass = 35.2795
E = mc2 atomic mass = 34.9689
Isotope 2: 17 protons 20 neutrons mass number = 37
17 x 1.00727 20 x 1.00866 atomic mass = 37.2968
E = mc2 atomic mass = 36.9659
the differences in these numbers are so small that in 'ordinary' chemistry calculations
there will be very little error in continuing to use the familiar whole numbers.
The mass spectrometer provides us with three important pieces of information:
average mass = (0.7577 x 34.9689) + (0.2423 x 36.9659) = 35.4528 amu molar mass = 35.4528 g In reality, you are not expected to calculate the average mass. Instead, you are expected to simply use the spectra to identify the 3 things listed above and also estimate the average mass. So, in the chlorine example above, it would be enough to recognise that the average would lie closer to 35 than 37, ideally that it would be about half way between 35 and 36. ISPS Chemistry Aug 2024 page 10 Atomic Structures & Properties AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.2 Practice Problems
The mass spectrum of the element Sb is most likely represented by which of the following? O A B C D
Which of the following elements has the
mass spectrum represented opposite?
O
A Nb B Mo
C U D Cf
The mass spectrum of a sample of a pure element is shown below. Based on the data, the peak at 26amu represents
an isotope of which of the following elements?
O
A Al with 13 neutrons
B Mg with 14 neutrons
C Fe with 26 neutrons
D Ti with 26 neutrons
ISPS Chemistry Aug 2024 page 11 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 4. The elements I and Te have similar
average atomic masses.
A sample that was believed to be a
mixture of I and Te was run through
a mass spectrometer, resulting in the
data opposite.
All of the following statements are true.
Which one would be the best basis for
concluding that the sample was pure Te?
A Te forms ions with a −2 charge, whereas I forms ions with a −1 charge.
B Te is more abundant than I in the universe.
O
C I consists of only one naturally occurring isotope with 74 neutrons, whereas Te has more than one isotope.
D I has a higher first ionization energy than Te does
The mass spectrum of element X is
presented in the diagram opposite.
Based on the spectrum, which of the
following can be concluded about
element X?
A X is a transition metal, and each peak represents an oxidation state of the metal. B X contains five electron sublevels.
C The atomic mass of X is 90.
O
D The atomic mass of X is between 90 and 92.
ISPS Chemistry Aug 2024 page 12 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.2 Quick Check FRQ
1.2 Quick Check FRQ
Using the information in the graph, determine the average atomic mass of Lv in the
Using the information in the graph, determine the average atomic mass of Lv in the sample to four significant figures.
sample to four significant figures.
Average atomic mass = 2/10( 291.2 ) + 3/10( 292.2 )+ 5/10 ( 293.2 )=292.5 amu
ISPS Chemistry Aug 2024 page 13 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.3 Elemental Composition of Pure Substances
ISPS Chemistry Aug 2024 page 14 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Molecular and Empirical Formulae
Even though the atom is the smallest representative sample of an
element, only the noble-gas elements are normally found in nature
as isolated atoms, (monatomic). Several elements are found in nature
in molecular form—two (diatomic) or more of the same type of atom
bound together.
Molecular formulae describe the actual numbers of atoms but they can often be simplified to their smallest ratio which is called the empirical formula.
Molecular
Formula Empirical Formula Mole Ratio Mass Ratio
H2O
H2O
H : O
2 : 1 H : O 1 : 8
H2O2
HO
H : O
1 : 1
H : O
1 : 16
Ethylene
Glycol, C2H6O2
C2H6O2
CH3O
C : H : O
1 : 3 : 1
C : H : O
12 : 3 : 16
Benzene, C6H6 C6H6 CH C : H 1 : 1 C : O 12 : 1
C2H4
CH2
C : H
1 : 2 C : H 6 : 1
CH4
CH4
C : H
1 : 4 C : H 3 : 1
Dinitrogen
Tetroxide,N2O4
N2O4
NO2
N : O
1 : 2
C : O
7 : 16
ISPS Chemistry Aug 2024 page 15 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Calculating Empirical Formulae
The key to calculating empirical formulaes is usually to start by determining accurate masses and then convert to numbers of moles before, finally, converting to mole ratios.
This can often be done using Gravimetric Analysis - accurate measurements of mass.
Example 1, when a 2.50 g sample of copper is heated, it forms 3.13 g of an oxide. What is its empirical formula?
mass of Cu = 2.50 g
mass of O = 3.13 - 2.50 = 0.63 g
moles of Cu = 2.50 / 63.55 = 0.393 mol
mass of O = 0.63 / 16.00 = 0.394 mol
mole ratio = 0.393 : 0.394 = 1 : 1 = CuO
Example 2, 1.285 g of copper chloride hydrate (CuxCly·nH2O) was heated in a crucible. After heating and then cooling, the final mass is 1.012 g of copper chloride, CuxCly.
The CuxCly sample was dissolved in 50 mL of deionized water and 0.2 g of fine aluminum mesh was added to the beaker.
