
- ✓

- B

- C

- DNone of these






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$B + {H_2}O \to C + HCl$
Compound $(A),\,(B)$ and $(C)$ will be respectively
($A$) can also be prepared by reaction of $\mathrm{P}_4$ and $\mathrm{HNO}_3$
($B$) is diamagnetic
($C$) contains one $\mathrm{N}-\mathrm{N}$ bond
($D$) reacts with $\mathrm{Na}$ metal producing a brown gas
$(i)$ Increase in concentration of reactant increases the rate of a zero order reaction
$(ii)$ Rate constant $k$ is equal to $'A'$ if $E_a = 0$
$(iii)$ Rate constant $k$ is equal to $'A'$ if $E_a = \infty$
$(iv)$ $log_e k$ vs $T$ is a straight line
$(v)$ $log_ek$ vs $1/T$ is straight line
Correct statement are
$\mathrm{I}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}\ \ \ \mathrm{E}^{\circ}=0.54 $
$\mathrm{Cl}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \ \ \ \mathrm{E}^{\circ}=1.36 $
$\mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} \ \ \ \mathrm{E}^{\circ}=1.50 $
$\mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} \ \ \ \mathrm{E}^{\circ}=0.77 $
$\mathrm{O}_2+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O} \ \ \ \mathrm{E}^{\circ}=1.23$
$1.$ Among the following, identify the correct statement.
$(A)$ Chloride ion is oxidized by $\mathrm{O}_2$
$(B)$ $\mathrm{Fe}^{2+}$ is oxidized by iodine
$(C)$ Iodide ion is oxidized by chlorine
$(D)$ $\mathrm{Mn}^{2+}$ is oxidized by chlorine
$2.$ While $\mathrm{Fe}^{3+}$ is stable, $\mathrm{Mn}^{3+}$ is not stable in acid solution because
$(A)$ $\mathrm{O}_2$ oxidises $\mathrm{Mn}^{2+}$ to $\mathrm{Mn}^{3+}$
$(B)$ $\mathrm{O}_2$ oxidises both $\mathrm{Mn}^{2+}$ and $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$
$(C)$ $\mathrm{Fe}^{3+}$ oxidizes $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$
$(D)$ $\mathrm{Mn}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$
$3.$ Sodium fusion extract, obtained from aniline, on treatment with iron $(II)$ sulphate and $\mathrm{H}_2 \mathrm{SO}_4$ in presence of air gives a Prussian blue precipitate. The blue colour is due to the formation of
$(A)$ $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$ $(B)$ $\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$
$(C)$ $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$ $(D)$ $\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$
Give the answer question $1,2$ and $3.$
$\text { [en }= H _2 NCH _2 CH _2 NH _2 \text { ] }$
Reason : Lower the activation energy, faster is the reaction.