91Ó°ÊÓ

The given data can be expressed below as:

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

This formula states that the dot product of the two given vectors is equal to the cosine 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 magnitude gives 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:

$\stackrel{→}{A}.\stackrel{→}{B}=\left|\stackrel{→}{A}\right|\left|\stackrel{→}{B}\right|\cos \mathrm{θ}$

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

Substituting the values in the above equation, we get:

$$\begin{gathered} \left(\hat{i}+\hat{j}+\hat{k}\right).\left(\hat{i}-\hat{j}-\hat{k}\right)=\left(\sqrt{1^2+1^2+1^2}\right)×\left(\sqrt{1^2+{\left(-1\right)}^2+{\left(-1\right)}^2}\right)\cos \mathrm{θ} \\ \left(1-1-1\right)=\left(\sqrt{3}\right)×\left(\sqrt{3}\right)×\cos \mathrm{θ} \\ \cos \mathrm{θ}=\frac{-1}{3} \\ \mathrm{θ}≈109.5° \end{gathered}$$

Thus, the angle between these two C-H bonds is approximate $109.5°$.