This is the second in a series of topics on Chemical Bonds.hibr />
This is intended  for use in general chemistry classes. More particularly, for non-chemistry major and high school students.
Why Hank's Campbell's Science 2.0 website? Well, the answer is because of its popularity, my expectations that members in here care to be of help through comments, and more so the publications here are protected by copyright. 

TOPIC OUTLINE

LESSON 1 REVIEW ON ATOMIC STRUCTURE AND PERIODIC TABLE
LESSON 2 PERIODIC TRENDS
LESSON 3 ELECTRON CONFIGURATIONS
LESSON 4 IONIC BONDS
LESSON 5 COVALENT BONDS
LESSON 6 LEWIS FORMULAS
LESSON 7 VSEPR AND MOLECULAR SHAPES
LESSON 8 METALLIC BONDS
LESSON 9 INTERPARTICLES ATTRACTIVE FORCES (IPAF)

LESSON 2

PERIODIC TRENDS

Suggested Time Frame: 

3 contact hours

Material: Periodic Table

Text

Elements in the periodic table are arranged according to their atomic number in order to group them according to similarity in properties. Actually, the original purpose is to group elements according to properties and as a consequence, they have to be arranged according to atomic number.

Periodic trend in atomic properties is another consequence of arranging the atoms according to atomic number. Remember that trends only give direction. It does not say that going to that direction is always smooth. It can be rough. Similarly, the periodic trends of atomic properties have exceptions. Meaning, if we say electron affinity increases across the periodic table, not all elements follow this, there are exceptions. These exceptions and their reasons for exceptions are not covered in this module. However, rest assured that the trends in periodic properties are good generalization because most atoms/elements follow the trends. So come on, enjoy the trip across the periodic table.

2.0 Ionization Energy (IE). The energy needed to remove an electron

Ionization energy is the required energy to remove an electron in a free atom in its gaseous phase. Electrons outside the nucleus are attracted to the positive nucleus of an atom, so that, to remove an electron needs the use of energy.

Formation of Cations (positively charged ion). A consequence of removing electrons

Cations are formed when an electrons are removed from a neutral atom.

Neutral Atom(g) + IE                             Cation (+) + e-

Mg(g) + IE Mg2+ + 2e-


2.01 First IE1, 2nd IE2 and so on… It's easy the first time but harder the next

The energy needed to remove the first outermost (valence) electron is called first ionization energy (IE1); for the next electron, it’s called the second ionization energy (IE2); the next is the 3rd (IE3); the 4th (IE4) and so on.

The electron farthest from the nucleus (valence electron) has the weakest attraction among the electrons to its nucleus making it the easiest to remove, thus, the IE is lowest. It appears that removal of the first outermost electron makes the nuclear charge(+) stronger with respect to its attraction towards the remaining electrons. This makes it harder to remove the next electron. It follows that the IE’s are arranged from low energy (IE1) to higher energy (IE2, IE3…). This means that to remove the first electron (IE1) at the outermost energy level is easier (requires lower energy) than the second electron (IE2), and so on... Indeed, it is easy the first time, but harder the next.

IE1  IE IE3   IE4…..

Increasing ionization energy


2.02 Stable Cations.   The cations comfort zone.

Cations are formed in their stable form. Stability of cations depends on the ionization energy, or on how much energy is needed to remove the electrons. In ordinary chemical reactions, the removal of electron stops at the point where the next electron requires very high energy to remove. That is, the point where there is a big gap in ionization energy (Table 2)


Stops at IE1

Look at the sodium (Na) in the table above. The data show that the first electron is easy to remove ( IE1 of 495.6), but the next electron is already hard to remove ( IE2 of 4562). There is a big gap between IE1 and IE2. Removal of an electron through ordinary chemical reaction stops there where one electron is removed, thus the stable cation for sodium is sodium(I) or sodium(1+) or Na1+ . That is, at IE1.

