The equation which is balanced and represents the correct product(s) is

(a) ${\mathrm{Li}}_2\mathrm{O}+2\mathrm{KCl}\to 2\mathrm{LiCl}+{\mathrm{K}}_2\mathrm{O}$

The above equation is incorrect because amine a stronger base cannot be generated by a weaker base, ${\mathrm{Li}}_2\mathrm{O}$

(b) ${\left[\mathrm{CoCl}{\left({\mathrm{NH}}_3\right)}_5\right]}^{+}+5{\mathrm{H}}^{+}\to {\mathrm{Co}}^{2+}+5{\mathrm{NH}}_4^{+}+{\mathrm{Cl}}^{-}$

The above equation is correct because amine complexes decompose under acidic medium. Thus, the complex [CoCl(NH₃)₅] decomposes under acidic conditions to give ammonium ions.

(c) ${\left[\mathrm{Mg}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}+{\left(\mathrm{EDTA}\right)}^{4-}\stackrel{\text{excess NaOH}}{\to }\left[\mathrm{Mg}{\left(\mathrm{EDTA}\right)}^{2+}+6{\mathrm{H}}_2\mathrm{O}\right]$

The above equation is incorrect because the product formed would be [Mg (EDTA)]^2-.

(d) ${\mathrm{CuSO}}_4+4\mathrm{KCN}\to {\mathrm{K}}_2\left[\mathrm{Cu}{\left(\mathrm{CN}\right)}_4+{\mathrm{K}}_2{\mathrm{SO}}_4\right]$

The above equation is incorrect. Thus, the correct equation is

$2\mathrm{KCN}+{\mathrm{CuSO}}_4\to {\mathrm{K}}_2{\mathrm{SO}}_4+\mathrm{Cu}+\underset{\left(\text{Cyanogen gas}\right)}{{\left(\mathrm{CN}\right)}_2}$