Problem Set 10 Periodic Trends PDF

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Chemistry NYA: Suggested Readings and Practice Problems Problem Set 10: Periodic Trends Chemistry A Molecular Approach, 4th edition, N. J. Tro Readings Chapter 2: 2.7 Chapter 9: 9.5, 9.6, 9.7 (Ionic Radii, Ionization Energy, Trends in First Ionization Energy, Exceptions to Trends in First Ionization Energy, Trends in Second and Successive Ionization Energies), 9.8, 9.9 Practice Problems Chapter 9: 26, 27, 28, 29, 32, 33, 34, 35, 36, 37, 55, 57, 59, 61, 67, 69, 71, 73, 75, 77, 91b, c, 93, 97, 98, 105, 106, 125

Additional Practice Problems 1. Arrange the following species in increasing order of size: O+ , O , O– 2. Arrange the following species in increasing order of first ionization energy (IE1): Ge , As , Se 3. Explain the trend in electron affinity:

Species EA (kJ •mol–1)

Al –44

Si –120

P –74

4.

For the following pairs of elements: N and O, choose the one with a: (a) more favourable electron affinity (b) higher first ionization energy (c) larger size

5.

Consider each set of successive ionization energies separately. (a) Explain which element in period 3 has the following ionization energies and write its electron configuration. IE kJ• mol–1

IE1 1012

IE2 1903

IE3 2910

IE4 4956

IE5 6278

IE6 22,230

(b) Explain (i) which element in period 3 has the following ionization energies and (ii) write its electron configuration. (iii) How many core electrons are present in this element? (iv) Calculate ‘Zeff’ for the first valence electron and then for the first core electron and explain the significance for each. IE kJ• mol–1

IE1 577

IE2 1816

IE3 2744

IE4 11,576

IE5 14,829

IE6 18,375

6.

Use the Bohr model to calculate the value of ‘Zeff’ for each species below given the ‘IE’ value for: (a) the one-electron species ‘H’, if IE1 is 1312 kJ/mol. (b) the multielectron species ‘Li’, if IE1 is 519 kJ/mol. (c) the multielectron species ‘Li’, if IE2 is 7298 kJ/mol. (The second ionization energy for ‘Li’ is removing a second mole of electrons from ‘Li’ which refers to removing one mole of electrons from ‘Li+’.) (d) the multielectron species ‘Li’, if IE3 is 11815 kJ/mol. (The third ionization energy for ‘Li’ is removing a third mole of electrons from ‘Li’ which refers to removing one mole of electrons from ‘Li+2’.)

7.

For the following pairs of elements: Mg or Al , explain which one has the larger: (i) IE1, (ii) IE2, (iii) IE3.

8.

Arrange the following species in increasing order of size: S–2 , Cl–1 , K+1 , Ca+2

9.

Consider the following situations involving the Group IA Alkali metals (Li, Na, K, Rb, Cs) and the Group IIA Alkaline Earth metals (Be, Mg, Ca, Sr, Ba, Ra) (a) In general, the ability to lose electrons (reducing ability) increases down the group, explain why? (b) Choose the smallest alkali metal (excluding hydrogen) in this group and write its balanced reaction with water. (c) Choose the alkali metal with the third highest ionization energy (including hydrogen) within the group and write the balanced reaction with hydrogen gas. (d) Choose the alkali metal with the most favourable electron affinity that is larger in size than sodium (Na) in that group and write the balanced reaction with the halogen whose anion would be isoelectronic with krypton (Kr). (e) Beryllium and Magnesium react with oxygen at high temperature to form their oxides. Write the balanced reaction for the larger of these two species. (f) Calcium, Strontium, and Barium react with water at room temperature. Write a balanced reaction with water for only the element that has the smallest ionization energy.

