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 bond order 1.5 or greater (x) and those with bond order less than 1.5 (y). Finally, we will calculate |x - y|.

### Step-by-Step Solution:

1. **Identify the Species and Their Bond Orders**:
- **NO₃⁻ (Nitrate Ion)**:
- Nitrogen has 5 valence electrons, and with one negative charge, it has 6 electrons.
- The structure has 3 bonds (1 double bond and 2 single bonds) with 3 oxygen atoms.
- Bond Order = Total Bonds / Number of Bonding Pairs = 4 / 3 = 1.33

- **SO₄²⁻ (Sulfate Ion)**:
- Sulfur has 6 valence electrons, and with two negative charges, it has 8 electrons.
- The structure has 4 bonds (1 double bond and 3 single bonds) with 4 oxygen atoms.
- Bond Order = 6 / 4 = 1.5

- **ClO₃⁻ (Chlorate Ion)**:
- Chlorine has 7 valence electrons, and with one negative charge, it has 8 electrons.
- The structure has 3 bonds (2 double bonds and 1 single bond) with 3 oxygen atoms.
- Bond Order = 5 / 3 = 1.67

- **SO₃ (Sulfur Trioxide)**:
- Sulfur has 6 valence electrons.
- The structure has 3 double bonds with 3 oxygen atoms.
- Bond Order = 6 / 3 = 2

- **PO₄³⁻ (Phosphate Ion)**:
- Phosphorus has 5 valence electrons, and with three negative charges, it has 8 electrons.
- The structure has 4 bonds (1 double bond and 3 single bonds) with 4 oxygen atoms.
- Bond Order = 4 / 4 = 1.0

- **XeO₃ (Xenon Trioxide)**:
- Xenon has 8 valence electrons.
- The structure has 3 double bonds with 3 oxygen atoms.
- Bond Order = 6 / 3 = 2

- **CO₃²⁻ (Carbonate Ion)**:
- Carbon has 4 valence electrons, and with two negative charges, it has 6 electrons.
- The structure has 4 bonds (1 double bond and 2 single bonds) with 3 oxygen atoms.
- Bond Order = 4 / 3 = 1.33

- **SO₃²⁻ (Sulfite Ion)**:
- Sulfur has 6 valence electrons, and with two negative charges, it has 8 electrons.
- The structure has 4 bonds (1 double bond and 2 single bonds) with 3 oxygen atoms.
- Bond Order = 4 / 3 = 1.33

2. **Categorize the Species**:
- **Species with Bond Order ≥ 1.5 (x)**:
- SO₄²⁻ (1.5)
- ClO₃⁻ (1.67)
- SO₃ (2)
- XeO₃ (2)
- **Total (x) = 4**

- **Species with Bond Order < 1.5 (y)**:
- NO₃⁻ (1.33)
- PO₄³⁻ (1.0)
- CO₃²⁻ (1.33)
- SO₃²⁻ (1.33)
- **Total (y) = 4**

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

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