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Match each species in List-I with its correct properties in List-II. List-I: (P) XeO2F2 (Q) SF4 (R) XeO6⁴- (S) [I(CN)2]- List-II: (1) The species has regular octahedral geometry (2) Five atoms of the species may lie in the same plane (3) The species uses d_(x²-y²) orbital in its bonding (4) The species uses d_(z²) orbital in its bonding (5) All atoms of the species lie in one plane

1. P -> 4; Q -> 4; R -> 1,2,3,4; S -> 2,4,5
2. P -> 4; Q -> 3,4; R -> 1,2; S -> 4,5
3. P -> 1,2,3; Q -> 1,3,4; R -> 1,2,5; S -> 1,5
4. P -> 2; Q -> 5; R -> 1,2; S -> 1,2,3,5

Correct answer: P -> 4; Q -> 3,4; R -> 1,2; S -> 4,5

Solution

XeO2F2: Xe has 5 electron pairs (2 bonds to O, 2 bonds to F, 1 lone pair) -> sp3d see-saw shape. sp3d uses d_(z²). In see-saw, 4 atoms can be in same plane (equatorial F and O atoms + center). P -> 4 (uses d_z²), and possibly 2 (5 atoms? No, only 5 atoms total including Xe, so 5 in plane is all atoms -> option 5? In see-saw: equatorial plane has 3 atoms + Xe = 4 atoms in one plane, not all 5). SF4: sp3d, see-saw shape. Uses d_z². -> 4. Also 4 (equatorial S, 2F equatorial, axial F... eq plane has S + 2 equatorial F = 3 in plane). d_(x²-y²) also involved in some descriptions of sp3d. XeO6⁴-: Xe has 6 bonds to O, no lone pair -> sp3d2 -> regular octahedral (option 1). 5 atoms in same plane (equatorial plane has 4 O + Xe = 5) -> option 2. [I(CN)2]-: I with 2 CN groups, I in +1. Hybridization sp3d (3 lone pairs + 2 bonds). Actually linear: sp3d with 3 equatorial lone pairs -> linear geometry with all 3 atoms in same plane (and all atoms collinear -> option 5). Uses d_z² -> option 4. S -> 4, 5. Standard answer: P->4; Q->3,4; R->1,2; S->4,5.

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