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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
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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).
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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).
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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,
ISPS Chemistry Aug 2024
m = 5.5 x 10-4 x 183.18
page 5
= 0.10 g = 100 mg
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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
to 18 g (1mole)
=
=
18.0 amu
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
ISPS Chemistry Aug 2024
and
n = number of particles / NA
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1.1 Practice Problems
1.
A solution is prepared by adding 16. g of CH3OH (molar mass 32 g) to 90. g of H2O (molar
mass 18 g). The mole fraction of CH3OH in this solution is closest to which of the
following?
A 0.1
O
2.
B 0.2
C 0.3
D 0.4E 0.6
A solution of methanol, CH3OH, in water is prepared by mixing together 128 g of
methanol and 108 g of water. The mole fraction of methanol in the solution is closest to
O
E 0.20
D 1.2 x 1024
E 6.0 x 1024
A 0.80 B 0.60 C 0.50 D 0.40
3.
How many carbon atoms are contained in 2.8 g of C2H4 ?
A 1.2 x 10 O
B 3.0 x 1023
23
4.
C 6.0 x 1023
In 1.00 mol of potassium zirconium sulfate trihydrate, K4Zr(SO4)4 • 3 H2O, there are
A 3 x 6.02 x 1023 hydrogen atoms
B 6.02 x 1023 sulfur atoms
C 4 x 6.02 x 10 potassium atoms
O
D 4 moles of oxygen atoms
23
E 4 moles of zirconium atoms
5.
A student has a 1g sample of each of the following compounds: NaCl, KBr, and KCl. Which
of the following lists the samples in order of increasing number of moles in the sample?
A NaCl < KCl < KBr
B NaCl < KBr < KCl
O
C KCl < NaCl < KBr D KBr < KCl < NaCl
6.
A student obtains a sample of a pure solid compound. In addition to Avogadro’s number, which of the following must the student know in order to determine how many molecules are in the sample?
A Mass of the sample, volume of the sample
B Mass of the sample, density of the sample
C Molar mass of the compound, mass of the sample
O
D Molar mass of the compound, density of the sample
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7.
Which of the following numerical expressions gives the number of particles in 2.0g of Ne ?
A B
O
C D
8.
(
)
(
)
According to the information in the table shown,
a 1.00 g sample of which of the following contains
the greatest mass of oxygen?
O
A Na2O B MgO
C K2O D CaO
1.1 Quick Check FRQ
1.
Answer the following questions related to the analysis of CaBr2.
a)
A student has a 10.0g sample of CaBr2 . Show the setup of the calculation to
determine the number of moles of CaBr2 in the sample. Include units in the setup. (You do not need to do any calculations.)
(The following or an equivalent variant)
10.00 g CaBr2 ÷ 199.88 g/mol
(Molar mass can be written with any number of significant figures)
b)
What number, in addition to the answer to part a), is needed to determine the
number of atoms of Ca in the sample?
The response gives the term “Avogadro’s number” or the value of Avogadro’s number,
(with any number of significant figures, units of or per mole are not required).
c)
A different student is given a 10.0g sample labeled CaBr2 that may contain an inert (nonreacting) impurity. Identify a quantity from the results of laboratory analysis that the student could use to determine whether the sample was pure.
One from
ISPS Chemistry Aug 2024
Mass of Ca in sample
Number of moles of Ca in sample
Mass of Br in sample
Number of moles of Br in sample
Mass or number of moles of element other than Ca or Br in sample
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1.2
Mass Spectra of Elements
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.
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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
17 x 1.00727 18 x 1.00866
E = mc2
mass number = 35
atomic mass = 35.2795
atomic mass = 34.9689
Isotope 2: 17 protons 20 neutrons
17 x 1.00727 20 x 1.00866
E = mc2
mass number = 37
atomic mass = 37.2968
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:
1.
2.
3.
the number of isotopes present, i
the atomic mass of each isotope the relative amount of each isotope
(2 peaks so 2 isotopes)
('35' and '37')
(75.77% and 24.23%)
From this we can calculate the average or relative atomic mass;
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.
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1.2 Practice Problems
1.
The mass spectrum of the element Sb is most likely represented by which of the
following?
O
A
B
CD
2.
Which of the following elements has the
mass spectrum represented opposite?
B
Mo
O
A
Nb C
U D
3.
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?
A
Al with 13 neutrons
O
B
Mg with 14 neutrons
C
Fe with 26 neutrons
D
Ti with 26 neutrons
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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.
D
I has a higher first ionization energy than Te does
C
I consists of only one naturally occurring isotope with 74 neutrons,
O
whereas Te has more than one isotope.
5.
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.
D
The atomic mass of X is between 90 and 92.
O
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1.2 Quick Check FRQ
1.
A new element with atomic number 116 was discovered in 2000. In 2012 it was named
livermorium, Lv. Although Lv is radioactive and short-lived, its chemical properties and
reactivity should follow periodic trends.
a)
Write the electron configuration for the valence electrons of Lv in the ground state.
7s2 7p4
b)
According to periodic properties, what would be the most likely formula for the product obtained when Lv reacts with H2(g)?
LvH2 (or H2Lv)
c)
The first ionization energy of polonium, Po, is 812 kJ/mol. Is the first ionization
energy of Lv expected to be greater than, less than, or equal to that of Po?
