Unit 1: Periodic Trends
Exploring the systematic variations in atomic properties across the Periodic Table.
1.7 Size & Ionization Energy
The properties of elements exhibit predictable patterns, or periodic trends, across periods and down groups in the Periodic Table. These trends are primarily influenced by three factors: nuclear charge, shielding effect, and electron-electron repulsion.
Atomic Size (Atomic Radius)
Atomic size refers to the distance from the nucleus to the outermost electrons.- Across a Period (Left to Right): Atomic size generally decreases. As you move across a period, the number of protons (nuclear charge) increases, pulling the outer electrons closer to the nucleus. The number of principal energy shells remains constant, so the shielding effect from inner electrons does not significantly increase. This stronger effective nuclear charge ($Z_{eff}$) pulls the electron cloud inward, resulting in a smaller atomic radius.
Example: In Period 3, Na is the largest atom, and Ar is the smallest. - Down a Group (Top to Bottom): Atomic size generally increases. As you move down a group, electrons occupy new principal energy shells that are successively further from the nucleus. This increase in the number of shells significantly increases the shielding effect, which outweighs the increase in nuclear charge. The outer electrons are less strongly attracted to the nucleus and are held further away, leading to a larger atomic radius.
Example: In Group 1, Li is smaller than Na, which is smaller than K.
Ionic Size
Ionic size refers to the radius of an ion.- Cations: Always smaller than their parent neutral atoms. When an atom loses electrons to form a cation, the remaining electrons experience less electron-electron repulsion, and the effective nuclear charge on them increases, pulling them closer to the nucleus. Additionally, the outermost electron shell may be completely removed.
Example: Na⁺ is smaller than Na. - Anions: Always larger than their parent neutral atoms. When an atom gains electrons to form an anion, the increased electron-electron repulsion among the added electrons causes the electron cloud to expand, leading to a larger ionic radius.
Example: Cl⁻ is larger than Cl.
First Ionization Energy ($IE_1$)
The first ionization energy is the minimum energy required to remove one electron from each atom in a mole of gaseous atoms to form a mole of gaseous 1+ ions. It is an endothermic process: $$ M(g) \rightarrow M^+(g) + e^- $$- Across a Period (Left to Right): First ionization energy generally increases. As the nuclear charge increases and the shielding effect remains relatively constant, the attraction between the nucleus and the outermost electrons increases. More energy is therefore required to remove an electron.
Exceptions:- Group 2 to Group 13 (e.g., Be to B, Mg to Al): There is a slight decrease. The electron being removed from Group 13 elements (e.g., Boron) is a p-electron, which is slightly higher in energy and experiences more shielding from the inner s-electrons compared to the s-electron being removed from Group 2 elements (e.g., Beryllium). This makes the p-electron easier to remove despite the increased nuclear charge.
- Group 15 to Group 16 (e.g., N to O, P to S): There is a slight decrease. The electron being removed from Group 16 elements (e.g., Oxygen) is a paired electron in a p-orbital. The electron-electron repulsion within this paired orbital makes it easier to remove one of these electrons compared to removing an unpaired electron from a half-filled p-orbital in Group 15 elements (e.g., Nitrogen).
- Down a Group (Top to Bottom): First ionization energy generally decreases. As you move down a group, the number of principal energy shells increases, leading to greater atomic size and a more significant shielding effect. Although the nuclear charge increases, the increased distance and shielding reduce the effective nuclear attraction on the outermost electrons, making them easier to remove.
Example: Li has a higher $IE_1$ than Na, which has a higher $IE_1$ than K.
Solved Examples:
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Explain why a Sodium (Na) atom is larger than a Chlorine (Cl) atom.
Solution: Both Na and Cl are in Period 3, meaning they have the same number of principal energy shells. However, Chlorine has a higher nuclear charge (17 protons) than Sodium (11 protons). This stronger nuclear attraction in Chlorine pulls its outer electrons closer to the nucleus, resulting in a smaller atomic radius compared to Sodium. -
Why is a Potassium (K) atom larger than a Sodium (Na) atom?
