To determine the basic strength of the given molecules, we need to consider the electron-donating and electron-withdrawing effects of substituents on the phenyl group.
- (A): \(\text{C}_6\text{H}_4\text{NH}_2\text{O}\) (Hydroxyl group is an electron-donating group, which increases the basicity of the amine group.)
- (B): \(\text{C}_6\text{H}_4\text{NH}_2\text{MeO}\) (Methoxy group is a strong electron-donating group, which further increases the basicity of the amine group.)
- (C): \(\text{C}_6\text{H}_4\text{NH}_2\text{NO}_2\) (Nitro group is a strong electron-withdrawing group, which decreases the basicity of the amine group.)
- (D): \(\text{C}_6\text{H}_4\text{NH}_2\text{CH}_3\) (Methyl group is a weak electron-donating group, but its effect is weaker than the hydroxyl or methoxy group.) The order of basic strength is determined by the electron-donating ability of the substituent groups. Therefore, the order is: \[ \text{B}>\text{A}>\text{D}>\text{C} \] Thus, the correct order is \( B>A>C>D \), which corresponds to option (4).
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,