To determine which of the given complexes shows maximum paramagnetism, we need to analyze the number of unpaired electrons in each complex. The more unpaired electrons a complex has, the more paramagnetic it is.
### Step-by-Step Solution:
1. **Identify the complexes and their ligands**:
- We have four complexes to analyze. Each complex contains different ligands which can affect the oxidation state of the central metal atom.
2. **Calculate the oxidation state of the central metal in each complex**:
- **Complex 1**: Cobalt complex with oxalate, ammonia, and chloride.
- Oxalate (C2O4) has a charge of -2, ammonia (NH3) is neutral (0), and chloride (Cl) has a charge of -1.
- Let the oxidation state of cobalt be \( x \).
- The equation becomes: \( x - 2 - 1 = 0 \) ⇒ \( x = +4 \).
- **Complex 2**: Iron complex with ethylenediamine, pyridine, and ammonia.
- All ligands are neutral.
- Let the oxidation state of iron be \( y \).
- The equation becomes: \( y = +2 \).
- **Complex 3**: Platinum complex with glycine and triphenylphosphine.
- Glycine has a charge of -1 and triphenylphosphine is neutral.
- Let the oxidation state of platinum be \( z \).
- The equation becomes: \( z - 1 = 0 \) ⇒ \( z = 0 \).
- **Complex 4**: Titanium complex with water.
- Water is neutral.
- Let the oxidation state of titanium be \( w \).
- The equation becomes: \( w = +3 \).
3. **Determine the electronic configuration of the metal in each oxidation state**:
- **Cobalt in +4 state**:
- Cobalt (Co) has an atomic number of 27, so its electronic configuration is \( [Ar] 3d^7 4s^2 \).
- In +4 state, it loses 4 electrons (2 from 4s and 2 from 3d), resulting in \( 3d^5 \) configuration.
- **Unpaired electrons**: 5.
- **Iron in +2 state**:
- Iron (Fe) has an atomic number of 26, so its electronic configuration is \( [Ar] 3d^6 4s^2 \).
- In +2 state, it loses 2 electrons (both from 4s), resulting in \( 3d^6 \) configuration.
- **Unpaired electrons**: 4.
- **Platinum in 0 state**:
- Platinum (Pt) has an atomic number of 78, so its electronic configuration is \( [Xe] 4f^{14} 5d^9 6s^1 \).
- In 0 state, it retains all its electrons, resulting in \( 4d^8 5s^2 \) configuration.
- **Unpaired electrons**: 2.
- **Titanium in +3 state**:
- Titanium (Ti) has an atomic number of 22, so its electronic configuration is \( [Ar] 3d^2 4s^2 \).
- In +3 state, it loses 3 electrons (2 from 4s and 1 from 3d), resulting in \( 3d^1 \) configuration.
- **Unpaired electrons**: 1.
4. **Compare the number of unpaired electrons**:
- Complex 1 (Cobalt +4): 5 unpaired electrons.
- Complex 2 (Iron +2): 4 unpaired electrons.
- Complex 3 (Platinum 0): 2 unpaired electrons.
- Complex 4 (Titanium +3): 1 unpaired electron.
5. **Conclusion**:
- The complex with the maximum number of unpaired electrons is the cobalt complex with 5 unpaired electrons. Therefore, it shows the maximum paramagnetism.
### Final Answer:
The complex that shows maximum paramagnetism is the **Cobalt complex** (Option 1).
