The shape of $H_3O^{+}$ ion is

Match the following:

$A$. $SnCl_2$	$I$. Trigonal planar
$B$. $NH_3$	$II$. Linear
$C$. $I_3^{-}$	$III$. Angular
$D$. $SO_3$	$IV$. Trigonal pyramidal

The correct answer is:

$XeF_4$ is square planar while $XeF_6$ has a distorted octahedral structure. What is the correct explanation for this observation?

Which of the following possess two lone pairs of electrons on the central atom and are square planar in shape?
$I. SF_4, II. XeO_4, III. XeF_4, IV. ICl_4^-$

The shape of the $BF_3$ molecule is:

The correct increasing order of adjacent bond angle among $BF_3$,$PF_3$,and $ClF_3$ is: