Question

In a planar tetra-atomic molecule $X{Y}_3$, $X$ is at the centroid of the equilateral triangle formed by the atoms, $Y$. If the $X-Y$ bond distance is $1A^{\circ }$. What is the distance between the centres of any two $Y$ atoms?

• A. $\frac{1}{\sqrt{3}}\stackrel{o}{A}$
• B. $\sqrt{2}\stackrel{o}{A}$
• C. $\sqrt{3}\stackrel{o}{A}$
• D. $\frac{1}{\sqrt{2}}\stackrel{o}{A}$

Answer 1

The correct option is C $\sqrt{3}\stackrel{o}{A}$

The given molecule $X{Y}_3$ can be depicted as-

Since $\Delta$ $ABD$ is equilateral and $C$ is
centroid, therefore, $\angle CBD={30}^{\circ }$
Now, in $\Delta BCE$,
$\frac{BE}{BC}=cos{30}^{\circ }$
Now, $BC=1\stackrel{o}{A}$ (as $X-Y$ bond distance = $1\stackrel{o}{A}$)
So, $BE=BC\times cos{30}^0=1A^{\circ }\times cos{30}^{\circ }=\frac{\sqrt{3}}{2}\stackrel{o}{A}$
Here,
$BD=2BE$ (As $E$ is mid point of $BD$)
$\therefore BD=\sqrt{3}\stackrel{o}{A}$
Since $\Delta$ $ABD$ is equilateral, so $AB=BD=DA=\sqrt{3}\stackrel{o}{A}$
Hence, distance between the centres of
any two $Y$ atoms is $\sqrt{3}\stackrel{o}{A}$