1. Given:
\[ \sin A = \sin^2 B \quad \text{and} \quad 2 \cos^2 A = 3 \cos^2 B. \]
2. Since \( A \) and \( B \) are acute:
\[ \sin A = \sin^2 B \implies A = \arcsin(\sin^2 B). \]
3. Substitute \( A = \frac{\pi}{6} \) and \( B = \frac{\pi}{4} \) into the second condition:
\[ 2 \cos^2 \frac{\pi}{6} = 3 \cos^2 \frac{\pi}{4}. \]
4. Verify:
- This holds true.
Thus, \( (A, B) = \left( \frac{\pi}{6}, \frac{\pi}{4} \right). \)
\( \text{M} \xrightarrow{\text{CH}_3\text{MgBr}} \text{N} + \text{CH}_4 \uparrow \xrightarrow{\text{H}^+} \text{CH}_3\text{COCH}_2\text{COCH}_3 \)
Identify the ion having 4f\(^6\) electronic configuration.