Question

The number of species from the following that are involved in $ sp^3d^2 $ hybridization is $ \text{[Co(NH}_3\text{)}_6\text{]}^{3+}, \text{SF}_6, \text{[CrF}_6\text{]}^{3-}, \text{[CoF}_6\text{]}^{3-}, \text{[Mn(CN)}_6\text{]}^{3-} $ and $ \text{[MnCl}_6\text{]}^{3-} $

• A. 3
• B. 4
• C. 6
• D. 5

Hint:

- Strong field ligands (\( \text{NH}_3 \), \( \text{CN}^- \)) cause inner orbital hybridization (\( d^2sp^3 \)) - Weak field ligands (\( \text{F}^- \), \( \text{Cl}^- \)) promote outer orbital hybridization (\( sp^3d^2 \)) - Non-metal compounds like \( \text{SF}_6 \) always use \( sp^3d^2 \)

Answer 1

The Correct Option is B

To determine the number of species among the given ones that involve \(sp^3d^2\) hybridization, we must analyze the electronic configuration and coordination of each complex. The \(sp^3d^2\) hybridization involves an octahedral geometry because it corresponds to six hybrid orbitals.

1. \([\text{Co}(\text{NH}_3)_6]^{3+}\): Cobalt in this complex is in the +3 oxidation state, resulting in a d⁶ electronic configuration. This complex is low-spin due to the strong field ligand \(\text{NH}_3\). This results in \(d^2sp^3\) hybridization, leading to an octahedral arrangement.
2. \(\text{SF}_6\): This molecule has a central sulfur atom with six identical fluorine atoms. It is a classic example of \(sp^3d^2\) hybridization, resulting in an octahedral shape.
3. \([\text{CrF}_6]^{3-}\): Chromium here is in the +3 oxidation state, equivalent to d³. With fluorine as a ligand, which is a weak field ligand, the complex involves \(sp^3d^2\) hybridization, resulting in an octahedral geometry.
4. \([\text{CoF}_6]^{3-}\): For Co³⁺ with fluorine ligands (weak field), the hybridization is \(sp^3d^2\). The resulting structure is octahedral as well.
5. \([\text{Mn(CN)}_6]^{3-}\): Manganese in +3 oxidation, together with CN^- ligands, leads to a low-spin configuration. The complex adopts \(d^2sp^3\) hybridization due to the strong field ligand CN^-, forming an octahedral geometry.
6. \([\text{MnCl}_6]^{3-}\): Manganese is in the +3 oxidation state and Cl^- being a weak field ligand results in \(sp^3d^2\) hybridization with an octahedral arrangement.

Thus, the species that exhibit \(sp^3d^2\) hybridization with an octahedral geometry are:

• \(\text{SF}_6\)
• [\text{CoF}_6]^{3-}
• \([\text{MnCl}_6]^{3-}\)

Therefore, the number of species involved in \(sp^3d^2\) hybridization is 4.

Answer 2

To determine the number of species involved in \(sp^3d^2\) hybridization from the given list, we need to analyze the electronic configurations and oxidation states of the central atoms for each compound. \(sp^3d^2\) hybridization occurs in octahedral complexes, typically involving d-orbitals from inner shells (inner d-orbitals).

1. \(\text{[Co(NH}_3\text{)}_6\text{]}^{3+}\):
   • \(\text{Co}^{3+}\) has the electronic configuration: \([Ar] \, 3d^6\).
   • Ammonia \((NH_3)\) is a strong field ligand, leading to pairing of electrons. This results in \(d^2sp^3\) hybridization, involving inner orbitals.
2. \(\text{SF}_6\):
   • Sulfur has an electronic configuration of \([Ne] \, 3s^2 \, 3p^4\).
   • For \(\text{SF}_6\), sulfur utilizes one 3s, three 3p, and two 3d orbitals for hybridization, which results in \(sp^3d^2\) hybridization (typical for \(\text{SF}_6\)).
3. \(\text{[CrF}_6\text{]}^{3-}\):
   • \(\text{Cr}^{3+}\) has the electronic configuration: \([Ar] \, 3d^3\).
   • Fluoride is a weak field ligand, resulting in \(sp^3d^2\) hybridization (utilizing outer orbitals).
4. \(\text{[CoF}_6\text{]}^{3-}\):
   • \(\text{Co}^{3+}\) has the electronic configuration: \([Ar] \, 3d^6\).
   • Fluoride is a weak field ligand, leading to electron promotion and \(sp^3d^2\) hybridization (outer orbitals used).
5. \(\text{[Mn(CN)}_6\text{]}^{3-}\):
   • \(\text{Mn}^{3+}\) has the electronic configuration: \([Ar] \, 3d^4\).
   • Cyanide \((CN^-)\) is a strong field ligand, which results in \(d^2sp^3\) hybridization (inner d-orbitals used).
6. \(\text{[MnCl}_6\text{]}^{3-}\):
   • \(\text{Mn}^{3+}\) has the electronic configuration: \([Ar] \, 3d^4\).
   • Chloride is a weak field ligand, resulting in promotion of electrons and \(sp^3d^2\) hybridization.

Based on the above analysis, the species that involve \(sp^3d^2\) hybridization are: \(\text{SF}_6, \text{[CrF}_6\text{]}^{3-}, \text{[CoF}_6\text{]}^{3-}, \text{[MnCl}_6\text{]}^{3-}\). Therefore, the number of species showing \(sp^3d^2\) hybridization is 4.