Electronegativity Difference and Bond Type Worksheet

Problem Overview:

The task involves determining the electronegativity difference (ΔEN) and the bond type for a series of chemical compounds using the Pauling Scale Periodic Table provided. The bond types are categorized as follows:
- Nonpolar covalent: ΔEN < 0.5
- Polar covalent: 0.5 ≤ ΔEN ≤ 2.0
- Ionic: ΔEN > 2.0

Solution Approach:

1. Identify the electronegativity values of the elements involved in each compound from the Pauling Scale Periodic Table.
2. Calculate the electronegativity difference (ΔEN):
\[
\Delta EN = | \text{Electronegativity of more electronegative element} - \text{Electronegativity of less electronegative element} |
\]
3. Determine the bond type based on the calculated ΔEN using the given criteria.

Step-by-Step Solution:

#### 1. NaCl
- Electronegativity of Na: 0.9
- Electronegativity of Cl: 3.0
- ΔEN = |3.0 - 0.9| = 2.1
- Bond type: Ionic (since ΔEN > 2.0)

#### 2. HBr
- Electronegativity of H: 2.1
- Electronegativity of Br: 2.8
- ΔEN = |2.8 - 2.1| = 0.7
- Bond type: Polar covalent (since 0.5 ≤ ΔEN ≤ 2.0)

#### 3. CH₄
- Electronegativity of C: 2.5
- Electronegativity of H: 2.1
- ΔEN = |2.5 - 2.1| = 0.4
- Bond type: Nonpolar covalent (since ΔEN < 0.5)

#### 4. Cl₂
- Electronegativity of Cl: 3.0
- Both atoms are the same, so ΔEN = 0
- Bond type: Nonpolar covalent (since ΔEN < 0.5)

#### 5. CO
- Electronegativity of C: 2.5
- Electronegativity of O: 3.5
- ΔEN = |3.5 - 2.5| = 1.0
- Bond type: Polar covalent (since 0.5 ≤ ΔEN ≤ 2.0)

#### 6. HCl
- Electronegativity of H: 2.1
- Electronegativity of Cl: 3.0
- ΔEN = |3.0 - 2.1| = 0.9
- Bond type: Polar covalent (since 0.5 ≤ ΔEN ≤ 2.0)

#### 7. HF
- Electronegativity of H: 2.1
- Electronegativity of F: 4.0
- ΔEN = |4.0 - 2.1| = 1.9
- Bond type: Polar covalent (since 0.5 ≤ ΔEN ≤ 2.0)

#### 8. K₂S
- Electronegativity of K: 0.8
- Electronegativity of S: 2.5
- ΔEN = |2.5 - 0.8| = 1.7
- Bond type: Ionic (since ΔEN > 2.0)

#### 9. LiF
- Electronegativity of Li: 1.0
- Electronegativity of F: 4.0
- ΔEN = |4.0 - 1.0| = 3.0
- Bond type: Ionic (since ΔEN > 2.0)

#### 10. H₂
- Electronegativity of H: 2.1
- Both atoms are the same, so ΔEN = 0
- Bond type: Nonpolar covalent (since ΔEN < 0.5)

#### 11. MgO
- Electronegativity of Mg: 1.2
- Electronegativity of O: 3.5
- ΔEN = |3.5 - 1.2| = 2.3
- Bond type: Ionic (since ΔEN > 2.0)

#### 12. HI
- Electronegativity of H: 2.1
- Electronegativity of I: 2.5
- ΔEN = |2.5 - 2.1| = 0.4
- Bond type: Nonpolar covalent (since ΔEN < 0.5)

#### 13. CuF
- Electronegativity of Cu: 1.9
- Electronegativity of F: 4.0
- ΔEN = |4.0 - 1.9| = 2.1
- Bond type: Ionic (since ΔEN > 2.0)

#### 14. NO
- Electronegativity of N: 3.0
- Electronegativity of O: 3.5
- ΔEN = |3.5 - 3.0| = 0.5
- Bond type: Polar covalent (since 0.5 ≤ ΔEN ≤ 2.0)

Final Answer:

\[
\begin{array}{|c|c|c|}
\hline
\text{Formula} & \text{Electronegativity Difference (ΔEN)} & \text{Bond Type} \\
\hline
\text{NaCl} & 2.1 & \text{Ionic} \\
\hline
\text{HBr} & 0.7 & \text{Polar covalent} \\
\hline
\text{CH}_4 & 0.4 & \text{Nonpolar covalent} \\
\hline
\text{Cl}_2 & 0 & \text{Nonpolar covalent} \\
\hline
\text{CO} & 1.0 & \text{Polar covalent} \\
\hline
\text{HCl} & 0.9 & \text{Polar covalent} \\
\hline
\text{HF} & 1.9 & \text{Polar covalent} \\
\hline
\text{K}_2\text{S} & 1.7 & \text{Ionic} \\
\hline
\text{LiF} & 3.0 & \text{Ionic} \\
\hline
\text{H}_2 & 0 & \text{Nonpolar covalent} \\
\hline
\text{MgO} & 2.3 & \text{Ionic} \\
\hline
\text{HI} & 0.4 & \text{Nonpolar covalent} \\
\hline
\text{CuF} & 2.1 & \text{Ionic} \\
\hline
\text{NO} & 0.5 & \text{Polar covalent} \\
\hline
\end{array}
\]

\boxed{
\begin{array}{|c|c|c|}
\hline
\text{Formula} & \text{Electronegativity Difference (ΔEN)} & \text{Bond Type} \\
\hline
\text{NaCl} & 2.1 & \text{Ionic} \\
\hline
\text{HBr} & 0.7 & \text{Polar covalent} \\
\hline
\text{CH}_4 & 0.4 & \text{Nonpolar covalent} \\
\hline
\text{Cl}_2 & 0 & \text{Nonpolar covalent} \\
\hline
\text{CO} & 1.0 & \text{Polar covalent} \\
\hline
\text{HCl} & 0.9 & \text{Polar covalent} \\
\hline
\text{HF} & 1.9 & \text{Polar covalent} \\
\hline
\text{K}_2\text{S} & 1.7 & \text{Ionic} \\
\hline
\text{LiF} & 3.0 & \text{Ionic} \\
\hline
\text{H}_2 & 0 & \text{Nonpolar covalent} \\
\hline
\text{MgO} & 2.3 & \text{Ionic} \\
\hline
\text{HI} & 0.4 & \text{Nonpolar covalent} \\
\hline
\text{CuF} & 2.1 & \text{Ionic} \\
\hline
\text{NO} & 0.5 & \text{Polar covalent} \\
\hline
\end{array}
}