Use the electronegativity values given to answer the following:
Electronegativity values: [tex]$C =2.5, H =2.1, N=3.0, O =3.5, Cl =3.0$[/tex]
For each bond below:
a) Determine the electronegativity difference
b) State the bond type (ionic, polar covalent, or non-polar covalent)
c) Predict whether the molecule is polar or non-polar (assume correct geometry)
| Molecule | [tex]$\Delta EN$[/tex] | Bond Type | Molecule Polarity |
|----------|------------------|-----------|-------------------|
| [tex]$H - Cl$[/tex] | | | |
| [tex]$C - H$[/tex] | | | |
| [tex]$N - H$[/tex] | | | |
| [tex]$O - H$[/tex] | | | |
Answer (2)
Calculate the electronegativity difference ( Δ EN ) for each bond using Δ EN = ∣ E N 1 − E N 2 ∣ .
Determine the bond type based on the calculated Δ EN values: non-polar covalent ( Δ EN < 0.4 ), polar covalent ( 0.4 ≤ Δ EN < 1.7 ), or ionic ( Δ EN ≥ 1.7 ).
Predict the molecular polarity based on the bond polarity (since the molecules are diatomic, the molecular polarity is the same as the bond polarity).
Fill in the table with the calculated Δ EN values, bond types, and molecular polarities: See table in step 17 .
Explanation
Understanding the Problem We want to determine the electronegativity difference, bond type, and molecular polarity for H-Cl, C-H, N-H, and O-H bonds.
Calculating Electronegativity Difference First, we calculate the electronegativity difference ( Δ EN ) for each bond using the formula Δ EN = ∣ E N 1 − E N 2 ∣ , where E N 1 and E N 2 are the electronegativity values of the two atoms in the bond.
H-Cl Electronegativity Difference For H-Cl: Δ EN = ∣2.1 − 3.0∣ = 0.9
C-H Electronegativity Difference For C-H: Δ EN = ∣2.5 − 2.1∣ = 0.4
N-H Electronegativity Difference For N-H: Δ EN = ∣3.0 − 2.1∣ = 0.9
O-H Electronegativity Difference For O-H: Δ EN = ∣3.5 − 2.1∣ = 1.4
Determining Bond Type Next, we determine the bond type based on the following rules:
If Δ EN < 0.4 , the bond is non-polar covalent.
If 0.4 ≤ Δ EN < 1.7 , the bond is polar covalent.
If Δ EN ≥ 1.7 , the bond is ionic.
H-Cl Bond Type For H-Cl: Δ EN = 0.9 , so the bond is polar covalent.
C-H Bond Type For C-H: Δ EN = 0.4 , so the bond is polar covalent.
N-H Bond Type For N-H: Δ EN = 0.9 , so the bond is polar covalent.
O-H Bond Type For O-H: Δ EN = 1.4 , so the bond is polar covalent.
Predicting Molecular Polarity Finally, we predict the molecular polarity. Since all the molecules are diatomic, the molecular polarity is the same as the bond polarity.
H-Cl Molecular Polarity H-Cl is polar covalent, so the molecule is polar.
C-H Molecular Polarity C-H is polar covalent, so the molecule is polar.
N-H Molecular Polarity N-H is polar covalent, so the molecule is polar.
O-H Molecular Polarity O-H is polar covalent, so the molecule is polar.
Final Answer Here's the completed table:
Molecule
Δ EN
Bond Type
Molecule Polarity
H - Cl
0.9
Polar Covalent
Polar
C - H
0.4
Polar Covalent
Polar
N - H
0.9
Polar Covalent
Polar
O - H
1.4
Polar Covalent
Polar
Examples
Understanding electronegativity differences and bond polarity is crucial in many real-world applications. For instance, in drug design, knowing the polarity of different bonds helps predict how a drug molecule will interact with its target in the body. Polar molecules tend to dissolve better in water-based environments like our bloodstream, while non-polar molecules might interact better with fatty tissues. This knowledge allows scientists to design drugs that are more effective and have fewer side effects. Similarly, in materials science, understanding bond polarity helps in designing polymers with specific properties, such as flexibility or strength.
The electronegativity differences for the H-Cl, C-H, N-H, and O-H bonds were calculated as 0.9, 0.4, 0.9, and 1.4, respectively. All bonds are classified as polar covalent, leading to the conclusion that the molecules are polar. Therefore, each bond is polar covalent, and the overall polarity of the molecules is also polar.
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