- ✓$2$
- B$-2$
- C$3$
- D$4$
or $x = + 6 - 4 = + 2$
Oxidation state of $Fe = + 2$
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$(1)$ $C{H_3}C{H_2}C{H_2}Cl + $ $\begin{array}{*{20}{c}}
{\,\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,C{H_3}} \\
{\,\,\,|}
\end{array}} \\
{C{H_3} - C - {O^ - }} \\
{\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$ $(2)$ $C{H_3}C{H_2}C{H_2}I + $$\begin{array}{*{20}{c}}
{\,\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,C{H_3}} \\
{\,\,\,|}
\end{array}} \\
{C{H_3} - C - {O^ - }} \\
{\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(3)$ $\begin{array}{*{20}{c}}
{{H_3}C - CH - C{H_3}} \\
{|\,\,\,\,} \\
{Br\,\,\,}
\end{array}$ $+CH_3O^-$ $(4)$ $\begin{array}{*{20}{c}}
{{H_3}C - CH - C{H_3}} \\
{|\,\,\,\,} \\
{Br\,\,\,}
\end{array}$ $+CH_3S^-$
$\,\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,COOH}\\
{\,\,\,\,\,\,\,\,\,|}\\
{\mathop {N{H_3}}\limits^{\,\, \oplus } - C - H}\\
{\,\,\,\,\,\,\,\,\,\,|}\\
{\,\,\,\,\,\,\,\,\,\,R}
\end{array}\,\,\,$
${C_2}{H_{4(g)}} + {H_{2(g)}} \rightleftharpoons {C_2}{H_{6(g)}}\,;$
$\Delta H = -32.7\,Kcal$ carried out is a vessel, the equilibrium concentration of $C_2H_4$ can be increased by
Reason : $NH_3$ is non-polar.