Answers: 1. O+ (smallest) , O , O– (largest) These are the same elements that differ by only the number of electrons. The number of protons is fixed at 8 p+. We expect that the positively charged oxygen to be smallest in size since it has one less electron. As a result it will have lower e- -e- repulsion which results in a larger p+-e- attraction. The negatively charged oxygen will be largest in size since it has one more electron. As a result it will have greater e- -e- repulsion which results in a smaller p+-eattraction. The neutral oxygen will have a size that is intermediate to the negatively and positively charged oxygen ions. In order to assess the level of attraction for these electrons in each species, we can calculate the ‘Zeff’ for each species. If we calculate the simplified ‘Zeff’ we would get the same value since we would be considering only the core electrons in the calculation. If we calculate the more rigorous ‘Zeff’ we can notice a difference in value and make a better assessment.

O+ :

1s22s22p3 (one electron has been removed) simplified: Zeff O+ = 8-2=6 more rigorous: Zeff O+ = 8-2-1/2(4)=4

O:

1s22s22p4 (neutral, no electrons removed) simplified: Zeff O = 8-2=6 more rigorous: Zeff O = 8-2-1/2(5)=3.5

O– :

1s22s22p5 (one electron has been added)

simplified: Zeff O- = 8-2=6 more rigorous: Zeff O- = 8-2-1/2(6)=3

In going from ‘O+’ to ‘O’ to ‘O–’, the more rigorous ‘Zeff’ is showing a decline in value. As ‘Zeff’ decreases, there is, on average, less attraction for outer electrons, and size increases. 2.

Ge (smallest) , Se , As (largest) ‘IE’ is the energy required to remove one mole of electrons from a gaseous element. These elements are all in the same period. Across a period, ‘Zeff’ increases, p+-e- attraction increases, and size decreases. As a result, the valence electrons, are held with greater attraction to the nucleus and are more difficult to remove. We expect the IEGe to be the smallest. In comparing the ‘IE’ values of Group 5A and 6A elements we note that there is an exception. Writing the spdf notation for each we get: As: 1s22s22p63s23p64s23d104p3

Se: 1s22s22p63s23p64s23d104p4

Draw the orbital (box) diagram for the orbital from which the electron will be removed. We can see that the box diagram for ‘As’ has a single electron in each orbital while the ‘Se’ shows a first electron pair. The electron that is likely to be removed is circled. We expect that there will be added e- -e- repulsion in ‘Se’ that will lower the attraction for the electron being removed and it will be easier to remove that electron. As a result, ‘IESe’ will be slightly lower than ‘IEAs’. This is an exception to the expected trend that occurs between Group5A and Group 6A elements for the first four periods. ↑

↑

4p 3.

↑

↑↓

↑

↑

4p

The first electron affinity (EA) is the energy change, per mol, that occurs when an electron is added to a neutral gaseous atom to form a gaseous ion. If the value of EA is negative this implies that the process is exothermic and produces a stable gaseous ion. In general, the EA values tend to get more negative in going from left to right across a period. This is a result of the effective nuclear charge (Zeff) that is increasing significantly when going across a period. Therefore, when an electron is added to form the gaseous ion, the electron undergoes an attraction to the nucleus that is large enough to form a stable gaseous ion. This helps to explain that the values from ‘Al’ to ‘P’ that become more negative across the period. In these particular series of values, there is a deviation in going from the element ‘Si’ to ‘P’ since in addition to attraction between protons and electrons, there are also repulsive forces occurring between electrons. Notice that the EA value for ‘P’ is less negative than that of ‘Si’. This can be explained by considering that the electron in the ‘P’ atom for EA is being added to the ‘3p’ subshell which is already half-filled. The electron in the ‘Si’ atom

for EA is being added to the ‘3p’ subshell which has an empty orbital. Therefore, when we consider the repulsive forces between the added electron in ‘Si’ and the added electron in ‘P’, the electron added in the ‘P’ atom is undergoing stronger repulsive forces since it forms the first electron pair within that subshell in comparison to ‘Si’, where the added electron causes the ‘3p’ subshell to become half-filled. The greater repulsion in forming the first electron pair in ‘P’ is reflected by a less negative EA value, and the ‘P– (g)’ ion is more difficult to form than the ‘Si– (g)’ ion. Si: P: ↑

