Question

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

• A. Both have same geometry with two lone pair of electrons each.
• B. Both have different geometry with one lone pair of electrons each.
• C. Both have same geometry with no lone pair of electrons each.
• D. Both have different geometry with no lone pair of electrons each.

Hint:

A molecule of type \(\text{AX}_4\) with no lone pair on the central atom always has tetrahedral geometry.

Answer 1

The Correct Option is C

Concept:
According to VSEPR theory, the geometry of a molecule depends on the number of bond pairs and lone pairs around the central atom. If the central atom has four bond pairs and no lone pair, the geometry is tetrahedral.

Step 1: Check \(\text{CH}_4\).
In \(\text{CH}_4\), carbon forms four bonds with hydrogen atoms. So, carbon has:
• four bond
• zero lone pairs Thus, geometry is tetrahedral.

Step 2: Check \(\text{SiCl}_4\).
In \(\text{SiCl}_4\), silicon forms four bonds with chlorine atoms. So, silicon also has:
• four bond pairs
• zero lone pairs Thus, geometry is also tetrahedral.

Step 3: Compare both molecules.
Both molecules have:
• same tetrahedral geometry
• no lone pair on the central atom Hence, the correct answer is:
\[ \boxed{(C)\ \text{Both have same geometry with no lone pair of electrons each}} \]