Question:medium

Identify the correct statement regarding geometry and lone pair of electrons present in \(\text{CH}_4\) and \(\text{SiCl}_4\).

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A molecule of type \(\text{AX}_4\) with no lone pair on the central atom always has tetrahedral geometry.
Updated On: May 14, 2026
  • Both have same geometry with two lone pair of electrons each.
  • Both have different geometry with one lone pair of electrons each.
  • Both have same geometry with no lone pair of electrons each.
  • Both have different geometry with no lone pair of electrons each.
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
The geometry of a molecule and the number of lone pairs on its central atom can be predicted using VSEPR (Valence Shell Electron Pair Repulsion) theory.
We need to determine the number of valence electrons of the central atom and how many bonds it forms.
Step 2: Key Formula or Approach:
Use VSEPR theory to determine geometry.
Count the valence electrons of the central atom, subtract electrons used in bonding to find lone pairs, and determine hybridization based on the steric number.
Step 3: Detailed Explanation:
Let's analyze the first molecule, \(\text{CH}_4\) (Methane):
The central atom is Carbon (\(\text{C}\)), which belongs to Group 14 and has 4 valence electrons.
It forms 4 single covalent bonds with 4 Hydrogen (\(\text{H}\)) atoms.
All 4 valence electrons are involved in bonding.
Therefore, the number of lone pairs on the central carbon atom = \(4 - 4 = 0\).
With 4 bond pairs and 0 lone pairs, the hybridization is \(\text{sp}^3\), and the geometry is tetrahedral.
Now let's analyze the second molecule, \(\text{SiCl}_4\) (Silicon tetrachloride):
The central atom is Silicon (\(\text{Si}\)), which is situated directly below Carbon in Group 14 and also has 4 valence electrons.
It forms 4 single covalent bonds with 4 Chlorine (\(\text{Cl}\)) atoms.
Similar to methane, all 4 valence electrons are utilized in bonding.
Therefore, the number of lone pairs on the central silicon atom = \(4 - 4 = 0\).
With 4 bond pairs and 0 lone pairs, its hybridization is also \(\text{sp}^3\), and its geometry is tetrahedral.
Comparing both molecules, they both possess the same tetrahedral geometry and neither central atom has any lone pairs of electrons.
Step 4: Final Answer:
Option (C) accurately describes both molecules.
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