Solution: Potassium is in Period 4, while Sodium is in Period 3. Moving down a group (from Na to K), a new principal energy shell is added. This additional shell increases the distance of the outermost electrons from the nucleus and enhances the shielding effect, making the Potassium atom larger than the Sodium atom. -
Compare the first ionization energies of Magnesium (Mg) and Aluminum (Al). Explain the trend.
Solution: Magnesium ($[Ne] 3s^2$) has a higher first ionization energy than Aluminum ($[Ne] 3s^2 3p^1$). Although Aluminum has a higher nuclear charge, the electron being removed from Aluminum is a 3p electron. This 3p electron is at a slightly higher energy and experiences more shielding from the 3s electrons compared to the 3s electron being removed from Magnesium, making it easier to remove. -
Why is the first ionization energy of Nitrogen (N) higher than that of Oxygen (O)?
Solution: Nitrogen ($1s^2 2s^2 2p^3$) has a half-filled 2p subshell, which is a relatively stable configuration. Oxygen ($1s^2 2s^2 2p^4$) has one paired electron in its 2p subshell. The electron-electron repulsion within this paired orbital makes it easier to remove one of these electrons compared to removing an unpaired electron from a half-filled 2p subshell in Nitrogen, despite Oxygen having a higher nuclear charge. -
Order the following elements by increasing atomic size: Li, K, F, C.
Solution: Atomic sizes generally decrease across a period and increase down a group. * F (Period 2, Group 17) is the smallest in Period 2. * C (Period 2, Group 14) is larger than F but smaller than Li. * Li (Period 2, Group 1) is the largest in Period 2. * K (Period 4, Group 1) is in a lower period than Li, so it is significantly larger. Therefore, the order of increasing atomic size is: F < C < Li < K.
1.8 Electron Affinity & Ions
First Electron Affinity ($EA_1$)
The first electron affinity is the energy change that occurs when one electron is added to each atom in a mole of gaseous atoms to form a mole of gaseous 1- ions. It is often an exothermic process (energy is released, so $EA_1$ is negative), but can be endothermic for some elements. $$ M(g) + e^- \rightarrow M^-(g) $$- Across a Period (Left to Right): First electron affinity generally becomes more negative (more energy is released, indicating a greater attraction for an incoming electron). As nuclear charge increases and shielding remains constant, the effective nuclear charge on the valence shell increases, making it easier for the atom to attract and accommodate an additional electron.
Exceptions:- Group 1 to Group 2 (e.g., Li to Be, Na to Mg): Electron affinity becomes less negative or even positive. For Group 2 elements, the incoming electron must enter a higher-energy p-orbital, which is shielded by the filled s-orbital. This makes it less favorable to add an electron.
- Group 14 to Group 15 (e.g., C to N, Si to P): Electron affinity becomes less negative or even positive. For Group 15 elements, the incoming electron must pair up with an existing electron in a half-filled p-orbital, leading to increased electron-electron repulsion. This makes the addition of an electron less favorable.
- Down a Group (Top to Bottom): First electron affinity generally becomes less negative (less energy is released, indicating a weaker attraction for an incoming electron). As you move down a group, the atomic size increases, and the outermost electrons are further from the nucleus. The increased distance and shielding reduce the attraction for an incoming electron, making the process less exothermic.
General Tendency to Form Ions
The trends in ionization energy and electron affinity collectively describe an atom's general tendency to gain or lose electrons:- Elements on the Left (Groups 1 & 2): Low ionization energies and less negative (or positive) electron affinities. They tend to lose electrons easily to form cations, achieving a stable noble gas configuration. These are the highly reactive metals.
- Elements on the Right (Groups 16 & 17): High ionization energies and highly negative electron affinities. They tend to gain electrons easily to form anions, also achieving a stable noble gas configuration. These are the highly reactive nonmetals.
- Noble Gases (Group 18): Very high ionization energies and positive electron affinities. They have a full valence shell, making them very stable and unlikely to gain or lose electrons. This explains their chemical inertness.
