The increasing order of `K _(b)` for the following compounds will be:
`H _(2) N- overset(NH) overset(||) underset(I)(C)-NH_(2), CH_(3) -overset(NH) overset(||)underset(II) (C)-NH_(2), underset(III)(CH_(2))-NH_(2)`

Options

A. `II lt I lt II`
B. `I lt IIlt III`
C. `IIIlt IIlt I`
D. `II lt I lt III`

Explanation

To determine the increasing order of \( K_b \) for the given compounds, we need to analyze the basicity of each compound based on their structures and the ability of their nitrogen atoms to donate electrons. The compounds are:

1. \( H_2N-N=I-C-NH_2 \) (Compound I)
2. \( CH_3-N=II-C-NH_2 \) (Compound II)
3. \( CH_2-NH_2 \) (Compound III)

### Step 1: Analyze Compound I
- In Compound I, the nitrogen atoms are involved in a conjugated system. The lone pair on the nitrogen can participate in resonance with the adjacent double bond (N=I), leading to delocalization of electrons.
- This delocalization makes the lone pair less available for donation, resulting in lower basicity.

### Step 2: Analyze Compound II
- In Compound II, there is also a nitrogen atom involved in a conjugated system, but there is only one double bond (N=II) adjacent to it.
- While there is still some delocalization, it is less extensive compared to Compound I, meaning the lone pair on the nitrogen is more available for donation than in Compound I, leading to higher basicity.

### Step 3: Analyze Compound III
- In Compound III, the nitrogen atom is not involved in any conjugated system. There are no adjacent double bonds to delocalize the lone pair.
- Therefore, the lone pair on the nitrogen is fully available for donation, resulting in the highest basicity among the three compounds.

### Conclusion
Based on the analysis:
- Compound I has the lowest basicity due to extensive electron delocalization.
- Compound II has moderate basicity due to limited delocalization.
- Compound III has the highest basicity as there is no delocalization.

Thus, the increasing order of \( K_b \) for the compounds is:
\[ K_b \text{ (Compound I)} < K_b \text{ (Compound II)} < K_b \text{ (Compound III)} \]

### Final Answer
The increasing order of \( K_b \) for the compounds is:
\[ K_b \text{ (I)} < K_b \text{ (II)} < K_b \text{ (III)} \]

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