\( \text{H}_2\text{N}-\text{NH}_2 < \text{NH}_3 < \text{CH}_3\text{CH}_2\text{NH}_2 < (\text{CH}_3\text{CH}_2)_3\text{N} < (\text{CH}_3\text{CH}_2)_2\text{NH} \)
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 two reactions A and B: 
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.
Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?

Consider the following reaction sequence: 
Given: Compound (x) has percentage composition \(76.6%\ \text{C}\), \(6.38%\ \text{H}\) and vapour density \(=47\). Compound (y) develops a characteristic colour with neutral \(\mathrm{FeCl_3}\) solution. Identify the {INCORRECT statement.}
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).

Method used for separation of mixture of products (B and C) obtained in the following reaction is: 