The number of unpaired electrons calculated in `[Co(NH_(3))_(6)]^(3+)` and `|CoF_(6)|^(3-)` are

Options

A. 4 and 4
B. 0 and 2
C. 2 and 4
D. 0 and 4

Answer

To determine the number of unpaired electrons in the complexes \([Co(NH_3)_6]^{3+}\) and \([CoF_6]^{3-}\), we will follow these steps:

### Step 1: Determine the oxidation state of cobalt in both complexes.
- For \([Co(NH_3)_6]^{3+}\):
- Ammonia (NH3) is a neutral ligand, so its contribution to the oxidation state is 0.
- Therefore, the oxidation state of cobalt (Co) is +3.

- For \([CoF_6]^{3-}\):
- Each fluorine (F) has an oxidation state of -1.
- Let the oxidation state of cobalt be \(x\). The equation is:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]
- Thus, cobalt is also +3 in this complex.

### Step 2: Write the electron configuration of cobalt in the +3 oxidation state.
- The atomic number of cobalt (Co) is 27.
- The electron configuration of neutral cobalt is:
\[ [Ar] 3d^7 4s^2 \]
- For \(Co^{3+}\), we remove 3 electrons (2 from 4s and 1 from 3d):
\[ Co^{3+}: [Ar] 3d^6 4s^0 \]

### Step 3: Analyze the ligand field strength and determine the splitting of d-orbitals.
- In \([Co(NH_3)_6]^{3+}\):
- Ammonia (NH3) is a strong field ligand, which causes pairing of electrons.
- The 3d orbitals split into \(t_{2g}\) and \(e_g\) levels.
- The 6 electrons will fill the \(t_{2g}\) orbitals completely:
- \(t_{2g}: \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow\) (6 electrons)
- \(e_g: \) empty
- Therefore, there are **0 unpaired electrons**.

- In \([CoF_6]^{3-}\):
- Fluorine (F) is a weak field ligand, which does not cause pairing of electrons.
- The 3d orbitals will also split into \(t_{2g}\) and \(e_g\) levels.
- The 6 electrons will fill the orbitals as follows:
- \(t_{2g}: \uparrow \uparrow \uparrow \uparrow \) (4 electrons)
- \(e_g: \uparrow \uparrow \) (2 electrons)
- Therefore, there are **4 unpaired electrons**.

### Final Answer:
- The number of unpaired electrons in \([Co(NH_3)_6]^{3+}\) is **0**.
- The number of unpaired electrons in \([CoF_6]^{3-}\) is **4**.

### Summary:
- \([Co(NH_3)_6]^{3+}\): 0 unpaired electrons
- \([CoF_6]^{3-}\): 4 unpaired electrons