VSEPR Theory and Shapes of Molecules

$BrO_3^-$: total valence electrons $= 7 + 3 \times 6 + 1 = 26$. The central Br has 3 bond pairs and 1 lone pair, so it is $sp^3$ hybridised and the ion is trigonal pyramidal.
$XeO_3$: total valence electrons $= 8 + 3 \times 6 = 26$. The central Xe has 3 bond pairs and 1 lone pair, so it is $sp^3$ hybridised and the molecule is trigonal pyramidal.
Both have 26 valence electrons and in both cases the structure is trigonal pyramidal, so they are isostructural.
$SO_3^{2-}$ is pyramidal (one lone pair on S), but $NO_3^-$ is trigonal planar (no lone pair on N).
$BF_3$ is trigonal planar ($sp^2$), but $NF_3$ is pyramidal ($sp^3$ with one lone pair).
$SF_4$ is see-saw shaped ($sp^3d$ with one lone pair), but $XeF_4$ is square planar ($sp^3d^2$ with two lone pairs).
So the isostructural pair is $BrO_3^-$ and $XeO_3$.

$NO_2^+$ (nitronium ion): the nitrogen has two double bonds and no lone pair, $O=\overset{+}{N}=O$. It is $sp$ hybridised, hence linear, with a bond angle of $180^\circ$. It is isoelectronic with $CO_2$.
$NO_2^-$: the nitrogen is $sp^2$ hybridised with one lone pair, so the ion is bent (about $115^\circ$).
$SO_2$: the sulphur is $sp^2$ hybridised with one lone pair, so the molecule is bent (about $119^\circ$).
$O_3$: the central oxygen is $sp^2$ hybridised with one lone pair, so the molecule is bent (about $117^\circ$).
So the species with a linear shape is $NO_2^+$.

$SiCl_4$: silicon has four bond pairs and no lone pair. It is $sp^3$ hybridised and tetrahedral.
$SO_4^{2-}$: sulphur has four sigma bond pairs and no lone pair; $sp^3$, tetrahedral.
$PO_4^{3-}$: phosphorus has four sigma bond pairs and no lone pair; $sp^3$, tetrahedral.
$NH_4^+$: nitrogen has four bond pairs and no lone pair; $sp^3$, tetrahedral.
$SCl_4$: sulphur has six valence electrons; four are used in the four S-Cl bonds and two remain as one lone pair. With five electron pairs it is $sp^3d$ hybridised, and the shape is see-saw (distorted tetrahedral), like $SF_4$.
So $SCl_4$ is not isostructural with $SiCl_4$.

$BF_3$: boron has three valence electrons and forms three B-F bonds with no lone pair. It is $sp^2$ hybridised, so the molecule is trigonal planar with bond angles of $120^\circ$.
$NH_3$: three bond pairs and one lone pair on nitrogen ($sp^3$), so it is trigonal pyramidal.
$PCl_3$: three bond pairs and one lone pair on phosphorus, so it is also trigonal pyramidal.
$IF_3$: three bond pairs and two lone pairs on iodine ($sp^3d$), so it is T-shaped.
So the molecule with trigonal planar geometry is $BF_3$.