AP e 1 Unit 1 - Atomic Structure & Properties

AP Chemistry Unit 1

Atomic Structures & PropertiesISPS Chemistry July 2020 page 1

Unit 1 - Atomic Structure& Properties

1.1 Moles and Molar Mass 1.2 Mass Spectroscopy 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

AP Chemistry Unit 1


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AP Chemistry Unit 1


1.1 Moles & Molar Mass

AP Chemistry Unit 1


The Atomic Mass Scale

Scientists of the nineteenth century were aware that atoms of different elements have differentmasses 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).

AP Chemistry Unit 1


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

AP Chemistry Unit 1


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

AP Chemistry Unit 1


1.1 Practice Problems1. 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 B 0.2 C 0.3 D 0.4 E 0.6

2. 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

A 0.80 B 0.60 C 0.50 D 0.40 E 0.20

3. How many carbon atoms are contained in 2.8 g of C2H4 ?

A 1.2 x 1023 B 3.0 x 1023 C 6.0 x 1023 D 1.2 x 1024 E 6.0 x 1024

4. 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 1023 potassium atoms D 4 moles of oxygen atoms

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

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

D Molar mass of the compound, density of the sample

O

O

O

O

O

O

AP Chemistry Unit 1


7. Which of the following numerical expressions gives the number of particles in 2.0g of Ne ?

A B

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?

A Na2O B MgO

C K2O D CaO

1.1 Quick Check FRQ1. 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 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

O

O

AP Chemistry Unit 1


1.2 Mass Spectroscopy of Elements

AP Chemistry Unit 1


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 calculationsthere 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. the number of isotopes present, i (2 peaks so 2 isotopes) 2. the atomic mass of each isotope ('35' and '37') 3. the relative amount of each isotope (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.

AP Chemistry Unit 1


1.2 Practice Problems1. The mass spectrum of the element Sb is most likely represented by which of the following?

A B

C D

2. Which of the following elements has the mass spectrum represented opposite?

A Nb B Mo

C U D Cf

3. 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

B Mg with 14 neutrons

C Fe with 26 neutrons

D Ti with 26 neutrons

O

O

O

AP Chemistry Unit 1


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.

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

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

O

AP Chemistry Unit 1


1.2 Quick Check FRQ1. 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

AP Chemistry Unit 1


1.3 Elemental Composition of Pure Substances

AP Chemistry Unit 1


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.

MolecularFormula

EmpiricalFormula

MoleRatio

MassRatio

H2O H2OH : O2 : 1

H : O1 : 8

H2O2 HO H : O1 : 1

H : O1 : 16

Ethylene Glycol, C2H6O2

C2H6O2 CH3OC : H : O1 : 3 : 1

C : H : O12 : 3 : 16

Benzene, C6H6

C6H6 CH C : H1 : 1

C : O12 : 1

C2H4 CH2C : H1 : 2

C : H6 : 1

CH4 CH4C : H1 : 4

C : H3 : 1

Dinitrogen Tetroxide,N2O4

N2O4 NO2N : O1 : 2

C : O7 : 16

AP Chemistry Unit 1


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 gtreat 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

AP Chemistry Unit 1


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 : 1multiply 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 moles of O moles of C moles of H ☞ empirical ⇧ ⇩ ⇩ mass of O ☜ mass of C mass of H

Calculation forhydrocarbons

Calculation for complex organics

subtract C and Hfrom sample

AP Chemistry Unit 1


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 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.

ethanol

dimethylether

AP Chemistry Unit 1


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.

copper(II) chlorideCuCl2

titanium(IV) oxideTiO2

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 : 1.5multiply to make whole numbers 2 : 3 Fe2O3 so iron(III) oxide

AP Chemistry Unit 1


1.3 Practice Problems1. 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 CH3O B C2H6O C C3H9O2 D C4H12O2

2. 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 KTeO B KTe2O C K2TeO3 D K2TeO6 E K4TeO6

3. A compound contains 30. percent sulfur and 70. percent fluorine by mass. The empirical formula of the compound is

A SF B SF2 C SF4 D SF6 E S2F

4. 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 SF2 B SF3 C SF4 D SF5 E SF6

5. 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 CH2N B CH5N C C2H5N D C3H3N2

6. 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. B The dehydrated sample absorbed moisture after heating.

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.

O

O

O

O

O

O

AP Chemistry Unit 1


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?