↑

↑

↑

↑

↑

3p

↓

3p

4. (a) ‘O’ has more favourable electron affinity (more negative value)… (b) ‘N’ has higher first ionization energy… (c) ‘N’ has larger size… 5. (a) Since there is a large increase (“jump”) in the ionization energy for this element in

going from IE5 to IE6, this indicates that the element has five valence electrons since the sixth electron removed required much more energy and corresponds to the removal of its first core electron in this sequence. We are given that the element belongs to Period 3, and therefore must correspond to: Phosphorous, P, with an electron configuration 1s22s22p63s23p3. (b) (i) Since there is a large increase (“jump”) in the ionization energy for this element in

going from IE3 to IE4, this indicates that the element has three valence electrons since the fourth electron removed required much more energy and corresponds to the removal of a first core electron in this sequence. We are given that the element belongs to Period 3, and therefore must correspond to: Aluminum, Al. (ii) Aluminum has an electron configuration 1s22s22p63s23p1. (iii) In Aluminum there are 10 core electrons. (iv) The ‘Zeff’ are as follows: First valence electron:

simplified: more rigorous:

Zeff=13-10=+3 Zeff=13-10-1/2(2)=+2

First core electron:

simplified: more rigorous:

Zeff=13-2=+11 Zeff=13-2-1/2(7)=+7.5

The value of ‘Zeff’ gives an indication on the level of attraction that a chosen electron has towards the nucleus in that atom. A larger ‘Zeff’ implies a larger attraction and vice versa. As we can see there is a large increase in the value of ‘Zeff’ in going from the first valence electron to the first core electron, and therefore a large increase in the level of attraction. When the attraction increases, the energy required to remove that electron also increases. Therefore the large increase in ‘Zeff’ that occurs from a valence electron to the first core electron, correspond to a large increase in the energy required to remove that electron and we observe a characteristic “jump” in successive IE values for a given element when removing its first core electron.

6. (a) When we consider ionization energy (kJ/mol), we remove one mole of electrons starting from a lower energy quantum state and we bring the electron far away from the nucleus, where nf→∞. Also, in order to make use of the Bohr model, which is a convenient model, we must work with a one-electron species. We may do so by reducing the nuclear charge in an atom (Z) to an effective nuclear charge (Zeff) for a multielectron species, where ‘Zeff=Z–S’. In doing so, we “remodel” the atom such that there is one electron that exists in an “average” attractive nuclear field given by an “average” number of protons ‘Zeff’, now surrounded by one electron. The spdf notation of ‘H’ is: 1s1 with a nuclear charge Z=1. In this situation, since ‘H’ is already a oneelectron species we expect that Z=Zeff since there is only one type of electrostatic interaction, that of p+–e– attraction. For ‘H’, the effective nuclear charge Zeff, is expected to equal to ‘1’ since: Zeff=Z–S=1–0=1. In order to show this more rigorously we can use the Bohr model, if we have an ionization energy, where ΔE= –2.178×10–18J (Z)2(1/nf2–1/ni2), which becomes ΔE = –2.178×10–18J (Zeff)2(0–1/ni2), which then becomes ΔE = +2.178×10–18J (Zeff/ni)2. In this problem, we convert 1312 kJ/mol to J/atom and get 2.179×10–18 J/atom. Let nf→∞, and ni=1 since the single electron in the ‘H’ atom has its valence electron in the first shell corresponding to ni=1. Solving for ‘Zeff’, we get Zeff=1, which is expected for ‘H’. (b) When we consider ionization energy (kJ/mol), we remove one mole of electrons starting from a lower energy quantum state and we bring the electron far away from the nucleus, where nf→∞. Also, in order to make use of the Bohr model, which is a convenient model, we must work with a one-electron species. We may do so by reducing the nuclear charge in an atom (Z) to an effective nuclear charge (Zeff) for a multielectron species. In doing so, we “remodel” the atom such that there is one electron that exists in an “average” attractive nuclear field given by an “average” number of protons ‘Zeff’, now surrounded by one electron. The spdf notation of ‘Li’ is: 1s22s1 with a nuclear charge Z=3. In this situation, ‘Li’ is not a one-electron species where there are 3 electrons, so we do not expect that Z=Zeff since there are two types of electrostatic interaction, that of p+–e– attraction and e––e– repulsion. The nuclear charge is reduced to an effective nuclear charge (Zeff), which we can estimate using (i) the equation ‘Zeff=Z–S’ or using a more rigorous approach, the (ii) the Bohr model, if we have an ionization energy. For ‘Li’, the effective nuclear charge Zeff, is not expected to equal to ‘3’ since: Zeff=Z–S≈3–2≈1. Using the Bohr model, if we have an ionization energy, ΔE= –2.178×10–18J (Z)2(1/nf2–1/ni2), which becomes ΔE = –2.178×10–18J (Zeff)2(0–1/ni2), which then becomes ΔE = +2.178×10–18J (Zeff/ni)2. Notice that ‘Z’ has been changd to ‘Zeff’ and in doing so remodel the ‘Li’ atom into a one-electron species with an average nucleus containing ‘Zeff’ number of protons. In this problem, we are given the ionization energy of ‘Li’, where IE1=519 kJ/mol. Convert this number to J/atom and get 8.618×10–19 J/atom. Let nf→∞, and ni=2 since we reduce the Li atom to a one-electron species where the last electron is in the second shell corresponding to ni=2. Solving for ‘Zeff’, we get Zeff=1.26, a value that is not too far from the estimated value of ‘1’ from ‘Zeff=Z–S’.

