Molecule ${\text{Xe}}{{\text{O}}_{\text{3}}}$ is isostructural with ${\text{BrO}}_3^ - $.
Answer whether the above statement is true or false.

Hint:To answer this question we have to find the geometry of both the given molecules. We will draw the Lewis structure of all the molecules to determine the electron pair of the central atom. Then by using valence shell electron pair repulsion theory the geometry can be determined.

Complete step-by-step answer:
We will write the Lewis structure as follows:
-First we will write the basic structure. Then we will decide the central atom around which we will write all atoms of the molecule. The least electronegative atom is the central atom.
-Then we will count total valence electrons.
-Two electrons are used in the formation of a bond.So, we will count the total electron used in bond formation.
-Then we will subtract the electrons used in bond formation from the total valence electrons.
-Then we will arrange the remaining electrons around each atom to complete the octet.

The valence shell electron pair repulsion theory is as follows:
Electron pair is the number of electron pairs present around the central atom in a molecule.
According to VSEPR the electron pairs present around the central atom repel each other. So, all the pairs get arranged to minimize the repulsion. Based on the number of electron pair the geometry can be determined as follows:



Lewis structure of the pair${\text{Xe}}{{\text{O}}_{\text{3}}}$ is as follows:

Total valence electrons in ${\text{Xe}}{{\text{O}}_{\text{3}}}$ are as follows:
$ = \,\left( {1 \times 8} \right) + \left( {6 \times 3} \right)$
$ = \,26$
The total sigma electron pair around xenon is four, so the geometry will be tetrahedral but all four electron pairs are not the same . Out of four three are bond pair and one is lone pair so, the shape or structure of${\text{Xe}}{{\text{O}}_{\text{3}}}$ is pyramidal.
The structure or shape of ${\text{Xe}}{{\text{O}}_{\text{3}}}$is shown as follows:

Lewis structure of the pair ${\text{BrO}}_3^ - $ is as follows:

Total valence electrons in ${\text{BrO}}_3^ - $ are as follows:
$ = \,\left( {7 \times 1} \right) + \left( {6\, \times 3} \right) + 1$
$ = \,26$
The total sigma electron pair around bromine is four, so the geometry will be tetrahedral but all four electron pairs are not the same . Out of four three are bond pairs and one is lone pair, so the shape or structure of${\text{BrO}}_3^ - $ is pyramidal.
The geometry of ${\text{BrO}}_3^ - $is shown as follows:

So, it is true that ${\text{Xe}}{{\text{O}}_{\text{3}}}$ is isostructural with ${\text{BrO}}_3^ - $.
Therefore, the statement is true.

Note: Geometry depends upon the lone pair as well as sigma bond pair. The shape or structure of the molecule depends upon the bond pairs only. To determine the total valence electrons of a molecule, we sum all the valence electrons of the atoms present in the molecule. We subtract one for every positive charge and add one for every negative charge. Geometry around the central atom is decided only on the basis of sigma bond pair and lone pair only. Pi bonds are not counted to determine the geometry. In tetrahedral geometry if the number of bond pairs will be three and lone pairs will be one then the shape will be pyramidal.