If the oxidation numbers of A, B, and C are + 2, +5, and –2 respectively, then the possible formula of the compound is:

1.	A2(BC2)2	2.	A3(BC4)2
3.	A2(BC3)2	4.	A3(B2C)2

Fluorine reacts with ice as per the following reaction

H₂O(s) + F₂(g) → HF(g) + HOF(g)

This reaction is a redox reaction because-

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

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

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 oxidising agent and reducing agent in the given reaction are

$5{\mathrm{P}}_{4\left(\mathrm{s}\right)}+12{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}+12{{\mathrm{HO}}^{-}}_{\left(\mathrm{aq}\right)}\to$
$8{\mathrm{PH}}_{3\left(\mathrm{g}\right)}+12{{\mathrm{HPO}}_2^{-}}_{\left(\mathrm{aq}\right)}$

1. Oxidising agent = ${\mathrm{P}}_4$; Reducing agent = ${\mathrm{P}}_4$

2. Oxidising agent = ${\mathrm{P}}_4$; Reducing agent = ${\mathrm{H}}_2\mathrm{O}$

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

4. None of the above

The balanced equation for the reaction between chlorine and sulphur dioxide in water is:

Consider the given data:

${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}=0.77;$ ${\mathrm{E}}_{{\mathrm{I}}^{-}/{\mathrm{I}}_2}=$ $-0.54$
${\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}=0.80;$ ${\mathrm{E}}_{\mathrm{Cu}/{\mathrm{Cu}}^{2+}}=$ $-0.34$
${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}=0.77;$ ${\mathrm{E}}_{\mathrm{Cu}/{\mathrm{Cu}}^{2+}}=$ $-0.34$
${\mathrm{E}}_{\mathrm{Ag}/{\mathrm{Ag}}^{+}}=-0.80;$ ${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}=0.77$

Using the electrode potential values given above, identify the reaction which is not feasible:

The correct statement about the electrolysis of an aqueous solution of ${\mathrm{AgNO}}_3$ with Pt electrode is:
 

The correct statement about electrolysis of an aqueous solution of ${\mathrm{CuCl}}_2$ with Pt electrode is-

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