There are two groups of compounds A and B. Groups A contains three compounds `Px_(4), Qy_(3), Rz_(2)`. Groups B also contains three compounds `Sx_(4), Ty_(3), Uz_(2)`. Hybridization of each central atom of group A compounds is same as that of iodine in `IBrCl^(-)` while in group B compounds it is same as that of iodine `IBrCl^(+)`. Substituents X, Y and Z exhibit covalency of one in ground state. Then find the value of x/y.
Where, x and y are total number of lone pair present at central atoms of compounds of group A and B respectively.

Answer

`2`

Explanation

`I_(3)^(-)rArrsp^(3)d`
Group A
`AsF_(4)^(+)rArrsp^(3)`
Group B