Question

how does ionization energy change across a period? 1. label your x-axis as atomic number and the y-axis as ionization energy 2. fill in the table below 3. use this link: http://bit.ly/462jofm
element name | atomic number | ionization energy
lithium | 3 | 51.93
boron | 5 | 800.6
oxygen | 8 | 1.317
fluorine | 9 |

1. plot on your grapha

Explanation:

To determine how ionization energy changes across a period (e.g., period 2 for Li, B, O, F):

1. Understand Periodic Trend: Ionization energy (IE) generally increases across a period (left to right). This is because:

• Atomic radius decreases (electrons are closer to the nucleus).
• Effective nuclear charge (\(Z_{\text{eff}}\)) increases (more protons pull electrons tighter).
• It becomes harder to remove an electron (higher IE).

1. Analyze Given Elements:

• Lithium (Li, atomic number 3): IE ≈ 519.9 kJ/mol (note: the handwritten “91.9” likely has a unit error; correct IE for Li is ~520 kJ/mol).
• Boron (B, atomic number 5): IE ≈ 800.6 kJ/mol.
• Oxygen (O, atomic number 8): IE ≈ 1314 kJ/mol (handwritten “1.31” likely missing “×10³” or unit error).
• Fluorine (F, atomic number 9): IE ≈ 1681 kJ/mol (not filled, but follows the trend).

1. Graphical Interpretation: Plotting atomic number (x-axis) vs. IE (y-axis) will show a rising trend (with minor exceptions, e.g., between Be/B or N/O, but not in this subset).

Answer:

Across a period (left to right), ionization energy generally increases (with minor exceptions). For the elements Lithium (Li), Boron (B), Oxygen (O), and Fluorine (F) (in period 2), the ionization energy values (after correcting units) show a clear upward trend as atomic number increases: \( \text{Li} < \text{B} < \text{O} < \text{F} \) in terms of ionization energy. This is due to decreasing atomic radius, increasing effective nuclear charge, and stronger electron-nucleus attraction across the period.