A) \[{{N}^{II}}\]
B) \[Electrical\text{ }conductances\propto \frac{1}{size\text{ }of\text{ }hydrated\text{ }cation}\]
C) \[(N{{H}_{3}})=14+3=17\]
D) \[(N{{H}_{3}})=14+14=28\]
Correct Answer: B
Solution :
Molecular mass of \[Ti{{O}_{2}}\Rightarrow T{{i}^{4+}}=[Ar]3{{d}^{0}},4{{s}^{0}}\] \[CuS{{O}_{4}}\Rightarrow C{{u}^{2+}}=[Ar]3{{d}^{9}},4{{s}^{0}}\] \[TiC{{l}_{3}}\Rightarrow T{{i}^{3+}}=[Ar]3{{d}^{1}},4{{s}^{0}}\] \[NiC{{l}_{2}}\Rightarrow N{{i}^{2+}}=[Ar]3{{d}^{8}},4{{s}^{0}}\] \[\alpha \] \[C{{H}_{3}}C{{H}_{2}}OH\] [\[C{{H}_{3}}COC{{H}_{3}}\]Radioactive changes follow 1st order kinetics.] or \[3Br{{O}^{-}}\xrightarrow{{}}BrO_{3}^{-}+2B{{r}^{-}}\] \[=\frac{-d[Br{{O}^{-}}]}{3dt}=+\frac{{{[Br{{O}^{-}}]}^{2}}}{dt}=+\frac{[B{{r}^{-}}]}{2dt}\] \[=\frac{{{k}_{1}}{{[Br{{O}^{-}}]}^{2}}}{3}={{k}_{2}}{{[Br{{O}^{-}}]}^{2}}=\frac{{{k}_{3}}{{[Br{{O}^{-}}]}^{2}}}{2}\]
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