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 determine the hybridization of each compound and classify them based on the number of p-orbitals used in the hybridization. We will then calculate the values of x, y, and z, where:

- x = number of compounds using all three p-orbitals
- y = number of compounds using only two p-orbitals
- z = number of compounds using only one p-orbital

Let's analyze each compound step by step:

### Step 1: Calculate Hybridization for Each Compound

1. **IF₅**
- Valence electrons of Iodine (I) = 7
- Monovalent atoms (F) = 5
- Charge = 0
- Hybridization = \( \frac{1}{2} (7 + 5 + 0) = \frac{12}{2} = 6 \) → **sp³d²**
- **p-orbitals used = 5**

2. **ClI₄⁻**
- Valence electrons of Chlorine (Cl) = 7
- Monovalent atoms (I) = 4
- Charge = +1 (due to negative charge)
- Hybridization = \( \frac{1}{2} (7 + 4 + 1) = \frac{12}{2} = 6 \) → **sp³d²**
- **p-orbitals used = 5**

3. **XeO₂F₂**
- Valence electrons of Xenon (Xe) = 8
- Monovalent atoms (F) = 2
- Charge = 0
- Hybridization = \( \frac{1}{2} (8 + 2 + 0) = \frac{10}{2} = 5 \) → **sp³d**
- **p-orbitals used = 4**

4. **NH₂⁻**
- Valence electrons of Nitrogen (N) = 5
- Monovalent atoms (H) = 2
- Charge = +1
- Hybridization = \( \frac{1}{2} (5 + 2 + 1) = \frac{8}{2} = 4 \) → **sp³**
- **p-orbitals used = 3**

5. **BCl₃**
- Valence electrons of Boron (B) = 3
- Monovalent atoms (Cl) = 3
- Charge = 0
- Hybridization = \( \frac{1}{2} (3 + 3 + 0) = \frac{6}{2} = 3 \) → **sp²**
- **p-orbitals used = 2**

6. **BeCl₂**
- Valence electrons of Beryllium (Be) = 2
- Monovalent atoms (Cl) = 2
- Charge = 0
- Hybridization = \( \frac{1}{2} (2 + 2 + 0) = \frac{4}{2} = 2 \) → **sp**
- **p-orbitals used = 1**

7. **AsCl₄⁺**
- Valence electrons of Arsenic (As) = 5
- Monovalent atoms (Cl) = 4
- Charge = -1 (due to positive charge)
- Hybridization = \( \frac{1}{2} (5 + 4 - 1) = \frac{8}{2} = 4 \) → **sp³**
- **p-orbitals used = 3**

8. **B(OH)₃**
- Valence electrons of Boron (B) = 3
- Monovalent atoms (OH) = 3
- Charge = 0
- Hybridization = \( \frac{1}{2} (3 + 3 + 0) = \frac{6}{2} = 3 \) → **sp²**
- **p-orbitals used = 2**

9. **NO₂⁻**
- Valence electrons of Nitrogen (N) = 5
- Monovalent atoms (O) = 2
- Charge = +1
- Hybridization = \( \frac{1}{2} (5 + 2 + 1) = \frac{8}{2} = 4 \) → **sp²**
- **p-orbitals used = 2**

10. **ClO₂⁺**
- Valence electrons of Chlorine (Cl) = 7
- Monovalent atoms (O) = 2
- Charge = -1 (due to positive charge)
- Hybridization = \( \frac{1}{2} (7 + 2 - 1) = \frac{8}{2} = 4 \) → **sp²**
- **p-orbitals used = 2**

### Step 2: Count the Number of Compounds for x, y, and z

- **x (using all 3 p-orbitals)**:
- IF₅, ClI₄⁻, XeO₂F₂, NH₂⁻, AsCl₄⁺ → **5 compounds**

- **y (using only 2 p-orbitals)**:
- BCl₃, B(OH)₃, NO₂⁻, ClO₂⁺ → **4 compounds**

- **z (using only 1 p-orbital)**:
- BeCl₂ → **1 compound**

### Step 3: Calculate x + y - z

Now we can compute the final value:
- \( x + y - z = 5 + 4 - 1 = 8 \)

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

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