If following molecules undergo dimerisation then find the value of `(YZ)/(X):`
(i) `CiO_(3)`
(ii) `OF`
(iii) `GaH_(3)`
(iv) `AlCl_(3)`
(v) `IC l_(3)`
(vi) `BeH_(2)`
(vi) `NO_(2)`

Where X=Number of molecules which are hypevalennt in dimeric form.
Y=Number of molecules which complete octet in dimer form
Z=Number of molecules which are hypovalent in dimeric form.

Explanation

To solve the problem, we need to analyze each of the given molecules and determine the values of X, Y, and Z based on their dimerization behavior.

### Step-by-Step Solution:

1. **Identify Hypervalent, Hypovalent, and Complete Octet Molecules:**
- **Hypervalent:** Molecules with more than 8 electrons in the valence shell.
- **Hypovalent:** Molecules with fewer than 8 electrons in the valence shell.
- **Complete Octet:** Molecules that have exactly 8 electrons in the valence shell.

2. **Analyze Each Molecule:**
- **(i) ClO₃:**
- Dimerizes to form Cl₂O₆.
- Each Cl atom has more than 8 electrons (hypervalent).
- **Count:** Hypervalent = 2, Complete Octet = 0, Hypovalent = 0.

- **(ii) OF:**
- Dimerizes to form O₂F₂.
- Each O atom has 8 electrons and each F has 8 electrons (complete octet).
- **Count:** Hypervalent = 0, Complete Octet = 2, Hypovalent = 0.

- **(iii) GaH₃:**
- Dimerizes to form Ga₂H₆.
- Each Ga atom has less than 8 electrons (hypovalent).
- **Count:** Hypervalent = 0, Complete Octet = 0, Hypovalent = 2.

- **(iv) AlCl₃:**
- Dimerizes to form Al₂Cl₆.
- Each Al atom has 8 electrons (complete octet).
- **Count:** Hypervalent = 0, Complete Octet = 2, Hypovalent = 0.

- **(v) ICl₃:**
- Dimerizes to form I₂Cl₆.
- Each I atom has more than 8 electrons (hypervalent).
- **Count:** Hypervalent = 2, Complete Octet = 0, Hypovalent = 0.

- **(vi) BeH₂:**
- Dimerizes to form Be₂H₄.
- Each Be atom has less than 8 electrons (hypovalent).
- **Count:** Hypervalent = 0, Complete Octet = 0, Hypovalent = 2.

- **(vii) NO₂:**
- Dimerizes to form N₂O₄.
- Each N atom has 8 electrons (complete octet).
- **Count:** Hypervalent = 0, Complete Octet = 2, Hypovalent = 0.

3. **Summarize the Counts:**
- **Hypervalent (X):** ClO₃ (2) + ICl₃ (2) = 4
- **Complete Octet (Y):** OF (2) + AlCl₃ (2) + NO₂ (2) = 6
- **Hypovalent (Z):** GaH₃ (2) + BeH₂ (2) = 4

4. **Calculate the Values of X, Y, and Z:**
- X = 4 (hypervalent)
- Y = 6 (complete octet)
- Z = 4 (hypovalent)

5. **Calculate the Value of (YZ)/(X):**
\[ \frac{YZ}{X} = \frac{(6)(4)}{4} = \frac{24}{4} = 6 \]

### Final Answer:
The value of \((YZ)/(X)\) is **6**.