Justify your answer in terms of Coulomb’s law.
Less
than that of Po. The two atoms have comparable effective nuclear charges, but the valence electrons in Lv would be at a greater distance from the nucleus than those in Po. By Coulomb’s law, the attractive force between the valence electrons and the nucleus decreases by the inverse square of the distance between them.
d)
Shown below is a hypothetical mass spectrum for a sample of Lv containing 10
atoms.
Using the information in the graph, determine the average atomic mass of Lv in the
sample to four significant figures.
Average atomic mass = 2/10( 291.2 ) + 3/10( 292.2 )+ 5/10 ( 293.2 )=292.5 amu
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1.3 Elemental Composition of Pure Substances
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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.
Ethylene
Glycol, C2H6O2
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
C2H6O2
CH3O
C:H:O
1:3:1
C:H:O
12 : 3 : 16
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
N2O4
NO2
N:O
1:2
C:O
7 : 16
Benzene, C6H6
Dinitrogen
Tetroxide,N2O4
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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 =
mass of O =
2.50 g
3.13 - 2.50 = 0.63 g
moles of Cu =
mass of O =
2.50 / 63.55 = 0.393 mol
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 =
mass of Cl =
treat H2O like an 'element'
mass of H2O =
0.479 g
1.012 - 0.479 = 0.533 g
1.285 - 1.012 = 0.273 g
moles of Cu =
moles of Cl =
moles of H2O =
0.479 / 63.55 = 7.53 x 10-3 mol
0.533 / 35.45 = 1.50 x 10-2 mol
0.273 / 18.01 = 1.51 x 10-2 mol
divide each by smallest
CuClH2O
7.53 x 10-3
7.53 x 10-3
1.50 x 10-2
7.53 x 10-3
1 :
2 :
CuCl2 . 2H2O
ISPS Chemistry Aug 2024
page 16
1.51 x 10-2
7.53 x 10-3
2
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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
mass of H
mass of O
=
=
=
moles of C =
moles of H =
mass of O =
40.92 g
4.58 g
54.50 g
40.92 / 12.01 = 3.407 mol
4.58 / 1.008 = 4.54 mol
54.50 / 16.00 = 3.406 mol
divide each by smallest
CHO
3.407
3.406
4.54
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
⇩
moles of CO2
⇩ x 1
moles of O
moles of C
⇧ ⇩
subtract C and H mass of O
☜ mass of C
from
sample
Calculation for complex organics
ISPS Chemistry Aug 2024
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mass of H2O
⇩
moles of H2O
⇩x2
moles of H
⇩
mass of H
Calculation for
hydrocarbons
☞ empirical
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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 O
mass of H = 13.5 x 2.016/18.02 = 1.51 g
= 11.5 - (6.0 + 1.51) = 4.0 g
moles of C =
moles of H =
mass of O =
6.0 / 12.01 = 0.50 mol
1.51 / 1.008 = 1.50 mol
4.0 / 16.00 = 0.25 mol
divide each by smallest
CHO
0.50
0.25
1.50
0.25
2 :
6 :
0.25
0.25
C2 H6 O
1
Empirical formulae, by themselves, are not
conclusive proof of the identity of a molecule.
ethanol
However, when used with other tools like
IR Spectroscopy - identifies functional groups, and Mass Spectroscopy - identifies mass of molecule
dimethylether
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.
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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
mass of O
=
=
moles of Fe =
mass of O =
34.97 g
15.03 g
34.97 / 55.85
15.03 / 16.00
= 0.6261 mol
= 0.9394 mol
divide each by smallest
FeO
0.6261
0.6261
0.9394
0.6261
1 :
multiply to make whole numbers
2 :
ISPS Chemistry Aug 2024
1.5
3 Fe2O3
page 19
so iron(III) oxide
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1.3 Practice Problems
1.
A 23.0g sample of a compound contains 12.0g of C, 3.0g of H, and 8.0g of O.
Which of the following is the empirical formula of the compound?
A
2.
C
C3H9O2
D
C4H12O2
KTeO
B
KTe2O
C K TeO O
2
3
D
K2TeO6
E
K4TeO6
SF B
O
SF2 C SF4 D
SF6 E
SF2 B
SF3 C SF4 D
S2F
O SF
SF5 E
6
A sample of a compound that contains only the elements C, H, and N is completely burned
in O2 to produce 44.0 g of CO2, 45.0 g of H2O, and some NO2 .
A possible empirical formula of the compound is
A
6.
6
A compound contains 3.21 g of sulfur and 11.4 g of fluorine.
Which of the following represents the empirical formula of the compound?
A
5.
2
A compound contains 30. percent sulfur and 70. percent fluorine by mass.
The empirical formula of the compound is
A
4.
B C H O
O
A compound contains 1.10 mol of K, 0.55 mol of Te, and 1.65 mol of O.
What is the simplest formula of this compound?
A
3.
CH3O
CH2N
B CH N
O
5
C
C2H5N
D
C3H3N2
After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of 38 percent. The correct value for the percentage of water in the hydrate is 51 percent. Which of the following is the most likely explanation for
this difference?
A
Strong initial heating caused some of the hydrate sample to spatter out of the crucible.
