Which of the following complex posses $sp^{3}d^{2}$ hybridization ?

Question

Accepted Answer

To determine which complex possesses $sp^{3}d^{2}$ hybridization, we need to analyze the electronic configuration and coordination environment of each complex.

1. **Option 1: $[CoF_{6}]^{3-}$**

- **Central Metal Ion:** Cobalt (Co) in this complex is in the +3 oxidation state, so it is $Co^{3+}$.
   - **Electronic Configuration of $Co^{3+}$:** The electronic configuration of Co is [Ar] 3d$^7$ 4s$^2$. For $Co^{3+}$, it becomes [Ar] 3d$^6$.
   - **Ligand Field:** Fluoride (F$^-$) is a weak field ligand, which does not cause pairing of electrons in the 3d orbitals.
   - **Hybridization:** In the presence of weak field ligands like F$^-$, the 3d electrons remain unpaired, and the complex uses outer orbitals for hybridization. Thus, the hybridization involves 4s, 4p, and 4d orbitals, resulting in $sp^{3}d^{2}$ hybridization.
   - **Geometry:** This hybridization leads to an octahedral geometry.

Therefore, $[CoF_{6}]^{3-}$ has $sp^{3}d^{2}$ hybridization, supporting the correct answer as Option 1.

2. **Option 2: $[Co(NH_{3})_{6}]^{3+}$**

- **Central Metal Ion:** Cobalt (Co) in the +3 oxidation state, $Co^{3+}$.
   - **Electronic Configuration of $Co^{3+}$:** [Ar] 3d$^6$.
   - **Ligand Field:** Ammonia (NH$_3$) is a strong field ligand, which causes pairing of electrons in the 3d orbitals.
   - **Hybridization:** Due to the strong field ligand, the electrons pair up, and the complex uses inner d orbitals for hybridization, resulting in $d^2sp^3$ hybridization.
   - **Geometry:** This leads to an octahedral geometry.

Therefore, $[Co(NH_{3})_{6}]^{3+}$ does not have $sp^{3}d^{2}$ hybridization.

3. **Option 3: $[Co(C_{2}O_{4})_{3}]^{3-}$**

- **Central Metal Ion:** Cobalt (Co) in the +3 oxidation state, $Co^{3+}$.
   - **Electronic Configuration of $Co^{3+}$:** [Ar] 3d$^6$.
   - **Ligand Field:** Oxalate (C$_2$O$_4^{2-}$) is a bidentate ligand and acts as a moderate field ligand.
   - **Hybridization:** The presence of moderate field ligands can lead to pairing of electrons, resulting in $d^2sp^3$ hybridization.
   - **Geometry:** This also leads to an octahedral geometry.

Therefore, $[Co(C_{2}O_{4})_{3}]^{3-}$ does not have $sp^{3}d^{2}$ hybridization.

4. **Option 4: $[Mn(CN)_{6}]^{3-}$**

- **Central Metal Ion:** Manganese (Mn) in the +3 oxidation state, $Mn^{3+}$.
   - **Electronic Configuration of $Mn^{3+}$:** [Ar] 3d$^4$.
   - **Ligand Field:** Cyanide (CN$^-$) is a strong field ligand, which causes pairing of electrons in the 3d orbitals.
   - **Hybridization:** Due to the strong field ligand, the electrons pair up, and the complex uses inner d orbitals for hybridization, resulting in $d^2sp^3$ hybridization.
   - **Geometry:** This leads to an octahedral geometry.

Therefore, $[Mn(CN)_{6}]^{3-}$ does not have $sp^{3}d^{2}$ hybridization.

In conclusion, the only complex that exhibits $sp^{3}d^{2}$ hybridization is $[CoF_{6}]^{3-}$, making Option 1 the correct answer.