An octahedral complex of `Co^(3+)` is diamagnetic . The hybridisation involved in the formation of the complex is :

Options

A. `sp^(3)d^(2)`
B. `dsp^(2)`
C. `d^(2)sp^(3)`
D. `sp^(3)d`

Answer

To determine the hybridization involved in the formation of an octahedral complex of `Co^(3+)` that is diamagnetic, we can follow these steps:

### Step 1: Determine the oxidation state and electronic configuration of Co
- Cobalt (Co) has an atomic number of 27. Its ground state electronic configuration is `[Ar] 3d^7 4s^2`.
- In the `Co^(3+)` oxidation state, cobalt loses three electrons. The two electrons from the 4s orbital and one electron from the 3d orbital are removed.
- Thus, the electronic configuration for `Co^(3+)` is `3d^6`.

**Hint:** Remember to account for the loss of electrons when determining the oxidation state.

### Step 2: Analyze the electron configuration for pairing
- The `3d^6` configuration means there are six electrons in the 3d subshell.
- In an octahedral field, the 3d orbitals split into two sets: the lower-energy t2g (three orbitals) and the higher-energy eg (two orbitals).
- To achieve a diamagnetic state, all electrons must be paired.

**Hint:** Consider how the splitting of d-orbitals in an octahedral field affects electron pairing.

### Step 3: Pair the electrons
- In the case of `Co^(3+)`, the six electrons will fill the t2g orbitals first, and then pair up in these orbitals.
- The pairing will occur as follows:
- The first three electrons will occupy the three t2g orbitals (one in each).
- The next three electrons will pair up in the t2g orbitals, resulting in a configuration of (↑↓)(↑↓)(↑↓) in the t2g orbitals.

**Hint:** Remember that pairing of electrons in lower energy orbitals is energetically favorable.

### Step 4: Identify the hybridization
- In an octahedral complex, the hybridization involves the mixing of the d, s, and p orbitals.
- For `Co^(3+)`, after pairing, we have:
- 2 electrons in the t2g orbitals (paired)
- 0 electrons in the eg orbitals
- The 4s and 4p orbitals are also involved in hybridization.
- The hybridization for an octahedral complex where 6 orbitals are involved (2 from d, 1 from s, and 3 from p) is `d^2sp^3`.

**Hint:** Recall that in octahedral complexes, the hybridization is typically `d^2sp^3`.

### Conclusion
The hybridization involved in the formation of the octahedral complex of `Co^(3+)` that is diamagnetic is **d²sp³**.

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