(c) We remove one mole of electrons starting from a lower energy quantum state and we bring the electron far away from the nucleus, where nf→∞. The spdf notation of ‘Li+’ is: 1s22s0 with a nuclear charge Z=3. ‘Li+’ is a multielectron species with 2 electrons, so we do not expect that Z=Zeff since there are two types of electrostatic interaction, p+– e– attraction and e––e– repulsion. We will (i) estimate and then (ii) calculate ‘Zeff’. (i) Zeff=Z–S ≈ 3 – 0 ≈ 3 since we expect that same orbital shielding is less significant and so we estimate S≈0. (ii) From the Bohr model: ΔE= –2.178×10–18J (Z)2(1/nf2–1/ni2) ΔE = –2.178×10–18J (Zeff)2(0–1/ni2) ΔE = +2.178×10–18J (Zeff/ni)2 We are given the second ionization energy of ‘Li’, where IE2=7298 kJ/mol. Convert this number to J/atom and get 1.2119×10–17 J/atom. Let nf→∞, and ni=1 since the electron being removed from ‘Li+’ is in the first shell (1s22s0) corresponding to ni=1. Solving for ‘Zeff’, we get Zeff=2.36, a value that is not too far from the estimated value of ‘3’ from ‘Zeff=Z–S’. We notice a “jump” in the Zeff of ‘Li’ when removing its first core electron. (d) From above, (i) Zeff=Z–S = 3 – 0 = 3, since we now only have one electron remaining in ‘Li’ when we are removing its third electron from ‘Li2+’. There is only one type of electrostatic interaction like in the hydrogen atom, with p+–e– attraction and S=0. (ii) From the Bohr model: ΔE= –2.178×10–18J (Z)2(1/nf2–1/ni2), ΔE = –2.178×10–18J (Zeff)2(0– 1/ni2), ΔE = +2.178×10–18J (Zeff/ni)2. We are given the third ionization energy of ‘Li’, where IE3=11815 kJ/mol. Convert this number to J/atom and get 1.96197×10–17 J/atom. Let nf→∞, and ni=1 since the electron being removed from ‘Li+2’ is in the first shell (1s12s0) corresponding to ni=1. Solving for ‘Zeff’, we get Zeff=3, which is equal to the ‘Z’ value for ‘Li’ as expected for a one–electron species when we are removing the last electron in this element. 7. (i) Al : When we remove the first electron (IE1) across a period we expect that the IE1 value increases and it becomes more difficult to remove an electron as we proceed from left to right. Here the trend is explained as a result of the increase in the effective nuclear charge (Zeff) which causes an increase in p+-e- attraction and a decrease in size such that electrons are held more tightly to the nucleus. However, these trends are general and there are some exceptions to note. For the first four periods (2, 3, 4) there is a slight decrease in the ‘IE1’ value between elements in Group IIA and Group IIIA. This is a result of the effective shielding that occurs from the electrons occupying the ‘s’ orbital. These electrons are known to penetrate closer to the nucleus, are lower in energy than the ‘p-orbital’ electrons and can lower the attraction for the ‘p’ electrons, making it easier to remove the outer valence electrons occupying the ‘p’ orbital. The trend for ‘IE1’ which involves an exception to the general period trend is: IE1 Al < IE1 Mg.