After reacting and dissolving the excess aluminum, 0.479 g of dried copper metal is recovered.
mass of Cu = 0.479 g
mass of Cl = 1.012 - 0.479 = 0.533 g
treat H2O like an 'element' mass of H2O = 1.285 - 1.012= 0.273 g
moles of Cu = 0.479 / 63.55 = 7.53 x 10-3 mol
moles of Cl = 0.533 / 35.45 = 1.50 x 10-2 mol
moles of H2O = 0.273 / 18.01 = 1.51 x 10-2 mol
divide each by smallest Cu Cl H2O 7.53 x 10-3 1.50 x 10-2 1.51 x 10-2
7.53 x 10-3 7.53 x 10-3 7.53 x 10-3
1 : 2 : 2
CuCl2 . 2H2O
ISPS Chemistry Aug 2024 page 16 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Alternatively we can use the results of Compositional Analysis - accurate determination of percentages. 'Trick' is to assume 100g sample so %'s convert directly to masses, and then proceed as before.
Example 3, Ascorbic acid (vitamin C) cures scurvy. It is composed of 40.92 % carbon (C), 4.58 % hydrogen (H), and 54.50 % oxygen (O) by mass. Determine its empirical formula. mass of C = 40.92 g mass of H = 4.58 g mass of O = 54.50 g moles of C = 40.92 / 12.01 = 3.407 mol moles of H = 4.58 / 1.008 = 4.54 mol mass of O = 54.50 / 16.00 = 3.406 mol divide each by smallest C H O 3.407 4.54 3.406 3.406 3.406 3.406 1 : 1.33 : 1 multiply to make whole numbers 3 4 3 C3 H4 O3 Finally we can use the results of Combustion Analysis - accurate determination of products of combustion.
mass of sample, CxHyOz mass of CO2 mass of H2O ⇩ ⇩ moles of CO2 moles of H2O ⇩ x 1 ⇩ x 2 Calculation for hydrocarbons moles of O moles of C moles of H ☞ empirical ⇧ ⇩ ⇩ mass of O ☜ mass of C mass of H subtract C and H from sample
Calculation for complex organics
ISPS Chemistry Aug 2024 page 17 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Example 4, The combustion of 11.5 g of a liquid produced 22.0 g of CO2 and 13.5 g of H2O. Determine its empirical formula.
mass of CO2 = 22.0 g ⇨ mass of C = 22.0 x 12.01/44.01 = 6.0 g mass of H2O = 13.5 g ⇨ mass of H = 13.5 x 2.016/18.02 = 1.51 g mass of O = 11.5 - (6.0 + 1.51) = 4.0 g
moles of C = 6.0 / 12.01 = 0.50 mol
moles of H = 1.51 / 1.008 = 1.50 mol
mass of O = 4.0 / 16.00 = 0.25 mol
divide each by smallest C H O
0.50 1.50 0.25
0.25 0.25 0.25
2 : 6 : 1 C2 H6 O
Empirical formulae, by themselves, are not
ethanol
dimethylether
conclusive proof of the identity of a molecule.
However, when used with other tools like IR Spectroscopy - identifies functional groups, and Mass Spectroscopy - identifies mass of molecule it can be an important first step.
Example 5, Combustion analysis of a hydrocarbon produced: 33.01g CO2 and 13.5g H2O. Determine its empirical formula.
mass of CO2 = 33.01 g ⇨ moles of CO2 = 33.01/44.01 = 0.75 mol moles of C = 0.75 x 1 = 0.75 mol
mass of H2O = 13.5 g ⇨ moles of H2O = 13.5/18.02 = 0.75 mol moles of H = 0.75 x 2 = 1.50 mol
C : H = 0.75 : 1.50 = 1 : 2 CH2 - empirical formula
Mass Spectroscopy reveals molecular mass = 56 amu so C4H8 - molecular formula IR Spectroscopy reveals presence of C = C bond so a butene rather than cyclobutane
nmr Spectroscopy would narrow down which isomer of butene,
but eventually differences in
physical properties such as
melting points might be needed
for final identification.
ISPS Chemistry Aug 2024 page 18 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Calculating Empirical Formulae for Non Molecular Substances
Nothing really changes except that our 'normal' formula for these
substances is really just an empirical formula, as it represents the
simplest ratio rather than an attempt to measure actual numbers.
So, copper(II) oxide is CuO - representing a 1:1 ratio, whereas
copper(I) oxide is Cu2O - representing a 2:1 ratio.
titanium(IV) oxide
TiO2
copper(II) chloride
CuCl2
Example 6, A sample of the black mineral hematite, an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?
mass of Fe = 34.97 g
mass of O = 15.03 g
moles of Fe = 34.97 / 55.85 = 0.6261 mol
mass of O = 15.03 / 16.00 = 0.9394 mol
divide each by smallest Fe O
0.6261 0.9394
0.6261 0.6261
1 : 1.5
multiply to make whole numbers 2 : 3 Fe2O3 so iron(III) oxide
ISPS Chemistry Aug 2024 page 19 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.3 Practice Problems
a) Explain why the student can correctly conclude that the hydrate was heated a
a) Explain why the student can correctly conclude that the hydrate was heated a sufficient number of times in the experiment.
sufficient number of times in the experiment.
No additional mass was lost during the third heating, indicating that all the water of hydration had been driven off.
b) Use the data above to
b) Use the data above to
i) calculate the total number of moles of water lost when the sample was heated, i) calculate the total number of moles of water lost when the sample was heated,
25.825 − 23.976 = 1.849 g 1.848 g H2O ÷ 18.02 g mol-1 = 0.1026 mol H2O and ii) determine the formula of the hydrated compound.
and ii) determine the formula of the hydrated compound.
