Aqueous solutions of two compounds M_1OH and | vedclass

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(A) The nature of the solution depends on the bond cleavage in $M-O-H$. If the $M-O$ bond breaks,it releases $OH-$,making it basic. If the $O-H$ bond

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Acidic,Basic

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(A) The nature of the solution depends on the bond cleavage in $M-O-H$. If the $M-O$ bond breaks,it releases $OH-$,making it basic. If the $O-H$ bond breaks,it releases $H $,making it acidic.Bond cleavage occurs where the difference in electronegativity $(\Delta EN)$ is higher.For $M_1OH$: $\Delta EN(M_1-O) = |3.4 - 3.5| = 0.1$ and $\Delta EN(O-H) = |3.5 - 2.1| = 1.4$. Since $\Delta EN(O-H) > \Delta EN(M_1-O)$,the $O-H$ bond breaks,releasing $H $,so the solution is acidic.For $M_2OH$: $\Delta EN(M_2-O) = |1.2 - 3.5| = 2.3$ and $\Delta EN(O-H) = |3.5 - 2.1| = 1.4$. Since $\Delta EN(M_2-O) > \Delta EN(O-H)$,the $M_2-O$ bond breaks,releasing $OH^-$,so the solution is basic.Therefore,the solutions are acidic and basic respectively.

Aqueous solutions of two compounds $M_1OH$ and $M_2OH$ are prepared in two different beakers. If the $EN$ of $M_1 = 3.4$,$EN$ of $M_2 = 1.2$,$EN$ of $O = 3.5$,and $EN$ of $H = 2.1$,then the nature of the two solutions will be:

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