The correct statement(s) about the given reaction is -

\(\mathrm{{X} eO_{6 (aq)}^{4 -} + 2 F^{- 1}_{(aq)} + 6 H^{+}_{(aq )} \rightarrow}~\mathrm{{X} eO_{3 (g)} + {F}_{2 (g)} + 3 H_{2} O_{(l)}}\)${}_{}$

1. ${\mathrm{XeO}}_6^{4-}$ oxidises F^-

2. The oxidation number of F increases from -1  to  zero

3. ${\mathrm{XeO}}_6^{4-}$ is a stronger oxidizing agent that ${\mathrm{F}}^{-}$

4. All of the above.

The oxidizing agent and reducing agent in the given reaction are :

$3{\mathrm{N}}_2{\mathrm{H}}_{4\left(\mathrm{l}\right)}+4{\mathrm{ClO}}_{3\left(\mathrm{aq}\right)}^{-}\to$
$6{\mathrm{NO}}_{\left(\mathrm{g}\right)}+4{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}+6{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$

1. Oxidising agent = ${\mathrm{N}}_2{\mathrm{H}}_4$; Reducing agent = ${\mathrm{ClO}}_3^{-}$

2. Oxidising agent = ${\mathrm{ClO}}_3^{-}$; Reducing agent = ${\mathrm{N}}_2{\mathrm{H}}_4$

3. Oxidising agent = ${\mathrm{N}}_2{\mathrm{H}}_4$ ; Reducing agent = ${\mathrm{N}}_2{\mathrm{H}}_4$

4. Oxidising agent = ${\mathrm{ClO}}_3^{-}$ ; Reducing agent = ${\mathrm{ClO}}_3^{-}$

The oxidising agent and reducing agent in the given reaction are :

${}_{}$${\mathrm{Cl}}_2{\mathrm{O}}_{7\left(\mathrm{g}\right)}+4{\mathrm{H}}_2{\mathrm{O}}_{2\left(\mathrm{aq}\right)}+2{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}\to$
$2{\mathrm{ClO}}_{2\left(\mathrm{aq}\right)}^{-}+4{\mathrm{O}}_{2\left(\mathrm{g}\right)}+5{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$${}_{}$

1. Oxidizing agent = H₂O₂; Reducing agent = ${\mathrm{Cl}}_2{\mathrm{O}}_7$

2. Oxidizing agent = ${\mathrm{Cl}}_2{\mathrm{O}}_7$; Reducing agent = ${\mathrm{H}}_2{\mathrm{O}}_2$

3. Oxidizing agent = H₂O₂; Reducing agent = H₂O₂

4. None of the above