1 point is earned for calculating the correct number of moles of anhydrous MgCl2. 23.977 − 22.347 = 1.630 g 1.630 g ÷ 95.2 g mol-1 = 0.01712 mol MgCl2
1 point is earned for writing the correct formula (with supporting calculations). 0.1026 mol H2O / 0.01712 mol MgCl2 = 5.993 ≈6 so MgCl2 . 6H2O c) A different student heats the hydrate in an uncovered crucible, and some of the solid
c) A different student heats the hydrate in an uncovered crucible, and some of the solid spatters out of the crucible. This spattering will have what effect on the calculated
spatters out of the crucible. This spattering will have what effect on the calculated mass of the water lost by the hydrate? Justify your answer.
mass of the water lost by the hydrate? Justify your answer.
The calculated mass (or moles) of water lost by the hydrate will be too large because the mass of the solid that was lost will be assumed to be water when it actually included some MgCl2 as well.
ISPS Chemistry Aug 2024 page 23 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Q1 contd.
Q1 contd.
In the second experiment, a student is given 2.94 g of a mixture containing anhydrous
In the second experiment, a student is given 2.94 g of a mixture containing anhydrous MgCl2 and KNO3 . To determine the percentage by mass of MgCl2 in the mixture,
MgCl2 and KNO3 . To determine the percentage by mass of MgCl2 in the mixture, the student uses excess AgNO3(aq) to precipitate the chloride ion as AgCl(s).
the student uses excess AgNO3(aq) to precipitate the chloride ion as AgCl(s). d) Starting with the 2.94 g sample of the mixture dissolved in water, briefly describe
d) Starting with the 2.94 g sample of the mixture dissolved in water, briefly describe the steps necessary to quantitatively determine the mass of the AgCl precipitate.
the steps necessary to quantitatively determine the mass of the AgCl precipitate.
2 points are point is earned for all three major steps: filtering the mixture, drying the precipitate, and determining the mass by difference.
1 point is earned for any two steps.
Add excess AgNO3 .– Separate the AgCl precipitate (by filtration).
– Wash the precipitate and dry the precipitate completely.
– Determine the mass of AgCl by difference
e) The student determines the mass of the AgCl precipitate to be 5.48 g. On the basis of
e) The student determines the mass of the AgCl precipitate to be 5.48 g. On the basis of this information, calculate each of the following.
this information, calculate each of the following.
i) The number of moles of MgCl2 in the original mixture
i) The number of moles of MgCl2 in the original mixture
1 point is earned for calculating the number of moles of AgCl. precipitate, and determining the mass by difference.
5.84 g AgCl ÷ 143.32 g mol-1 = 0.0382 mol AgCl
1 point is earned for conversion to moles of MgCl2.
2 mol AgCl → 1 mol MgCl2 0.0382 mol AgCl → 0.0191 mol MgCl2 ii) The percent by mass of MgCl2 in the original mixture
ii) The percent by mass of MgCl2 in the original mixture
0.0191 mol MgCl2 x 95.20 g mol-1 = 1.82 g MgCl2
(1.82 g MgCl2 / 2.94 g sample) x 100% = 61.9% MgCl2 by mass
ISPS Chemistry Aug 2024 page 24 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.4 Composition of Mixtures
Working with Mixtures
As long as a mixture is homogeneous, we can usually find a way of separating and determining at least one of the components and determine the purity. For example, the active ingredient in aspirin. acetyl salicylate, can be extrated from the 'filler' material and analysed to determine the purity of a tablet of known mass.
ISPS Chemistry Aug 2024 page 25 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Separating Mixtures
To separate the different components in a mixture we
often need to make use of differences in their properties -
particularly solubility.
A mixture of two compounds, one of which is soluble
(CuCl2) while the other is insoluble (CuCl) would simply
require the mixture to be added to water, stirred and then
the insoluble CuCl would be separated by filtration.
The accuracy of this method would depend on the different solubilities but with CuCl2 having a solubility of 76 g per 100 ml of water and CuCl being 3.91 mg per 100 ml of water, it should be possible to accurately determine the CuCl as long as the quantity was not too low to begin with.
Determining a mixture of two soluble compounds, such as CuCl2
and NaCl, would need a different approach. Adding a third
chemical such as NaOH would cause Cu to precipitate out
as solid Cu(OH)2. This could be filtered, dried to constant mass
and weighed, allowing the amount of Cu (and hence CuCl2) to be
determined.
Another alternative would be to add a more reactive metal, such
as aluminium, that would cause solid copper to precipitate out
(a displacement reaction). The excess Al could be reacted away with acid, before the copper was filtered, dried to constant mass and weighed.
An important method of separating the
components of a homogeneous mixture is
distillation, a process that depends on the
different abilities of substances to form
vapours.
The amount of ethanol in an alcoholic drink
or the amount of salt in a sample of sea water
could both be determined using a form of
distillation.
Whatever methods of analysis are used, the
ability to use mole relationships linked to an
understanding of
chemical formulae
and percentages
will be crucial.
ISPS Chemistry Aug 2024 page 26 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.4 Practice Problems
A sample of carbonate rock is a mixture of CaCO3 and MgCO3 . The rock is analyzed in a laboratory, and the results are recorded in the table above.
Which columns in the table provide all the information necessary to determine the mole ratio of Ca to Mg in the rock?
O
A 1, 2, 5 B 2, 5, 6 C 3, 4, 6, 7 D 2, 3, 4, 5
4. A 5.0g sample of MgCl2 may contain measurable amounts of other compounds as impurities. Which of the following quantities is (are) needed to determine that the sample is pure MgCl2 ?
