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 compound to determine the hybridization of the central atom and categorize them based on the number of p-orbitals used in the hybridization. We will denote:

- **x**: Total number of compounds where the central atom uses all three p-orbitals.
- **y**: Total number of compounds where the central atom uses only two p-orbitals.
- **z**: Total number of compounds where the central atom uses only one p-orbital.

Let's analyze each compound step by step:

### Step 1: Analyze IF₅
- **Valence Electrons**: Iodine (I) has 7 valence electrons. There are 5 fluorine atoms contributing 5 electrons.
- **Steric Number Calculation**: (7 + 5) / 2 = 6
- **Hybridization**: sp³d²
- **p-orbitals Used**: All three p-orbitals are used. (Count: 1)

### Step 2: Analyze ClI₄⁻
- **Valence Electrons**: Chlorine (Cl) has 7, 4 iodine atoms contribute 4, and the negative charge adds 1.
- **Steric Number Calculation**: (7 + 4 + 1) / 2 = 6
- **Hybridization**: sp³d²
- **p-orbitals Used**: All three p-orbitals are used. (Count: 2)

### Step 3: Analyze XeO₂F₂
- **Valence Electrons**: Xenon (Xe) has 8, 2 oxygen atoms contribute 4, and 2 fluorine atoms contribute 2.
- **Steric Number Calculation**: (8 + 4 + 2) / 2 = 7
- **Hybridization**: sp³d
- **p-orbitals Used**: All three p-orbitals are used. (Count: 3)

### Step 4: Analyze NH₂⁻
- **Valence Electrons**: Nitrogen (N) has 5, and the negative charge adds 1.
- **Steric Number Calculation**: (5 + 2 + 1) / 2 = 4
- **Hybridization**: sp³
- **p-orbitals Used**: All three p-orbitals are used. (Count: 4)

### Step 5: Analyze BCl₃
- **Valence Electrons**: Boron (B) has 3, and 3 chlorine atoms contribute 3.
- **Steric Number Calculation**: (3 + 3) / 2 = 3
- **Hybridization**: sp²
- **p-orbitals Used**: Only two p-orbitals are used. (Count: 1)

### Step 6: Analyze BeCl₂
- **Valence Electrons**: Beryllium (Be) has 2, and 2 chlorine atoms contribute 2.
- **Steric Number Calculation**: (2 + 2) / 2 = 2
- **Hybridization**: sp
- **p-orbitals Used**: Only one p-orbital is used. (Count: 1)

### Step 7: Analyze AsCl₄⁺
- **Valence Electrons**: Arsenic (As) has 5, 4 chlorine atoms contribute 4, and the positive charge subtracts 1.
- **Steric Number Calculation**: (5 + 4 - 1) / 2 = 4
- **Hybridization**: sp³
- **p-orbitals Used**: All three p-orbitals are used. (Count: 5)

### Step 8: Analyze B(OH)₃
- **Valence Electrons**: Boron (B) has 3, and 3 hydroxyl groups contribute 3.
- **Steric Number Calculation**: (3 + 3) / 2 = 3
- **Hybridization**: sp²
- **p-orbitals Used**: Only two p-orbitals are used. (Count: 2)

### Step 9: Analyze NO₂⁻
- **Valence Electrons**: Nitrogen (N) has 5, 2 oxygen atoms contribute 4, and the negative charge adds 1.
- **Steric Number Calculation**: (5 + 4 + 1) / 2 = 5
- **Hybridization**: sp²
- **p-orbitals Used**: Only two p-orbitals are used. (Count: 3)

### Step 10: Analyze ClO₂⁺
- **Valence Electrons**: Chlorine (Cl) has 7, 2 oxygen atoms contribute 4, and the positive charge subtracts 1.
- **Steric Number Calculation**: (7 + 4 - 1) / 2 = 5
- **Hybridization**: sp²
- **p-orbitals Used**: Only two p-orbitals are used. (Count: 4)

### Summary of Counts
- **x (3 p-orbitals used)**: 5 compounds (IF₅, ClI₄⁻, XeO₂F₂, NH₂⁻, AsCl₄⁺)
- **y (2 p-orbitals used)**: 4 compounds (BCl₃, B(OH)₃, NO₂⁻, ClO₂⁺)
- **z (1 p-orbital used)**: 1 compound (BeCl₂)

### Final Calculation
Now we calculate the value of "x + y - z":
- x = 5
- y = 4
- z = 1

So, \( x + y - z = 5 + 4 - 1 = 8 \).

### Final Answer
The value of \( x + y - z \) is **8**.