Aqueous solutions 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 $\mathrm{M}_{1}=3.4$, $\mathrm{M}_{2}=1.2, \mathrm{O}=3.5$ and $\mathrm{H}=2.1$ then the nature of two solutions…

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Aqueous solutions 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 $\mathrm{M}_{1}=3.4$, $\mathrm{M}_{2}=1.2, \mathrm{O}=3.5$ and $\mathrm{H}=2.1$ then the nature of two solutions will be respectively :

Aicharya · Accepted Answer

The electronegativity difference between $\mathrm{M}_{1}$ and $\mathrm{O}$ is $0.1$, which indicates $\mathrm{M}_{1}-\mathrm{O}$ bond will be covalent, since $\mathrm{O}-\mathrm{H}$ bond having more ionic character thus bond will break and $\mathrm{H}^{+}$ ions gets release and acidic solution is formed and whereas difference between electronegativity of $\mathrm{M}_{2} \mathrm{O}$ is $2.3$, thus, $\mathrm{M}_{2}-\mathrm{OH}$ bond will break. Hence, solution will be basic in nature. ,

Aqueous solutions of two compounds $\mathrm{M}_{1}-\mathrm{O}-\mathrm{H}$ and…

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