A The color and density of the sample B The mass of Mg in the sample only
C The number of moles of Cl in the sample only
O
D The mass of Mg and the mass of Cl in the sample
5. The mass percent of carbon in pure glucose, C6H12O6, is 40.0 percent. A chemist analyzes an impure sample of glucose and determines that the mass percent of carbon is 38.2 %. Which of the following impurities could account for the low mass percent of carbon in the sample?
O
A Water, H2O B Ribose, C5H10O5 C Fructose, C6H12O6, an isomer of glucose D Sucrose, C12H22O11
ISPS Chemistry Aug 2024 page 27 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 6. The percentage of silver in a solid sample is determined gravimetrically by converting the silver to Ag+(aq) and precipitating it as silver chloride. Failure to do which of the following could cause errors in the analysis?
I Account for the mass of the weighing paper when determining the mass of the sample II Measure the temperature during the precipitation reaction
III Wash the precipitate
IV Heat the AgCl precipitate to constant mass
O
A I only B I and II C I and IV D II and III E I, III and IV
7. To gravimetrically analyze the silver content of a piece of jewelry made from an alloy of Ag and Cu, a student dissolves a small preweighed sample in HNO3(aq).
Ag+(aq) and Cu2+(aq) ions form in the solution. Which of the following should be the next step in the analytical process?
A Centrifuging the solution to isolate the heavier ions
B Evaporating the solution to recover the dissolved nitrates
C Adding enough base solution to bring the pH up to 7.0
O
D Adding a solution containing an anion that forms an insoluble salt with only one of the metal ions
ISPS Chemistry Aug 2024 page 28 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.4 Quick Check FRQ - same question was used earlier
1.4 Quick Check FRQ - same question was used earlier
Balancing the forces of repulsion and attraction that
can exist between electrons and between electrons
and the positive nucleus required the creation
of a brand new set of maths, Quantum Mechanics -
particularly once it was realised that electrons have
both the properties of charged particles and
wave properties.
One use of Quantum Mechanics is to plot
potential positions for a single electron.
The region in space where there is a greater
than 90% probability of finding the electron
is called an orbital. The simplest shape formed
is spherical and is called an s orbital.
Repulsions between electrons means that a maximum of
2 electrons can occupy the same orbital.
These repulsions also force electrons to move further away
from the nucleus to occupy higher energy levels called
shells.
Shells are represened by a principal quantum number, n, where
n = 1 (1st shell), n = 2 (2nd shell), n = 3 (3rd shell) etc.
There is an s orbital within each shell, so 1s, 2s, 3s etc.
Notice that the absolute position of an orbital
can change.
The 1s orbital, for example, only exists, by
itself, in a Hydrogen (1+) or Helium (2+)
atom.
Coulombs Law, F ∝ (q1 x q2) / r2
explains why the 1s orbital will be closer to
the nucleus in a Helium (2+) compared to a
Hydrogen (1+) atom and, therefore, that a
He atom will be smaller than a H atom.
In larger atoms, the higher charges on
the nuclei pull the 1s orbital even closer.
ISPS Chemistry Aug 2024 page 32 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. The first shell is small and only has room for the
1s orbital.
The second shell is larger and there is room for a 2s orbital and a set of 3 2p orbitals which are dumb-bell shape.
The third shell is even larger and there is room for a 3s orbital, a set of
3p orbitals and a set of 5 3d orbitals which are double dumb-bell shape.
The fourth shell is even larger and there is room for a 4s orbital, a set of 4p orbitals, a set of 4d orbitals and a set of 7 4f orbitals which have complex shapes.
Once electrons start occupying these orbitals, the complex
balance of attractive and repulsive forces means that the
energies of these orbitals are continually shifting, depending on
the occupancy of each orbital.
In particular, there are times when the 4s orbital is at a lower energy
than the 3d orbital.
The filling of orbitals are
governed by a number of rules.
The first of these is called the
Aufbau principle which states
that an electron occupies orbitals
in order from lowest energy to
highest.
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The Aufbau principle ensures that
the 1s orbital must be filled before
an electron can be placed in a
higher energy 2s orbital
Our second rule is called the Pauli Exclusion Principle which states that no two electrons can have the same set of quantum numbers. That is, no two electrons can be in the same state.
Electrons can be in the same shell, same n - 2 in 1st shell
rising to 32 by 4th shell
Electrons can be in the same shape/type of orbital, same l,
2 in an s-orbital rising to 14 in the f-orbital set
Electrons can be in the same type of orbital with same
orientation, same m, both in a px or dxy orbital
So the 2 electrons in a 2px orbital,
for example, would be in the
same shell (2nd), same shaped
orbital (p) with the same
orientation (along x-axis)
so they would have to have
opposite spins.
ISPS Chemistry Aug 2024 page 34 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Orbitals within the same set, e.g. 2p,
are all of equal energy (described
as being degenerate) so the Aufbau
Principle would not differentiate
between them. Our third rule,
Hund's Rule has to be applied.
Hund's Rule of Maximum Multiplicity states that every orbital in a sublevel is singly occupied before any orbital is doubly occupied, and that all of the electrons in singly occupied orbitals have the same spin (to maximize total spin).
Once each orbital is occupied, then
the Pauli Exclusion Principle ensures
that the second electron has the
opposite spin to the original single
electrons.
We have two main methods for describing
the electron configuration of an atom. The
first method is referred to as standard
notation and describes in terms of shells
and sub-shells.