C
The amount of the hydrate sample used was too small.
D
The crucible was not heated to constant mass before use.
E
Excess heating caused the dehydrated sample to decompose.
B The dehydrated sample absorbed moisture after heating.
O
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7.
Two different ionic compounds each contain only copper and chlorine. Both compounds are powders, one white and one brown. An elemental analysis is performed on each
powder. Which of the following questions about the compounds is most likely to be
answered by the results of the analysis?
A
What is the density of each pure compound?
C
What is the chemical reactivity of each compound?
D
Which of the two compounds is more soluble in water?
B What is the formula unit of each compound?
O
8.
Complete combustion of a sample of a hydrocarbon in excess oxygen produces equimolar quantities of carbon dioxide and water. Which of the following could be the
molecular formula of the compound?
A
9.
B
C2H6
C C H D C H
O
4
8
6
6
In which of the following compounds is the mass ratio of chromium to oxygen closest to 1.62 to 1.00?
A
10.
C2H2
CrO3
B CrO C CrO D Cr O O
2
2
E
Cr2O3
A student has two samples of NaCl , each one from a different source. Assume that the only potential contaminant in each sample is KCl . The student runs an experiment to
determine the percent by mass of chlorine in each sample. From the results of this
experiment alone, which of the following questions is most likely to be answered?
A Which sample has the higher purity?
O
11.
B
Which sample has the higher density?
C
What is the source of the contaminants present in each of the samples?
D
Which sample came from a salt mine, and which sample came from the ocean?
What is the empirical formula of an oxide of chromium that is 48 percent oxygen by mass?
A
12.
CrO
B
CrO2
C CrO D Cr O O
3
2
E
Cr2O3
What number of moles of O2 is needed to produce 14.2 grams of P4O10 from P?
(Molecular weight P4O10 = 284)
A
0.0500
ISPS Chemistry Aug 2024
B
0.0625
C
0.125
page 21
D 0.250 O
E
0.500
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13.
A student has samples of two pure compounds, XClO3 and ZClO3 , which contain
unknown alkali metals X and Z . The student measures the mass of each sample and then strongly heats the samples to drive off all the oxygen, leaving solid residues of XCl and ZCl
The student measures the mass of the solid residue from each sample. Which of the
following questions can be answered from the results of the experiment?
A Which has the greater molar mass, X or Z ?
O
14.
B
Which has the higher boiling point, X or Z ?
C
Which has the higher melting point, XCl or ZCl ?
D
Which has the greater density, XCl or ZCl ?
M+ is an unknown metal cation with a +1 charge. A student dissolves the chloride of the unknown metal, MCl, in enough water to make 100.0 mL of solution.
The student then mixes the solution with excessAgNO3solution, causingAgClto precipitate.
The student collects the precipitate by filtration, dries it, and records the data shown below.
(The molar mass of AgCl is 143 g/mol.)
mass of unknown chloride, MCl
0.74 g
mass of filter paper plus AgCl precipitate 2.23 g
mass of filter paper
0.80 g
What is the identity of the metal chloride?
A
15.
NaCl
B KCl C CuCl D LiCl
O
To determine the percentage of water in a hydrated salt, a student heated a 1.2346 g sample
of the salt for 30 minutes; when cooled to room temperature, the sample weighed 1.1857 g.
After the sample was heated for an additional 10 minutes and again cooled to room
temperature, the sample weighed 1.1632 g. Which of the following should the student do next?
A
Use the smallest mass value to calculate the percentage of water in the hydrated salt.
B
Repeat the experiment with a new sample of the same mass and average the results.
C
Repeat the experiment with a new sample that has a different mass.
E
Use the average of the mass values obtained after the two heatings to calculate the
percentage of water in the hydrated salt.
D Reheat the sample until its mass is constant.
O
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1.3 Quick Check FRQ
1.
Answer the following questions relating to gravimetric analysis.
In the first of two experiments, a student is assigned the task of determining the number of
moles of water in one mole of MgCl2 .nH2O. The student collects the data shown in the
following table.
a)
Explain why the student can correctly conclude that the hydrate was heated a
sufficient number of times in the experiment.
additional mass was lost during the third heating, indicating that all the water
No
of hydration had been driven off.
b)
Use the data above to
i)
calculate the total number of moles of water lost when the sample was heated,
25.825 − 23.976 = 1.849 g
and
ii)
1.848 g H2O ÷ 18.02 g mol-1 = 0.1026 mol H2O
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
c)
so MgCl2 . 6H2O
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 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
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Q1 contd.
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, 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
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
this information, calculate each of the following.
i)
The number of moles of MgCl2 in the original mixture
point is earned for calculating the number of moles of AgCl. precipitate, and
1
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
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
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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.
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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.
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1.4 Practice Problems
1.
A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, the percent of CaCO3 in the sample is
A
2.
B
30%
40%
C
70%
D
75%
O
E
100%
A student is given two 10g samples, each a mixture of only NaCl(s) and KCl(s) but in
different proportions. Which of the following pieces of information could be used to
determine which mixture has the higher proportion of KCl(s) ?
O The mass of Cl in each mixture
A
The volume of each mixtureB
C
The number of isotopes of Na and K
D
The reaction of each mixture with water
3.
