Question

In $C{H}_4$ molecule, there are four $C-H$ bonds. If two adjacent bonds are in $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}-\hat{k}$ direction, then find the angle between these bonds.

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

Answer 1

Answer: (D)

${\cos }^{-1}\left(\frac{-1}{3}\right)$

Let,
$H_1=\hat{i}+\hat{j}+\hat{k}$
$H_2=\hat{i}-\hat{j}-\hat{k}$
$\therefore \overrightarrow{H_1}.\overrightarrow{H_2}=H_1{H}_{2}cos\theta$

$\therefore \cos \theta =\frac{(\hat{i}+\hat{j}+\hat{k}).(\hat{i}-\hat{j}-\hat{k})}{\sqrt{3}.\sqrt{3}}=\frac{-1}{3}$

$\therefore \theta ={\cos }^{-1}\left(\frac{-1}{3}\right)$