Explanation of Assertion (A):
In aqueous solutions, $\text{Cr}^{2+}$ acts as a reducing agent and is oxidised to $\text{Cr}^{3+}$. This is because $\text{Cr}^{3+}$ has a stable $d^3$ electronic configuration. Conversely, $\text{Mn}^{3+}$ acts as an oxidising agent and is reduced to $\text{Mn}^{2+}$, which has a stable half-filled $d^5$ electronic configuration.
Explanation of Reason (R):
The half-filled electronic configuration provides extra stability due to symmetrical distribution of electrons and exchange energy. This explains why $\text{Cr}^{3+}$ and $\text{Mn}^{2+}$ are more stable compared to their respective other oxidation states.
Conclusion:
Both Assertion (A) and Reason (R) are true. The reason given (R) correctly explains why $\text{Cr}^{2+}$ is reducing and $\text{Mn}^{3+}$ is oxidising, as it is related to the stability of the resulting electronic configurations.
Given below are two statements:
Statement (I): The first ionization energy of Pb is greater than that of Sn.
Statement (II): The first ionization energy of Ge is greater than that of Si.
In light of the above statements, choose the correct answer from the options given below:
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32