Question

How many right angles are there in a $Xe{F_5}^ +$ ion?

Answer 1

Bases on the valence electrons of all atoms in a molecule, the molecular geometry and electron geometry will be expected. $Xe{F_5}^ +$ has octahedral as electron geometry and square pyramidal as molecular geometry. Based on the molecular geometry, right angles can be calculated.

Complete answer:
$Xe{F_5}^ +$ is a molecule consisting of Xenon as a central metal atom, and five fluorine atoms. The valence electrons on Xenon are $8$ , and five fluorine atoms have $5 \times 7 = 35$ electrons, as each fluorine atom has $7$ valence electrons. Thus, the total valence electrons on the given molecule will be $8 + 35 - 1 = 42$ electrons.
The eight electrons on a xenon atom are involved in bond formation with five fluorine atoms and one electron was lost as it is a cation, the remaining two electrons will exist as a lone pair of electrons. Thus, the hybridization is $s{p^3}{d^2}$ . Thus, the electron geometry is octahedral, and the molecular geometry is square pyramidal due to the presence of one lone pair of electrons.
The structure of $Xe{F_5}^ +$ is

The five fluorine atoms were considered as $1,2,3,4,$ and $5$ . Based on the above structure the right angles were:
$3 - Xe - 1$
$3 - Xe - 2$
$3 - Xe - 4$
$3 - Xe - 5$
$1 - Xe - 2$
$1 - Xe - 5$
$4 - Xe - 5$
$2 - Xe - 4$
Thus, there is a total of eight right angles in $Xe{F_5}^ +$

Note:
Right angle means the angle must be equal to ${90^0}$ . In the above structure, there were eight right angles, meaning there were eight bonds in which the bond angle equal to ${90^0}$ . The right angles can be considered based on the molecular geometry but not electron geometry.