This question involves matching hybrid orbitals (from List-I) with the correct coordination complexes (from List-II). Hybridization in coordination compounds is determined by the geometry of the complex, which in turn depends on the coordination number and the nature of the ligands (weak field or strong field).

Key Concepts:

Hybridization and Geometry:

• dsp² hybridization gives square planar geometry (coordination number 4).
• sp³ hybridization gives tetrahedral geometry (coordination number 4).
• sp³d² hybridization gives octahedral geometry (outer orbital complex, high spin).
• d²sp³ hybridization gives octahedral geometry (inner orbital complex, low spin).

Strong field vs Weak field ligands: Strong field ligands (e.g., CN^-, CO, NH₃) cause pairing of electrons (low spin complexes), while weak field ligands (e.g., F^-, Cl^-, H₂O) do not (high spin complexes).

Step-by-Step Matching:

(P) dsp²: This hybridization is for square planar complexes. Look for a 4-coordinate complex that is square planar. ${\left[\mathrm{Ni}\right(\mathrm{CN}\left)4\right]}^{2-}$ (option 6) is square planar because CN^- is a strong field ligand, causing pairing of Ni²⁺ electrons (d⁸ configuration).

(Q) sp³: This hybridization is for tetrahedral complexes. Look for a 4-coordinate complex that is tetrahedral. ${\left[\mathrm{FeCl}4\right]}^{2-}$ (option 4) is tetrahedral because Cl^- is a weak field ligand, and Fe is in +2 oxidation state (d⁶, high spin). Also, Ni(CO)₄ (option 5) is tetrahedral (CO is strong field but Ni is in 0 oxidation state, d¹⁰ configuration, which is always tetrahedral). So, (Q) matches with both (4) and (5).

(R) sp³d²: This is outer orbital octahedral hybridization (high spin). Look for an octahedral complex with weak field ligands. ${\left[\mathrm{FeF}6\right]}^{4-}$ (option 1) has F^- (weak field), so it is high spin and uses outer orbitals (sp³d²). Also, [Ti(H₂O)₃Cl₃] (option 2) is octahedral. Ti is in +3 oxidation state (d¹), which is always high spin regardless of ligand, so it also uses sp³d². So, (R) matches with both (1) and (2).

(S) d²sp³: This is inner orbital octahedral hybridization (low spin). Look for an octahedral complex with strong field ligands. ${\left[\mathrm{Cr}\right(\mathrm{NH}3\left)6\right]}^{3+}$ (option 3) has NH₃ (strong field), and Cr³⁺ (d³) is always low spin, so it uses d²sp³ hybridization.

Final Matching:

• P → 6
• Q → 4, 5
• R → 1, 2
• S → 3

Related Topics & Formulae:

Valence Bond Theory (VBT): Explains hybridization in coordination compounds. The type of hybridization dictates the geometry.

Crystal Field Theory (CFT): Explains the splitting of d-orbitals and helps determine whether a complex is high spin or low spin, which is crucial for choosing between inner (d²sp³) and outer (sp³d²) orbital complexes for octahedral geometry.

Common Geometries:

• Coordination Number 4: Tetrahedral (sp³) or Square Planar (dsp²)
• Coordination Number 6: Octahedral (sp³d² or d²sp³)