Which of the underlined atoms in the molecules shown below have sp-hybridization ?
(u) `ulCH_2CHCH_3` (v) `CH_2ulC"CHCl"` (w) `CH_3ulCH_(2)^(+)` (x) `H-C-=C-H`
(y) `CH_3ulCN` (z) `(CH_3)_2CulN"NH"_2`

Options

A. x and z
B. x , y and z
C. u , w and x
D. v ,x and y

Explanation

To determine which of the underlined atoms in the given molecules have sp-hybridization, we will calculate the steric number for each relevant carbon atom. The steric number is calculated as the sum of the number of sigma bonds and lone pairs around the atom. The hybridization type can be inferred from the steric number as follows:

- Steric number of 2 corresponds to sp hybridization.
- Steric number of 3 corresponds to sp² hybridization.
- Steric number of 4 corresponds to sp³ hybridization.

Let's analyze each molecule step by step:

### Step 1: Analyze molecule (u) `CH2=CH-CH3`
- The underlined carbon is the first carbon in the double bond.
- It forms 3 sigma bonds (2 with hydrogen and 1 with the second carbon) and has no lone pairs.
- **Steric number = 3** → **Hybridization = sp²** (not sp).

### Step 2: Analyze molecule (v) `CH2=C(CHCl)-CH3`
- The underlined carbon is the second carbon in the double bond.
- It forms 2 sigma bonds (1 with the first carbon and 1 with the chlorine) and has 2 pi bonds (from the double bond).
- **Steric number = 2** → **Hybridization = sp** (this is sp).

### Step 3: Analyze molecule (w) `CH3-CH2^+`
- The underlined carbon is the second carbon in the chain.
- It forms 3 sigma bonds (2 with hydrogen and 1 with the first carbon) and has no lone pairs.
- **Steric number = 3** → **Hybridization = sp²** (not sp).

### Step 4: Analyze molecule (x) `H-C≡C-H`
- The underlined carbon is one of the carbons in the triple bond.
- It forms 1 sigma bond (with hydrogen) and has 2 pi bonds (from the triple bond).
- **Steric number = 1** → **Hybridization = sp** (this is sp).

### Step 5: Analyze molecule (y) `CH3-C≡N`
- The underlined carbon is the carbon in the triple bond with nitrogen.
- It forms 1 sigma bond (with the nitrogen) and has 2 pi bonds (from the triple bond).
- **Steric number = 1** → **Hybridization = sp** (this is sp).

### Step 6: Analyze molecule (z) `(CH3)2C=N-NH2`
- The underlined carbon is the central carbon.
- It forms 3 sigma bonds (2 with methyl groups and 1 with nitrogen) and has no lone pairs.
- **Steric number = 3** → **Hybridization = sp²** (not sp).

### Summary of Results:
- (u) sp²
- (v) sp
- (w) sp²
- (x) sp
- (y) sp
- (z) sp²

### Final Answer:
The underlined atoms with sp-hybridization are in molecules (v), (x), and (y).

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