Question

The tetra atomic molecules/ions with different shapes are

• A. \(TeCl_4,\ SeF_4\)
• B. \(CH_4,\ PCl_4^+\)
• C. \(SF_4,\ NH_4^+\)
• D. \(SiH_4,\ CCl_4\)

Hint:

According to VSEPR theory: \[ AX_4 \rightarrow \text{tetrahedral} \] and \[ AX_4E \rightarrow \text{see-saw} \] where \(E\) represents a lone pair.

Answer 1

The Correct Option is C

Step 1: Apply VSEPR to SF₄.
S in SF₄: Valence electrons of S = 6. Four bonds to F use 4 electrons; one lone pair remains. Electron geometry = trigonal bipyramidal (5 electron pairs). The lone pair occupies an equatorial position. Molecular shape = see-saw (or disphenoidal).
Step 2: Apply VSEPR to NH₄⁺.
N in NH₄⁺: Valence electrons of N = 5; losing 1 electron (+ charge) gives 4 effective pairs. All 4 are bond pairs (no lone pairs). Electron geometry = molecular geometry = tetrahedral.
Step 3: Compare the two shapes.
SF₄ is see-saw (not a regular polyhedron due to lone pair). NH₄⁺ is tetrahedral. These are different shapes, confirming the pair given in option 3.
Step 4: Analyze other options for comparison.
Options including molecules with the same shape (e.g., two tetrahedral species, or two linear species) would be incorrect. Here SF₄ (see-saw) and NH₄⁺ (tetrahedral) are genuinely different.
Step 5: Explain why the shapes differ.
SF₄ has 4 bonding pairs + 1 lone pair; the lone pair distorts the geometry from tetrahedral to see-saw. NH₄⁺ has 4 bonding pairs and no lone pairs; perfect tetrahedral symmetry.
Step 6: Conclusion.
SF₄ (see-saw) and NH₄⁺ (tetrahedral) are a valid pair of tetraatomic species with different shapes.
\[ \boxed{SF_4 \text{ (see-saw) and } NH_4^+ \text{ (tetrahedral)}} \]