Aqueous solutions of two compounds $\mathrm{M}-\mathrm{O}-\mathrm{H}$ and $M^{\prime}-\mathrm{O}-\ma

Question

Aqueous solutions of two compounds $\mathrm{M}-\mathrm{O}-\mathrm{H}$ and $M^{\prime}-\mathrm{O}-\mathrm{H}$ have been prepared in two different beakers. If the electronegativity of $M=3.5$, $M^{\prime}=1.72, \mathrm{O}=3.0$ and $\mathrm{H}=2.1$, then the solutions respectively are\

Professor PC Thomas Classes · Answer

(b): In the compound $M-\mathrm{O}-\mathrm{H}$, if $I E$ or $E . N$. of $M$ is low, the compound will act as a base and if the $I E$ or E.N. of $M$ is high, the compound will behave as an acid. Therefore, $M-\mathrm{O}-\mathrm{H}$ will act as an acid as E.N. of $M$ is high (3.5) and $M^{\prime}-\mathrm{O}-\mathrm{H}$ will act as a base as E.N. of $M^{\prime}$ is low (1.72).