Solved Examples:
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Explain why the first electron affinity of Fluorine (F) is more negative than that of Oxygen (O).
Solution: Fluorine (Z=9) has a higher nuclear charge than Oxygen (Z=8). As you move from Oxygen to Fluorine across Period 2, the effective nuclear charge increases, and the shielding remains similar. This stronger attraction makes Fluorine more readily accept an additional electron into its valence shell, releasing more energy (more negative electron affinity). -
Why is the first electron affinity of Chlorine (Cl) less negative than that of Fluorine (F)?
Solution: Both are in Group 17. Moving down the group from F to Cl, the atomic size increases due to the addition of a new principal energy shell. The incoming electron in Chlorine is added to a 3p orbital, which is further from the nucleus and experiences more shielding than the 2p orbital in Fluorine. This reduced attraction makes the electron addition less exothermic for Chlorine. -
Predict whether Magnesium (Mg) or Sulfur (S) would more readily gain an electron. Explain.
Solution: Sulfur (Group 16) would more readily gain an electron than Magnesium (Mg) (Group 2). Sulfur is a nonmetal with a strong tendency to gain two electrons to achieve a stable octet, resulting in a negative electron affinity. Magnesium is a metal that tends to lose electrons to form a cation, and its electron affinity is positive, meaning it requires energy to gain an electron. -
Why do noble gases like Neon (Ne) have positive (or very slightly negative) electron affinities?
Solution: Noble gases have a completely filled valence electron shell, which is a highly stable electronic configuration. Adding an extra electron would require placing it in a new, higher-energy principal shell, where it would experience significant repulsion from the already filled inner shells and be poorly attracted to the nucleus. This process is energetically unfavorable, hence the positive electron affinity. -
WAEC-style: Which of the following elements would have the highest tendency to form an anion: Na, Al, S, Ar? Justify your answer.
Solution: Sulfur (S). Elements that readily form anions are typically nonmetals with high electron affinities (tendency to gain electrons) and relatively high ionization energies. Among the given options, Sulfur is a nonmetal (Group 16) that needs to gain only two electrons to achieve a stable octet, making it have a strong tendency to form an anion (S²⁻). Na is a metal that forms cations. Al is a metal that forms cations. Ar is a noble gas and does not readily form ions.
1.9 Electronegativity & Metallic Character
Electronegativity
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons in a chemical bond. It is a relative scale (Pauling scale is common) and does not have units.- Across a Period (Left to Right): Electronegativity generally increases. As the nuclear charge increases and atomic size decreases across a period, the nucleus exerts a stronger attractive force on electrons in chemical bonds, pulling them closer.
- Down a Group (Top to Bottom): Electronegativity generally decreases. As atomic size increases down a group due to additional electron shells and increased shielding, the attractive force of the nucleus on bonding electrons is diminished.
Metallic and Nonmetallic Character
The terms metallic character and nonmetallic character describe the set of chemical properties associated with metals and nonmetals, respectively.- Metallic Character: Associated with the ease of losing electrons (low ionization energy, low electronegativity). Metals are typically shiny, malleable, ductile, and good conductors of heat and electricity.
- Across a Period (Left to Right): Metallic character decreases. Elements transition from metals to metalloids to nonmetals. This is because ionization energy increases, and atoms become less likely to lose electrons.
- Down a Group (Top to Bottom): Metallic character increases. Atomic size increases, and ionization energy decreases, making it easier for atoms to lose their valence electrons.
- Nonmetallic Character: Associated with the ease of gaining electrons (high electron affinity, high electronegativity). Nonmetals are typically dull, brittle, poor conductors (insulators), and tend to form anions.
- Across a Period (Left to Right): Nonmetallic character increases. Atoms become more likely to gain electrons.
- Down a Group (Top to Bottom): Nonmetallic character decreases. Atoms become less likely to gain electrons.
Solved Examples:
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Which element is more electronegative: Oxygen (O) or Sulfur (S)? Explain.