B What is the formula unit of each compound?

C What is the chemical reactivity of each compound?

D Which of the two compounds is more soluble in water?

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 C2H2 B C2H6 C C4H8 D C6H6

9. In which of the following compounds is the mass ratio of chromium to oxygen closest to 1.62 to 1.00?

A CrO3 B CrO2 C CrO D Cr2O E Cr2O3

10. 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?

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?

11. What is the empirical formula of an oxide of chromium that is 48 percent oxygen by mass?

A CrO B CrO2 C CrO3 D Cr2O E Cr2O3

12. What number of moles of O2 is needed to produce 14.2 grams of P4O10 from P? (Molecular weight P4O10 = 284)

A 0.0500 B 0.0625 C 0.125 D 0.250 E 0.500

O

O

O

O

O

O

AP Chemistry Unit 1


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 ?

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 ?

14. 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 0.80 g

mass of filter paper plus AgCl precipitate 2.23 g

What is the identity of the metal chloride?

A NaCl B KCl C CuCl D LiCl

15. 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.

D Reheat the sample until its mass is constant.

E Use the average of the mass values obtained after the two heatings to calculate the percentage of water in the hydrated salt.

O

O

O

AP Chemistry Unit 1


1.3 Quick Check FRQ1. 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.

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

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.

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 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.

AP Chemistry Unit 1


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

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

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

AP Chemistry Unit 1


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.

AP Chemistry Unit 1


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.

AP Chemistry Unit 1


1.4 Practice Problems1. 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 30% B 40% C 70% D 75% E 100%

2. 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) ?

A The volume of each mixture B The mass of Cl in each mixture

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 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

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?

A Water, H2O B Ribose, C5H10O5

C Fructose, C6H12O6, an isomer of glucose D Sucrose, C12H22O11

O

O

O

O

O

AP Chemistry Unit 1


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 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

D Adding a solution containing an anion that forms an insoluble salt with only one of the metal ions

O

O

AP Chemistry Unit 1


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 that a Cl atom has the same number of valence electrons as a Br atom.

AP Chemistry Unit 1


1.5 Atomic Structure & Electron Configuration

AP Chemistry Unit 1


Electromagnetic ForcesThere 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.

AP Chemistry Unit 1


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.

AP Chemistry Unit 1


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.

AP Chemistry Unit 1


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.

AP Chemistry Unit 1


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, thenthe 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.

AP Chemistry Unit 1


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

AP Chemistry Unit 1


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

AP Chemistry Unit 1


MagnetismThe 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 ensuresthat paired electrons have opposite spins so their magneticproperties cancel out.

Most magnetic properties areshown by transition metals asHund's Rule of MaximumMultiplicity ensures that up to5d orbitals can be occupied bysingle 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. Thiscan best be achieved with iron atoms. Repeated application of amagnetic field (and heat) will leave the iron atoms lined up and apermanent magnetic effect can be produced - ferromagnetism.

AP Chemistry Unit 1


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 S B Ca C Ga D Sb E Br 2. The atom that contains exactly two unpaired electrons 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? A 1s22s1 B 1s22s22p2 C 1s22s22p6 D 1s22s22p63s1

4. Which of the following is the ground-state electron configurations of the F— ion ? 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

C 1s22s22p63s23p64s23d104p4 D 1s22s22p63s23p64s23d104p5

6. How many protons, neutrons, and electrons are in an 5266 Fe atom?

Protons Neutrons Electrons

A 26 30 26

B 26 56 26

C 30 26 30

D 56 26 26

E 56 82 56

O

O

O

O

O

O

AP Chemistry Unit 1


7. Of the following electron configurations of neutral atoms, which represents an atom in an excited state?

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.

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

D 1s22s22p63s1 E 1s22s22p63s13p1

10. Which of the following represents a pair of iso topes?

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 D 7 13 7 14

E 8 10 16 20

O

O

O

O

AP Chemistry Unit 1


11. Which of the following represents the ground state electron configuration for the Mn3+ ion? (Atomic number Mn = 25)

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

B 55 79 55

C 55 79 79

D 79 55 79

E 134 55 134

O

O

AP Chemistry Unit 1


1.5 Quick Check FRQ1. 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

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.

c) 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.

1 point is earned for indicating that Cl2 and Br2 are both nonpolar and/or have only London 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.

AP Chemistry Unit 1


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).