A variation is to describe the inner shells
by reference to the equivalent noble gas and
then describe the outer shell in detail.
I: 1s22s22p63s23p64s23d104p65s24d105p5
becomes I: [Kr]5s24d105p5
The other method is often referred to as 'electrons in boxes'
and has the advantage of being able to show spins.
Both methods, however, will require you to remember that the
overlapping of shells means that, for example, the 4s orbital is
filled before the 3d orbitals due to the fact that the 4s orbital is at
a lower energy and, by the Aufbau Principle, must be filled before the higher energy 3d orbitals.
However, once filled, the 4s orbital is higher than the 3d so, when forming ions, electrons are lost from the 4s orbital first before electrons are lost from 3d orbitals.
ISPS Chemistry Aug 2024 page 35 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Effective Nuclear Charge
Coulombs Law, F ∝ (q1 x q2) / r2 describes how the force of attraction will be affected by the both the charge on an electron (q1) and the
charge on the nucleus (q2).
The charge on the nucleus can be calculated
on the basis of the number of protons, so a
sodium atom should have a charge of 11+.
However, the outer electron (the valence
electron) will actually experience a
significantly smaller charge.
The 10 electrons in the inner orbitals
(the core electrons) will be shielding the outer
electron so that it effectively experiences an overall charge closer to 11 - 10 = 1+. Atoms in the same group would have the same effective nuclear charge ...... Li: 3 - 2 = 1+ Be: 4 - 2 = 2+ O: 8 - 2 = 6+ F: 9 - 2 = 7+ Na: 11 - 10 = 1+ Mg: 12 - 10 = 2+ S: 16 - 10 = 6+ Cl: 17 - 10 = 7+ K: 19 - 18 = 1+ Ca: 20 - 18 = 2+ Se: 34 - 28 = 6+ Br: 35 - 28 = 7+ ..... whilst the effective nuclear charge increases as you go across a row in the Periodic Table.
In reality, the maths is more
complicated, so effective nuclear
charges are rarely exact whole numbers
but the same relationships apply.
In reality the effective nuclear charge
increases slightly as you go down a
Group, but this has a minimal effect
compared with the extra distance, r ,
caused by the extra electron shell.
Atomic radius decreases as you go across a Period because effective nuclear charge increases so outer shell is pulled closer - F ∝ (q1 x q2) / r2 and q is increasing
Atomic radius increases as you go down a Group because an extra shell is being added so pull weakens - F ∝ (q1 x q2) / r2 and r is increasing
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AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International.
Ionisation energy decreases as you go down a
Group because an extra shell is being added so
attraction weakens - F ∝ (q1 x q2) / r2 and r is
increasing - so it is easier to remove an electron.
Ionisation energy increases as you go across a
Period because effective nuclear charge increases
so outer shell is attracted more strongly and pulled
closer - F ∝ (q1 x q2) / r2 and q is increasing - so it
is harder to remove an electron.
Another very important property that can be explained using Coulomb's Law is electronegativity.
Electronegativity measures an atoms ability to attract the electrons in a shared pair - in a covalent bond, eg. P—H in PH3 . If one atom is sufficiently more electronegative than the other then the bond will be polar covalent, eg. O��-—H��+ in H2O . If one atom is much more electronegative than the other then the bond will become ionic, eg. Na+ Cl—.
Electronegativity decreases as you go down a Group because an extra shell is being added so attraction weakens - F ∝ (q1 x q2) / r2 and r is increasing - so it is harder to attract an electron.
Electronegativity increases as you go across a Period because effective nuclear charge increases so shared electrons are attracted more strongly - F ∝ (q1 x q2) / r2 and q is increasing
ISPS Chemistry Aug 2024 page 37 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Magnetism
The way a substance behaves in a magnetic field can in some cases provide insight into the arrangements of its electrons. Molecules with one or more unpaired electrons are attracted to a magnetic field. The more unpaired electrons in a species, the stronger the attractive force. This type of magnetic behavior is called paramagnetism.
Substances with no unpaired electrons are weakly
repelled by a magnetic field. This property is called
diamagnetism. The Pauli Exclusion Principle ensures
that paired electrons have opposite spins so their magnetic
properties cancel out.
Most magnetic properties are
shown by transition metals as
Hund's Rule of Maximum
Multiplicity ensures that up to
5d orbitals can be occupied by
single electrons.
The effects are small but can be measured by
balancing a tube containing a sample of your
substance against masses that are sitting on a
very accurate (at least 4 dp) balance.
Switching on the electromagnets will have no effect on a
diamagnetic substance. A paramagnetic substance, however, will
be pulled down by the magnetic field.
The mass being recorded by the balance will decrease slightly.
We would detect this with a tube filled with an Fe2+ compound,
and even more with an Fe3+ compound
The effect is weak because the iron ions will start off with
all possible orientations. Once in a magnetic field they will
try and line up with the magnetic field but only a small
number will succeed.
The effect can be strengthened by helping the particles line up. This
can best be achieved with iron atoms. Repeated application of a
magnetic field (and heat) will leave the iron atoms lined up and a
permanent magnetic effect can be produced - ferromagnetism.
ISPS Chemistry Aug 2024 page 38 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.5 Practice Problems
Consider atoms of the following elements. Assume that the atoms are in the ground state. 1. The atom that contains only one electron in the highest occupied energy sublevel O
A S B Ca C Ga D Sb E Br
2. The atom that contains exactly two unpaired electrons
O
A S B Ca C Ga D Sb E Br
3. Which of the following ground-state electron configurations represents the atom that has the lowest first-ionization energy?
