Consider the following three compounds `(i)AX_(2n)^(n-)`, (ii)`AX_(3n)` and (ii)`AX_(4n)^(n+)`, where central atom A is 15th group element and their maximum covalency is 3n. If total number of proton in surrounding atom X is n and value of n is one, then calculate value of `x^(3)+y^(2)+z^(2)` . (where x, y and z are total number of lone pair at central atom in compounds (i), (ii) and (iii) respectively.

Explanation

To solve the problem, we will analyze the three compounds given and calculate the number of lone pairs of electrons on the central atom A for each compound. The compounds are:

1. \( AX_{2n}^{n-} \)
2. \( AX_{3n} \)
3. \( AX_{4n}^{n+} \)

Where A is a 15th group element (specifically nitrogen, as we will see), and \( n \) is given as 1.

### Step 1: Identify the central atom A and surrounding atom X
Since A is a 15th group element and we are considering the first member of this group, we take A as nitrogen (N). The surrounding atom X is hydrogen (H) since it has 1 proton (as \( n = 1 \)).

### Step 2: Write the compounds with \( n = 1 \)
1. For \( AX_{2n}^{n-} \):
- \( n = 1 \) gives us \( AX_{2} \) with a charge of -1.
- This becomes \( NH_2^{-} \) (ammonium ion).

2. For \( AX_{3n} \):
- \( n = 1 \) gives us \( AX_{3} \).
- This becomes \( NH_{3} \) (ammonia).

3. For \( AX_{4n}^{n+} \):
- \( n = 1 \) gives us \( AX_{4} \) with a charge of +1.
- This becomes \( NH_{4}^{+} \) (ammonium ion).

### Step 3: Determine the lone pairs of electrons for each compound
1. **For \( NH_2^{-} \)**:
- Nitrogen has 5 valence electrons. In \( NH_2^{-} \), it forms 2 bonds with hydrogen and has an extra electron due to the negative charge.
- Total electrons = 5 (from N) + 1 (extra) = 6.
- Number of lone pairs = \( \frac{6 - 2 \text{ (bonds)}}{2} = 2 \).
- So, \( x = 2 \).

2. **For \( NH_{3} \)**:
- Nitrogen has 5 valence electrons and forms 3 bonds with hydrogen.
- Total electrons = 5 (from N).
- Number of lone pairs = \( \frac{5 - 3}{2} = 1 \).
- So, \( y = 1 \).

3. **For \( NH_{4}^{+} \)**:
- Nitrogen has 5 valence electrons and forms 4 bonds with hydrogen.
- Total electrons = 5 (from N) - 1 (due to positive charge) = 4.
- Number of lone pairs = \( \frac{4 - 4}{2} = 0 \).
- So, \( z = 0 \).

### Step 4: Calculate \( x^3 + y^2 + z^2 \)
Now we can substitute the values of \( x \), \( y \), and \( z \) into the expression:

\[ x^3 + y^2 + z^2 = 2^3 + 1^2 + 0^2 = 8 + 1 + 0 = 9 \]

### Final Answer
The value of \( x^3 + y^2 + z^2 \) is **9**.