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⁻

 Question:

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⁻

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Short Solution (Text):

Key idea (VBT quick check)

• Octahedral complexes can be:

  • d²sp³ (inner-orbital) when strong-field ligands pair electrons in (n−1)d (low spin).

  • sp³d² (outer-orbital) when weak-field ligands don’t pair (high spin), so central atom uses the outer nd orbitals.

• SF₆ (molecular, not a complex): classic sp³d² around S.

Evaluate each species

1. [Co(NH₃)₆]³⁺
   Co³⁺: d⁶; NH₃ ~ intermediate/weak-to-borderline but for Co³⁺ it gives low spin (pairing favored) → inner (d²sp³). ❌

2. SF₆
   S uses 3s, three 3p, two 3d → sp³d² (octahedral). ✅

3. [CrF₆]³⁻
   Cr³⁺: d³; F⁻ weak field, but d³ doesn’t need pairing; inner (n−1)d are available → d²sp³ (inner). ❌

4. [CoF₆]³⁻
   Co³⁺: d⁶; F⁻ weak field → high spin, no pairing in (n−1)d → uses outer nd → sp³d². ✅

5. [Mn(CN)₆]³⁻
   Mn³⁺: d⁴; CN⁻ strong field → low spin (pairing) → d²sp³ (inner). ❌

6. [MnCl₆]³⁻
   Mn³⁺: d⁴; Cl⁻ weak field → high spin → sp³d² (outer). ✅

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✅ Final Answer:
Number of species with sp³d² hybridization = 3
(They are SF₆, [CoF₆]³⁻, [MnCl₆]³⁻)

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Conclusion – Video Solution:

For octahedral species, weak-field ligands → high spin → sp³d² (outer orbital), while strong-field ligands → low spin → d²sp³ (inner orbital). Applying this plus the known sp³d² geometry of SF₆ gives a total of 3 species with sp³d² hybridization.