If the formula of basic Beryllium nitrate is `[Be_(n)O(NO_(3))_(6)]`. What is the value of 'n' here

Answer

To find the value of 'n' in the formula of basic Beryllium nitrate, \[ Be_nO(NO_3)_6 \], we can follow these steps:

### Step 1: Identify the oxidation states
- Beryllium (Be) has an oxidation state of +2.
- Oxygen (O) in the oxide form has an oxidation state of -2.
- The nitrate ion \((NO_3^-)\) has an oxidation state of -1.

### Step 2: Write the charge balance equation
The overall charge of the compound must equal zero. Therefore, we can set up the equation based on the oxidation states:
\[ \text{Total charge} = \text{Charge from Be} + \text{Charge from O} + \text{Charge from NO}_3 \]
This can be expressed mathematically as:
\[ 2n + (-2) + (-1 \times 6) = 0 \]

### Step 3: Simplify the equation
Now, simplify the equation:
\[ 2n - 2 - 6 = 0 \]
This simplifies to:
\[ 2n - 8 = 0 \]

### Step 4: Solve for 'n'
Now, we can solve for 'n':
\[ 2n = 8 \]
\[ n = \frac{8}{2} = 4 \]

### Conclusion
The value of 'n' is 4, so the formula of basic Beryllium nitrate is \[ Be_4O(NO_3)_6 \].

---