- A) \( {[MnBr}_4]^{2-} \): The \( {Mn}^{2+} \) ion has an \( 3d^5 \) configuration, leading to an \( sp^3 \) hybridization with a diamagnetic behavior due to the pairing of electrons. Hence, it corresponds to (III).
- B) \( {[FeF}_6]^{3-} \): The \( {Fe}^{3+} \) ion has a \( 3d^5 \) configuration, resulting in \( sp^2d^2 \) hybridization with paramagnetic behavior. Hence, it corresponds to (II).
- C) \( {[Co(C}_2{O}_4)_3]^{3-} \): The \( {Co}^{3+} \) ion leads to a \( 3d^6 \) configuration, requiring \( d^2sp^3 \) hybridization with diamagnetic behavior. Hence, it corresponds to (I).
- D) \( [Ni(CO)_4] \): The \( {Ni}^{2+} \) ion has a \( 3d^8 \) configuration with \( sp^3 \) hybridization and paramagnetic behavior. Hence, it corresponds to (IV). Thus, the correct matching is (A)-(III), (B)-(II), (C)-(I), (D)-(IV).
If the system of equations \[ (\lambda - 1)x + (\lambda - 4)y + \lambda z = 5 \] \[ \lambda x + (\lambda - 1)y + (\lambda - 4)z = 7 \] \[ (\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9 \] has infinitely many solutions, then \( \lambda^2 + \lambda \) is equal to:
Let \( ABCD \) be a tetrahedron such that the edges \( AB \), \( AC \), and \( AD \) are mutually perpendicular. Let the areas of the triangles \( ABC \), \( ACD \), and \( ADB \) be 5, 6, and 7 square units respectively. Then the area (in square units) of the \( \triangle BCD \) is equal to: