Nodal planes of `pi`-bonds in `CH_(2)=C=C=CH_(2)` are located in,

Options

A. All are in molecular plane
B. Two in molecular plane and one in a plane perpendicular to molecular plane which contains C-C sigma-bond
C. On is molecular plane and two in plane perpendicular to molecular plane which contains C-C sigma-bonds
D. Two in molecular plane and one in a perpendicular to molecular plane which bisects C-C sigma-bonds at right angle.

Explanation

To determine the nodal planes of π-bonds in the molecule \( CH_2=C=C=CH_2 \), we will analyze the hybridization of the carbon atoms and the orientation of the π-bonds.

### Step 1: Draw the Structure
First, we need to draw the structure of the molecule \( CH_2=C=C=CH_2 \). This molecule consists of four carbon atoms connected by double bonds. The structure can be represented as:

```
H H
\ /
C
// \
C C
\ /
H H
```

### Step 2: Identify Hybridization
Next, we identify the hybridization of each carbon atom:
- The terminal carbon atoms (the ones bonded to \( H_2 \)) are \( sp^2 \) hybridized because they are involved in one double bond and are bonded to two hydrogen atoms.
- The central carbon atoms (the ones involved in double bonds with each other) are \( sp \) hybridized because they are involved in two double bonds.

### Step 3: Determine the Orbital Orientation
- For \( sp^2 \) hybridized carbons, the geometry is trigonal planar, and the \( p \) orbitals (which are involved in π-bonding) are oriented perpendicular to the plane of the molecule.
- For \( sp \) hybridized carbons, the geometry is linear, and the \( p \) orbitals are oriented along the axis of the molecule.

### Step 4: Analyze the π-Bonds
In the molecule, there are three π-bonds:
1. Between the first \( sp^2 \) carbon and the first \( sp \) carbon.
2. Between the first \( sp \) carbon and the second \( sp \) carbon.
3. Between the second \( sp \) carbon and the second \( sp^2 \) carbon.

### Step 5: Identify Nodal Planes
- The nodal plane of a π-bond is a plane where the electron density is zero. For the π-bonds formed by \( sp^2 \) and \( sp \) hybridized carbons, the nodal planes will be:
- For the first π-bond (between \( sp^2 \) and \( sp \)): The nodal plane is in the \( xy \)-plane (the molecular plane).
- For the second π-bond (between the two \( sp \) carbons): The nodal plane is in the \( xz \)-plane (perpendicular to the molecular plane).
- For the third π-bond (between the second \( sp \) carbon and the terminal \( sp^2 \) carbon): The nodal plane is again in the \( xy \)-plane.

### Conclusion
Thus, we conclude that:
- There are 2 π-bonds in the molecular plane (the \( xy \)-plane).
- There is 1 π-bond in a plane perpendicular to the molecular plane (the \( xz \)-plane).

The correct answer is:
**Option B: 2 in the molecular plane and 1 in the plane perpendicular to the molecular plane which contains the C-C σ bond.**