Question

In a methane $\left(C{H}_4\right)$ 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 $C-H$ bonds is in the direction of $\hat{i}+\hat{j}+\hat{k}$ , an adjacent $C-H$ bond in the $\hat{i}-\hat{j}-\hat{k}$ direction. Then angle between these two bonds is.

• A. ${\cos }^{-1}(-\frac{2}{3})$
• B. ${\cos }^{-1}(-\frac{2}{3})$
• C. ${\cos }^{-1}(-\frac{1}{3})$
• D. ${\cos }^{-1}\left(\frac{1}{3}\right)$

Answer 1

The correct option is C ${\cos }^{-1}(-\frac{1}{3})$

Given,
$\overrightarrow{A}=\hat{i}+\hat{j}+\hat{k}$
$\overrightarrow{B}=\hat{i}-\hat{j}-\hat{k}$
using,
$\cos \theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|}$
Substitute the values, we get
$\cos \theta =\frac{(\hat{i}+\hat{j}+\hat{k})(\hat{i}-\hat{j}-\hat{k})}{\sqrt{3}\sqrt{3}}=\frac{1-1-1}{3}$
$=-\frac{1}{3}$
Final Answer: (C)