Answer
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Hint: To solve this question, we must know the molecular structures of the reactants. Then from that, we can predict the making and breaking of bonds to form new products. The formation of products depends on the reactivity of reactant molecules.
Complete Step-by-Step Answer:
Before we move forward with the solution of the given reaction, let us first understand some important basic concepts.
The chemical name of \[Xe{F_4}\] is xenon tetrafluoride. The molecule of xenon tetrafluoride has a square planar geometric structure. This geometric structure consists of Xenon being the central atom with 4 sigma bonds and 2 lone pairs. The geometric arrangement of the molecule is a square planar structure where Xe is the central atom and 4 F atoms are arranged around the central atom. Each of these Fluorine atoms form 1 sigma bond each and have 3 lone pairs.
On the other hand, the molecular structure of \[{O_2}{F_2}\] can be understood as a single chain compound with dihedral angle whose values approach almost \[90^\circ \] . The oxygen atoms are placed between the fluorine atoms.
When \[Xe{F_4}\] reacts with \[{O_2}{F_2}\] , the two \[O-F\] bonds present in \[{O_2}{F_2}\] break. Now, the two Fluorine ions thus formed get attached to the \[Xe{F_4}\] . The fluorine atoms get attached at the sites on the central atom where we have lone pairs of electrons. Once these bonds are formed, the final structure obtained would have Xe as the central atom with 6 sigma bonds, which help the Xe atom to bond with 6 fluorine atoms. This results in leaving just 1 lone pair on the central atom. The \[{O_2}{F_2}\] molecule from where the fluorine atoms got removed, now forms the\[{O_2}\] molecule. Hence the final chemical equation for this reaction can be given as:
\[Xe{F_4} + {O_2}{F_2} \to Xe{F_6} + {O_2}\]
Hence, the compound A can be identified as \[Xe{F_6}\] .
Note: Xenon hexafluoride is a noble gas compound with the formula \[Xe{F_6}\] . It is a colourless solid that readily sublimes into intensely yellow vapours.
Complete Step-by-Step Answer:
Before we move forward with the solution of the given reaction, let us first understand some important basic concepts.
The chemical name of \[Xe{F_4}\] is xenon tetrafluoride. The molecule of xenon tetrafluoride has a square planar geometric structure. This geometric structure consists of Xenon being the central atom with 4 sigma bonds and 2 lone pairs. The geometric arrangement of the molecule is a square planar structure where Xe is the central atom and 4 F atoms are arranged around the central atom. Each of these Fluorine atoms form 1 sigma bond each and have 3 lone pairs.
On the other hand, the molecular structure of \[{O_2}{F_2}\] can be understood as a single chain compound with dihedral angle whose values approach almost \[90^\circ \] . The oxygen atoms are placed between the fluorine atoms.
When \[Xe{F_4}\] reacts with \[{O_2}{F_2}\] , the two \[O-F\] bonds present in \[{O_2}{F_2}\] break. Now, the two Fluorine ions thus formed get attached to the \[Xe{F_4}\] . The fluorine atoms get attached at the sites on the central atom where we have lone pairs of electrons. Once these bonds are formed, the final structure obtained would have Xe as the central atom with 6 sigma bonds, which help the Xe atom to bond with 6 fluorine atoms. This results in leaving just 1 lone pair on the central atom. The \[{O_2}{F_2}\] molecule from where the fluorine atoms got removed, now forms the\[{O_2}\] molecule. Hence the final chemical equation for this reaction can be given as:
\[Xe{F_4} + {O_2}{F_2} \to Xe{F_6} + {O_2}\]
Hence, the compound A can be identified as \[Xe{F_6}\] .
Note: Xenon hexafluoride is a noble gas compound with the formula \[Xe{F_6}\] . It is a colourless solid that readily sublimes into intensely yellow vapours.
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