AP Chemistry Unit 1


1.6 Photoelectron Spectroscopy

Photoelectron spectroscopy (PES)Photoelectron spectroscopy (PES) was formerly known as X-ray electron spectroscopy or XES. Itinvolves 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.

AP Chemistry Unit 1


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 electronsp-orbitals - 1 - 6 electronsd-orbitals - 1 - 10 electons

The highest binding energy (on left) hasto belong to the orbital closest to the nucleus so 1s2

This also establishes the height that isequivalent 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

AP Chemistry Unit 1


Explaining RelationshipsCoulomb'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.

AP Chemistry Unit 1


1.6 Practice Problems1.

The complete photoelectron spectrum of an element is given above. Which of the following electron configurations is consistent with the spectrum? 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.

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 .

O

O

AP Chemistry Unit 1


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.

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.

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.

O

O

AP Chemistry Unit 1


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 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.

D Nitrogen atoms have a smaller nuclear charge than oxygen atoms.

O

O

AP Chemistry Unit 1


1.6 Quick Check FRQ1. 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 or [Ar] 4s2

and ii) identify the element.

the element is Ca

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 = h c / 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)

AP Chemistry Unit 1


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

AP Chemistry Unit 1


1.7 Periodic Trends

AP Chemistry Unit 1


Periodic TrendsCoulomb'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

AP Chemistry Unit 1


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 rmust 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 C to N 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

AP Chemistry Unit 1


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.


As you go across a Period - negative ions only - the 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.

AP Chemistry Unit 1


1.7 Practice Problems1. 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 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 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 Cs B Ag C Pb D Br E Se

4. Which of the following elements has the largest atomic radius?


5. Which of the following elements has the lowest first-ionisation energy?


6. 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 ns2 B ns2np1 C ns2np2 D ns2np3

O

O

O

O

O

O

AP Chemistry Unit 1


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

B H+

C K+

D Cl—

8. The elements in which of the following have most nearly the same atomic radius?

A Be, B, C, N B Ne, Ar, Kr, Xe C Mg, Ca, Sr, Ba

D C, P, Se, I E Cr, Mn, Fe, Co

9. 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 B, because it has five core electrons

B 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

10. 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

B 242 pm 419 kJ/mol

C 120 pm 633 kJ/mol

D 120 pm 419 kJ/mol

O

O

O

O

AP Chemistry Unit 1


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.

C There is repulsion between paired electrons in oxygen’s 2p orbitals.

D There is attraction between paired electrons 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 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.

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.

O

O

O

AP Chemistry Unit 1


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

15. Which of the following elements has the largest first ionization energy?

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?

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

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.

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

O

O

O

O

O

AP Chemistry Unit 1


1.7 Quick Check FRQ1. 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.

1 point is earned for choosing an appropriate pair of compounds (LiI/KI, LiI/LiF, or 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 above.

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. ClF

AP Chemistry Unit 1


1.8 Valence Electrons & Ionic Compounds

Bonding & EnergyRegardless 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.

AP Chemistry Unit 1


Type of BondAll 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 probablyelectronegativity.

As atoms approach and overlap it will be differences in their ability to attract electronsthat will largely determine the type of bond thatforms.

However, this is not 'black & white' - bonding is acontinuum 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 Ielectronegativity 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 AlI3Formula

Molecular Formula Al2Cl6 Al2Br6 Al2I6

AP Chemistry Unit 1


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 becomemore 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 (⇡)

AP Chemistry Unit 1


1.8 Practice Problems1. 1s2 2s2 2p6 3s2 3p3 Atoms of an element, X, have the electronic con figu ration shown. The compound most likely formed with magnesium, Mg, is

A MgX B Mg2X C MgX2 D MgX3 E Mg3X2

2. 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 Li B Ba C Al D Cl E Ne

3. Which of the following forms monatomic ions with 2- charge in solutions

A F B S C Mg D Ar E Mn

4. 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.

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 remove

D The radii of atoms of alkaline earth metals increase moving down the group from Be to Ra.

O

O

O

O

O

AP Chemistry Unit 1


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 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

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.

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

C X2O3 D X2O5

10. Which of the following ions has the same number of electrons as Br— ?

A Ca2+ B K+ C Sr2+ D I— E Cl—

O

O

O

O

O

AP Chemistry Unit 1


1.8 Quick Check FRQ1. 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) .

AP e 1 Unit 1 - Atomic Structure & Properties

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