Solution: Oxygen is more electronegative than Sulfur. Both are in Group 16. As you move down a group, electronegativity decreases because the atomic size increases, and the valence electrons are further from the nucleus, experiencing less attraction from the nucleus. -
Compare the electronegativity of Carbon (C) and Fluorine (F). Explain the difference.
Solution: Fluorine is significantly more electronegative than Carbon. Both are in Period 2. As you move across a period from left to right, the nuclear charge increases while shielding remains relatively constant. This leads to a stronger attraction for bonding electrons, increasing electronegativity. -
Predict which element would be a better conductor of electricity: Sodium (Na) or Phosphorus (P). Justify your answer based on periodic trends.
Solution: Sodium (Na) would be a better conductor of electricity. Sodium is a metal (Group 1, s-block) with low ionization energy and readily loses its valence electron to form a "sea" of delocalized electrons, which is characteristic of good electrical conductors. Phosphorus (P) is a nonmetal (Group 15, p-block) and is generally a poor conductor of electricity. This reflects the decrease in metallic character across a period. -
WAEC-style: Explain why metallic character increases down Group 1.
Solution: Metallic character is associated with the ease of losing valence electrons. As you move down Group 1, the atomic size increases due to the addition of more electron shells. This increased distance from the nucleus and greater shielding reduce the attraction between the nucleus and the outermost electron, making it easier to remove that electron. Therefore, the metallic character increases down the group. -
Which element, Nitrogen (N) or Oxygen (O), has a greater tendency to attract electrons in a bond? Explain using electronegativity.
Solution: Oxygen (O) has a greater tendency to attract electrons in a bond. Oxygen is to the right of Nitrogen in Period 2. As you move across a period, the effective nuclear charge increases, and atomic size decreases, leading to a stronger pull on bonding electrons. Therefore, Oxygen is more electronegative than Nitrogen.
Knowledge Check (20 Questions)
Answer: It generally decreases.
Answer: The minimum energy required to remove one electron from each atom in a mole of gaseous atoms to form a mole of gaseous 1+ ions.
Answer: Smaller. When an atom loses electrons, electron-electron repulsion decreases, and the remaining electrons are pulled closer to the nucleus.
Answer: Down a group, atomic size increases due to more electron shells, leading to increased shielding. This reduces the effective nuclear attraction on the outermost electrons, making them easier to remove.
Answer: The energy change that occurs when one electron is added to each atom in a mole of gaseous atoms to form a mole of gaseous 1- ions.
Answer: It generally decreases.
Answer: Potassium (K) is more metallic. Metallic character increases down a group and decreases across a period. K is to the left of Ca in Period 4, and Group 1 elements are more metallic than Group 2.
Answer: Oxygen has a paired electron in one of its 2p orbitals, leading to electron-electron repulsion that makes it easier to remove compared to Nitrogen's stable half-filled 2p subshell.
Answer: Cl < Mg < Na < K. (Atomic radius decreases across a period, increases down a group).
Answer: They have a stable, full valence shell. Adding an electron would require placing it in a new, higher-energy shell, which is energetically unfavorable due to repulsion.
Answer: Nitrogen (N). Electronegativity increases across a period.
Answer: It generally decreases.
Answer: Larger. When an atom gains electrons, electron-electron repulsion increases, causing the electron cloud to expand.
Answer: Rb. First ionization energy decreases down a group.
Answer: The electron removed from Al is a 3p electron, which is at a slightly higher energy and experiences more shielding from the 3s electrons than the 3s electron removed from Mg.
Answer: Bromine (Br). Electron affinity becomes less negative down a group.
Answer: Higher electronegativity indicates a greater tendency to attract electrons in a bond.
Answer: Increased nuclear charge pulls outer electrons closer, while shielding remains relatively constant.
Answer: Mg (Magnesium). Magnesium is in Group 2 and readily loses two electrons to form a stable Mg²⁺ ion.
Answer: Inner shell electrons repelling outer shell electrons, reducing the nuclear attraction experienced by the outer electrons.