Aqueous solution of two compounds ${\mathrm{M}}_1-\mathrm{O}-\mathrm{H}$ and ${\mathrm{M}}_2-\mathrm{O}-\mathrm{H}$ are prepared in two different beakers. If the electronegativity of ${\mathrm{M}}_1=3.4\mathrm{ },\mathrm{ }{\mathrm{M}}_2=1.2,\mathrm{ }\mathrm{O}=3.5$ and $\mathrm{H}=2.1$ , then the nature of two solution will be respectively -

Arrange as directed:

$\mathrm{S} \left(16\right), \mathrm{O} \left(8\right), \mathrm{Te} \left(52\right), \mathrm{Se} \left(34\right)$ belonging to Group 16 of the long periodic table according to increasing order of electronegativity. 

(The atomic numbers have been given within first brackets after the symbols of the elements) 

In the reaction $\mathrm{B}(\mathrm{OH}{)}_3+2{\mathrm{H}}_2\mathrm{O}⟶{\left[\mathrm{B}(\mathrm{OH}{)}_4\right]}^{-}+{\mathrm{H}}_3\mathrm{O}$

$\mathrm{B}(\mathrm{OH}{)}_3$ functions as