In carprolactum if 'a' is number of lone pairs of `e^(-)` and 'b' is number of `sp^(3)` hybridised atoms. What is the vlaue of `a+b`?

Answer

To solve the problem, we need to determine the values of 'a' (the number of lone pairs of electrons) and 'b' (the number of sp³ hybridized atoms) in caprolactam. Let's break this down step by step.

### Step 1: Draw the Structure of Caprolactam
Caprolactam is a cyclic amide with the following structure:

```
O
||
H2N-C-CH2-CH2-CH2-CH2
```

### Step 2: Identify Lone Pairs of Electrons
- **Oxygen (O)**: Oxygen typically has 2 lone pairs of electrons.
- **Nitrogen (N)**: Nitrogen typically has 1 lone pair of electrons.

Now, we can calculate the total number of lone pairs:
- Lone pairs from Oxygen = 2
- Lone pairs from Nitrogen = 1

Thus, the total number of lone pairs (a) = 2 + 1 = **3**.

### Step 3: Identify sp³ Hybridized Atoms
Next, we need to determine the number of sp³ hybridized atoms:
- **Carbon (C)**: Each carbon atom in caprolactam is connected to four other atoms (either other carbons or hydrogens), which means they are sp³ hybridized.
- In the structure, we have:
- 5 carbon atoms in the chain (each sp³ hybridized).
- 1 carbon in the carbonyl group (C=O) that is not sp³ hybridized.

Thus, the total number of sp³ hybridized carbon atoms (b) = 5.

### Step 4: Calculate a + b
Now, we can calculate the final value:
- a (lone pairs) = 3
- b (sp³ hybridized atoms) = 5

So, a + b = 3 + 5 = **8**.

### Final Answer
The value of a + b is **8**.

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