Consider following four compounds:
(i) `C_(x) O_(y)`
(ii) `C_(x)O_(y+1)`
(iii) `C_(x+2)O_(y+1)` and (iv) `C_(x+11)O_(y+8)`,
if "x=y=1", then calculate the vlaue of |p-q|, where p and q are total number of `sp^(2)` and hybridized carbon atoms respectively in given four compounds.

Explanation

To solve the problem, we need to analyze each of the four compounds given the values \( x = 1 \) and \( y = 1 \). We will determine the number of sp² and sp hybridized carbon atoms in each compound and then calculate \( |p - q| \), where \( p \) is the total number of sp² hybridized carbon atoms and \( q \) is the total number of sp hybridized carbon atoms.

### Step 1: Analyze each compound

1. **Compound (i): \( C_{x}O_{y} \)**
- Substituting \( x = 1 \) and \( y = 1 \), we have \( CO \).
- The structure of \( CO \) can be represented as \( C \equiv O \) (a triple bond between carbon and oxygen).
- Hybridization:
- The carbon in \( CO \) is sp hybridized (due to the triple bond).
- Count:
- sp² carbon = 0
- sp carbon = 1

2. **Compound (ii): \( C_{x}O_{y+1} \)**
- Substituting \( x = 1 \) and \( y = 1 \), we have \( CO_{2} \).
- The structure of \( CO_{2} \) is \( O=C=O \) (a double bond between carbon and each oxygen).
- Hybridization:
- The carbon in \( CO_{2} \) is sp² hybridized.
- Count:
- sp² carbon = 1
- sp carbon = 0

3. **Compound (iii): \( C_{x+2}O_{y+1} \)**
- Substituting \( x = 1 \) and \( y = 1 \), we have \( C_{3}O_{2} \).
- The structure can be represented as \( O=C=C=C=O \) (with double bonds between the carbon atoms and double bonds to oxygen).
- Hybridization:
- Each of the three carbon atoms is sp² hybridized.
- Count:
- sp² carbon = 3
- sp carbon = 0

4. **Compound (iv): \( C_{x+11}O_{y+8} \)**
- Substituting \( x = 1 \) and \( y = 1 \), we have \( C_{12}O_{9} \).
- The structure can be complex, but it can be represented as a chain of carbon atoms with alternating single and double bonds to oxygen.
- Hybridization:
- The carbon atoms can be arranged such that 12 of them are sp² hybridized due to the presence of double bonds.
- Count:
- sp² carbon = 12
- sp carbon = 0

### Step 2: Total counts of sp² and sp hybridized carbons

Now we will sum up the counts from each compound:

- Total sp² carbon (\( p \)):
- From (i): 0
- From (ii): 1
- From (iii): 3
- From (iv): 12
- Total \( p = 0 + 1 + 3 + 12 = 16 \)

- Total sp carbon (\( q \)):
- From (i): 1
- From (ii): 0
- From (iii): 0
- From (iv): 0
- Total \( q = 1 + 0 + 0 + 0 = 1 \)

### Step 3: Calculate \( |p - q| \)

Now we can calculate \( |p - q| \):

\[ |p - q| = |16 - 1| = |15| = 15 \]

### Final Answer

The final answer is \( 15 \).