Question

1. ionization energy

is the energy required to remove an electron from a neutral atom.
trend across periods:
as you move from left to right across a period, ionization energy will ______.
explain why:

trends down a group:
as you move down a group, ionization energy will ______.
explain why:

1. electronegativity

is the relative ability of an atom to attract electrons in a chemical bond.
trend across periods:
as you move from left to right across a period, electronegativity will ______.
explain why:

trends down a group:
as you move down a group, electronegativity will ______.
explain why:

periodic trends practice

1. draw bohr models for boron, carbon, and nitrogen. which trend is shown?

1. draw bohr models for oxygen and sulfur. which trend is shown?

1. draw a bohr model of a neutral aluminum atom and an aluminum ion. which trend is shown?

Answer:

3. IONIZATION ENERGY

Trend across periods:

Step 1: Determine the trend direction
As you move from left to right across a period, ionization energy will increase.

Step 2: Explain why
Across a period, the number of protons in the nucleus increases (atomic number increases), while the electrons are added to the same energy level (shell). This increases the effective nuclear charge ($Z_{\text{eff}}$) experienced by the outermost electrons. A higher $Z_{\text{eff}}$ pulls the electrons closer to the nucleus, making them more tightly bound. Thus, more energy is required to remove an electron (ionization energy increases).

Trend down a group:

Step 1: Determine the trend direction
As you move down a group, ionization energy will decrease.

Step 2: Explain why
Down a group, electrons are added to new, larger energy levels (shells) that are farther from the nucleus. The outermost electrons are shielded from the nucleus by inner electron shells, reducing the effective nuclear charge they experience. Additionally, the distance between the nucleus and the outermost electrons increases. Both factors make it easier to remove an electron, so ionization energy decreases.

4. ELECTRONEGATIVITY

Trend across periods:

Step 1: Determine the trend direction
As you move from left to right across a period, electronegativity will increase.

Step 2: Explain why
Across a period, the effective nuclear charge ($Z_{\text{eff}}$) increases (more protons, same energy level for valence electrons). A higher $Z_{\text{eff}}$ means the nucleus has a stronger pull on bonding electrons. Thus, the atom’s ability to attract electrons in a chemical bond (electronegativity) increases.

Trend down a group:

Step 1: Determine the trend direction
As you move down a group, electronegativity will decrease.

Step 2: Explain why
Down a group, valence electrons occupy larger, more shielded energy levels (farther from the nucleus). The effective nuclear charge on these electrons is reduced, and the distance from the nucleus increases. Both factors weaken the nucleus’ ability to attract bonding electrons, so electronegativity decreases.

PERIODIC TRENDS PRACTICE

1. Boron, Carbon, Nitrogen

• Bohr Models:
• Boron (B): Atomic number 5. Nucleus: 5 protons, 6 neutrons (most common isotope). Energy levels: 2 electrons in n=1, 3 electrons in n=2.
• Carbon (C): Atomic number 6. Nucleus: 6 protons, 6 neutrons. Energy levels: 2 electrons in n=1, 4 electrons in n=2.
• Nitrogen (N): Atomic number 7. Nucleus: 7 protons, 7 neutrons. Energy levels: 2 electrons in n=1, 5 electrons in n=2.
• Trend Shown: These elements are in Period 2 (same period), moving left to right. The trend is increasing ionization energy and electronegativity (consistent with period trends: $B < C < N$ for both).

2. Oxygen, Sulfur

• Bohr Models:
• Oxygen (O): Atomic number 8. Nucleus: 8 protons, 8 neutrons. Energy levels: 2 electrons in n=1, 6 electrons in n=2.
• Sulfur (S): Atomic number 16. Nucleus: 16 protons, 16 neutrons. Energy levels: 2 electrons in n=1, 8 in n=2, 6 in n=3.
• Trend Shown: These elements are in Group 16 (same group), moving down. The trend is decreasing ionization energy and electronegativity (consistent with group trends: $O > S$ for both).

3. Aluminum Atom and Ion

• Neutral Aluminum (Al): Atomic number 13. Nucleus: 13 protons, 14 neutrons (most common isotope). Energy levels: 2 in n=1, 8 in n=2, 3 in n=3.
• Aluminum Ion (Al³⁺): Loses 3 valence electrons (from n=3). Energy levels: 2 in n=1, 8 in n=2 (stable octet).
• Trend Shown: The ion is smaller than the atom (ionization removes valence electrons, shrinking the electron cloud). This illustrates the trend of decreasing atomic radius with ionization (or the stability of noble gas electron configurations).

Final Answers (Fill-in-Blanks):

1. IONIZATION ENERGY

• Across periods: $\boldsymbol{\text{increase}}$ (explained above).
• Down a group: $\boldsymbol{\text{decrease}}$ (explained above).

1. ELECTRONEGATIVITY

• Across periods: $\boldsymbol{\text{increase}}$ (explained above).
• Down a group: $\boldsymbol{\text{decrease}}$ (explained above).