Question

use a model to determine bond character
9 use your electronegativity periodic table to complete the data table in part a of modeling molecular polarity
modeling molecular polarity
part a:
hydrogen fluoride
hf
nitrogen gas
n₂
carbon dioxide
co₂
chemical formula
molecular structure
electronegativity difference
bond character
a) non - polar covalent
b) polar covalent
c) ionic
a) non - polar covalent
b) polar covalent
c) ionic
a) non - polar covalent
b) polar covalent
c) ionic

Explanation:

Step1: Find electronegativity values

From the electronegativity table, \( \text{EN}_H = 2.1\) and \( \text{EN}_F=3.98\)

Step2: Calculate electronegativity difference

\(\Delta \text{EN}=\vert\text{EN}_F - \text{EN}_H\vert=\vert3.98 - 2.1\vert = 1.88\)

Step3: Determine bond character

Since \(0.5<\Delta \text{EN}=1.88 < 1.7\), the bond is polar covalent.

Answer:

Electronegativity difference for \(HF\): \(1.88\), Bond character: b) polar covalent.

For \(N_2\):

Step1: Find electronegativity values

\(\text{EN}_N = 3.04\) (for both \(N\) atoms in \(N_2\))

Step2: Calculate electronegativity difference

\(\Delta \text{EN}=\vert\text{EN}_N-\text{EN}_N\vert = 0\)

Step3: Determine bond character

Since \(\Delta \text{EN} = 0\), the bond is non - polar covalent.