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?
A
4.
B
1, 2, 5
2, 5, 6
C
3, 4, 6, 7 D
O
2, 3, 4, 5
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
C
The number of moles of Cl in the sample only
The mass of Mg in the sample only
D
The mass of Mg and the mass of Cl in the sample
O
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?
A
Water, H O
O
2
C
B
Fructose, C6H12O6, an isomer of glucose
ISPS Chemistry Aug 2024
page 27
D
Ribose, C5H10O5
Sucrose, C12H22O11
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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
A I only
7.
B I and II
C I and IV
D II and III
E I, III and IV
O
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
D Adding a solution containing an anion that forms an insoluble salt with only one of O
the metal ions
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1.4 Quick Check FRQ - same question was used earlier
1.
Answer the following questions related to the analysis of CaBr2.
a)
A student has a 10g sample of CaBr2 . Show the setup of the calculation to determine
the number of moles of CaBr2 in the sample. Include units in the setup.
(You do not need to do any calculations.)
mass ÷ molar mass
10 g CaBr2 ÷ 199.88 g mol-1 = number of moles of CaBr2
Note: Molar mass can be written with any number of significant figures.
b)
What number, in addition to the answer to part a), is needed to determine the
number of atoms of in the sample?
The
response gives the term “Avogadro’s number” or the value of Avogadro’s number,
6.022 x 1023 (with any number of significant figures, units of mol-1 or per mole are not
required).
c)
A different student is given a 10g sample labeled CaBr2 that may contain an inert
(nonreacting) impurity. Identify a quantity from the results of laboratory analysis that the student could use to determine whether the sample was pure.
The response gives an appropriate result (such as one of the following):
Mass of Ca in sample
Number of moles of Ca in sample
Mass of Br in sample
Number of moles of Br in sample
Mass or number of moles of element other than Ca or Br in sample
d)
Explain why CaCl2 is likely to have properties similar to those of CaBr2 .
The
response indicates that Cl is in the same group/family/column of the
periodic table as Br
OR
ISPS Chemistry Aug 2024
that a Cl atom has the same number of valence electrons as a Br atom.
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1.5 Atomic Structure &
Electron Configuration
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Electromagnetic Forces
There is not much that goes on in Chemistry that is not governed by the forces that exist between
charged particles.
The forces of repulsion and attraction are
enormously important in determining the stability and properties of the atoms that make up
everything in the universe.
The strength of these electromagnetic forces are determined by two factors:
*
the amount of charge , q1 and q2
(positive or negative)
F ∝ q1 x q2
*
the distance between charges, r
F ∝ 1 / r1 x r2
This can all be combined in what is known as Coulombs Law:
F ∝ q1 x q2
r2
Whenever possible, AP Chemistry requires explanations based on fundamental mathematical relationships such as Coulombs
Law.
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Electron Configurations
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.
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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.
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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.
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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 ......
4 - 2 = 2+
O:
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+
Li:
3 - 2 = 1+
Be:
8 - 2 = 6+
F:
9 - 2 = 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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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. O𝛿-—H𝛿+ 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
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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.
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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
A
2.
3.
A S
O
B
Ca C
Ga D
Sb E
Br
Which of the following ground-state electron configurations represents the atom that has the lowest first-ionization energy?
1s22s1
B
1s22s22p2
C
D 1s 2s 2p 3s
O
1s22s22p6
2
2
6
1
Which of the following is the ground-state electron configurations of the F— ion ?
A
5.
Ca C
The atom that contains exactly two unpaired electrons
A
4.
O Ga D Sb E Br
B
S
B
1s22s22p4
1s22s22p5
C 1s 2s 2p
O
2
2
D
6
1s22s22p63s23p6
Which of the following best represents the ground-state electron configuration for an atom
of selenium?
A
1s22s22p63s23p3
B
1s22s22p63s23p4
C 1s 2s 2p 3s 3p 4s 3d 4p D 1s 2s 2p 3s 3p 4s 3d 4p
O
2
6.
2
6
2
6
2
10
4
2
2
6
2
6
2
10
5
How many protons, neutrons, and electrons are in an 5266 Fe atom?
ProtonsNeutronsElectrons
A O
26
30
26
B
26
56
26
C
30
26
30
D
56
26
26
E
56
82
56
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7.
Of the following electron configurations of neutral atoms, which represents an atom in an excited state?
A
1s22s22p5
8.
D
B 1s 2s 2p 3s C 1s 2s 2p 3s
O
2
2
5
2
2
1s22s22p63s23p2 E
2
6
1
1s22s22p63s23p5
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
B
O
9.
Na has a greater density at standard conditions than Ne.
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.
Which of the following is the electron configuration of an excited atom that is likely to
emit a quantum of energy?
A
1s22s22p63s23p1
10.
D
B
C
1s22s22p63s23p5
1s22s22p63s1
1s22s22p63s2
E 1s 2s 2p 3s 3p
O
2
2
6
1
1
Which of the following represents a pair of iso­topes?
Atom 1Atom 2
Atomic Number
Mass Number
Atomic Number
Mass Number
A
614
714
B
67
1414
C 614
D 713
O
E 810
ISPS Chemistry Aug 2024
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714
1620
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11.
