Match List I with List II and select the correct answer using the codes given below the lists `:`
`{:("List "I,,"List "II),(a.B Br_(3),,i. "Dimer"),(b.Tl_(2)O,,ii."Trigonol planar"),(c. B(OH)_(3),,iii."Basic"),(d.AlCl_(3),,iv. "Monobasic acid"):}`

Options

A. `(x)/(2)+y`
B. `(2x+y)/(2)`
C. `(2x-y)/(2)`
D. `(y-2x)/(2)`

Explanation

To solve the matching question between List I and List II, we will analyze each compound in List I and determine its corresponding property from List II.

### Step-by-Step Solution:

1. **Identify the first compound: BBr₃**
- Boron (B) has a valency of 3 and forms three bonds with bromine (Br), which is monovalent.
- The geometry of BBr₃ is trigonal planar due to the three bonding pairs and no lone pairs on the boron atom.
- **Match:** BBr₃ → ii. "Trigonol planar"

2. **Identify the second compound: Tl₂O**
- Thallium (Tl) in Tl₂O can exhibit a +1 oxidation state. The oxygen atom has lone pairs and can act as a Lewis base.
- The compound is basic in nature due to the presence of oxygen which can donate lone pairs.
- **Match:** Tl₂O → iii. "Basic"

3. **Identify the third compound: B(OH)₃**
- Boric acid (B(OH)₃) is known to be a monobasic acid. It can accept hydroxide ions (OH⁻) to form B(OH)₄⁻, indicating its monobasic nature.
- **Match:** B(OH)₃ → iv. "Monobasic acid"

4. **Identify the fourth compound: AlCl₃**
- Aluminum chloride (AlCl₃) dimerizes to form Al₂Cl₆. In the dimer, aluminum accepts electron pairs from chlorine, forming a three-center two-electron bond.
- **Match:** AlCl₃ → i. "Dimer"

### Final Matches:
- a → ii (BBr₃ → "Trigonol planar")
- b → iii (Tl₂O → "Basic")
- c → iv (B(OH)₃ → "Monobasic acid")
- d → i (AlCl₃ → "Dimer")

### Answer Codes:
The correct answer using the codes given below the lists is:
- a: ii
- b: iii
- c: iv
- d: i