- A${N_2}$
- ✓$Cl$
- C$Na$
- D$C{O_2}$
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Reason : Oxidation state of $Cr$ in $Na_2CrO_4$ is $+VI$
$\mathop {\begin{array}{*{20}{c}}
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{C{H_2} = C - C{H_2} - C - C{H_3}}
\end{array}}\limits_{(I)} \rightleftharpoons $
$\mathop {\begin{array}{*{20}{c}}
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{C{H_3} - C - C{H_2} - C - C{H_3}}
\end{array}}\limits_{(II)} \rightleftharpoons $
$\mathop {\begin{array}{*{20}{c}}
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{C{H_3} - C = CH - C - C{H_3}}
\end{array}}\limits_{(III)} $
reagents ' $X$ ' and ' $Y$ ' respectively are:
| Exp. | $[A]$ | $[B]$ | Rate of reaction |
| $1$ | $0.1$ | $0.1$ | $2 \times {10^{ - 3}}\,mol\,{L^{ - 1}}\,{\sec ^{ - 1}}$ |
| $2$ | $0.4$ | $0.1$ | $0.4 \times {10^{ - 2}}\,mol\,{L^{ - 1}}\,{\sec ^{ - 1}}$ |
| $3$ | $0.1$ | $0.2$ | $1.4 \times {10^{ - 2}}\,mol\,{L^{ - 1}}\,{\sec ^{ - 1}}$ |