Na + 495.6 J Na1+ + 1e- IE1 = 495.6

Stops at IE2

Look at magnesium (Mg) in the table above. The first and second electrons are easy to remove (IE1 of 737.7 and IE2 of 1451), but the next electron is already hard to remove (IE3 of 7733). There is a big gap between IE2 and IE3. Thus, the stable cation for magnesium is magnesium(II) or magnesium(2+) or Mg2+.  That is at IE2..

Mg + 737.7 J Mg1+ + 1e- IE1 = 737.7

Mg1+ + 1451 J Mg2+ + 1e- IE2 = 1451


Sample Task 2.1

Now, get your periodic table, look at the ionization energy values of the elements in the table and infer its trend by answering  the following:

1. What is the trend in ionization energy within a column that is, from top to bottom( increasing or decreasing)?

Answer: ________________________

2. What is the trend in ionization energy within a row, from left to right (increasing or decreasing)?

Answer:_______________________

3. Of the list of pairs of atoms below, ring the one with the lower first ionization energy:

a. Magnesium (Mg) or Phosphorus (P)

b. Beryllium (Be) or Barium (Ba)

c. Carbon (C) or Fluorine (F)

d. Nitrogen (N) or Arsenic (As)

2.03 Atomic Size and Ionization Energy

Atomic size refers to the distance of the center of the nucleus to the valence electron (radius, r).


The truth is, it is hard to define atomic size because we only talk about the probability of finding electrons. Thus, atomic size is determined by the distance of two nuclei of two bonded atoms. How this works will not be taken up here, but in the next level chemistry course.

Periods and atomic size. Recall that Period number is the number of energy levels in an atom. The more energy levels, the wider the space where electrons in that atom can be found thus, the bigger the atom.

2.04 Relating Ionization Energy (IE) with Atomic Size.

Now, what is the relation of ionization energy with atomic size? Take note that the higher the period number, the bigger the size, the farther the valence electrons from the nucleus. It follows that the bigger the atom the lesser its attraction of the valence electrons to the nucleus compared to smaller atom.

Table 3 Relating Ionization Energy with Atomic Size

Atom

Period Number/Number of energy levels

Sketch of the atom representing its size

compare size the 3 atoms

Compare the IE of the 3 atoms


Lithium (Li)


2


Li is the smallest among the three


highest

Potassium (K)

4


K is smaller

higher

Cesium (Cs)

6


Cs is small

high

Sample Task 2.2

Complete the table below by writing the Period number, sketching the size, describing as to small, smaller, smallest, and describing the IE as to high, higher, highest.

Atom

Period Number/Number of energy levels

Sketch of the atom representing its size

compare the size of the 3 atoms

Compare the IE of the 3 atoms

Bromine (Br)






Iodine ( I )






Fluorine (F)






II. Infer the trend of atomic size in your periodic table.

What is the periodic trend of the atoms from top to bottom in the periodic table, increasing or decreasing?

1. atomic size Answer:_________________________

2. ionization energy Answer:____________________

2.05 Atominc Number (Z) and Ionization Energy

Recall that atomic number (Z) is the number of protons (positively charged particles). Recall that protons (+ charge) and neutrons ( no charge) are in the nucleus, making the nuclear charge or the charge of nucleus positive.

Which do you think has stronger nuclear charge, sodium (Na) or aluminium (Al) atom?

To answer this question, you have to look for the atomic numbers of sodium and aluminium (in the periodic table) and then decide which one has more positive charges in its nucleus.

Yes, you are right, aluminum with Z =13 has stronger nuclear charge compared to sodium with Z=11, thus aluminum has stronger nuclear charge compared to sodium. It follows that the more positive charges are in the nucleus, the greater the nuclear charge, the greater the attraction of the nucleus to the valence electrons, pulling the electrons towards the nucleus, so that the atom shrinks a little.

Now, that you know which of the two atoms (Al or Na) has stronger nuclear charge,

can you infer which of the two (Al or Na) has higher ionization energy?

To answer this, note that Al and Na belong to the same period, however one of them has more protons in the nucleus.

