Since conductance measurements show two ions per formula unit, all the ammonia ligands are present in the coordination sphere. Hence, the structural formula is $\left[ CoBr _{2}\left( NH _{3}\right)_{4}\right] Cl$
$ {\left[ CoBr _{2}\left( NH _{3}\right)_{4}\right] Cl + AgNO _{3} \rightleftharpoons\left[ CoBr _{2}\left( NH _{3}\right)_{4}\right]+ NO _{3}^{-}+ AgCl }$
${\left[ CoBr _{2}\left( NH _{3}\right)_{4}\right] Cl\rightleftharpoons\left[ CoBr _{2}\left( NH _{3}\right)_{4}\right]+Cl ^{-}}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{Z}^{-}$$\xrightarrow[{{\text{Sublimation}}}]{{{k_s}}} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Z}+\mathrm{Br}^{-}$
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{Z}^{-}$$\xrightarrow[{{\text{elimination}}}]{{{k_e}}}\mathrm{CH}_{3} \mathrm{CH}= \mathrm{CH}_{2} +\mathrm{HZ}+\mathrm{Br}^{-}$
where
$\mathrm{Z}^{-}=\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}(\mathrm{A})$ or $\begin{array}{*{20}{c}}
{\,C{H_3}} \\
{|\,\,\,\,\,} \\
{C{H_3} - C - {O^ - }(B)} \\
{|\,\,\,\,} \\
{\,\,C{H_3}}
\end{array}$
$\mathrm{k}_{\mathrm{s}}$ and $\mathrm{k}_{\mathrm{e}},$ are $,$ respectively, the rate constants for the substitution and elimination, and $\mu=\frac{\mathrm{k}_{\mathrm{s}}}{\mathrm{k}_{\mathrm{e}}},$ the correct options is
$A+B\xrightarrow{K}C$: