Which of the following facts about the complex `[Cr(NH_(3))_(6)]Cl_(3)` is wrong ?

Options

A. The complex involves `d^(2) sp^(3)` hybridisation and is octahedral in shape .
B. The complex is paramagnetic .
C. The complex is an outer orbital complex .
D. The complex gives white precipitate with silver nitrate solution .

Explanation

To determine which of the provided facts about the complex \([Cr(NH_3)_6]Cl_3\) is incorrect, let's analyze the options step by step.

### Step 1: Identify the oxidation state of chromium in the complex
The complex \([Cr(NH_3)_6]Cl_3\) contains six ammonia (NH₃) ligands and three chloride (Cl⁻) ions. Since NH₃ is a neutral ligand, it does not contribute to the charge. The three chloride ions contribute a total charge of -3. Therefore, to balance this, chromium must have an oxidation state of +3.

**Hint:** Remember that the charge of the complex must equal the sum of the charges of the ligands and the metal ion.

### Step 2: Determine the electronic configuration of Cr³⁺
The ground state electronic configuration of chromium (Cr) is \([Ar] 3d^5 4s^1\). When chromium loses three electrons to form Cr³⁺, it loses the 4s electron and two of the 3d electrons. Thus, the electronic configuration of Cr³⁺ is \([Ar] 3d^3\).

**Hint:** Keep in mind that transition metals lose their outermost electrons first, which are usually from the 4s subshell before the 3d subshell.

### Step 3: Analyze the hybridization of the complex
In the case of \([Cr(NH_3)_6]^{3+}\), the chromium ion is in the +3 oxidation state and has a configuration of 3d³. The ammonia ligands are strong field ligands and will cause pairing of the electrons. The hybridization involves the d-orbitals, specifically the inner d-orbitals (3d), along with the s and p orbitals from the same principal energy level (n=3). Therefore, the hybridization is \(d^2sp^3\), leading to an octahedral geometry.

**Hint:** Remember that the type of hybridization can indicate whether the complex is an inner or outer orbital complex.

### Step 4: Determine the magnetic properties
Since Cr³⁺ has three unpaired electrons in the 3d orbitals, the complex is paramagnetic. Paramagnetic complexes have unpaired electrons, which contribute to their magnetic properties.

**Hint:** Check the number of unpaired electrons to determine if the complex is paramagnetic or diamagnetic.

### Step 5: Evaluate the options
1. **The complex involves \(d^2sp^3\) hybridization and is octahedral in shape.** - This statement is correct.
2. **The complex is paramagnetic.** - This statement is also correct.
3. **The complex is an outer orbital complex.** - This statement is incorrect; it is an inner orbital complex.
4. **The complex gives a white precipitate with silver nitrate solution.** - This statement is correct as chloride ions will precipitate with silver ions.

Based on the analysis, the incorrect statement is that the complex is an outer orbital complex.

### Final Answer
The wrong fact about the complex \([Cr(NH_3)_6]Cl_3\) is that **the complex is an outer orbital complex**.