O
A 1s22s1 B 1s22s22p2 C 1s22s22p6 D 1s22s22p63s1
4. Which of the following is the ground-state electron configurations of the F— ion ? O
A 1s22s22p4 B 1s22s22p5 C 1s22s22p6 D 1s22s22p63s23p6
5. Which of the following best represents the ground-state electron configuration for an atom of selenium?
A 1s22s22p63s23p3 B 1s22s22p63s23p4
O
C 1s22s22p63s23p64s23d104p4 D 1s22s22p63s23p64s23d104p5
6. How many protons, neutrons, and electrons are in an 5266 Fe atom?
Protons Neutrons Electrons
O
A 26 30 26
B 26 56 26
C 30 26 30
D 56 26 26
E 56 82 56
ISPS Chemistry Aug 2024 page 39 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 7. Of the following electron configurations of neutral atoms, which represents an atom in an excited state?
O
A 1s22s22p5 B 1s22s22p53s2 C 1s22s22p63s1
D 1s22s22p63s23p2 E 1s22s22p63s23p5
8. The effective nuclear charge experienced by the outermost electron of Na is different than the effective nuclear charge experienced by the outermost electron of Ne. This difference best accounts for which of the following?
A Na has a greater density at standard conditions than Ne.
O
B Na has a lower first ionization energy than Ne.
C Na has a higher melting point than Ne.
D Na has a higher neutron-to-proton ratio than Ne.
E Na has fewer naturally occurring isotopes than Ne.
9. Which of the following is the electron configuration of an excited atom that is likely to emit a quantum of energy?
A 1s22s22p63s23p1 B 1s22s22p63s23p5 C 1s22s22p63s2 O
D 1s22s22p63s1 E 1s22s22p63s13p1
10. Which of the following represents a pair of isotopes?
Atom 1 Atom 2
Atomic Number Mass Number Atomic Number Mass Number A 6 14 7 14
B 6 7 14 14
C 6 14 14 28 O
D 7 13 7 14 E 8 10 16 20
ISPS Chemistry Aug 2024 page 40 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 11. Which of the following represents the ground state electron configuration for the Mn3+ ion? (Atomic number Mn = 25)
O
A 1s22s22p63s23p63d4 B 1s22s22p63s23p63d54s2
C 1s22s22p63s23p63d24s2 D 1s22s22p63s23p63d84s2
E 1s22s22p63s23p63d34s1
12. Which of the following shows the correct number of protons, neutrons, and electrons in a neutral cesium-134 atom?
Protons Neutrons Electrons
A 55 55 55
O
B 55 79 55
C 55 79 79
D 79 55 79
E 134 55 134
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AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.5 Quick Check FRQ
1.5 Quick Check FRQ
Photoelectron spectroscopy (PES)
Photoelectron spectroscopy (PES) was formerly known as X-ray electron spectroscopy or XES. It involves the use of a device that focuses a monochromatic beam of X-rays at a solid sample.
The process is similar to the situation that would exist if you needed to kick some balls out of some ditches. To help you, you have a highly trained kicker who can consistently kick a ball with exactly 100 J of energy.
One ball emerges with 70J of Kinetic Energy
so we know that it required 30J to escape.
from the ditch - escape energy = 30J
Two balls emerge with only 55J of KE, so we
know that it required 45J to escape - they
were in a deeper ditch.
In photoelectron spectroscopy, electrons are 'kicked' by X-rays
of equal energy. Some electrons (from higher orbitals) emerge
with more KE while others (from lower orbitals) needed more of
the energy to simply escape so they emerge with less KE.
It is the 'escape energy' or ionisation energy that is recorded.
ISPS Chemistry Aug 2024 page 44 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. This process is virtually identical to the
one used to determine the first
ionization energy of an atom except that
we are only able to remove one electron
from the valence shell.
Using very high energy UV or X-ray
photons we are able to remove core
electrons as well and build up a complete
'picture' of the electronic configuration.
High ionisation/binding/escape energies
correspond to orbitals closer to the
nucleus.
The height of a peak corresponds to the number of electrons present in an orbital set:
s-orbitals - 1 - 2 electrons
p-orbitals - 1 - 6 electrons
d-orbitals - 1 - 10 electons
The highest binding energy (on left) has
to belong to the orbital closest to the
nucleus so 1s2
This also establishes the height that is
equivalent to 2 electrons.
Moving left to right
The next signal has to be for the 2s orbital and the height confirms 2 electrons so - 2s2 The next signal has to be for the 2p orbitals and the height confirms 6 electrons so - 2p6 The next signal has to be for the 3s orbital and the height confirms 2 electrons so - 3s2 The final signal has to be for the 3p orbitals and the height confirms 3 electrons so - 3p3
So complete electron configuration - 1s2 2s2 2p6 3s2 3p3 15 electrons in total so Phosphorus
In this example, we again start with the
highest binding energy peak on the left
which will always be the 1s orbital.
If there are electrons further out then
the 1s must be filled so 1s2.
