Xenon reacts with fluorine under different conditions to form binary fluorides, i.e., \(\mathrm{XeF}_{2}, \mathrm{XeF}_{4}\) and \(\mathrm{XeF}_{6}\). The products formed depends on ratio of \(\mathrm{Xe}\) and \(\mathrm{F}_{2}\).
AIIMS - 2017
CHXI11:THE P-BLOCK ELEMENTS
316376
Xenon hexafluoride reacts with silica to form a xenon compound \({\rm{X}}\). The oxidation state of xenon in \({\rm{X}}\) is
1 +2
2 +4
3 +6
4 0
Explanation:
Xenon hexafluoride reacts with silica to form \(\mathrm{XeOF}_{4}\) as \(2 \mathrm{XeF}_{6}+\mathrm{SiO}_{2} \longrightarrow 2 \mathrm{XeOF}_{4}+\mathrm{SiF}_{4}\) The oxidation state of xenon in \(\mathrm{XeOF}_{4}\) is +6 as calculated below: Let the oxidation state of \(\mathrm{Xe}\) is \(a\). \(\begin{aligned}\therefore \quad & a+1 \times(-2)+4 \times(-1)=0 \\& a-2-4=0 \quad a=+6\end{aligned}\)
CHXI11:THE P-BLOCK ELEMENTS
316377
\(\mathrm{XeF}_{4}\) on partial hydrolysis produces
Xenon reacts with fluorine under different conditions to form binary fluorides, i.e., \(\mathrm{XeF}_{2}, \mathrm{XeF}_{4}\) and \(\mathrm{XeF}_{6}\). The products formed depends on ratio of \(\mathrm{Xe}\) and \(\mathrm{F}_{2}\).
AIIMS - 2017
CHXI11:THE P-BLOCK ELEMENTS
316376
Xenon hexafluoride reacts with silica to form a xenon compound \({\rm{X}}\). The oxidation state of xenon in \({\rm{X}}\) is
1 +2
2 +4
3 +6
4 0
Explanation:
Xenon hexafluoride reacts with silica to form \(\mathrm{XeOF}_{4}\) as \(2 \mathrm{XeF}_{6}+\mathrm{SiO}_{2} \longrightarrow 2 \mathrm{XeOF}_{4}+\mathrm{SiF}_{4}\) The oxidation state of xenon in \(\mathrm{XeOF}_{4}\) is +6 as calculated below: Let the oxidation state of \(\mathrm{Xe}\) is \(a\). \(\begin{aligned}\therefore \quad & a+1 \times(-2)+4 \times(-1)=0 \\& a-2-4=0 \quad a=+6\end{aligned}\)
CHXI11:THE P-BLOCK ELEMENTS
316377
\(\mathrm{XeF}_{4}\) on partial hydrolysis produces
Xenon reacts with fluorine under different conditions to form binary fluorides, i.e., \(\mathrm{XeF}_{2}, \mathrm{XeF}_{4}\) and \(\mathrm{XeF}_{6}\). The products formed depends on ratio of \(\mathrm{Xe}\) and \(\mathrm{F}_{2}\).
AIIMS - 2017
CHXI11:THE P-BLOCK ELEMENTS
316376
Xenon hexafluoride reacts with silica to form a xenon compound \({\rm{X}}\). The oxidation state of xenon in \({\rm{X}}\) is
1 +2
2 +4
3 +6
4 0
Explanation:
Xenon hexafluoride reacts with silica to form \(\mathrm{XeOF}_{4}\) as \(2 \mathrm{XeF}_{6}+\mathrm{SiO}_{2} \longrightarrow 2 \mathrm{XeOF}_{4}+\mathrm{SiF}_{4}\) The oxidation state of xenon in \(\mathrm{XeOF}_{4}\) is +6 as calculated below: Let the oxidation state of \(\mathrm{Xe}\) is \(a\). \(\begin{aligned}\therefore \quad & a+1 \times(-2)+4 \times(-1)=0 \\& a-2-4=0 \quad a=+6\end{aligned}\)
CHXI11:THE P-BLOCK ELEMENTS
316377
\(\mathrm{XeF}_{4}\) on partial hydrolysis produces
Xenon reacts with fluorine under different conditions to form binary fluorides, i.e., \(\mathrm{XeF}_{2}, \mathrm{XeF}_{4}\) and \(\mathrm{XeF}_{6}\). The products formed depends on ratio of \(\mathrm{Xe}\) and \(\mathrm{F}_{2}\).
AIIMS - 2017
CHXI11:THE P-BLOCK ELEMENTS
316376
Xenon hexafluoride reacts with silica to form a xenon compound \({\rm{X}}\). The oxidation state of xenon in \({\rm{X}}\) is
1 +2
2 +4
3 +6
4 0
Explanation:
Xenon hexafluoride reacts with silica to form \(\mathrm{XeOF}_{4}\) as \(2 \mathrm{XeF}_{6}+\mathrm{SiO}_{2} \longrightarrow 2 \mathrm{XeOF}_{4}+\mathrm{SiF}_{4}\) The oxidation state of xenon in \(\mathrm{XeOF}_{4}\) is +6 as calculated below: Let the oxidation state of \(\mathrm{Xe}\) is \(a\). \(\begin{aligned}\therefore \quad & a+1 \times(-2)+4 \times(-1)=0 \\& a-2-4=0 \quad a=+6\end{aligned}\)
CHXI11:THE P-BLOCK ELEMENTS
316377
\(\mathrm{XeF}_{4}\) on partial hydrolysis produces
Xenon reacts with fluorine under different conditions to form binary fluorides, i.e., \(\mathrm{XeF}_{2}, \mathrm{XeF}_{4}\) and \(\mathrm{XeF}_{6}\). The products formed depends on ratio of \(\mathrm{Xe}\) and \(\mathrm{F}_{2}\).
AIIMS - 2017
CHXI11:THE P-BLOCK ELEMENTS
316376
Xenon hexafluoride reacts with silica to form a xenon compound \({\rm{X}}\). The oxidation state of xenon in \({\rm{X}}\) is
1 +2
2 +4
3 +6
4 0
Explanation:
Xenon hexafluoride reacts with silica to form \(\mathrm{XeOF}_{4}\) as \(2 \mathrm{XeF}_{6}+\mathrm{SiO}_{2} \longrightarrow 2 \mathrm{XeOF}_{4}+\mathrm{SiF}_{4}\) The oxidation state of xenon in \(\mathrm{XeOF}_{4}\) is +6 as calculated below: Let the oxidation state of \(\mathrm{Xe}\) is \(a\). \(\begin{aligned}\therefore \quad & a+1 \times(-2)+4 \times(-1)=0 \\& a-2-4=0 \quad a=+6\end{aligned}\)
CHXI11:THE P-BLOCK ELEMENTS
316377
\(\mathrm{XeF}_{4}\) on partial hydrolysis produces
