For the following molecules :
`PCl_(5), BrF_(3), IC l_(2)^(-), XeF_(5)^(-),NO_(3)^(-),XeO_(2)F_(2),PCl_(4)^(+),CH_(3)^(+)`
Calculate the value of `(a+b)/(c )`
a = Number of species having `sp^(3)` d-hybridisation
b= Number of species which are planar
c= Number of species which are non-planar

Explanation

To solve the problem, we need to determine the values of \( a \), \( b \), and \( c \) based on the given molecules. Let's break it down step by step.

### Step 1: Identify Hybridization of Each Species
We need to determine the hybridization of each molecule listed:

1. **PCl₅**: The hybridization is \( sp^3d \).
2. **BrF₃**: The hybridization is \( sp^3d \).
3. **ICl₂⁻**: The hybridization is \( sp^3 \).
4. **XeF₅⁻**: The hybridization is \( sp^3d^2 \).
5. **NO₃⁻**: The hybridization is \( sp^2 \).
6. **XeO₂F₂**: The hybridization is \( sp^3d \).
7. **PCl₄⁺**: The hybridization is \( sp^3d \).
8. **CH₃⁺**: The hybridization is \( sp^3 \).

### Step 2: Count the Species with \( sp^3 \) Hybridization
From the above analysis, the species with \( sp^3 \) hybridization are:
- ICl₂⁻
- CH₃⁺

Thus, \( a = 2 \).

### Step 3: Identify Planarity of Each Species
Next, we need to determine which species are planar:
1. **PCl₅**: Non-planar.
2. **BrF₃**: Non-planar.
3. **ICl₂⁻**: Planar.
4. **XeF₅⁻**: Non-planar.
5. **NO₃⁻**: Planar.
6. **XeO₂F₂**: Non-planar.
7. **PCl₄⁺**: Planar.
8. **CH₃⁺**: Planar.

Counting the planar species gives us:
- ICl₂⁻
- NO₃⁻
- PCl₄⁺
- CH₃⁺

Thus, \( b = 4 \).

### Step 4: Count the Non-Planar Species
The non-planar species are:
- PCl₅
- BrF₃
- XeF₅⁻
- XeO₂F₂

Thus, \( c = 4 \).

### Step 5: Calculate \( \frac{(a+b)}{c} \)
Now we can substitute the values into the equation:
\[ \frac{(a+b)}{c} = \frac{(2+4)}{4} = \frac{6}{4} = 1.5 \]

### Final Answer
The value of \( \frac{(a+b)}{c} \) is \( 1.5 \).

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