Vsepr worksheet - Free Printable

To solve the problem, we need to analyze each molecule using the VSEPR (Valence Shell Electron Pair Repulsion) theory. This involves determining the number of bonding pairs and lone pairs around the central atom and then predicting the molecular shape based on these electron pair arrangements.

Step-by-Step Analysis:

#### 1. Molecule: \( \text{NCl}_3 \)

- Central Atom: Nitrogen (N)
- Lewis Structure:
- N has 5 valence electrons.
- Each Cl has 7 valence electrons.
- Total valence electrons = \( 5 + 3 \times 7 = 26 \).
- N forms 3 single bonds with Cl atoms.
- N has 1 lone pair (remaining 2 electrons).

- Number of Bonding Pairs: 3
- Number of Lone Pairs: 1
- Molecular Shape: Trigonal pyramidal

#### 2. Molecule: \( \text{BCl}_3 \)

- Central Atom: Boron (B)
- Lewis Structure:
- B has 3 valence electrons.
- Each Cl has 7 valence electrons.
- Total valence electrons = \( 3 + 3 \times 7 = 24 \).
- B forms 3 single bonds with Cl atoms.
- B has no lone pairs.

- Number of Bonding Pairs: 3
- Number of Lone Pairs: 0
- Molecular Shape: Trigonal planar

#### 3. Molecule: \( \text{ClF}_6^- \)

- Central Atom: Chlorine (Cl)
- Lewis Structure:
- Cl has 7 valence electrons.
- Each F has 7 valence electrons.
- Total valence electrons = \( 7 + 6 \times 7 + 1 = 50 \).
- Cl forms 6 single bonds with F atoms.
- Cl has no lone pairs.

- Number of Bonding Pairs: 6
- Number of Lone Pairs: 0
- Molecular Shape: Octahedral

#### 4. Molecule: \( \text{NO}_3^- \)

- Central Atom: Nitrogen (N)
- Lewis Structure:
- N has 5 valence electrons.
- Each O has 6 valence electrons.
- Total valence electrons = \( 5 + 3 \times 6 + 1 = 24 \).
- N forms 3 double bonds with O atoms.
- N has no lone pairs.

- Number of Bonding Pairs: 3
- Number of Lone Pairs: 0
- Molecular Shape: Trigonal planar

#### 5. Molecule: \( \text{CF}_4 \)

- Central Atom: Carbon (C)
- Lewis Structure:
- C has 4 valence electrons.
- Each F has 7 valence electrons.
- Total valence electrons = \( 4 + 4 \times 7 = 32 \).
- C forms 4 single bonds with F atoms.
- C has no lone pairs.

- Number of Bonding Pairs: 4
- Number of Lone Pairs: 0
- Molecular Shape: Tetrahedral

#### 6. Molecule: \( \text{XeCl}_6^{2+} \)

- Central Atom: Xenon (Xe)
- Lewis Structure:
- Xe has 8 valence electrons.
- Each Cl has 7 valence electrons.
- Total valence electrons = \( 8 + 6 \times 7 - 2 = 50 \).
- Xe forms 6 single bonds with Cl atoms.
- Xe has no lone pairs.

- Number of Bonding Pairs: 6
- Number of Lone Pairs: 0
- Molecular Shape: Octahedral

Final Answer:

| LEWIS STRUCTURES | No of Bonding Pairs | No of Lone Pairs | Molecular Shape |
|------------------|---------------------|------------------|-----------------|
| \( \text{NCl}_3 \) | 3 | 1 | Trigonal pyramidal |
| \( \text{BCl}_3 \) | 3 | 0 | Trigonal planar |
| \( \text{ClF}_6^- \) | 6 | 0 | Octahedral |
| \( \text{NO}_3^- \) | 3 | 0 | Trigonal planar |
| \( \text{CF}_4 \) | 4 | 0 | Tetrahedral |
| \( \text{XeCl}_6^{2+} \) | 6 | 0 | Octahedral |

\boxed{
\begin{array}{|c|c|c|c|}
\hline
\text{LEWIS STRUCTURES} & \text{No of Bonding Pairs} & \text{No of Lone Pairs} & \text{Molecular Shape} \\
\hline
\text{NCl}_3 & 3 & 1 & \text{Trigonal pyramidal} \\
\hline
\text{BCl}_3 & 3 & 0 & \text{Trigonal planar} \\
\hline
\text{ClF}_6^- & 6 & 0 & \text{Octahedral} \\
\hline
\text{NO}_3^- & 3 & 0 & \text{Trigonal planar} \\
\hline
\text{CF}_4 & 4 & 0 & \text{Tetrahedral} \\
\hline
\text{XeCl}_6^{2+} & 6 & 0 & \text{Octahedral} \\
\hline
\end{array}
}