Polarity of Molecules worksheet - Free Printable

To solve the problem, we need to analyze each molecule step by step. We will determine the electronegativity difference (Δ EN) between the atoms in each bond, identify the bond polarity, determine the molecular geometry, and finally assess the polarity of the molecule based on its Lewis structure and molecular shape.

Step-by-Step Solution:

#### 1. PCl₅
- Molecule: PCl₅
- Atoms Involved: P (Phosphorus) and Cl (Chlorine)
- Electronegativity Values:
- P: 2.19
- Cl: 3.16
- Electronegativity Difference (Δ EN):
\[
\Delta \text{EN} = 3.16 - 2.19 = 0.97
\]
- Bond Polarity: Since Δ EN > 0, the bond is polar.
- Molecular Geometry: Phosphorus pentachloride (PCl₅) has a trigonal bipyramidal geometry.
- Polarity of Molecule: Despite having polar bonds, the molecule is nonpolar because the dipole moments of the five P-Cl bonds cancel each other out due to the symmetrical trigonal bipyramidal shape.

#### 2. BeCl₂
- Molecule: BeCl₂
- Atoms Involved: Be (Beryllium) and Cl (Chlorine)
- Electronegativity Values:
- Be: 1.57
- Cl: 3.16
- Electronegativity Difference (Δ EN):
\[
\Delta \text{EN} = 3.16 - 1.57 = 1.59
\]
- Bond Polarity: Since Δ EN > 0, the bond is polar.
- Molecular Geometry: Beryllium chloride (BeCl₂) has a linear geometry.
- Polarity of Molecule: The molecule is polar because the dipole moments of the two Be-Cl bonds do not cancel each other out due to the linear but unsymmetrical arrangement of electron pairs around the central Be atom.

#### 3. CH₄
- Molecule: CH₄
- Atoms Involved: C (Carbon) and H (Hydrogen)
- Electronegativity Values:
- C: 2.55
- H: 2.20
- Electronegativity Difference (Δ EN):
\[
\Delta \text{EN} = 2.55 - 2.20 = 0.35
\]
- Bond Polarity: Since Δ EN > 0, the bond is polar.
- Molecular Geometry: Methane (CH₄) has a tetrahedral geometry.
- Polarity of Molecule: Despite having polar bonds, the molecule is nonpolar because the dipole moments of the four C-H bonds cancel each other out due to the symmetrical tetrahedral shape.

#### 4. OF₂
- Molecule: OF₂
- Atoms Involved: O (Oxygen) and F (Fluorine)
- Electronegativity Values:
- O: 3.44
- F: 3.98
- Electronegativity Difference (Δ EN):
\[
\Delta \text{EN} = 3.98 - 3.44 = 0.54
\]
- Bond Polarity: Since Δ EN > 0, the bond is polar.
- Molecular Geometry: Oxygen difluoride (OF₂) has a bent (V-shaped) geometry.
- Polarity of Molecule: The molecule is polar because the dipole moments of the two O-F bonds do not cancel each other out due to the bent shape, which results in a net dipole moment.

#### 5. SF₆
- Molecule: SF₆
- Atoms Involved: S (Sulfur) and F (Fluorine)
- Electronegativity Values:
- S: 2.58
- F: 3.98
- Electronegativity Difference (Δ EN):
\[
\Delta \text{EN} = 3.98 - 2.58 = 1.40
\]
- Bond Polarity: Since Δ EN > 0, the bond is polar.
- Molecular Geometry: Sulfur hexafluoride (SF₆) has an octahedral geometry.
- Polarity of Molecule: Despite having polar bonds, the molecule is nonpolar because the dipole moments of the six S-F bonds cancel each other out due to the symmetrical octahedral shape.

Final Table:

| Molecule | Δ EN | Bond Polarity | Molecular Geometry | Polarity of Molecule |
|----------|------|---------------|--------------------|-----------------------|
| 1. PCl₅ | 0.97 | Polar | Trigonal Bipyramidal | Nonpolar |
| 2. BeCl₂ | 1.59 | Polar | Linear | Polar |
| 3. CH₄ | 0.35 | Polar | Tetrahedral | Nonpolar |
| 4. OF₂ | 0.54 | Polar | Bent (V-shaped) | Polar |
| 5. SF₆ | 1.40 | Polar | Octahedral | Nonpolar |

Final Answer:

\boxed{
\begin{array}{|c|c|c|c|c|}
\hline
\text{Molecule} & \Delta \text{EN} & \text{Bond Polarity} & \text{Molecular Geometry} & \text{Polarity of Molecule} \\
\hline
1. \text{PCl}_5 & 0.97 & \text{Polar} & \text{Trigonal Bipyramidal} & \text{Nonpolar} \\
\hline
2. \text{BeCl}_2 & 1.59 & \text{Polar} & \text{Linear} & \text{Polar} \\
\hline
3. \text{CH}_4 & 0.35 & \text{Polar} & \text{Tetrahedral} & \text{Nonpolar} \\
\hline
4. \text{OF}_2 & 0.54 & \text{Polar} & \text{Bent (V-shaped)} & \text{Polar} \\
\hline
5. \text{SF}_6 & 1.40 & \text{Polar} & \text{Octahedral} & \text{Nonpolar} \\
\hline
\end{array}
}