Which of the following represents the ground state electron configuration for the Mn3+ ion? (Atomic number Mn = 25)
A 1s 2s 2p 3s 3p 3d O
B
1s22s22p63s23p63d54s2
C
1s22s22p63s23p63d24s2
D
1s22s22p63s23p63d84s2 E
1s22s22p63s23p63d34s1
2
12.
2
6
2
6
4
Which of the following shows the correct number of protons, neutrons, and electrons in a neutral cesium-134 atom?
ProtonsNeutronsElectrons
A
B O
55
55
55
55
79
55
C
55
79
79
D
79
55
79
E
134
55
134
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1.5 Quick Check FRQ
1.
Use the principles of atomic structure and/or chemical bonding to explain each of the
following. In each part, your answers must include references to both substances.
a)
The atomic radius of Li is larger than that of Be.
1 point is earned for indicating that Be has more protons than Li
1 point is earned for indicating that since the electrons are at about the same distance from the nucleus,there is more attraction in Be as a result of the largernumber of protons
Both Li and Be have their outer electrons in the same shell (and/or they have the same number
of innercore electrons shielding the valence electrons from the nucleus). However, Be has four protons and Li has only three protons. Therefore, the effective nuclear charge experienced
(attraction experienced) bythe valence (outer) electrons is greater in Be than in Li, so Be has a smaller atomic radius.
b)
The second ionization energy of K is greater than the second ionization energy of Ca.
1 point is earned for saying that electrons are removed from an inner (third) level in potassium but
one level higher, (fourth level) in calcium
1 point is earned for saying that the distance to the nucleus is less for the third level, so attraction is
greater and more energy is needed to remove an electron
c)
The second electron removed from a potassium atom comes from the third level (inner core). The
second electron removed from a calcium atom comes from the fourth level (valence level). The electrons in the third level are closer to the nucleus so the attraction is much greater than for
electrons in the fourth level.
The carbon-to-carbon bond energy in C2H4 is greater than it is in C2H6.
1 point is earned for indicating that C2H4 has a double bond and C2H6 has a single bond
1 point is earned for indicating that the carbon-carbon double bond in C2H4 required more
energy to break (is stronger) than the carbon-carbon bond in C2H6
C2H4 has a double bond between the two carbon atoms, whereas C2H6 has a carbon-carbon single bond. More energy is required to break a double bond in C2H4 than to break a single bond in C2H6; therefore, the carbon-to-carbon bond energy in C2H4 is greater.
d)
The boiling point of Cl2 is lower than the boiling point of Br2.
point is earned for indicating that Cl2 and Br2 are both nonpolar and/or have only London
1
dispersion forces (or van der waals).
1 point for indicating that the more electrons, the more polarizable, the greater the dispersion forces, and the higher the boiling point.
Cl2 has less electrons than Br2 and they are closer to the nucleus (3 shells) than Br2 (4 shells) so Cl2
is less polarisable than Br2. Less energy will be required to overcome the London Dispersion forces
between Cl2 molecules those between the Br2 molecules; therefore, the boiling point of Cl2 is lower
than the boiling point of Br2.
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2.
Answer the following questions related to sulfur and one of its compounds.
a)
Consider the two chemical species S and S2-.
i)
Write the electron configuration (e.g., 1s2 2s2 . . .) of each species.
1 point is earned for the correct configuration for S.
S: 1s2 2s2 2p6 3s2 3p4
1 point is earned for the correct configuration for S2–.S2−: 1s2 2s2 2p6 3s2 3p6
Note: Replacement of 1s2 2s2 2p6 by [Ne] is acceptable.
ii)
Explain why the radius of the S2− ion is larger than the radius of the S atom.
1 point is earned for a correct explanation.
The nuclear charge is the same for both species, but the eight valence electrons in the sulfide ion
experience a greater amount of electron-electron repulsion than do the six valence electrons in the
neutral sulfur atom. This extra repulsion in the sulfide ion increases the average distance between
the valence electrons, so the electron cloud around the sulfide ion has the greater radius.
iii)
Which of the two species would be attracted into a magnetic field? Explain.
1 point is earned for the correct answer with a correct explanation.
The sulfur atom would be attracted into a magnetic field. Sulfur has two unpaired p electrons, which results in a net magnetic moment for the atom. This net magnetic moment would interact with an external magnetic field, causing a net attraction into the field.
The sulfide ion would not be attracted into a magnetic field because all the electrons in the species
are paired, meaning that their individual magnetic moments would cancel each other.
b)
The S2- ion is isoelectronic with the Ar atom. From which species, S2- or Ar, is it easier to remove an electron? Explain.
1 point is earned for the correct answer with a correct explanation.
It requires less energy to remove an electron from a sulfide ion than from an argon atom.
A valence electron in the sulfide ion is less attracted to the nucleus (charge +16) than is a valence electron in the argon atom (charge +18).
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1.6 Photoelectron Spectroscopy
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.
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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 rightThe 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
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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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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?
B 1s 2s 2p 3s 3p
O
A
1s22s22p1
C
1s22s22p63s23p6D
2
2
6
2
3
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.
D
In its compounds, the element tends to form ions with a charge of +3 .
C
In its compounds, the element tends to form ions with a charge of +1 .
