\( \text{H}_2\text{N}-\text{NH}_2 < \text{NH}_3 < \text{CH}_3\text{CH}_2\text{NH}_2 < (\text{CH}_3\text{CH}_2)_3\text{N} < (\text{CH}_3\text{CH}_2)_2\text{NH} \)
This problem asks for the correct increasing order of basic strength for a set of nitrogen-containing compounds in an aqueous solution. The basicity of these compounds depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton from water.
The basicity of amines and related compounds in an aqueous solution is governed by a combination of several factors:
The final observed order is the net result of these competing effects.
Step 1: Analyze Hydrazine (\( \text{H}_2\text{N}-\text{NH}_2 \))
In hydrazine, two electronegative nitrogen atoms are bonded to each other. The lone pair on one nitrogen atom is pulled by the electron-withdrawing inductive effect (-I effect) of the adjacent -NH₂ group. This significantly reduces the availability of the lone pair for protonation, making hydrazine a very weak base, even weaker than ammonia. Therefore, hydrazine is the least basic compound in the list.
Step 2: Compare Hydrazine and Ammonia (\( \text{NH}_3 \))
Ammonia has no electron-donating or electron-withdrawing groups attached. As established in Step 1, due to the -I effect in hydrazine, ammonia is a stronger base than hydrazine.
\[ \text{H}_2\text{N}-\text{NH}_2 < \text{NH}_3 \]
Step 3: Compare Ammonia and Ethylamines
Ethylamine (\(\text{CH}_3\text{CH}_2\text{NH}_2\)), Diethylamine (\((\text{CH}_3\text{CH}_2)_2\text{NH}\)), and Triethylamine (\((\text{CH}_3\text{CH}_2)_3\text{N}\)) all have electron-donating ethyl groups. The +I effect of these alkyl groups increases the electron density on the nitrogen atom, making them all stronger bases than ammonia.
Step 4: Determine the Order of Basicity for Ethylamines (1°, 2°, 3°)
Here, the interplay between the inductive effect, solvation, and steric hindrance is crucial.
For ethylamines in an aqueous solution, the experimentally determined order of basic strength is: Secondary > Tertiary > Primary. The powerful combined +I effect in triethylamine outweighs its poor solvation enough to make it more basic than ethylamine.
\[ \text{CH}_3\text{CH}_2\text{NH}_2 < (\text{CH}_3\text{CH}_2)_3\text{N} < (\text{CH}_3\text{CH}_2)_2\text{NH} \]
Step 5: Combine All Compounds into the Final Order
Assembling the comparisons from the previous steps gives the final increasing order of basic nature:
\[ \text{H}_2\text{N}-\text{NH}_2 < \text{NH}_3 < \text{CH}_3\text{CH}_2\text{NH}_2 < (\text{CH}_3\text{CH}_2)_3\text{N} < (\text{CH}_3\text{CH}_2)_2\text{NH} \]
This corresponds to option (4) in the list.
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
Predict the major product $ P $ in the following sequence of reactions:
(i) HBr, benzoyl peroxide
(ii) KCN
(iii) Na(Hg), $C_{2}H_{5}OH$

Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.
The magnitude of heat exchanged by a system for the given cyclic process ABC (as shown in the figure) is (in SI units):
