Which of the following complex compounds will exhibit highest magnetic behaviour ?
`(At.No. Ti = 22, Cr = 24, Co = 27, Zn = 30)`

Options

A. `[Co(NH_(3))_(6)]^(3+)`
B. `[Zn(NH_(3))_(6)]^(2+)`
C. `[[Ti(NH_(3))_(6)]^(3+)`
D. `[cr(NH_(3))_(6)]^(3+)`

Answer

To determine which of the given complex compounds exhibits the highest magnetic behavior, we need to analyze the oxidation states and the number of unpaired electrons in each complex. The presence of unpaired electrons is what contributes to paramagnetism.

### Step-by-Step Solution:

1. **Identify the Complexes and Their Metal Ions:**
- The complexes are based on the following metals: Cobalt (Co), Zinc (Zn), Titanium (Ti), and Chromium (Cr).
- Their atomic numbers are: Ti = 22, Cr = 24, Co = 27, Zn = 30.

2. **Calculate the Oxidation States:**
- For each complex, we assume the metal ion's oxidation state based on the overall charge of the complex and the fact that the amine ligands are neutral.
- **Cobalt Complex:** Assume oxidation state of Co = x. If the overall charge is +3, then x + 0 = +3, so Co is +3.
- **Zinc Complex:** Zn is typically +2 in complexes. Therefore, Zn = +2.
- **Titanium Complex:** Assume oxidation state of Ti = x. If the overall charge is +3, then x + 0 = +3, so Ti is +3.
- **Chromium Complex:** Assume oxidation state of Cr = x. If the overall charge is +3, then x + 0 = +3, so Cr is +3.

3. **Determine Electron Configurations:**
- **Cobalt (Co +3):** Atomic number 27. Configuration: [Ar] 4s0 3d6. After losing 3 electrons, we have 3d6.
- **Zinc (Zn +2):** Atomic number 30. Configuration: [Ar] 4s0 3d10. After losing 2 electrons, we have 3d10 (fully filled).
- **Titanium (Ti +3):** Atomic number 22. Configuration: [Ar] 4s0 3d2. After losing 3 electrons, we have 3d1.
- **Chromium (Cr +3):** Atomic number 24. Configuration: [Ar] 4s0 3d5. After losing 3 electrons, we have 3d3.

4. **Count the Number of Unpaired Electrons:**
- **Cobalt (3d6):** In a strong field, all six electrons will pair up, resulting in **0 unpaired electrons**.
- **Zinc (3d10):** Fully filled d-orbitals mean **0 unpaired electrons**.
- **Titanium (3d1):** There is **1 unpaired electron**.
- **Chromium (3d3):** There are **3 unpaired electrons**.

5. **Calculate the Bohr Magneton Values:**
- The formula for Bohr magneton (μ) is given by:
\[ \mu = n(n + 2)^{1/2} \]
where n is the number of unpaired electrons.
- **Cobalt:** μ = 0 (0 unpaired electrons).
- **Zinc:** μ = 0 (0 unpaired electrons).
- **Titanium:** μ = 1(1 + 2)^{1/2} = √3.
- **Chromium:** μ = 3(3 + 2)^{1/2} = √15.

6. **Conclusion:**
- Comparing the values, Chromium (Cr) with 3 unpaired electrons has the highest magnetic behavior, as it has the highest Bohr magneton value.

### Final Answer:
**The complex compound that will exhibit the highest magnetic behavior is the Chromium complex.**