The following acid have arrange in the order of decreasing strength. Identify the correct order. `CIOH(I) BrOH(II) IOH(III)`

Options

A. `I gt II gt III`
B. `II gt I gt III`
C. `III gt II gt I`
D. `I gt III gt II`

Answer

To determine the order of decreasing strength of the acids ClOH (I), BrOH (II), and IOH (III), we can follow these steps:

### Step 1: Understand the acidity trend in the group
Acidity generally increases as we move down a group in the periodic table. This is due to the increasing size of the atoms, which leads to a weaker bond between the hydrogen atom and the halogen atom, making it easier for the acid to donate a proton (H⁺).

### Step 2: Identify the acids and their corresponding halogens
The acids in question are:
- ClOH (I) - Hypochlorous acid
- BrOH (II) - Hypobromous acid
- IOH (III) - Hypoiodous acid

### Step 3: Analyze the halogens
The halogens in these acids are:
- Chlorine (Cl)
- Bromine (Br)
- Iodine (I)

As we move from chlorine to bromine to iodine, we are moving down the group in the periodic table.

### Step 4: Apply the trend to the acids
Following the trend:
- ClOH (I) will be the weakest acid because chlorine is at the top of the group.
- BrOH (II) will be stronger than ClOH because bromine is below chlorine.
- IOH (III) will be the strongest acid because iodine is below bromine.

### Step 5: Write the order of acidity
Based on the analysis, the order of decreasing strength of the acids is:
BrOH (II) > IOH (III) > ClOH (I)

### Final Order
The correct order of decreasing strength of the acids is:
**BrOH (II) > IOH (III) > ClOH (I)**