O
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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.
D
The electrons in the 2p sublevel have the smallest binding energy.
O
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.
C
Peak Y, because F is more electronegative than C is.
D
Peak Y, because F has a smaller atomic radius than C has.
B
Peak X, because F has a greater nuclear charge than C has.
O
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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
D
Ca , because its nucleus has two more protons than the nucleus of Ar has
O
2+
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.
D
Nitrogen atoms have a smaller nuclear charge than oxygen atoms.
O
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1.6 Quick Check FRQ
1.
The complete photoelectron spectrum of an element in its ground state is represented
below.
a)
Based on the spectrum,
i)
write the ground-state electron configuration of the element,
1s2 2s2 2p6 3s2 3p6 4s2
and
ii)
or [Ar] 4s2
identify the element.
element is Ca
the
b)
Calculate the wavelength, in meters, of electromagnetic radiation needed to remove
an electron from the valence shell of an atom of the element.
The
response meets both of the following criteria.
The response indicates that the energy required is 0.980 x 10-18 J .
The response shows a calculation similar to the following.
E = h𝒗 = h c/λ
so λ = hc / E
λ = (6.6.26 x 10-34 J ) (2.998 x 108 ms-1 ) / 0.980 x 10-18 J
λ = 2.03 x 10-7 m (203 nm)
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2.
The photoelectron spectrum for an unknown element is shown below.
a)
Based on the photoelectron spectrum, identify the unknown element and write its electron configuration.
1s2 2s2 2p6 3s2 3p3
or [Ne] 3s2 3p3
the element is phosphorus, P
b)
Consider the element in the periodic table that is directly to the right of the element
identified in part a). Would the peak of this element appear to the left of, the right of, or in the same position as the peak of the element in part a)?
Explain your reasoning.
The response indicates that the peak would be to the left of the peak in the spectrum
shown.
The response indicates that the electron is in a lower energy state (and thus has a larger binding energy) because the nucleus of the element that is directly to the right
in the periodic table would have more protons (16) than the number of protons (15)
in the element corresponding to the given spectrum
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1.7 ISPS Chemistry Aug 2024
Periodic Trends
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Periodic Trends
Coulomb's Law can again be used to explain some aspects of Periodic Trends.
F ∝ (q1 x q2) / r2
Ionisation Energies:- The overall general trends are best explained using Coulombs Law.
As you go down a Group - Li - Fr or He - Rn - the ionisation energy decreases.
With each extra shell added, the distance r increases, so F decreases meaning that
it will become easier to remove an electron so ionisation energy decreases.
(Though charge on nucleus is increasing, this is cancelled out by the extra screening, so we consider atoms in the same group to have similar effective nuclear charge.)
As you go across a Period - Li - Ne or Na - Ar - the ionisation energy increases.
With each extra proton added, the effective nuclear charge q increases, so F increases
meaning that it will become harder to remove an electron so ionisation energy
increases.
However, the increase is not regular and we need to consider our shell model as well as Coulomb's law to fully explain the trend
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The increase from Li to Be is
expected and is explained by
increase in effective nuclear
charge.
The decrease from Be to B is
not expected, but is explained by
the fact that 2p orbitals are
further out and increase in r
must be having a bigger effect
than the increase in effective
nuclear charge.
The increase from B to C is
expected and is explained by
increase in effective nuclear
charge.
The decrease from N to O is not expected and cannot be fully explained simply using Coulomb's
Law. The effective nuclear charge, q has increased and we are in the same 2p orbitals so r should
be unchanged.
Instead, we explain the reduction in ionisation energy by assuming that the extra repulsions
caused by the first pairing of electrons is enough to counteract the increase in effective nuclear
charge.
The increase from O to F is expected and is explained by increase in effective nuclear charge.
Atomic & Ionic Radii:And again, the general trends are best explained using Coulombs Law.
Positive ions (cations) are always smaller than the original atom as the remaining electrons experience a
higher relative charge, q increases so
F increases which pulls the remaining electrons closer so the radius decreases.
With simple ions, all of the electrons in the outer shell are lost so the ion has
one shell less and will much smaller as
a result.
Na 1s2 2s2 2p6 3s1 ⇨ Na+ 1s2 2s2 2p6
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Negative ions (anions) are always larger than the original atom as the increased number of electrons
experience a lower relative charge,
q decreases so F decreases which
allows the remaining electrons to move further away so the radius increases.
O 1s2 2s2 2p6 ⇨ O2- 1s2 2s2 2p8
This time there is no change in the number of electron shells, just the same number of protons having to
attract more electrons.
As you go down a Group - for both atoms and ions - the radius increases.
With each extra shell added, the distance r increases.
As you go across a Period - atoms - the radius decreases.
With each extra proton added, the effective nuclear charge q increases, so F increases
meaning that it the outer shell will be pulled closer to nucleus so radius decreases.
As you go across a Period - positive ions only - the radius decreases.
With each extra proton added, the effective nuclear charge q increases, so F increases
meaning that it the outer shell will be pulled closer to nucleus so radius decreases.
As you go across a Period - negative ions only - the radius decreases.
With each extra proton added, the effective nuclear charge q increases, so F increases
meaning that it the outer shell will be pulled closer to nucleus so radius decreases.
