Question

Identify a pair of molecules having similar shapes of both members.

• A. $\text{NH}_{3}, \text{SO}_{2}$
• B. $\text{XeF}_{4}, \text{SF}_{4}$
• C. $\text{H}_{2}\text{O},\text{SCl}_{2}$
• D. $\text{PCl}_{5}, \text{BrF}_{5}$

Hint:

Molecules belonging to the same generic formula $\text{AX}_n\text{E}_m$ (where X is a bonded atom and E is a lone pair) will always have identically named shapes! Both $\text{H}_2\text{O}$ and $\text{SCl}_2$ are $\text{AX}_2\text{E}_2$ systems.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Molecular shape depends on the number of bonding pairs and lone pairs around the central atom (VSEPR theory).

Step 2: Formula Application:
Check steric numbers: - $H_2O$: 2 Bond Pairs + 2 Lone Pairs = Bent/V-shape. - $SCl_2$: 2 Bond Pairs + 2 Lone Pairs = Bent/V-shape.

Step 3: Explanation:
Both Oxygen (in $H_2O$) and Sulfur (in $SCl_2$) belong to Group 16. They have 6 valence electrons, use 2 for bonding with monovalent atoms, leaving 4 electrons (2 lone pairs). This results in an identical Bent geometry for both.

Step 4: Final Answer:
The pair with similar shapes is $H_2O$ and $SCl_2$.