The difference in the number of unpaired electrons in `Co^(2+)` ion in its high - spin and low - spin octahedral complexes is -----

Answer

To solve the question regarding the difference in the number of unpaired electrons in the `Co^(2+)` ion in its high-spin and low-spin octahedral complexes, we can follow these steps:

### Step 1: Determine the Atomic Number and Electronic Configuration of Cobalt
- The atomic number of cobalt (Co) is 27.
- The electronic configuration of neutral cobalt is:
\[ \text{Co: } [\text{Ar}] 3d^7 4s^2 \]

### Step 2: Determine the Electronic Configuration of `Co^(2+)`
- When cobalt loses 2 electrons to form `Co^(2+)`, the electronic configuration becomes:
\[ \text{Co}^{2+}: [\text{Ar}] 3d^7 4s^0 \]
- This means we have 7 electrons in the 3d subshell.

### Step 3: Understand the Concept of High-Spin and Low-Spin Complexes
- In octahedral complexes, the d-orbitals split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy).
- High-spin complexes occur with weak field ligands, where the splitting energy (\(\Delta\)) is less than the pairing energy, leading to maximum unpaired electrons.
- Low-spin complexes occur with strong field ligands, where \(\Delta\) is greater than the pairing energy, leading to pairing of electrons.

### Step 4: Fill the d-Orbitals for High-Spin Complex
- For the high-spin case (weak field ligands), the 7 electrons will fill the orbitals as follows:
- The \(t_{2g}\) orbitals will be filled first, followed by the \(e_g\) orbitals.
- The filling will look like this:
- \(t_{2g}\): ↑ ↑ ↑ (3 electrons)
- \(e_g\): ↑ ↑ (2 electrons)
- Total unpaired electrons in high-spin: **3 unpaired electrons**.

### Step 5: Fill the d-Orbitals for Low-Spin Complex
- For the low-spin case (strong field ligands), the 7 electrons will fill the orbitals as follows:
- The filling will look like this:
- \(t_{2g}\): ↑↑ ↑↑ ↑ (6 electrons, 3 pairs)
- \(e_g\): ↑ (1 electron)
- Total unpaired electrons in low-spin: **1 unpaired electron**.

### Step 6: Calculate the Difference in Unpaired Electrons
- The difference in the number of unpaired electrons between high-spin and low-spin is:
\[ \text{Difference} = \text{Unpaired (high-spin)} - \text{Unpaired (low-spin)} = 3 - 1 = 2 \]

### Final Answer
The difference in the number of unpaired electrons in `Co^(2+)` ion in its high-spin and low-spin octahedral complexes is **2**.

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