Molecular Shapes Worksheet with 3D models of PBr3, H2S, and CH4, including electron dot structure spaces 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.

---

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 = \( 5 + 3 \times 7 = 26 \).
- Phosphorus is the central atom because it is less electronegative than bromine.
- Each Br atom forms a single bond with P, using 6 electrons (3 bonds × 2 electrons per bond = 6 electrons).
- Remaining electrons = \( 26 - 6 = 20 \), which are distributed as lone pairs on the Br atoms (each Br gets 3 lone pairs, totaling 6 electrons per Br).

Electron dot structure:
```
:Br
|
P---Br
|
:Br
```

##### Step 2: Molecular Geometry
- The central phosphorus atom 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 the central phosphorus atom causes 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 = \( 6 + 2 \times 1 = 8 \).
- Sulfur is the central atom because it is less electronegative than hydrogen.
- Each H atom forms a single bond with S, using 2 electrons (2 bonds × 1 electron per bond = 2 electrons).
- Remaining electrons = \( 8 - 2 = 6 \), which are distributed as 3 lone pairs on the sulfur atom.

Electron dot structure:
```
:H
|
S---H
```

##### Step 2: Molecular Geometry
- The central sulfur atom 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 the sulfur atom cause 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 = \( 4 + 4 \times 1 = 8 \).
- Carbon is the central atom because it is less electronegative than hydrogen.
- Each H atom forms a single bond with C, using 4 electrons (4 bonds × 1 electron per bond = 4 electrons).
- No remaining electrons; all electrons are used in bonding.

Electron dot structure:
```
H
\
C---H
/
H
/
H
```

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

##### Step 3: Polarity
- The molecule is nonpolar because:
- All C-H bonds are identical, and the tetrahedral shape ensures that the dipole moments of the bonds cancel out symmetrically.
- There are no lone pairs on the central carbon atom to disrupt symmetry.

##### 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:
\[
\boxed{
\begin{array}{ll}
\text{PBr₃:} & \text{Trigonal pyramidal, Polar} \\
\text{H₂S:} & \text{Bent (angular), Polar} \\
\text{CH₄:} & \text{Tetrahedral, Nonpolar}
\end{array}
}
\]