Q78P - University Physics with Modern Physics

Question

In the methane molecule, ${\mathbf{CH}}_{\mathbf{4}}$, each hydrogen atom is at a corner of a regular tetrahedron with the carbon atom at the center. In coordinates for which one of the $\mathbf{C}\mathbf{-}\mathbf{H}$bonds is in the direction of $\hat{\mathbf{i}}\mathbf{+}\hat{\mathbf{j}}\mathbf{+}\hat{\mathbf{k}}$, an adjacent $\mathbf{C}\mathbf{-}\mathbf{H}$bond is in the $\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$direction. Calculate the angle between these two bonds.

Step-by-Step Solution

Answer

The angle between these two C-H bonds is approximately $109.5^{°}$

Step 1: Identification of the given data

The given data can be expressed below as:

• One of the C-H bonds is in the direction of $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$.
• Another C-H bond is in the direction of $\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$.

Step 2: Significance of the formula of the dot product to calculate the angle

This formula states that the dot product of the two given vectors is equal to the cosecant of the vector’s angle and the product of the two vector’s magnitude.

 

The dot product of the two vectors divided by the product of the two vectors gives the angle between the bonds.

Step 3: Determination of the angle between the bonds

From the rule of the dot product of the vectors, the angle between the two bonds can be expressed as:

$\stackrel{\rightharpoonup }{A}.\stackrel{\rightharpoonup }{B}=\left|\stackrel{\rightharpoonup }{A}\right|\left|\stackrel{\rightharpoonup }{B}\right|\cos \theta$

Here, $\stackrel{\rightharpoonup }{\mathrm{A}}\mathrm{and} \stackrel{\rightharpoonup }{\mathrm{B}}$are the direction of the first and the second C-H bond which are $\hat{i}+\hat{j}+\hat{k} \mathrm{and} \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$ respectively, and $\theta$is the angle between two vectors.

 

Substituting the values in the above equation, we get, 

$$\begin{gathered} \left(\hat{i}+\hat{j}+\hat{k}\right).\left(\hat{i}-\hat{j}-\hat{k}\right)=\left(\sqrt{1^2+1^2+1^2}\right)\times \left(\sqrt{1^2+{\left(-1\right)}^2+{\left(-1\right)}^2}\right)\cos \theta \\ \left(1-1-1\right)=\left(\sqrt{3}\right)\times \left(\sqrt{3}\right)\times \cos \theta \\ \cos \theta =\frac{-1}{3} \\ \theta \approx 109.5^{°} \\ \mathrm{Thus}, \mathrm{the} \mathrm{angle} \mathrm{between} \mathrm{these} \mathrm{two} \mathrm{C}-\mathrm{H} \mathrm{bonds} \mathrm{is} \mathrm{approximate} 109.5^{°} . \end{gathered}$$