Yes, you are correct. Aluminium (Al) has higher ionization energy compared to sodium (Na) because the nucleus of Al has more protons,and thus the greater is the (+) charge. It follows that the greater the nuclear charge (+), the greater is the attraction of the nucleus towards the valence electrons(-), thus the valence electrons are harder to remove, the greater the ionization energy. As to atomic size, aluminium is smaller compared to sodium because the strong attraction of the nucleus to the valence electrons makes the atom shrinks a little.

Sample Task 2.3

Complete the table below by writing the period number, atomic number (Z), describing the size (as to small, smaller, smallest), and describing the ionization energy, IE ( as to high, higher, highest).

Atom

Z

Period Number/ Number of energy levels

Sketch the size of the atom

compare the size of the 3 atoms

compare the IE of the 3 atoms

Siicon (Si)

14

3



Smaller

higher

Chorine ( Cl )


17

3



smallest

highest

Magnesium(Mg)



12

3


small

high

II. Infer the trend of atomic size and ionization energy in your periodic table.

1.The periodic trend on ionization energy across the periodic table from left to right across is _____________________ ( increasing/decreasing) while the atomic size is __________________ (increasing/decreasing)

2. The periodic trend on ionization energy from top to bottom of each vertical column is ______________________(increasing/decreasing) while the atomic size is ______________________(increasing/decreasing).

2.2 Metallic Properties

Look at the right side of your periodic table.

Can you see a sharp, bold line that looks like stairs?

Hint: it starts below boron (B), below silicon (Si), below arsenic(As), below tellurium(Te), to below astatine (At) down at the right side of Uus Or it starts at Al, Ge, Sb, Po, and ends at Uus.

Yes that one! That stairs separate the metals from the non-metals. The metals are at the left side of that line and the non-metals are at the right side of it. It shows that the metallic properties decreases from left going to the right across the periodic table. Somewhere at the stairs are the metalloids ( in yellow color) such as Si, Ge, As, Sb, and Te


Refer to your periodic table.

I. Which of the pairs of atoms is more metallic? Encircle your answer.

1. Gold (Au) or Barium (Ba) 2. Silver (Ag) or Strontium (Sr)

II. Which of the atoms below is the least metallic? Encircle your answer.

P, Cl, Si, Mg, S, Al

III. Which of the atoms below is most metallic? Encircle your answer.

P, Cl, Si, Mg, S, Al

2.21 What makes a metal a metal?

The answer lies on its ability to lose its valence electrons. The easier the atom loses electron the more metallic it is. Recall that the lower the IE the easier for the atom to lose electron. It follows that the lower the IE the more metallic the atom is. IE and metallic properties are inversely related. The higher the IE the less metallic, the lower the IE the more metallic.

General properties of metals ( the consequences of losing electron easily)

1. malleable (can be hammered into thin sheet without breaking)

2. Luster (it reflects light)

3. Good conductor of heat

4. Good conductor of electricity


2.30 Electron Affinity (EA)

Electrons cannot feel, but just to explain, let us put it this way, as an electron(e-). approaches an atom, it feels attracted to the atom, particularly, the nucleus (+) of the atom as the target. As the incoming electron gets closer to the atom, it encounters the electrons (e-) of the atom, pushing the incoming electron away from the atom (repulsion). However, if the nucleus (+) is strong enough, that is, the nucleus is highly positive enough, the attraction outweighs the repulsion thus, that incoming electron is successfully added as one of the valence electrons of the atom. As a consequence energy is released from the atom. This energy released is a net change in energy (energy final minus energy initial) called electron affinity.

Electron affinity is a change in energy due to gain of electron by a neutral gaseous atom.

X(g) + 1 e- X1-

Neutral gaseous atom + 1e- to form a negatively charged ion called anion.

To compare with, IE is the energy associated with the removal of electrons, while EA is the energy change associated with the gaining of electrons of an atom.

Take note, noble gases have, practically, no tendency of gaining or losing electrons. Thus, there are no electron affinity and ionization energy values yet assigned to them.

