Molecular Shapes Worksheet with 3D molecular models and questions on geometry and polarity.

Problem Analysis:

The worksheet asks us to analyze the molecular geometry and polarity of three molecules: PBr₃, H₂S, and CH₄. We need to determine their electron dot structures, molecular geometries, and whether they are polar or nonpolar.

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Solution:

#### 1. PBr₃ (Phosphorus Tribromide)

##### Step 1: Electron Dot Structure
- Phosphorus (P) is in Group 15 and has 5 valence electrons.
- Bromine (Br) is in Group 17 and has 7 valence electrons.
- Total valence electrons = \( 5 + 3 \times 7 = 26 \).

The electron dot structure for PBr₃ can be drawn as follows:
```
Br
\
P---Br
/
Br
```
Phosphorus forms three single bonds with bromine atoms, and it has one lone pair of electrons.

##### Step 2: Molecular Geometry
- The central atom (P) has 4 regions of electron density: 3 bonding pairs and 1 lone pair.
- According to VSEPR theory, this arrangement corresponds to a trigonal pyramidal geometry.

##### Step 3: Polarity
- The molecule is polar because:
- The lone pair on phosphorus creates an asymmetrical distribution of charge.
- The dipole moments of the P-Br bonds do not cancel out due to the trigonal pyramidal shape.

##### Final Answer for PBr₃:
- Geometry: Trigonal pyramidal
- Polarity: Polar

---

#### 2. H₂S (Hydrogen Sulfide)

##### Step 1: Electron Dot Structure
- Sulfur (S) is in Group 16 and has 6 valence electrons.
- Hydrogen (H) is in Group 1 and has 1 valence electron.
- Total valence electrons = \( 6 + 2 \times 1 = 8 \).

The electron dot structure for H₂S can be drawn as follows:
```
H
\
S---H
```
Sulfur forms two single bonds with hydrogen atoms and has two lone pairs of electrons.

##### Step 2: Molecular Geometry
- The central atom (S) has 4 regions of electron density: 2 bonding pairs and 2 lone pairs.
- According to VSEPR theory, this arrangement corresponds to a bent (or angular) geometry.

##### Step 3: Polarity
- The molecule is polar because:
- The lone pairs on sulfur create an asymmetrical distribution of charge.
- The dipole moments of the S-H bonds do not cancel out due to the bent shape.

##### Final Answer for H₂S:
- Geometry: Bent (angular)
- Polarity: Polar

---

#### 3. CH₄ (Methane)

##### Step 1: Electron Dot Structure
- Carbon (C) is in Group 14 and has 4 valence electrons.
- Hydrogen (H) is in Group 1 and has 1 valence electron.
- Total valence electrons = \( 4 + 4 \times 1 = 8 \).

The electron dot structure for CH₄ can be drawn as follows:
```
H
\
C---H
/ \
H H
```
Carbon forms four single bonds with hydrogen atoms, and there are no lone pairs on carbon.

##### Step 2: Molecular Geometry
- The central atom (C) has 4 regions of electron density: 4 bonding pairs and 0 lone pairs.
- According to VSEPR theory, this arrangement corresponds to a tetrahedral geometry.

##### Step 3: Polarity
- The molecule is nonpolar because:
- All bond dipoles (C-H) are identical and symmetrically arranged in a tetrahedral shape.
- The dipole moments cancel each other out.

##### Final Answer for CH₄:
- Geometry: Tetrahedral
- Polarity: Nonpolar

---

Final Answers:

1. PBr₃:
- Geometry: Trigonal pyramidal
- Polarity: Polar

2. H₂S:
- Geometry: Bent (angular)
- Polarity: Polar

3. CH₄:
- Geometry: Tetrahedral
- Polarity: Nonpolar

Boxed Final Answer:
\[
\boxed{
\begin{array}{ll}
\text{PBr}_3 & \text{Trigonal pyramidal, Polar} \\
\text{H}_2\text{S} & \text{Bent (angular), Polar} \\
\text{CH}_4 & \text{Tetrahedral, Nonpolar}
\end{array}
}
\]