Consider the following compounds :
`(i) IF_(5) " " (ii) ClI_(4)^(-) " " (iii) XeO_(2)F_(2) " " (iv) NH_(2)^(-) `
`(v) BCl_(3) " " (vi) BeCl_(2) " " (vii) AsCl_(4)^(+) " " (viii) B(OH)_(3)`
(ix) `NO_(2)^(-) " " (x) ClO_(2)^(+)`
Then calculate value of "x+y-z", here, x,y and z are total number of compounds in given compounds in which central atom used their all three p-orbitals, only two p-orbitals and only one p-orbital in hybridisation respectively .

Explanation

To solve the problem, we need to analyze each of the given compounds to determine the hybridization of the central atom and identify how many p-orbitals are used in each case. We will categorize them into three groups: compounds using all three p-orbitals (x), compounds using only two p-orbitals (y), and compounds using only one p-orbital (z). Finally, we will calculate the value of "x + y - z".

### Step-by-Step Solution:

1. **Identify the Central Atom and Count Valence Electrons**:
- For each compound, determine the central atom and count its valence electrons.

2. **Determine the Steric Number**:
- The steric number is calculated as the sum of the number of sigma bonds and lone pairs around the central atom.

3. **Determine Hybridization**:
- Based on the steric number:
- Steric number 2 → sp hybridization (uses 1 p-orbital)
- Steric number 3 → sp² hybridization (uses 2 p-orbitals)
- Steric number 4 → sp³ hybridization (uses 3 p-orbitals)
- Steric number 5 → sp³d hybridization (uses 3 p-orbitals and 1 d-orbital)
- Steric number 6 → sp³d² hybridization (uses 3 p-orbitals and 2 d-orbitals)

4. **Analyze Each Compound**:
- (i) IF₅: Iodine (7 valence electrons) + 5 bonds + 1 lone pair = Steric number 6 → sp³d² (uses 3 p-orbitals)
- (ii) ClI₄⁻: Iodine (7 + 1 for charge) + 4 bonds + 2 lone pairs = Steric number 6 → sp³d² (uses 3 p-orbitals)
- (iii) XeO₂F₂: Xenon (8 valence electrons) + 4 bonds + 1 lone pair = Steric number 5 → sp³d (uses 3 p-orbitals)
- (iv) NH₂⁻: Nitrogen (5 + 1 for charge) + 2 bonds + 2 lone pairs = Steric number 4 → sp³ (uses 3 p-orbitals)
- (v) BCl₃: Boron (3 valence electrons) + 3 bonds = Steric number 3 → sp² (uses 2 p-orbitals)
- (vi) BeCl₂: Beryllium (2 valence electrons) + 2 bonds = Steric number 2 → sp (uses 1 p-orbital)
- (vii) AsCl₄⁺: Arsenic (5 - 1 for charge) + 4 bonds = Steric number 4 → sp³ (uses 3 p-orbitals)
- (viii) B(OH)₃: Boron (3 valence electrons) + 3 bonds = Steric number 3 → sp² (uses 2 p-orbitals)
- (ix) NO₂⁻: Nitrogen (5 + 1 for charge) + 2 bonds + 1 lone pair = Steric number 3 → sp² (uses 2 p-orbitals)
- (x) ClO₂⁺: Chlorine (7 - 1 for charge) + 2 bonds + 1 lone pair = Steric number 3 → sp² (uses 2 p-orbitals)

5. **Count x, y, z**:
- **x** (compounds using all three p-orbitals): IF₅, ClI₄⁻, XeO₂F₂, NH₂⁻, AsCl₄⁺ → Total = 5
- **y** (compounds using only two p-orbitals): BCl₃, B(OH)₃, NO₂⁻, ClO₂⁺ → Total = 4
- **z** (compounds using only one p-orbital): BeCl₂ → Total = 1

6. **Calculate x + y - z**:
- x + y - z = 5 + 4 - 1 = 8

### Final Answer:
The value of "x + y - z" is **8**.