Match List-I with List-II and select the correct answer using the code given below the Lists: List-I (Compound/Molecule) List-II (Shape of Molecule) A. CH 3F 1. Trigonal planar B. HCHO 2. Tetrahedral C. HCN 3. Trigonal pyramidal D. NH 3 4. Linear
A-2, B-1, C-4, D-3
Understanding Molecular Shapes: A VSEPR Approach
Molecular shape is a crucial property that influences a molecule's physical and chemical characteristics. The Valence Shell Electron Pair Repulsion (VSEPR) theory is a common method used to predict the geometry of molecules based on the repulsion between electron pairs around the central atom.
To determine the shape of a molecule, we typically follow these steps for the central atom:
- Determine the number of valence electrons.
- Determine the number of atoms bonded to the central atom.
- Calculate the number of lone pairs on the central atom.
- Calculate the steric number (number of bonded atoms + number of lone pairs).
- Use the steric number and the number of lone pairs to predict the electron geometry and molecular shape.
Let's apply this to the molecules given in List-I:
|
List-I (Compound/Molecule) |
List-II (Shape of Molecule) |
|
A. CH <sub>3</sub>F |
1. Trigonal planar |
|
B. HCHO |
2. Tetrahedral |
|
C. HCN |
3. Trigonal pyramidal |
|
D. NH <sub>3</sub> |
4. Linear |
Analyzing Each Molecule's Shape
Let's analyze the shape of each compound/molecule from List-I:
- A. CH<sub>3</sub>F (Methyl fluoride):
- Central atom: Carbon (C)
- Valence electrons on C: 4
- Atoms bonded to C: 3 hydrogens (H) + 1 fluorine (F) = 4 atoms
- Lone pairs on C: \(\frac{1}{2} (4 - 4 \times 1) = 0\) lone pairs
- Steric number: 4 (4 bonded atoms + 0 lone pairs)
- Electron geometry: Tetrahedral
- Molecular shape: With 0 lone pairs, the molecular shape is the same as the electron geometry. Thus, CH<sub>3<sub>F has a Tetrahedral shape.
- B. HCHO (Formaldehyde):
- Central atom: Carbon (C)
- Valence electrons on C: 4
- Atoms bonded to C: 2 hydrogens (H) + 1 oxygen (O) = 3 atoms (The double bond to O is treated as one electron group for geometry)
- Lone pairs on C: \(\frac{1}{2} (4 - 3 \times 1) = \frac{1}{2}(4-4) = 0\) lone pairs (Correct calculation: \(4 - (\text{bonds to H} + \text{double bonds to O}) = 4 - (2 \times 1 + 1 \times 2 \text{ equivalent}) = 4 - 4 = 0\). Or simpler, carbon forms 4 bonds total, no lone pairs needed)
- Steric number: 3 (3 bonded atoms + 0 lone pairs)
- Electron geometry: Trigonal planar
- Molecular shape: With 0 lone pairs, the molecular shape is the same as the electron geometry. Thus, HCHO has a Trigonal planar shape.
- C. HCN (Hydrogen cyanide):
- Central atom: Carbon (C)
- Valence electrons on C: 4
- Atoms bonded to C: 1 hydrogen (H) + 1 nitrogen (N) = 2 atoms (The triple bond to N is treated as one electron group)
- Lone pairs on C: \(\frac{1}{2} (4 - 1 \times 1 - 1 \times 3) = \frac{1}{2}(4-4) = 0\) lone pairs
- Steric number: 2 (2 bonded atoms + 0 lone pairs)
- Electron geometry: Linear
- Molecular shape: With 0 lone pairs, the molecular shape is the same as the electron geometry. Thus, HCN has a Linear shape.
- D. NH<sub>3</sub> (Ammonia):
- Central atom: Nitrogen (N)
- Valence electrons on N: 5
- Atoms bonded to N: 3 hydrogens (H) = 3 atoms
- Lone pairs on N: \(\frac{1}{2} (5 - 3 \times 1) = \frac{1}{2}(2) = 1\) lone pair
- Steric number: 4 (3 bonded atoms + 1 lone pair)
- Electron geometry: Tetrahedral
- Molecular shape: With 1 lone pair and 3 bonded atoms, the molecular shape is Trigonal pyramidal.
Matching the Molecules to Shapes
Based on our analysis, the correct matching pairs are:
- A. CH<sub>3</sub>F <--> 2. Tetrahedral
- B. HCHO <--> 1. Trigonal planar
- C. HCN <--> 4. Linear
- D. NH<sub>3</sub> <--> 3. Trigonal pyramidal
This corresponds to the option A-2, B-1, C-4, D-3.
Revision Table: Common Shapes and Steric Numbers
| Steric Number | Bonded Atoms | Lone Pairs | Electron Geometry | Molecular Shape | Example |
| 2 | 2 | 0 | Linear | Linear | CO<sub>2</sub>, HCN |
| 3 | 3 | 0 | Trigonal Planar | Trigonal Planar | BF<sub>3</sub>, HCHO |
| 3 | 2 | 1 | Trigonal Planar | Bent (or Angular) | SO<sub>2</sub>, O<sub>3</sub> |
| 4 | 4 | 0 | Tetrahedral | Tetrahedral | CH<sub>4</sub>, CH<sub>3<sub>F |
| 4 | 3 | 1 | Tetrahedral | Trigonal Pyramidal | NH<sub>3</sub> |
| 4 | 2 | 2 | Tetrahedral | Bent (or Angular) | H<sub>2</sub>O |
Additional Information on Molecular Geometry and VSEPR
Molecular geometry describes the 3D arrangement of the atoms in a molecule. It is different from electron geometry, which describes the arrangement of all electron groups (both bonding pairs and lone pairs) around the central atom.
VSEPR theory is based on the idea that electron groups (single bonds, multiple bonds, lone pairs) repel each other and will arrange themselves as far apart as possible to minimize repulsion. Lone pairs generally exert more repulsion than bonding pairs, which can affect the final molecular shape.
Understanding the shapes of molecules helps predict their properties, such as polarity, reactivity, and intermolecular forces.
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