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 need to analyze the three compounds given, which are based on a central atom A from the 15th group of the periodic table, and determine the number of lone pairs of electrons around the central atom in each compound.

Let's break down the solution step-by-step:

### Step 1: Identify the central atom A and surrounding atom X
The central atom A is a 15th group element. The elements in this group include nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), and bismuth (Bi). Given the maximum covalency is 3n, and n is 1, we find that the maximum covalency is 3. The only 15th group element that can exhibit a maximum covalency of 3 is nitrogen (N).

The surrounding atom X has a total number of protons equal to n, which is given as 1. Therefore, X must be hydrogen (H).

### Step 2: Write the formulas for the compounds
Using the information from Step 1, we can write the three compounds:
1. Compound (i): AX₂N⁻ → NH₂⁻ (Amide ion)
2. Compound (ii): AX₃N → NH₃ (Ammonia)
3. Compound (iii): AX₄N⁺ → NH₄⁺ (Ammonium ion)

### Step 3: Determine the number of lone pairs for each compound
Next, we will determine the number of lone pairs on the central atom (N) in each compound.

**For NH₂⁻ (Compound i)**:
- Nitrogen has 5 valence electrons.
- It forms 2 bonds with hydrogen (2 electrons used).
- There is an additional electron due to the negative charge.
- Lone pairs = 5 (valence electrons) - 2 (bonding electrons) - 1 (extra electron) = 2 lone pairs.
- Thus, **x = 1**.

**For NH₃ (Compound ii)**:
- Nitrogen has 5 valence electrons.
- It forms 3 bonds with hydrogen (3 electrons used).
- Lone pairs = 5 (valence electrons) - 3 (bonding electrons) = 2 lone pairs.
- Thus, **y = 1**.

**For NH₄⁺ (Compound iii)**:
- Nitrogen has 5 valence electrons.
- It forms 4 bonds with hydrogen (4 electrons used).
- There is a positive charge, which means it has lost one electron.
- Lone pairs = 5 (valence electrons) - 4 (bonding electrons) - 1 (lost electron) = 0 lone pairs.
- Thus, **z = 0**.

### Step 4: Calculate x³ + y² + z²
Now we can substitute the values of x, y, and z into the expression:
- x = 1, y = 1, z = 0
- Calculate:
\[ x^3 + y^2 + z^2 = 1^3 + 1^2 + 0^2 = 1 + 1 + 0 = 2 \]

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