question_answer

In a planar tetra-atomic molecule\[P{{Q}_{3}},\]P is at the centroid of the equivalent triangle formed by the atoms, Q. If the P-Q bond distance is 2\[\overset{{}^\circ }{\mathop{\Alpha }}\,\], what is the distance between the centres of any two Q atoms?

A)  \[2\sqrt{3}\overset{{}^\circ }{\mathop{\Alpha }}\,\]                     

B)  \[\sqrt{3}\overset{{}^\circ }{\mathop{\Alpha }}\,\]

C)  \[4\sqrt{3}\overset{{}^\circ }{\mathop{\Alpha }}\,\]                     

D)  \[\frac{1}{\sqrt{3}}\overset{{}^\circ }{\mathop{\Alpha }}\,\]

Correct Answer: A

Solution :

                                                 [C = centeroid] \[\frac{BE}{BC}=\cos 30{}^\circ =\frac{\sqrt{3}}{2},\]\[\frac{BE}{2}=\frac{\sqrt{3}}{2}\Rightarrow BE=\sqrt{3}\] Distance between the centres of two Q atoms \[=2\times BE=2\times \sqrt{3}=2\sqrt{3}\overset{{}^\circ }{\mathop{\Alpha }}\,\]