The shape of ammonia molecule is

1. Analyze the Question

The question asks for the molecular geometry (shape) of the ammonia molecule ($NH_3$). To determine this, we must look at the central atom (Nitrogen) and how its valence electrons are arranged.

2. Explain the Concept

According to the Valence Shell Electron Pair Repulsion (VSEPR) theory:

• Nitrogen has 5 valence electrons. In $NH_3$, it forms 3 covalent bonds with 3 Hydrogen atoms.
• This leaves 1 lone pair of electrons on the Nitrogen atom.
• The total number of electron domains around the Nitrogen is 4 (3 bonding pairs + 1 lone pair).
• While the electron geometry of 4 domains is tetrahedral, the molecular shape is determined only by the positions of the atoms.
• The lone pair exerts a stronger repulsive force than the bonding pairs, pushing the three Hydrogen atoms closer together, resulting in a trigonal pyramidal shape.

3. Show the Solution

• Step 1: Identify the central atom: Nitrogen ($N$).
• Step 2: Count the electron pairs: 3 bonding pairs ($N-H$) and 1 lone pair.
• Step 3: Apply VSEPR theory: 4 electron pairs suggest a tetrahedral arrangement of electrons. However, because one of those positions is occupied by a lone pair (which is invisible in the molecular shape), the atoms form a pyramid with a triangular base.
• Step 4: Evaluate the options: • A (V-shaped) describes water ($H_2O$). • B (Tetrahedral) describes methane ($CH_4$). • C (Co-planar) describes molecules like $BF_3$. • D (Trigonal) refers to the trigonal pyramidal structure of ammonia.

4. Identify the Answer

Therefore, the correct option is D.

5. Exam Tip

To remember molecular shapes easily, use this quick guide:

• 2 bonding pairs, 0 lone pairs: Linear ($CO_2$)
• 3 bonding pairs, 0 lone pairs: Trigonal planar ($BF_3$)
• 2 bonding pairs, 2 lone pairs: V-shaped/Bent ($H_2O$)
• 3 bonding pairs, 1 lone pair: Trigonal pyramidal ($NH_3$)
• 4 bonding pairs, 0 lone pairs: Tetrahedral ($CH_4$)