Aqueous solutions of two compounds M_1-OH and | vedclass

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(A) The nature of a hydroxide $M-OH$ depends on the electronegativity difference between the metal/non-metal $M$ and oxygen $O$,or the electronegativi

Vedclass · Accepted Answer

Acidic,basic

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(A) The nature of a hydroxide $M-OH$ depends on the electronegativity difference between the metal/non-metal $M$ and oxygen $O$,or the electronegativity of $M$ itself.According to the rule,if the electronegativity of $M$ is greater than $1.7$,the $O-H$ bond breaks,making the solution acidic.If the electronegativity of $M$ is less than $1.7$,the $M-O$ bond breaks,releasing $OH^-$ ions,making the solution basic.For $M_1-OH$: Electronegativity of $M_1 = 3.4 > 1.7$,so it is acidic.For $M_2-OH$: Electronegativity of $M_2 = 1.2 Therefore,the nature of the solutions is 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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