Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

In a non polar $AX_2$, the bond moment of $AX$ is 0.56D then the bond angle of $AX_2$ is

$\begin{array}{1 1}(a)\;180^{\large\circ}&(b)\; 90^{\large\circ}\\(c)\;45^{\large\circ}&(d)\;30^{\large\circ}\end{array}$

Can you answer this question?

1 Answer

0 votes
Net dipolemoment $\mu$=2(bond moment)$\times \cos \large\frac{\theta}{2}$
But given $AX_2$ is a non polar molecule.Hence $\mu=0$
$0=2(0.56)D\times \cos\large\frac{\theta}{2}$
Hence (a) is the correct answer.
answered Mar 7, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App