`[Cr(NH_3)_6]^(3+)` ion is:

Options

A. (a) diamagnetic
B. (b) square planar
C. (c) paramagnetic
D. (d) none of these

Explanation

To determine the nature of the `[Cr(NH_3)_6]^(3+)` ion, we will analyze the oxidation state of chromium, the electronic configuration, and the magnetic properties of the complex.

### Step-by-Step Solution:

1. **Identify the oxidation state of chromium:**
- The complex ion is `[Cr(NH_3)_6]^(3+)`.
- Ammonia (NH₃) is a neutral ligand, contributing 0 charge.
- Therefore, the oxidation state of chromium (Cr) can be calculated as follows:
\[ \text{Charge of the complex} = \text{Oxidation state of Cr} + \text{Charge from ligands} \]
\[ 3+ = x + 0 \implies x = +3 \]
- Thus, the oxidation state of chromium in this complex is +3.

2. **Determine the electronic configuration of Cr(III):**
- The atomic number of chromium (Cr) is 24. The ground state electronic configuration is:
\[ \text{Cr: } [Ar] 3d^5 4s^1 \]
- When chromium is in the +3 oxidation state, it loses three electrons (two from 4s and one from 3d):
\[ \text{Cr}^{3+}: [Ar] 3d^3 \]

3. **Analyze the hybridization:**
- The complex has six ammonia ligands, indicating an octahedral geometry.
- In an octahedral field, the 3d orbitals split into two sets: \(t_{2g}\) and \(e_g\).
- For \(3d^3\), the three electrons will occupy the lower energy \(t_{2g}\) orbitals according to Hund's rule, resulting in:
\[ t_{2g}^3 \quad (e_g^0) \]

4. **Determine the magnetic properties:**
- Since all three electrons in the \(t_{2g}\) orbitals are unpaired, the complex will exhibit paramagnetism.
- A complex is paramagnetic if it has unpaired electrons.

### Conclusion:
The `[Cr(NH_3)_6]^(3+)` ion is **paramagnetic** in nature.

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