\( \text{H}_2\text{N}-\text{NH}_2 < \text{NH}_3 < \text{CH}_3\text{CH}_2\text{NH}_2 < (\text{CH}_3\text{CH}_2)_3\text{N} < (\text{CH}_3\text{CH}_2)_2\text{NH} \)
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.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,