Question

prepwork part 1: atomic radius oer practice
one of the ap chemistry students approaches you after hearing you learned about effective nuclear charge and atomic radius today. he knows he should be able to explain this trend in terms of the atomic structure of atoms, but the ap chemistry student cannot remember the concept! for the scenario below, explain the trend using coulombs law.

a. **magnesium has a smaller atomic radius than barium
magnesium and barium are found in the same _, which means they have the same _ but different _
the relationship between _ and coulombic force is _ proportional, which means _

b. fluorine has a smaller atomic radius than carbon

Explanation:

• For part a:
• Magnesium (Mg) and barium (Ba) are in the same group (Group 2 - alkaline earth metals). Elements in the same group have the same number of valence electrons. But they have different numbers of electron shells (Ba has more electron shells as it is lower in the group).
• Coulomb's Law is \(F = k\frac{q_1q_2}{r^2}\), where \(F\) is the electrostatic force, \(k\) is a constant, \(q_1\) and \(q_2\) are the charges (in an atom, \(q_1\) can be considered as the effective nuclear charge \(Z_{eff}\) and \(q_2\) as the charge of the electron), and \(r\) is the distance (related to the number of electron shells). The relationship between the distance (\(r\), related to the number of electron shells) and the coulombic force is inversely proportional. As the number of electron shells (distance \(r\)) increases (from Mg to Ba), the coulombic attraction between the nucleus and the valence electrons decreases. So, the atomic radius increases (Mg has fewer electron shells, so smaller radius).
• For part b:
• Fluorine (F) and carbon (C) are in the same period. In a period, as we move from left to right, the atomic number increases. The number of electron shells is the same for elements in the same period. But the effective nuclear charge (\(Z_{eff}\)) increases (for F, \(Z = 9\); for C, \(Z=6\), and after shielding, \(Z_{eff}\) of F is higher than that of C).
• From Coulomb's Law \(F = k\frac{q_1q_2}{r^2}\), when the distance (\(r\), number of electron shells) is the same (same period), the relationship between the effective nuclear charge (\(Z_{eff}\), \(q_1\)) and the coulombic force is directly proportional. A higher \(Z_{eff}\) (in F compared to C) means a stronger coulombic attraction between the nucleus and the valence electrons. So, the electrons are pulled closer to the nucleus, resulting in a smaller atomic radius for F compared to C.

Answer:

a. Magnesium and barium are found in the same group, which means they have the same number of valence electrons but different number of electron shells. The relationship between distance (number of electron shells) and coulombic force is inversely proportional, which means as the number of electron shells (distance) increases (from Mg to Ba), the coulombic attraction between the nucleus and valence electrons decreases, so atomic radius increases (Mg has smaller radius).
b. Fluorine and carbon are in the same period (same number of electron shells). The relationship between effective nuclear charge and coulombic force is directly proportional. Since F has a higher effective nuclear charge (compared to C in the same period), the coulombic attraction between its nucleus and valence electrons is stronger, pulling electrons closer to the nucleus, resulting in a smaller atomic radius for F.