The next peak must be 2s but height
tells us 2s1 - 1s2 2s1 - Lithium
ISPS Chemistry Aug 2024 page 45 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Explaining Relationships
Coulomb's Law can again be used to explain some aspects of Photoelectron Spectroscopy. F ∝ (q1 x q2) / r2
Both hydrogen, 1s1 , and helium, 1s2 , have very similar electron configurations but you should notice, and be able to explain, the fact that it takes slightly more energy to remove from helium than from hydrogen.
Whilst you might predict that the extra repulsions caused by having two electrons in the 1s orbital would cause the orbital to expand further out from the nucleus, increasing r and, therefore decreasing F, the opposite is happening. F must be increasing as more energy is needed to remove the electrons.
So it must be due to the fact that the Helium nucleus is 2+ whereas the Hydrogen is only 1+. Increasing q will cause F to increase and will explain why more energy is needed to remove the electrons.
With lithium and beryllium, we now have both 1s and 2s orbitals. The 2s orbitals are further from the nucleus so the increased r means that F has decreased, making it easier to remove 2s electrons than 1s electrons.
Notice, however, that each time the energy needed to remove from the 1s orbital is increasing. As the nuclear charge increases to 3+ and then 4+. Increasing q will cause F to increase and will explain why more energy is needed to remove the electrons.
With boron, we now have 2p orbitals. The
2p orbitals are further from the nucleus so
the increased r means that F has decreased,
making it easier to remove 2p electrons than
2s and 1s electrons.
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AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.6 Practice Problems
1.
The complete photoelectron spectrum of an element is given above. Which of the following electron configurations is consistent with the spectrum?
O
A 1s22s22p1 B 1s22s22p63s23p3
C 1s22s22p63s23p6 D 1s22s22p63s23p64s23d5
2.
The complete photoelectron spectrum for an element is shown above. Which of the following observations would provide evidence that the spectrum is consistent with the atomic model of the element?
A A neutral atom of the element contains exactly two electrons.
B The element does not react with other elements to form compounds. O
C In its compounds, the element tends to form ions with a charge of +1 . D In its compounds, the element tends to form ions with a charge of +3 .
ISPS Chemistry Aug 2024 page 47 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 3.
The photoelectron spectrum for the element nitrogen is represented above. Which of the following best explains how the spectrum is consistent with the electron shell model of the atom?
A The leftmost peak represents the valence electrons.
B The two peaks at the right represent a total of three electrons.
C The electrons in the 1s sublevel have the smallest binding energy.
O
D The electrons in the 2p sublevel have the smallest binding energy.
4. A sample containing atoms
of C and F was analyzed
using x-ray photoelectron
spectroscopy.
The portion of the spectrum
showing the 1s peaks for
atoms of the two elements
is shown opposite.
Which of the following
correctly identifies the 1s
peak for the F atoms and
provides an appropriate
explanation?
A Peak X, because F has a smaller first ionization energy than C has.
O
B Peak X, because F has a greater nuclear charge than C has.
C Peak Y, because F is more electronegative than C is.
D Peak Y, because F has a smaller atomic radius than C has.
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AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 5.
The photoelectron spectra of the 1s electrons
of two isoelectronic species, Ca2+ and Ar, are
shown opposite.
Which of the following correctly identifies the
species associated with peak X and provides a
valid justification?
A Ar, because it has completely filled energy levels
B Ar, because its radius is smaller than the radius of Ca2+
C Ca2+, because its nuclear mass is greater than that of Ar
O
D Ca2+, because its nucleus has two more protons than the nucleus of Ar has
6.
The photoelectron spectra
opposite show the energy
required to remove a 1s
electron from a nitrogen
atom and from an oxygen
atom.
Which of the following
statements best accounts
for the peak in the upper
spectrum being to the right
of the peak in the lower
spectrum?
A Nitrogen atoms have a half-filled p subshell.
B There are more electron-electron repulsions in oxygen atoms than in nitrogen atoms.
C Electrons in the p subshell of oxygen atoms provide more shielding than electrons in the p subshell of nitrogen atoms.
O
D Nitrogen atoms have a smaller nuclear charge than oxygen atoms.
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AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.6 Quick Check FRQ
1.6 Quick Check FRQ
On the basis of the data, element X is most likely to be
O
A Na B Mg C Al D Si E P
13. Which of the following best helps to account for the fact that the F— ion is smaller than the O2- ion?
A F— has a larger nuclear mass than O2- has.
O
B F— has a larger nuclear charge than O2- has.
C F— has more electrons than O2- has.
D F— is more electronegative than O2- is.
E F— is more polarizable than O2- is.
ISPS Chemistry Aug 2024 page 59 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 14. Which of the following correctly identifies which has the higher first-ionization energy, Cl or Ar, and supplies the best justification?
A Cl, because of its higher electronegativity
B Cl, because of its higher electron affinity
C Ar, because of its completely filled valence shell
O
D Ar, because of its higher effective nuclear charge
15. Which of the following elements has the largest first ionization energy? O
A Li B Be C B D C E N
16. Which of the following lists Mg, P, and Cl in order of increasing atomic radius? O
A Cl < P < Mg B Cl < Mg< P C Mg< P < Cl
D Mg < Cl < P E P < Cl < Mg
17. Which of the following properties generally decreases across the periodic table from sodium to chlorine?
A First ionization energy B Atomic mass C Electronegativity O
D Maximum value of oxidation number E Atomic radius
18.
For element X represented above, which of the following is the most likely explanation for the large difference between the second and third ionization energies?
A The effective nuclear charge decreases with successive ionizations.
