Consider the following species `NO_(3)^(-), SO_(4)^(2-), ClO_(3)^(-),SO_(3), PO_(4)^(3-), XeO_(3),CO_(3)^(2-), SO_(3)^(2-)`
Then calculate value of |x-y|, where
x : Total number of species which have bond order 1.5 or greater than 1.5
y : Total number of species which have bond order less than 1.5

Explanation

To solve the problem, we need to determine the bond order of each species listed and categorize them into two groups: those with a bond order of 1.5 or greater (x) and those with a bond order less than 1.5 (y). Finally, we will calculate |x - y|.

### Step-by-Step Solution:

1. **Identify the species and calculate their total valence electrons:**
- **NO₃⁻**: N (5) + 3O (6) + 1 (for the negative charge) = 24 electrons
- **SO₄²⁻**: S (6) + 4O (6) + 2 (for the negative charge) = 32 electrons
- **ClO₃⁻**: Cl (7) + 3O (6) + 1 (for the negative charge) = 26 electrons
- **SO₃**: S (6) + 3O (6) = 24 electrons
- **PO₄³⁻**: P (5) + 4O (6) + 3 (for the negative charge) = 32 electrons
- **XeO₃**: Xe (8) + 3O (6) = 26 electrons
- **CO₃²⁻**: C (4) + 3O (6) + 2 (for the negative charge) = 24 electrons
- **SO₃²⁻**: S (6) + 3O (6) + 2 (for the negative charge) = 26 electrons

2. **Draw the Lewis structures and calculate bond orders:**
- **NO₃⁻**: Bond order = 4 bonds / 3 oxygen = 1.33 (y)
- **SO₄²⁻**: Bond order = 6 bonds / 4 oxygen = 1.5 (x)
- **ClO₃⁻**: Bond order = 5 bonds / 3 oxygen = 1.67 (x)
- **SO₃**: Bond order = 6 bonds / 3 oxygen = 2 (x)
- **PO₄³⁻**: Bond order = 5 bonds / 4 oxygen = 1.25 (y)
- **XeO₃**: Bond order = 6 bonds / 3 oxygen = 2 (x)
- **CO₃²⁻**: Bond order = 4 bonds / 3 oxygen = 1.33 (y)
- **SO₃²⁻**: Bond order = 4 bonds / 3 oxygen = 1.33 (y)

3. **Count the number of species in each category:**
- **x (bond order ≥ 1.5)**: SO₄²⁻, ClO₃⁻, SO₃, XeO₃ (4 species)
- **y (bond order < 1.5)**: NO₃⁻, PO₄³⁻, CO₃²⁻, SO₃²⁻ (4 species)

4. **Calculate |x - y|:**
- x = 4
- y = 4
- |x - y| = |4 - 4| = 0

### Final Answer:
The value of |x - y| is **0**.