Atomic numbers of `Cr` and `Fe` are respectively `24` and `26`. Which of the following is paramagnetic with the spin of the electron?

Options

A. `[Cr(CO)_(6)]`
B. `[Fe(CO)_(5)]`
C. `[Fe(CN)_(6)]^(4-)`
D. `[Cr(NH_(3))_(6)]^(3+)`

Explanation

To determine which of the given coordination compounds is paramagnetic and to find the spin of the electrons, we can follow these steps:

### Step 1: Identify the oxidation states of chromium in the complexes.
- For the complex \( \text{Cr(CO)}_6 \), since CO is a neutral ligand, the oxidation state of chromium (Cr) is 0.
- For the complex \( \text{Cr(NH}_3)_6^{3+} \), we can set up the equation:
\[ x + 0 = +3 \quad (\text{where } x \text{ is the oxidation state of Cr}) \]
Therefore, \( x = +3 \).

### Step 2: Write the electron configuration for chromium in different oxidation states.
- The atomic number of chromium (Cr) is 24. The ground state electron configuration of Cr is:
\[ \text{Cr: } [\text{Ar}] 4s^1 3d^5 \]
- For \( \text{Cr}^{0} \) (in \( \text{Cr(CO)}_6 \)):
- The configuration remains \( [\text{Ar}] 4s^1 3d^5 \).
- For \( \text{Cr}^{3+} \) (in \( \text{Cr(NH}_3)_6^{3+} \)):
- We remove 3 electrons (2 from 4s and 1 from 3d):
\[ \text{Cr}^{3+}: [\text{Ar}] 3d^3 \]

### Step 3: Determine the electron configuration in the presence of ligands.
- In the presence of strong field ligands like CO, the \( 3d \) electrons will pair up.
- For \( \text{Cr(CO)}_6 \):
- The \( 3d^5 \) configuration will lead to pairing of electrons, resulting in no unpaired electrons. Thus, it is **not paramagnetic**.

- For \( \text{Cr(NH}_3)_6^{3+} \):
- The \( 3d^3 \) configuration will have three unpaired electrons (since NH3 is a weak field ligand and does not cause pairing).
- Therefore, it is **paramagnetic**.

### Step 4: Determine the spin of the electrons.
- The spin of an electron is given by the formula \( \text{Spin} = \frac{n}{2} \), where \( n \) is the number of unpaired electrons.
- For \( \text{Cr(NH}_3)_6^{3+} \):
- There are 3 unpaired electrons, so:
\[ \text{Spin} = \frac{3}{2} = 1.5 \]

### Conclusion:
- The complex \( \text{Cr(NH}_3)_6^{3+} \) is paramagnetic with a spin of 1.5.