The number of species from the following that are involved in sp³d² hybridization is :
[Co(NH₃)₆]³⁺, SF₆, [CrF₆]³⁻, [CoF₆]³⁻, [Mn(CN)₆]³⁻, and [MnCl₆]³⁻

Solution

Chemistry ma’am · 3 steps

For each complex, ask whether two inner 3d orbitals are left empty. If not, the outer 4d orbitals are used.

1. The deciding test. Every species here is octahedral, so each uses six hybrid orbitals. If two inner 3d orbitals are free after the metal's own d electrons are placed, you get d²sp³ (inner orbital). If not, the metal must use 4d orbitals, giving sp³d² (outer orbital).
2. Ligand strength. Strong-field ligands like CN⁻, and NH₃ with Co³⁺, pair up the d electrons and free inner orbitals. Weak-field F⁻ and Cl⁻ cause no pairing. A d³ ion has two empty 3d orbitals even without pairing.

   Species	Central atom	Empty 3d orbitals	Hybridisation
   [Co(NH₃)₆]³⁺	Co³⁺, 3d⁶, paired by NH₃	2	d²sp³
   SF₆	S: 3s¹ 3p³ 3d² when excited	uses its own 3d	sp³d²
   [CrF₆]³⁻	Cr³⁺, 3d³	2, even with a weak ligand	d²sp³
   [CoF₆]³⁻	Co³⁺, 3d⁶, high spin	0	sp³d²
   [Mn(CN)₆]³⁻	Mn³⁺, 3d⁴, paired by CN⁻	2	d²sp³
   [MnCl₆]³⁻	Mn³⁺, 3d⁴, high spin	1	sp³d²

3. So. Only SF₆, [CoF₆]³⁻ and [MnCl₆]³⁻ are sp³d². That makes 3, option (b).

The idea to keep. Octahedral does not automatically mean sp³d². Count the empty inner 3d orbitals: two free gives d²sp³. A d³ ion always has two free, whatever the ligand.

Answer from the NTA final answer key. Solution by PrepWiser.