Which of the following complexes exhibits the highest paramagnetic behaviour?
where gly=glycine, en=ethylenediamine and bipy =bipyridyl
(At. no. `Ti=22, V=23, Fe=26, Co=27`)

Options

A. `[V(gly)_(2)(OH)_(2)(NH_(3))_(2)]^(2+)`
B. `[Fe(en)(py)(NH_(3))_(2)]^(2+)`
C. `[Co("ox")_(2)(OH)_(2)]^(-)`
D. `[Ti(NH_(3))_(6)]^(3+)`

Explanation

To determine which of the given complexes exhibits the highest paramagnetic behavior, we need to analyze each complex based on the oxidation state of the central metal ion and the number of unpaired electrons present in each case. The more unpaired electrons a complex has, the more paramagnetic it will be.

### Step-by-Step Solution:

1. **Identify the complexes and their central metal ions:**
- Complex 1: Vanadium complex with glycine (Gly)
- Complex 2: Iron complex with ethylenediamine (en) and bipyridyl (bipy)
- Complex 3: Cobalt complex with oxalate (Ox) and hydroxide (OH)
- Complex 4: Titanium complex with ammonia (NH3)

2. **Calculate the oxidation state of the central metal ion in each complex:**
- **Complex 1 (Vanadium):**
- Let the oxidation state of Vanadium be \( X \).
- The overall charge of the complex is +2.
- \( X + 2(0) + 2(-1) = +2 \) (Gly is neutral, OH is -1)
- Therefore, \( X - 2 = 2 \) ⇒ \( X = +4 \).

- **Complex 2 (Iron):**
- Let the oxidation state of Iron be \( X \).
- The overall charge is +2.
- \( X + 1(0) + 1(0) + 2(0) = +2 \) (en and bipy are neutral)
- Therefore, \( X = +2 \).

- **Complex 3 (Cobalt):**
- Let the oxidation state of Cobalt be \( X \).
- The overall charge is -1.
- \( X + 2(-2) + 2(-1) = -1 \) (Ox is -2, OH is -1)
- Therefore, \( X - 6 = -1 \) ⇒ \( X = +5 \).

- **Complex 4 (Titanium):**
- Let the oxidation state of Titanium be \( X \).
- The overall charge is +3.
- \( X + 6(0) = +3 \) (NH3 is neutral)
- Therefore, \( X = +3 \).

3. **Determine the electronic configuration and number of unpaired electrons:**
- **Complex 1 (Vanadium +4):**
- Atomic number of Vanadium = 23, configuration: \( [Ar] 4s^2 3d^3 \)
- For Vanadium +4: \( 4s^0 3d^3 \) ⇒ 3 unpaired electrons.

- **Complex 2 (Iron +2):**
- Atomic number of Iron = 26, configuration: \( [Ar] 4s^2 3d^6 \)
- For Iron +2: \( 4s^0 3d^6 \)
- Strong field ligands (en, bipy) cause pairing ⇒ 0 unpaired electrons (diamagnetic).

- **Complex 3 (Cobalt +5):**
- Atomic number of Cobalt = 27, configuration: \( [Ar] 4s^2 3d^7 \)
- For Cobalt +5: \( 4s^0 3d^6 \) ⇒ 0 unpaired electrons (diamagnetic).

- **Complex 4 (Titanium +3):**
- Atomic number of Titanium = 22, configuration: \( [Ar] 4s^2 3d^2 \)
- For Titanium +3: \( 4s^0 3d^2 \) ⇒ 2 unpaired electrons.

4. **Compare the number of unpaired electrons:**
- Complex 1 (Vanadium +4): 3 unpaired electrons (paramagnetic)
- Complex 2 (Iron +2): 0 unpaired electrons (diamagnetic)
- Complex 3 (Cobalt +5): 0 unpaired electrons (diamagnetic)
- Complex 4 (Titanium +3): 2 unpaired electrons (paramagnetic)

5. **Conclusion:**
- The complex with the highest paramagnetic behavior is the Vanadium complex, as it has the most unpaired electrons (3).

### Final Answer:
The complex that exhibits the highest paramagnetic behavior is the Vanadium complex.