Two different beakers contain M₁-O-H, and M₂-O-H solutions separately. Find the nature of the two solutions if the electronegativity of M₁ = 3.4, M₂ = 1.2, O = 3.5, H = 2.1

The correct option is acidic, basic.

To determine the nature (acidic or basic) of the M-O-H type solutions, we need to examine which bond is more polar and likely to break when dissolved in water. The bond with the larger electronegativity (EN) difference is more polar and has higher ionic character, making it easier to cleave.

First, let us calculate the electronegativity difference between oxygen (O) and hydrogen (H):

$$\Delta \text{EN(O-H)}=\text{EN(O)}-\text{EN(H)}=3.5-2.1=1.4$$

1. For the M₁-O-H solution:
The electronegativity of M₁ is 3.4. The electronegativity difference between oxygen (O) and M₁ is:

$$\Delta \text{EN(O-}{\text{M}}_1$$

Comparing the two electronegativity differences:
Since $\Delta \text{EN(O-H)}=1.4$ is much greater than $\Delta \text{EN(O-}{\text{M}}_1\text{)}=0.1$, the O-H bond is more polar than the M₁-O bond. As a result, the O-H bond cleaves in water to release H⁺ ions, making the solution acidic:

$${\text{M}}_1\text{-O-H}\to {\text{M}}_1{\text{-O}}^{-}+{\text{H}}^{+}$$

2. For the M₂-O-H solution:
The electronegativity of M₂ is 1.2. The electronegativity difference between oxygen (O) and M₂ is:

$$\Delta \text{EN(O-}{\text{M}}_2\text{)}=\text{EN(O)}-\text{EN(}{\text{M}}_2\text{)}=3.5-1.2=2.3$$

Comparing the two electronegativity differences:
Since $\Delta \text{EN(O-}{\text{M}}_2\text{)}=2.3$ is greater than $\Delta \text{EN(O-H)}=1.4$, the M₂-O bond is more polar (more ionic) than the O-H bond. As a result, the M₂-O bond cleaves in water to release OH^- ions, making the solution basic:

$${\text{M}}_2\text{-O-H}\to {\text{M}}_2^{+}+{\text{OH}}^{-}$$

Therefore, the nature of the M₁-O-H solution is acidic and the M₂-O-H solution is basic.