(ii) When we try to remove the second electron from either species, the ‘IE2’ value corresponds to removal of an outer valence electron for both. Remember that any successive ‘IE’ value is higher than the previous one for any element. It becomes exceedingly more difficult to remove the next electron for each species. Therefore we know that the ‘IE2’ value for Mg and Al are higher than their ‘IE1’ values. The second electrons that are removed from either of these elements are found in the ‘3s’ orbital for both species. For Mg we are removing the last of the electrons in the ‘3s’ orbital while for Al we are removing the second of the third electron in that orbital. We expect that electrons in the same orbital do not shield each other effectively and so the remaining electron in the ‘3s’ orbital for Al does not cause a significant shielding effect. We may compare the ‘Zeff’ value for each species and see that the attraction for an electron is larger in ‘Al’ than it is in ‘Mg’: IE2:

From Al+ to Al2+:

Second valence e–:

IE2:

From Mg+ to Mg2+:

Second (last) valence e–:

simplified: more rigorous:

Zeff=13-10=+3 Zeff=13-10-1/2(1)=+2.5

simplified: Zeff=12-10=+2 more rigorous: Zeff=12-10-1/2(0)=+2

The trend for ‘IE2’ goes back to the regular period trend without the exception as expected for the ‘IE1’ of these elements and we get: IE2 Mg < IE2 Al. From Zumdahl (IE values in kJ/mol): IE1 Mg=735 , IE1 Al=580 IE2Mg=1445 , IE2Al=1815

IE1 Mg > IE2 Al IE2 Mg < IE2 Al

(iii) When we try to remove the third electron from either species, the ‘IE3’ value corresponds to removal of an outer valence electron for ‘Al’ but to a first core electron for ‘Mg’. It becomes more difficult to remove the first core electron since core electrons are held with greater attraction to the nucleus for any species. We can calculate the ‘Zeff’ value for each species to compare the level of attraction. IE1:

From Al to Al+:

First valence e–:

simplified, Zeff=13-10=+3 more rigorous, Zeff=13-10-1/2(2)=+2

IE2:

From Al+ to Al2+:

Second valence e–:

simplified, Zeff=13-10=+3 more rigorous, Zeff=13-10-1/2(1)=+2.5

IE3:

From Al2+ to Al3+:

Third (last) valence e–: simplified, Zeff=13-10=+3 more rigorous, Zeff=13-10-1/2(0)=+3

IE1:

From Mg to Mg+:

First valence e–:

IE2:

From Mg+ to Mg2+:

Second (last) valence e–:

IE3:

From Mg2+ to Mg3+:

First core e–: simplified, Zeff=12-2=+10 more rigorous, Zeff=12-2-1/2(7)=+6.5

simplified, Zeff=12-10=+2 more rigorous, Zeff=12-10-1/2(...