- 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).
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: