Aqueous solutions of two compounds M_1-OH and | vedclass

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(A) The nature of the $M-OH$ bond depends on the electronegativity difference between $M$ and $O$. If the electronegativity difference between $M$ and

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acidic,basic

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(A) The nature of the $M-OH$ bond depends on the electronegativity difference between $M$ and $O$. If the electronegativity difference between $M$ and $O$ is small,the $M-O$ bond is covalent,and the $O-H$ bond breaks,releasing $H^{+}$ ions,making the solution acidic. If the electronegativity difference between $M$ and $O$ is large,the $M-O$ bond is ionic,and the $M-OH$ bond breaks,releasing $OH^{-}$ ions,making the solution basic. For $M_1-OH$: Electronegativity difference $= |3.5 - 3.4| = 0.1$. Since this is small,the $O-H$ bond breaks,resulting in an acidic solution. For $M_2-OH$: Electronegativity difference $= |3.5 - 1.2| = 2.3$. Since this is large,the $M_2-OH$ bond breaks,resulting in a basic solution. Therefore,the solutions are acidic and basic respectively.

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

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