In a methane `(CH_(4)` molecule each hydrogen atom is at a corner of a regular tetrahedron with the carbon atom at the centre. In coordinates where one of the `C-H` bond in the `hat(i)-hat(j)-hat(k)`, an adjacent `C-H` bond in the `hat(i)-hat(j)-hat(k)` direction. Then angle between these two bonds.

Options

A. `cos^(-1)""(-(2)/(3))`
B. `cos^(-1)""((2)/(3))`
C. `cos^(-1)""(-(1)/(3))`
D. `cos^(-1)""((1)/(3))`

Answer

To find the angle between the two C-H bonds in a methane molecule, we will follow these steps:

### Step 1: Define the Vectors
We need to define the vectors representing the two C-H bonds in the methane molecule.

- Let vector **A** represent one C-H bond in the direction of \( \hat{i} - \hat{j} - \hat{k} \).
- Let vector **B** represent the adjacent C-H bond in the direction of \( \hat{i} + \hat{j} + \hat{k} \).

Thus, we have:
- \( \mathbf{A} = \hat{i} - \hat{j} - \hat{k} \)
- \( \mathbf{B} = \hat{i} + \hat{j} + \hat{k} \)

### Step 2: Calculate the Dot Product
The dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z \]
Substituting the components of vectors **A** and **B**:
\[ \mathbf{A} \cdot \mathbf{B} = (1)(1) + (-1)(1) + (-1)(1) = 1 - 1 - 1 = -1 \]

### Step 3: Calculate the Magnitudes of the Vectors
The magnitude of vector **A** is:
\[ |\mathbf{A}| = \sqrt{(1)^2 + (-1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

The magnitude of vector **B** is:
\[ |\mathbf{B}| = \sqrt{(1)^2 + (1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 4: Use the Dot Product to Find the Cosine of the Angle
Using the formula for the cosine of the angle \( \theta \) between two vectors:
\[ \cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \]
Substituting the values we calculated:
\[ \cos(\theta) = \frac{-1}{\sqrt{3} \cdot \sqrt{3}} = \frac{-1}{3} \]

### Step 5: Calculate the Angle
To find the angle \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(-\frac{1}{3}\right) \]

### Final Answer
The angle between the two C-H bonds in the methane molecule is:
\[ \theta = \cos^{-1}\left(-\frac{1}{3}\right) \]