
Oxygen can show positive oxidation state in its compound with
A) Fluorine
B) Nitrogen
C) Hydrogen
D) Sulphur
Answer
563.4k+ views
Hint:Fluorine is more electronegative than oxygen. compounds of fluorine and oxygen are called fluorides. Oxygen inhibits positive oxidation state in compounds with fluorine because fluorine is more electronegative than oxygen.
Complete step by step answer:
-Fluorine is the most reactive element because of its small atomic size, highly electronegativity, low bond energy and high standard reduction potential.
-Fluorine is the most electronegative element in the periodic table. Oxygen is the second electronegative element in the periodic table. Fluorine has the electronegativity \[4.0\] and oxygen has the electronegativity \[3.5\].
-Fluorine is more electronegative than oxygen. Thus \[O{F_2}\] is not fluorine oxide but it is oxygen fluoride.
-Oxygen fluorides are compounds of elements of oxygen and fluorine with general formula \[{O_n}{F_2}\] where \[n = 1\] to \[6\]. Many different oxygen fluorides are known as-
-Oxygen difluoride $\left( {O{F_2}} \right)$
-Dioxygen difluoride $\left( {{O_2}{F_2}} \right)$
-Tetra oxygen difluoride $\left( {{O_4}{F_2}} \right)$
-Dioxygen monofluoride $\left( {{O_2}F} \right)$
-Oxygen fluorides are strong oxidising agents with high energy and can release their energy either instantaneously or at a controlled rate. Thus, these compounds attracted much attention as potential fuels in jet propulsion systems.
Preparation of oxygen fluorides-
\[{O_2} + {F_2} \rightleftharpoons {O_2}{F_2}\]
\[3{O_2} + 2{F_2} \rightleftharpoons 2{O_3}{F_2}\]
\[2{O_2} + {F_2} \rightleftharpoons {O_4}{F_2}\]
-Oxygen difluoride is the chemical compound with the formula \[O{F_2}\]. As predicted by VSEPR theory (Valence shell electron pair repulsion theory). Which is based on
\[{l_p} - {l_p} > {l_p} - {B_p} > {B_p} - {B_p}\]
Where \[{l_p}\] stands for lone pair and \[{B_p}\] stands for bond pair.
Note:
We have to remember that Fluoride is more electronegative than oxygen. Thus, it would attract towards itself during complete transfer. It would result in a \[ + 2\] charge on oxygen. Thus, oxygen has \[ + 2\] oxidation number.
Complete step by step answer:
-Fluorine is the most reactive element because of its small atomic size, highly electronegativity, low bond energy and high standard reduction potential.
-Fluorine is the most electronegative element in the periodic table. Oxygen is the second electronegative element in the periodic table. Fluorine has the electronegativity \[4.0\] and oxygen has the electronegativity \[3.5\].
-Fluorine is more electronegative than oxygen. Thus \[O{F_2}\] is not fluorine oxide but it is oxygen fluoride.
-Oxygen fluorides are compounds of elements of oxygen and fluorine with general formula \[{O_n}{F_2}\] where \[n = 1\] to \[6\]. Many different oxygen fluorides are known as-
-Oxygen difluoride $\left( {O{F_2}} \right)$
-Dioxygen difluoride $\left( {{O_2}{F_2}} \right)$
-Tetra oxygen difluoride $\left( {{O_4}{F_2}} \right)$
-Dioxygen monofluoride $\left( {{O_2}F} \right)$
-Oxygen fluorides are strong oxidising agents with high energy and can release their energy either instantaneously or at a controlled rate. Thus, these compounds attracted much attention as potential fuels in jet propulsion systems.
Preparation of oxygen fluorides-
\[{O_2} + {F_2} \rightleftharpoons {O_2}{F_2}\]
\[3{O_2} + 2{F_2} \rightleftharpoons 2{O_3}{F_2}\]
\[2{O_2} + {F_2} \rightleftharpoons {O_4}{F_2}\]
-Oxygen difluoride is the chemical compound with the formula \[O{F_2}\]. As predicted by VSEPR theory (Valence shell electron pair repulsion theory). Which is based on
\[{l_p} - {l_p} > {l_p} - {B_p} > {B_p} - {B_p}\]
Where \[{l_p}\] stands for lone pair and \[{B_p}\] stands for bond pair.
Note:
We have to remember that Fluoride is more electronegative than oxygen. Thus, it would attract towards itself during complete transfer. It would result in a \[ + 2\] charge on oxygen. Thus, oxygen has \[ + 2\] oxidation number.
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