In the compound M-O-H, the M-O bond will be broken if:

Options

A. `Delta(EgtN.)` of M and O `lt Delta` (E.N.) of O and H
B. `Delta` (E.N.) of M and O=`Delta` (E.N.) of O and H
C. `Delta` (E.N.) of M and `O gt Delta` (E.N.) of O and H
D. Cannot be predicted according `Delta(E.N.)` data

Answer

To determine when the M-O bond in the compound M-O-H will be broken, we need to analyze the electronegativity of the elements involved. Here's a step-by-step solution:

### Step 1: Understand the Bonding in M-O-H
In the compound M-O-H, we have:
- M: a metal or a non-metal
- O: oxygen
- H: hydrogen

The bonds present are M-O and O-H.

### Step 2: Analyze Electronegativity
Electronegativity is a measure of an atom's ability to attract and hold onto electrons. The bond will break if the attraction between the atoms is weaker than the attraction within the other bonds.

### Step 3: Compare Electronegativity Values
We need to compare the electronegativities of:
- M and O (M-O bond)
- O and H (O-H bond)

### Step 4: Determine Conditions for Bond Breaking
For the M-O bond to break:
- The electronegativity of the M-O bond must be greater than the electronegativity of the O-H bond. This means that the M-O bond holds the electrons more tightly than the O-H bond.

### Step 5: Conclusion
Thus, the M-O bond will break if the electronegativity of M and O is greater than the electronegativity of O and H. Therefore, the correct answer is:
**Electronegativity of M and O is greater than electronegativity of O and H.**

### Final Answer:
The M-O bond will be broken if the electronegativity of M and O is greater than the electronegativity of O and H.

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