Question

Match List - I with List - II. Match the shapes of the molecules on the bases of VSEPR : 

Choose the correct answer from the options given below : 
 

• A. A-I, B-III, C-II, D-IV
• B. A-I, B-II, C-III, D-IV
• C. A-IV, B-III, C-I, D-II
• D. A-III, B-I, C-IV, D-II

Hint:

Be careful to distinguish between "Molecular Geometry" (shape of the atoms only) and "Electron Geometry" (arrangement of all electron domains). In this question, $XeF_4$ has a square planar molecular shape, but an octahedral electron geometry.

Answer 1

The Correct Option is D

Step 1: Concept:
The question asks us to match molecules with their corresponding electron-domain geometries based on the Valence Shell Electron Pair Repulsion (VSEPR) theory. Note that the List II contains the base "electron geometries" (determined by hybridization) rather than the final molecular shapes.

Step 2: Key Formula or Approach:
To find the electron geometry, calculate the Steric Number (SN):
\[ \text{SN} = (\text{Number of bonding pairs}) + (\text{Number of lone pairs on the central atom}) \]
- SN = 3 $\rightarrow$ Trigonal Planar (Plane triangle)
- SN = 4 $\rightarrow$ Tetrahedral
- SN = 5 $\rightarrow$ Trigonal Bipyramidal (Triangular dipyramid)
- SN = 6 $\rightarrow$ Octahedral

Step 3: Step-by-step Explanation:

• A. $ClF_3$: Chlorine (Group 17) has 7 valence electrons. It forms 3 single bonds with Fluorine, leaving 4 non-bonding electrons (2 lone pairs).
SN = $3 \text{ (bonds)} + 2 \text{ (lone pairs)} = 5$.
Electron geometry: Triangular dipyramid (III).

• B. $BF_3$: Boron (Group 13) has 3 valence electrons. It forms 3 single bonds with Fluorine and has 0 lone pairs.
SN = $3 \text{ (bonds)} + 0 \text{ (lone pairs)} = 3$.
Electron geometry: Plane triangle (I).

• C. $XeF_4$: Xenon (Group 18) has 8 valence electrons. It forms 4 single bonds with Fluorine, leaving 4 non-bonding electrons (2 lone pairs).
SN = $4 \text{ (bonds)} + 2 \text{ (lone pairs)} = 6$.
Electron geometry: Octahedral (IV).

• D. $CH_4$: Carbon (Group 14) has 4 valence electrons. It forms 4 single bonds with Hydrogen and has 0 lone pairs.
SN = $4 \text{ (bonds)} + 0 \text{ (lone pairs)} = 4$.
Electron geometry: Tetrahedral (II).
Matching them up gives A-III, B-I, C-IV, D-II.

Step 4: Final Answer:
The correct match is found in option (D).