B The shielding of outer electrons increases with successive ionizations. O
C The electron removed during the third ionization is, on average, much closer to the nucleus than the first two electrons removed were.
D The ionic radius increases with successive ionizations
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AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.7 Quick Check FRQ
1.7 Quick Check FRQ
A student examines the data in the table and poses the following hypothesis:
A student examines the data in the table and poses the following hypothesis: the number of F atoms that will bond to a nonmetal is always equal to 8 minus the
the number of F atoms that will bond to a nonmetal is always equal to 8 minus the number of valence electrons in the nonmetal atom.
number of valence electrons in the nonmetal atom.
Based on the student’s hypothesis, what should be the formula of the compound that
Based on the student’s hypothesis, what should be the formula of the compound that forms between chlorine and fluorine?
forms between chlorine and fluorine?
1 point is earned for the correct formula. ClF
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AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.8 Valence Electrons & Ionic Compounds
Bonding & Energy Regardless of the nature of the bond being formed, the driving force is always the extra stability that comes from moving to a lower energy level.
Forces of attraction (F ∝ (q1 x q2) / r2) between shared electrons and positive nuclei or between positive and negative ions will be balanced by the forces of repulsion (F ∝ (q1 x q2) / r2) between electrons and between nuclei or between positive ions or between negative ions.
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AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. Type of Bond
All of the properties discussed previously (size of atoms and ions, effective nuclear charge, ionisation energies, electronegativity and electron affinities) can play a part in determining the type of bond that forms.
The most useful property, however, is probably
electronegativity.
As atoms approach and overlap it will be
differences in their ability to attract electrons
that will largely determine the type of bond that
forms.
However, this is not 'black & white' - bonding is a
continuum with many 'shades of grey'
For example, aluminium forms analogous compounds with the halogens - AlF3 , AlCl3 , AlBr3 etc.
Al F Al Cl Al Br Al I
electronegativity 1.6 3.98 1.6 3.16 1.6 2.96 1.6 2.66 ∆EN = 2.38 = 1.56 = 1.36 = 1.06 prediction ionic polar covalent polar covalent polar covalent MPt (°C) 1290 193 98 188 Structure network molecular molecular molecular Empirical AlF3 AlCl3 AlBr3 AlI3
Formula
Molecular Formula Al2Cl6 Al2Br6 Al2I6
ISPS Chemistry Aug 2024 page 63 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. So, though a compound between a metal and a non-metal, AlCl3 has a low melting point and a molecular structure which is typical of a covalent compound. However, when molten, AlCl3 is a good electrolyte which means that it does form Al3+ and Cl— ions and can also be considered an ionic compound.
Normally, it is safe enough to assume that atoms near the edges of the Periodic Table will behave 'normally' but properties can be a bit more mixed towards the middle.
Many atoms gain or lose electrons to become
more stable. Having the the same number of
electrons as the noble gas closest to them in
the periodic table can be part of that stability.
Same But Different - Elements in the same
Group will have many similar properties,
such as same charge on ion, same effective
charge (Zeff or q ) and, same formulae.
However, as number of shells / distance
from nucleus (r) increases, expect other
properties to be changing:-
atomic radius (⇡), electronegativity (⇣)
ionisation energy (⇣), ionic character (⇡)
Same But Different - Elements in the same Period/Row have the same number of shells / similar distance from nucleus (r).
However, as effective charge (Zeff or q ) increases, expect other properties to be changing:- atomic radius (⇣) electronegativity (⇡) ionisation energy (⇡)
ISPS Chemistry Aug 2024 page 64 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.8 Practice Problems
All the chlorides of the alkaline earth metals have similar empirical formulas, as shown in the table above. Which of the following best helps to explain this observation?
A Cl2(g) reacts with metal atoms to form strong, covalent double bonds.
B Cl has a much greater electronegativity than any of the alkaline earth metals. O
C The two valence electrons of alkaline earth metal atoms are relatively easy to remove
D The radii of atoms of alkaline earth metals increase moving down the group from Be to Ra.
ISPS Chemistry Aug 2024 page 65 Atomic Structures & Properties
AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 6. RbCl has a high boiling point. Which of the following compounds is also likely to have a high boiling point, and why?
A NO , because its elements are in the same period of the periodic table B ClF , because its elements are in the same group of the periodic table.
C Cl2O , because its elements have similar electronegativities and it is a covalent compound.
O
D CsCl , because its elements have very different electronegativities and it is an ionic compound
7. Which of the following best describes solid ethyl alcohol C2H5OH A A network solid with covalent bonding
B A molecular solid with zero dipole moment
O
C A molecular solid with hydrogen bonding
D An ionic solid E A metallic solid
8. Based on the information opposite and
periodic trends, which of the following
is the best hypothesis regarding the
oxide(s) formed by Rb?
A Rb will form only Rb2O.
B Rb will form only Rb2O2.
O
C Rb will form only Rb2O and Rb2O2. D Rb will form Rb2O, Rb2O2 and RbO2.
9. Based on the ionization energies of element X
given in the table above, which of the following
is most likely the empirical formula of an oxide
of element X?
A XO2 B X2O
O
C X2O3 D X2O5
10. Which of the following ions has the same number of electrons as Br— ? O
A Ca2+ B K+ C Sr2+ D I— E Cl—
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AP Chemistry Notes © 2020 by Gordon Watson is licensed under Attribution-NonCommercial-ShareAlike 4.0 International. 1.8 Quick Check FRQ
1.8 Quick Check FRQ