Question

Xenon hexafluoride on partial hydrolysis produces compounds ‘X’ and ‘Y’. Compounds ‘X’ and ‘Y’ and the oxidation state of $Xe$ are respectively ………………….
A.$XeO{F_4}$($+ 6$) and $Xe{O_2}{F_2}$($+ 6$)
B.$XeO{F_4}$($+ 6$) and $Xe{O_3}$($+ 6$)
C.$Xe{O_2}{F_2}$($+ 6$) and $Xe{O_2}$($+ 4$)
D.$Xe{O_2}$($+ 4$) and $Xe{O_3}$($+ 6$)

Answer 1

Partial hydrolysis is when a limited amount of water is added to another molecule to break the molecule in different parts. Oxidation state is to check which element get oxidized, that is the total number of electrons removed from an element.

Complete step by step solution:
So first we need to check the partial hydrolysis of Xenon hexafluoride to know the compound “X”:
$Xe{F_6} + {H_2}O \to XeO{F_4} + 2HF$
So as we see this equation when Xenon hexafluoride reacts with water it results in formation of xenon oxyfluoride and hydrogen fluoride.
So the “X” compound is $XeO{F_4}$.
Now the partial hydrolysis of xenon oxyfluoride to get “Y” compound:
$XeO{F_4} + {H_2}O \to Xe{O_2}{F_2} + 2HF$
So as we see this equation when xenon oxyfluoride reacts with water it results in formation of Xenon Dioxide Difluoride and hydrogen fluoride.
So the “Y” compound is $Xe{O_2}{F_2}$.
Now we have to check the oxidation state of xenon in each compound.
First, $XeO{F_4}$:
Oxidation state of Xenon + Oxidation state of Oxygen+ $4 \times$ (Oxidation state of Fluorine) =0
$Xe + {{ }}( - 2){{ }} + 4 \times ( - 1){{ }} = 0$
(Oxidation state of Oxygen=-$2$, Oxidation state of Fluorine=$- 1$)
$Xe$= $+ 6$
Now for $Xe{O_2}{F_2}$:
Oxidation state of Xenon + $2 \times$(Oxidation state of Oxygen) + $2 \times$ (Oxidation state of Fluorine) =$0$
$Xe + 2 \times ( - 2) + 2 \times ( - 1) = 0$
$Xe$= $+ 6$

So, option A is the correct.

Additional Information:
To check the oxidation of an atom we need to keep some points in mind. Those are:
For ionic bonding the oxidation number of atoms is the charge that exists on the atom.
For simple ions, the oxidation number of atoms is charged on ions.
For compound molecules, the oxidation number of an atom is the sum of the oxidation numbers of the constituent atoms.
Atoms having d-shell have several oxidation numbers.
In complex molecules, the oxidation number of an atom can be calculated by assuming the oxidation number of other atoms are fixed.
Oxidation number of metals which have more than one oxidation state will be represented by Roman numerals.

Note: Hydrolysis is adding water molecules to break the atom and to form other molecules. Oxidation states of compound molecules are calculated by adding the oxidation state of its constituent atoms linked with atoms.