Bond energies of `(H - H), (O = O) and (O - H)` are 105, 120 and 220 kcal/mol respectively, then magnitude of `Delta H` in the reaction in kcal is :
`2H_(2)(g) + O_(2)(g) rarr 2H_(2)O(l)`

Explanation

To calculate the change in enthalpy (ΔH) for the reaction:

\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]

we will use the bond energies provided:

- Bond energy of \( H-H \) = 105 kcal/mol
- Bond energy of \( O=O \) = 120 kcal/mol
- Bond energy of \( O-H \) = 220 kcal/mol

### Step-by-Step Solution:

1. **Identify Bonds Broken in Reactants:**
- In the reactants \( 2H_2(g) + O_2(g) \):
- Each \( H_2 \) has 1 \( H-H \) bond, and there are 2 \( H_2 \) molecules:
\[ \text{Bonds broken: } 2 \times (1 \text{ bond}) = 2 \text{ } H-H \text{ bonds} \]
- The \( O_2 \) molecule has 1 \( O=O \) bond:
\[ \text{Bonds broken: } 1 \text{ } O=O \text{ bond} \]

2. **Identify Bonds Formed in Products:**
- In the products \( 2H_2O(l) \):
- Each \( H_2O \) has 2 \( O-H \) bonds, and there are 2 \( H_2O \) molecules:
\[ \text{Bonds formed: } 2 \times (2 \text{ bonds}) = 4 \text{ } O-H \text{ bonds} \]

3. **Calculate Total Bond Energies:**
- **Total energy of bonds broken:**
- For \( H-H \): \( 2 \times 105 \text{ kcal/mol} = 210 \text{ kcal} \)
- For \( O=O \): \( 1 \times 120 \text{ kcal/mol} = 120 \text{ kcal} \)
- Total:
\[ \text{Total energy of bonds broken} = 210 + 120 = 330 \text{ kcal} \]

- **Total energy of bonds formed:**
- For \( O-H \): \( 4 \times 220 \text{ kcal/mol} = 880 \text{ kcal} \)

4. **Calculate ΔH:**
- Using the formula:
\[ \Delta H = \text{(Energy of bonds broken)} - \text{(Energy of bonds formed)} \]
- Substitute the values:
\[ \Delta H = 330 \text{ kcal} - 880 \text{ kcal} = -550 \text{ kcal} \]

### Final Answer:
The magnitude of \( \Delta H \) for the reaction is:
\[ \Delta H = -550 \text{ kcal} \]