Electron Affinities:-
The electron affinity (EA) of an element is the energy change that
occurs when an electron is added to a gaseous atom to give an anion.
E(g) + e −→ E−(g)
While ionisation energies and electronegativities have quite definite trends which can, therefore,
be 'easily' explained, electron affinities are harder to describe and explain.
In general, elements with the most negative electron affinities (the highest affinity for an added
electron) are those with the smallest size and highest ionization energies and are located in the
upper right corner of the periodic table.
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Taking the halogens for example, the increasing negative value from At to Cl can be explained by
the fact that the electron gained is entering a shell that is closer to the nucleus and so would be
attracted more strongly.
With Fluorine, however, the electron is entering the 2nd shell, which is significantly smaller than
the 3rd shell (for Chlorine), and the extra repulsions mean that, despite being closer to the
nucleus, the electron affinity is less negative for fluorine then chlorine.
Electronegativity:-
measures an atoms ability to attract the electrons in a shared pair
- in a covalent bond
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
The properties discussed in this section (size of atoms and ions, effective nuclear charge,
ionisation energies, and electron affinities) are central to understanding chemical reactivity.
For example, because fluorine has an energetically favorable EA and a large energy barrier to
ionisation (IE), it is much easier to form fluorine anions (F—) than cations (F+).
Metallic properties depend on having electrons that can be removed easily. Similarly for oxidising
& reducing agents.
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1.7 Practice Problems
1.
Which of the following best helps to explain why the electron affinity of Br has a greater magnitude than that of I ?
A
Br has a lower electronegativity than I does.
B
Br has a lower ionization energy than I does.
C
An added electron would go into a new shell in Br but not in I.
D
There is a greater attraction between an added electron and the nucleus in Br
O
than in I.
2.
Which of the following best helps to explain why the electronegativity of Cl is less than
that of F ?
A
The mass of the Cl atom is greater than the mass of the F atom.
B
The Cl nucleus contains more protons than the F nucleus contains.
C
When Cl and F form bonds with other atoms, the Cl bonding electrons are more O
shielded from the positive Cl nucleus than the F bonding electrons are shielded
from the positive F nucleus.
D
Because Cl is larger than F, the repulsions among electrons in the valence shell of Cl
are less than the repulsions among electrons in the valence shell of F.
3.
Which of the following elements has the highest electronegativity?
A
4.
5.
6.
Cs B
Ag C
O Br E Se
Pb D
Which of the following elements has the largest atomic radius?
A Cs B Ag C Pb D Br E Se
O
Which of the following elements has the lowest first-ionisation energy?
A Cs B Ag C Pb D Br E Se
O
Which of the following represents an electron configuration that corresponds to the
valence electrons of an element for which there is an especially large jump between the second and third ionization energies? (Note: n represents a principal quantum number equal to or greater than 2.)
A ns B ns np O
2
ISPS Chemistry Aug 2024
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1
C
ns2np2
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D
ns2np3
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7.
Zn(s) is used to reduce other compounds in chemical reactions. If a chemist needs a
substance that is more effective in its reducing ability, which of the following species would be the best choice?
A
Na
O
8.
9.
B
H+
C
K+
D
Cl—
The elements in which of the following have most nearly the same atomic radius?
A
Be, B, C, N
D
C, P, Se, I
B
Ne, Ar, Kr, Xe
E
Cr, Mn, Fe, Co
O
C
Mg, Ca, Sr, Ba
The first five ionization energies of a second-period
element are listed in the table opposite.
Which of the following correctly identifies the element
and best explains the data in the table?
A
OB
10.
B, because it has five core electrons
B, because it has three valence electrons
C
N, because it has five valence electrons
D
N, because it has three electrons in the p sublevel
Based on periodic trends and the data in the
table opposite, which of the following are the
most probable values of the atomic radius and
the first ionization energy for potassium,
respectively?
A
242 pm
633 kJ/mol
242 pm
419 kJ/mol
C
120 pm
633 kJ/mol
D
120 pm
419 kJ/mol
OB
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11.
The table opposite shows the first ionization energy and atomic radius of several elements.
Which of the following best helps to explain the
deviation of the first ionization energy of oxygen
from the overall trend?
A
The atomic radius of oxygen is greater than
the atomic radius of fluorine.
B
The atomic radius of oxygen is less than
the atomic radius of nitrogen.
D
There is attraction between paired electrons
in oxygen’s 2p orbitals.
C
There is repulsion between paired electrons
O
in oxygen’s 2p orbitals.
12.
The ionization energies for element X are listed in the table below.
On the basis of the data, element X is most likely to be
A
13.
Na
B
Mg
C
Al
O
D
Si
E
P
Which of the following best helps to account for the fact that the F­— ion is smaller than
the O2- ion?
A
OB
F­— has a larger nuclear mass than O2- has.
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.
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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
D
Ar, because of its higher effective nuclear charge
O
15.
Which of the following elements has the largest first ionization energy?
A
16.
C
Be
B
D
E
O
C
N
Which of the following lists Mg, P, and Cl in order of increasing atomic radius?
A
Cl < P < Mg
O
D
17.
