In a planar tetraatomic molecule, `AB_(3)`, A is at the centroid of the equilateral triangle formed by the atoms, B. If the A-B bond distance is `1Å`, what is the distance between the centres of any two B atoms? 

Options

A. `1//sqrt3Å`
B. `sqrt2Å`
C. `sqrt3Å`
D. `1//sqrt2Å`

Answer

C

Explanation

Ccntroid is at a distance of `2/3` of h, `therefore 2/3 xx h = 1 A^(0)`
`h=1.5 A^(0)`. The distance b/w centres of two .B. atoms is 2r. Here `2r^(2) = h^(2) + r^(2)` , i.e. `4r^(2) - r^(2) = h^(2), 3r^(2) = 1.5 xx 1.5, r= sqrt(3)/2 A^(0), therefore 2r = sqrt(3)A^(0)`