The shape and magnetic property of `[Cr(NH_3)_6]^(3+)` ion is :

Options

A. octahedral, paramagnetic
B. octahedral, diamagnetic
C. square planar, paramagnetic
D. square planar, diamagnetic

Explanation

To determine the shape and magnetic property of the complex ion \([Cr(NH_3)_6]^{3+}\), we can follow these steps:

### Step 1: Determine the oxidation state of chromium
The complex is \([Cr(NH_3)_6]^{3+}\). Since ammonia (NH₃) is a neutral ligand, it does not contribute to the charge. Therefore, the oxidation state of chromium (Cr) can be calculated as follows:

\[ \text{Oxidation state of Cr} + 0 = +3 \]

Thus, the oxidation state of chromium in this complex is +3.

### Step 2: Write the electronic configuration of chromium
Chromium has an atomic number of 24. The ground state electronic configuration of chromium is:

\[ \text{Cr: } [Ar] \, 4s^1 \, 3d^5 \]

When chromium is in the +3 oxidation state, it loses three electrons. The electrons are removed first from the 4s orbital and then from the 3d orbital. Therefore, the electronic configuration for \([Cr]^{3+}\) is:

\[ \text{Cr}^{3+}: [Ar] \, 3d^3 \]

### Step 3: Determine the hybridization
In the complex \([Cr(NH_3)_6]^{3+}\), there are six ammonia ligands. Ammonia is a strong field ligand, which means it will cause pairing of electrons in the d-orbitals.

The 3d orbitals will be filled as follows:

- The 3d orbitals will hold 3 electrons (3d³).
- Since we have 6 ligands, we will use the 3d, 4s, and 4p orbitals to accommodate the six pairs of electrons.

The hybridization can be determined as follows:

\[ \text{Hybridization: } d^2sp^3 \]

### Step 4: Determine the geometry
The hybridization \(d^2sp^3\) corresponds to an octahedral geometry. In an octahedral arrangement, the six ligands are symmetrically arranged around the central metal ion.

### Step 5: Determine the magnetic property
To determine the magnetic property, we check the number of unpaired electrons. In the case of \([Cr(NH_3)_6]^{3+}\):

- The 3d configuration has 3 electrons, which means there are 3 unpaired electrons.

Since there are unpaired electrons, the complex is **paramagnetic**.

### Final Answer
The shape of \([Cr(NH_3)_6]^{3+}\) is **octahedral** and it is **paramagnetic**.

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