Provide the shapes of $H_2O$,$NH_3$,and $SO_2$ including their lone pairs of electrons.

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(N/A) The shapes of the molecules are determined by the $VSEPR$ theory,considering both bonding and lone pairs of electrons:
$1$. $H_2O$: The central oxygen atom has $2$ bond pairs and $2$ lone pairs. The geometry is tetrahedral,but the shape is bent (or $V$-shaped) with a bond angle of $104.5^{\circ}$.
$2$. $NH_3$: The central nitrogen atom has $3$ bond pairs and $1$ lone pair. The geometry is tetrahedral,but the shape is trigonal pyramidal with a bond angle of $107^{\circ}$.
$3$. $SO_2$: The central sulfur atom has $2$ bond pairs (double bonds) and $1$ lone pair. The shape is bent (or $V$-shaped) with a bond angle of $119.5^{\circ}$.

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The incorrect order of bond angle is:

Based on the $VSEPR$ model,match the xenon compounds given in List-$I$ with the corresponding geometries and the number of lone pairs on xenon given in List-$II$ and choose the correct option.
List-$I$List-$II$
$(P)$ $XeF_2$$(5)$ Trigonal bipyramidal and three lone pairs of electrons
$(Q)$ $XeF_4$$(3)$ Square planar and two lone pairs of electrons
$(R)$ $XeO_3$$(2)$ Pyramidal and one lone pair of electrons
$(S)$ $XeO_3F_2$$(4)$ Trigonal bipyramidal and no lone pairs of electrons

Which of the following represents pairs of isostructural species?

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The shape of $XeF_{5}^{-}$ will be

Match List-$I$ with List-$II$ :
List-$I$ (Compound / Species) List-$II$ (Shape / Geometry)
$A$. $SF_4$ $I$. Tetrahedral
$B$. $BrF_3$ $II$. Pyramidal
$C$. $BrO_3^{-}$ $III$. See-saw
$D$. $NH_4^{+}$ $IV$. Bent $T$-shape

Choose the correct answer from the options given below :

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