Li B
Mg < Cl < P
B
Cl < Mg< P
E
P < Cl < Mg
C
Mg< P < Cl
Which of the following properties generally decreases across the periodic table from
sodium to chlorine?
A
First ionization energy
B
D
Maximum value of oxidation number
Atomic mass
OE
C
Electronegativity
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.
D
The ionic radius increases with successive ionizations
C
The electron removed during the third ionization is, on average, much closer to
O
the nucleus than the first two electrons removed were.
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1.7 Quick Check FRQ
1.
A student learns that ionic compounds have
significant covalent character when a cation
has a polarizing effect on a large anion.
As a result, the student hypothesizes that salts composed of small cations and large
anions should have relatively low melting
points.
a)
Select two compounds from the table and explain how the data support the student’s
hypothesis.
point is earned for choosing an appropriate pair of compounds (LiI/KI, LiI/LiF, or
1
LiI/NaF).
1 point is earned for an explanation that supports the hypothesis.
e.g.
LiI and KI. LiI has a small cation and a large anion and KI has a large cation and the same large anion.
The melting point of LiI (with its smaller cation) is lower than that of KI.
b)
Identify a compound from the table that can be dissolved in water to produce a basic
solution. Write the net ionic equation for the reaction that occurs to cause the
solution to be basic.
1 point is earned for choosing one of the correct compounds. Either LiF or NaF.
1 point is earned for writing a correct balanced equation. F- + H2O ⇄ HF + OH
2.
Some binary compounds that form between fluorine and various nonmetals are listed in the table below.
Nonmetal
C
N
O
Ne
Si
P
S
Ar
Formula of
Compound
CF4
NF3
OF2 compound SiF4
PF3
SF2
No
compound
No
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
number of valence electrons in the nonmetal atom.
Based on the student’s hypothesis, what should be the formula of the compound that
forms between chlorine and fluorine?
1 point is earned for the correct formula.
ISPS Chemistry Aug 2024
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ClF
Atomic Structures & Properties
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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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Atomic Structures & Properties
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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.
electronegativity
Al
1.6
F
3.98
∆EN
= 2.38
prediction
ionic
MPt (°C)
1290
Al
1.6
Cl
3.16
Al
1.6
Br
2.96
= 1.56
= 1.36
polar covalent
polar covalent
193
98
Al
1.6
I
2.66
= 1.06
polar covalent
188
Structurenetworkmolecularmolecularmolecular
Empirical
Formula
AlF3
AlCl3
AlBr3
AlI3
Molecular Formula
Al2Cl6
Al2Br6
Al2I6
ISPS Chemistry Aug 2024
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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 (⇣) ISPS Chemistry Aug 2024
electronegativity (⇡)
page 64
ionisation energy (⇡)
Atomic Structures & Properties
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1.8 Practice Problems
1.
1s2 2s2 2p6 3s2 3p3 Atoms of an element, X, have the electronic con­figu­ration shown. The compound most likely formed with magnesium, Mg, is
A
2.
Mg2X
C
MgX2
D
MgX3
E Mg X
O
3
2
O Ba C Al D Cl E Ne
LiB
Which of the following forms monatomic ions with 2- charge in solutions
A
4.
B
Atoms of Mg combine with atoms of F to form a compound. Atoms of which of the
following elements combine with atoms of F in the same ratio?
A
3.
MgX
F
B S
O
C
Mg D
Ar E
Mn
If Na reacts with chlorine to form NaCl , which of the following elements reacts with Na
to form an ionic compound in a one-to-one ratio, and why?
A
K, because it is in the same group as Na.
B
Mg, because its mass is similar to that of Na.
C
Ar, because its mass is similar to that of Cl.
D
Br, because it has the same number of valence electrons as Cl.
O
5.
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.
C
The two valence electrons of alkaline earth metal atoms are relatively easy to
O
remove
D
The radii of atoms of alkaline earth metals increase moving down the group from Be to Ra.
ISPS Chemistry Aug 2024
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Atomic Structures & Properties
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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.
D
CsCl , because its elements have very different electronegativities and it is an ionic O
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
D
An ionic solid
C
A molecular solid with hydrogen bonding
O
8.
9.
E
A metallic solid
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.
C
Rb will form only Rb2O and Rb2O2.
D Rb will form Rb O, Rb O and RbO .
O
2
2
2
2
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
C X O D X O
O
2
10.
3
2
5
Which of the following ions has the same number of electrons as Br— ?
A
Ca2+ B
ISPS Chemistry Aug 2024
O Sr D I E Cl
K+ C
page 66
2+
—
—
Atomic Structures & Properties
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1.8 Quick Check FRQ
1.
a)
What type of chemical bond is present in the Cl2 molecule?
The response indicates that the bond is covalent.
(No explanation is required, however, the bond is covalent because the two atoms have the same electronegativity and will form a nonpolar covalent bond).
b)
Cl2 reacts with the element Sr to form an ionic compound. Based on periodic
properties, identify a molecule, X2 , that is likely to to react with Sr in a way similar to how Cl2 reacts with Sr. Justify your choice.
The response meets both of the criteria below:
F2 , Br2 , I2 or At2 is written.
the element chosen is in the same group or family as Cl (or Cl2) .
ISPS Chemistry Aug 2024
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Atomic Structures & Properties