2.31 Positive and Negative EA

The key phrase here is change in energy. This change in energy associated by gaining electrons is brought about by the competing forces: repulsion and attractive forces. Nucleus (+) attracts the incoming electron (-) while the electrons already present in the atom repel the incoming electron ( negative particle repel another negative particle). These repulsion and attractive forces happen simultaneously as the incoming electron tries to join the atom.

Positive and negative signs are conventional (or traditional but generally followed). A negative electron affinity simply means the energy is released as the incoming electron approaches the atom. This negative energy usually happens with the first electron affinity (EA1) and is spontaneous (or can happen naturally).

A positive electron affinity simply means that the addition of electron to the atom is not spontaneous (do not happen naturally). You have to force the incoming electron to join the atom (or you need to supply energy to make the incoming electron join the atom). This usually happens with 2nd electron affinity (EA2). In EA1 the neutral atom has gained electron and it is already negatively charged, so that addition of another electron will experience greater repulsion (from the combined forces of the already negatively charged atom and the electrons present in the atom itself) requiring an input of energy. Thus, the EA2 is positive.

Ex. O + 1e- O1- EA = -141.0 KJ/mol

O1- + 1e- O2- EA = + 744.0 KJ/mol

Negative EA means energy is released when electron is gained by the neutral atom. This

process is spontaneous, meaning this happens without supplying energy.

Positive EA means energy need to be supplied to force the neutral atom to gain electron. This

process is NOT spontaneous.


Atomic Size and EA


It looks like it is easier to add an electron to a smaller neutral atom compared to a bigger one, but this is not always the case. Considering the individual EA values of atoms in the periodic table, it is indeed hard to make generalization for electron affiinity EA. There are a lot of exceptions. Well, logically we can say that in smaller atom the valence shell where the electron is to be added is closer to the positive nucleus, so the attraction towards the atom is stronger and gaining of electron is easier, thus the electron affinity is a larger negative value. It follows that the bigger the negative value of EA, the easier for the neutral atom to gain electron forming a negative ion (anion).

2.32 Metals, Nonmetals and EA

Metals have lesser tendency to gain electron compared to nonmetals. Nonmetals tend to gain electrons while metals tend to lose electrons. Nonmetals tend to form negative ions, while metals tend to form positive ion. Generally, we can say that metals have lower electron affinity while nonmetals have higher electron affinity. Metals have lower ionization energy while non-metals have higher ionization energy.


2.40 Electronegativity


Electronegativity is the attraction of a bonded atom towards a shared electrons. Nonmetals have high electronegativity while metals have low electronegativity.

More discussion on electronegativity in Lesson __, chemical bonding and polarity.




IA IIA IIIA IVA VA VIA VIIA VIIIA

1


















2

3

4











5

6

7

8

9

10


11

12

13
























































































118


Representative elements are those belonging to groups A (IA, IIA, IIIA,…VIIIA)


























Name:_________________________________ Score:______________

Class Schedule:_________________________ Date:_______________





A. Encircle the atom that is asked for in each question. Refer to your periodic table.

Question

Choices

1. Which one is bigger?

(1) Ca or Zn (2) F or Br

2. Which one has greater ionization energy?

(1) Na or Rb (2) N or As

3. Which one is more metallic?

(1) Ag or Au (2) Fe or Zn

4. Which one has greater electron affinity?

(1) B or F (2) Na or Cl

5. Which one is more electronegative?

(1) Cl or I (2) Al or S

6. Which one is the smallest?

(1) Ca, Sr, Ba (2) K, Cr, Ge

7. Which one has the greatest tendency to lose a valence electron?

(1) K, Fe, Zn (2) Cu, Ag, Au

8. Which one has the greatest tendency to gain electron?

(1) B, C, O (2) F, Cl, Br

9. Which one has the least attraction towards shared electrons?

(1) Br, Cl, F (2) Si, P, S

10. Which one is least metallic?

(1) Mg, Ca, Sr (2) Rb, Sr, Sb

11. Which one is more likely to form a negative ion (anion)?

(1) Li or S (2) Ba or O

12. Which one is more likely to form a cation?

(1) Be or Br (2) Cr or Cl