能正确表示下列反应的离子方程式的是$(\quad\ \ \ \ )$

将$\\text{S}{{\\text{O}}_{2}}$通入$\\text{NaClO}$溶液中：$\\text{S}{{\\text{O}}_{2}}+{{\\text{H}}_{2}}\\text{O}+\\text{Cl}{{\\text{O}}^{-}}=\\text{HClO}+\\text{HSO}_{3}^{-}$

向${{\\text{H}}_{2}}{{\\text{C}}_{2}}{{\\text{O}}_{4}}$溶液中滴入少量$\\text{NaOH}$溶液：$\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}+2 \\mathrm{OH}^{-}=\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}+2 \\mathrm{H}_{2} \\mathrm{O}$

用惰性电极电解饱和$\\text{MgC}{{\\text{l}}_{2}}$溶液：$2\\text{C}{{\\text{l}}^{-}}+2{{\\text{H}}_{2}}\\text{O}\\begin{matrix} \\underline{\\underline{通电}} \\\\ {} \\\\\\end{matrix}\\text{C}{{\\text{l}}_{2}}\\uparrow +{{\\text{H}}_{2}}\\uparrow +2\\text{O}{{\\text{H}}^{-}}$

$\\text{NiS}{{\\text{O}}_{4}}$溶于氨水形成硫酸六氨合镍溶液：$\\text{N}{{\\text{i}}^{2+}}+6\\text{N}{{\\text{H}}_{3}}\\cdot {{\\text{H}}_{2}}\\text{O}={{\\left[ \\text{Ni}{{\\left( \\text{N}{{\\text{H}}_{3}} \\right)}_{6}} \\right]}^{2+}}+6{{\\text{H}}_{2}}\\text{O}$

$\rm A$．$\text{S}{{\text{O}}_{2}}$通入$\text{NaClO}$溶液中，反应氧化还原反应生成氯离子和硫酸根离子，$\text{Cl}{{\text{O}}^{-}}+\text{S}{{\text{O}}_{\text{2}}}+\text{ }{{\text{H}}_{\text{2}}}\text{O }=\text{SO}_{\text{4}}^{\text{2-}}+\text{C}{{\text{l}}^{-}}+\text{2}{{\text{H}}^{+}}\,\,$，$\rm A$错误；

$\rm B$．${{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}$为二元弱酸，${{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}$溶液中滴入少量$\text{NaOH}$溶液生成$\text{H}{{\text{C}}_{\text{2}}}\text{O}_{\text{4}}^{-}$：${{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{+O}{{\text{H}}^{-}}\text{=H}{{\text{C}}_{\text{2}}}\text{O}_{\text{4}}^{-}+{{\text{H}}_{\text{2}}}\text{O}$，$\rm B$错误；

$\rm C$．惰性电极电解饱和$\text{MgC}{{\text{l}}_{2}}$溶液会生成氢气、氯气和氢氧化镁沉淀，$\text{2C}{{\text{l}}^{-}}\text{+2}{{\text{H}}_{\text{2}}}\text{O+M}{{\text{g}}^{\text{2+}}}\begin{matrix} \underline{\underline{通电}} \\ {} \\\end{matrix}\text{C}{{\text{l}}_{\text{2}}}\uparrow +{{\text{H}}_{2}}\uparrow +\text{Mg}{{\left( \text{OH} \right)}_{2}}\downarrow $，$\rm C$错误；

$\rm D$．$\text{NiS}{{\text{O}}_{4}}$溶于氨水形成硫酸六氨合镍溶液，方程式正确，$\rm D$正确；

故选：$\rm D$