Question

Identify the pair of species with same geometries

• A. \(XeF_4,\ PCl_4^+\)
• B. \(NH_3,\ ClF_3\)
• C. \(SF_4,\ NH_4^+\)
• D. \(XeF_2,\ I_3^-\)

Hint:

Species with type \[ AX_2E_3 \] always have linear geometry because three lone pairs occupy equatorial positions in trigonal bipyramidal arrangement.

Answer 1

The Correct Option is D

Step 1: Recall VSEPR notation.
The molecular geometry depends on the number of bond pairs (BP) and lone pairs (LP) around the central atom. We use the notation $ AX_nE_m $ where $ X $ = bonding pair and $ E $ = lone pair. The geometric shape is determined by both BP and LP.
Step 2: Analyze XeF2 and I3-.
For $ XeF_2 $: Xe has 8 valence electrons, forms 2 bonds with F, leaving $ (8-2)/2 = 3 $ lone pairs. Type: $ AX_2E_3 $. Geometry: linear (the 3 LPs occupy equatorial positions in a trigonal bipyramid). For $ I_3^- $: Central I has 7 + 1 (charge) = 8 electrons, forms 2 bonds, leaving $ (8-2)/2 = 3 $ lone pairs. Type: $ AX_2E_3 $. Geometry: linear. Both are linear - same geometry!
Step 3: Verify option 1 - XeF4 and PCl4+.
$ XeF_4 $: $ AX_4E_2 $ = square planar. $ PCl_4^+ $: $ AX_4 $ = tetrahedral. Different geometries.
Step 4: Verify option 2 - NH3 and ClF3.
$ NH_3 $: $ AX_3E_1 $ = trigonal pyramidal. $ ClF_3 $: $ AX_3E_2 $ = T-shaped. Different geometries.
Step 5: Verify option 3 - SF4 and NH4+.
$ SF_4 $: $ AX_4E_1 $ = see-saw shape. $ NH_4^+ $: $ AX_4 $ = tetrahedral. Different geometries.
Step 6: Final answer.
Only $ XeF_2 $ and $ I_3^- $ share the same molecular geometry (linear), making option (4) correct. \[ \boxed{XeF_2\ \text{and}\ I_3^-\ \text{are both linear}} \]