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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Chemistry
Chapter
p Block Elements (Group 15, 16, 17 & 18)

JIPMER 2018 · Chemistry · p Block Elements (Group 15, 16, 17 & 18)

Xenon hexafluoride reacts with silica to form a xenon compound X ....

Xenon hexafluoride reacts with silica to form a xenon compound X . The oxidation state of xenon in X is

Options

  1. A+2+2
  2. B+4+4
  3. C+6+6
  4. D00

Correct answer

C. +6+6

Step-by-step solution

Xenon hexa fluoride reacts with silica to form XeOF4\mathrm{XeOF}_4 as 2XeF6+SiO2⟶2XeOF4+SiF42 \mathrm{XeF}_6+\mathrm{SiO}_2 \longrightarrow 2 \mathrm{XeOF}_4+\mathrm{SiF}_4 The oxidation state of xenon in XeOF4\mathrm{XeOF}_4 is +6 as calculated below: Let the oxidation state of Xe is a ∴a+1×(−2)+4×(−1)=0a−2−4=0a=+6\begin{aligned} \therefore \quad \mathrm{a}+1 \times(-2)+4 \times(-1) & =0 \\ \mathrm{a}-2-4 & =0 \\ \mathrm{a} & =+6\end{aligned}
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