Question

1. use coulomb’s law to find the force on an electron in oxygen’s valence shell. then, use coulomb’s law to find the force on an electron in fluorine’s valence shell. the atomic radius of fluorine is 135 pm and the atomic radius of oxygen is 152 pm.

coulomb’s law: $f_e = k \frac{q_1 q_2}{r^2}$

$k = 0.000231 \\, \text{n} \cdot \text{pm}^2/\text{e}^2$

1. use your answers from question 1 to explain why a fluorine atom has a higher electronegativity than an oxygen atom.

Explanation:

Step1: Determine charges for oxygen

Oxygen has a nuclear charge \( q_1 = +8e \) (since atomic number is 8) and the valence electron has \( q_2 = -e \). The distance \( r_{O} = 152 \, \text{pm} \). Using Coulomb's law \( F_e = k\frac{q_1q_2}{r^2} \), substitute values: \( k = 0.000231 \, \text{N·pm}^2/\text{e}^2 \), \( q_1 = 8e \), \( q_2 = -e \), \( r = 152 \, \text{pm} \). So \( F_{O} = 0.000231 \times \frac{(8e)(-e)}{(152)^2} \). Calculate numerator: \( 8e \times -e = -8e^2 \). Denominator: \( 152^2 = 23104 \). Then \( F_{O} = 0.000231 \times \frac{-8e^2}{23104} \approx -0.000231 \times \frac{8}{23104} e^2 \approx -8.0 \times 10^{-9} e^2 \, \text{N} \) (magnitude calculation: \( 0.000231 \times 8 / 23104 \approx 8.0 \times 10^{-9} \)).

Step2: Determine charges for fluorine

Fluorine has nuclear charge \( q_1 = +9e \) (atomic number 9) and valence electron \( q_2 = -e \). Distance \( r_{F} = 135 \, \text{pm} \). Using Coulomb's law: \( F_{F} = 0.000231 \times \frac{(9e)(-e)}{(135)^2} \). Numerator: \( -9e^2 \), denominator: \( 135^2 = 18225 \). \( F_{F} = 0.000231 \times \frac{-9e^2}{18225} \approx -0.000231 \times \frac{9}{18225} e^2 \approx -1.15 \times 10^{-8} e^2 \, \text{N} \) (magnitude: \( 0.000231 \times 9 / 18225 \approx 1.15 \times 10^{-8} \)).

Step3: Analyze electronegativity (for part 2)

Electronegativity relates to the attraction for electrons. The magnitude of the force on fluorine's valence electron (\( |F_F| \)) is greater than on oxygen's (\( |F_O| \)) because \( F_F \) has a larger nuclear charge (9e vs 8e) and smaller radius (135 pm vs 152 pm), leading to a stronger attractive force. A stronger attractive force means fluorine more strongly attracts bonding electrons, so higher electronegativity.

Answer:

(for part 1 forces):
Force on oxygen's electron: \( \approx -8.0 \times 10^{-9} e^2 \, \text{N} \) (attractive, negative sign), Force on fluorine's electron: \( \approx -1.15 \times 10^{-8} e^2 \, \text{N} \) (more attractive).